Electrical Network Theory
Electrical Network Theory
Electrical Network Theory
NETWORK
THEORY
Norman BALABANIAN
Theodore A. BICKART
Sundaram Seshu
ELECTRICAL
NETWORK
THEORY
NORMAN BALABANIAN
THEODORE A. BICKART
Syracuse University
with contributions
SUNDARAM
SESHU
University of Illinois
LONDON
SYDNEY
TORONTO
PREFACE
vi
PREFACE
Matrix analysis is not treated all in one place but some of it is introduced
at the time it is required. Thus introductory considerations are discussed
in Chapter 1 but functions of a matrix are introduced in Chapter 4 in
which a solution of the vector state equation is sought. Similarly, equi
valence, canonic forms of a matrix, and quadratic forms are discussed in
Chapter 7, preparatory to the development of analytic properties of net
work functions.
The analysis of networks starts in Chapter 2 with a precise formulation
of the fundamental relationships of Kirchhoff, developed through the
application of graph theory. The classical methods of loop, node, nodepair, and mixed-variable equations are presented on a topological base.
In Chapter 3 the port description and the terminal description of
multiterminal networks are discussed. The usual two-port parameters are
introduced, but also discussed are multiport networks. The indefinite
admittance and indefinite impedance matrices and their properties make
their appearance here. The chapter ends with a discussion of formulas
for the calculation of network functions by topological concepts.
The state formulation of network equations is introduced in Chapter 4.
Procedures for writing the state equations for passive and active and
reciprocal and nonreciprocal networks include an approach that requires
calculation of multiport parameters of only a resistive network (which
may be active and nonreciprocal). An extensive discussion of the timedomain solution of the vector state equation is provided.
Chapter 5 deals with integral methods of solution, which include the
convolution integral and superposition integrals. Numerical methods of
evaluating the transition matrix, as well as the problem of errors in
numerical solutions, are discussed.
Chapters 6 and 7 provide a transition from analysis to synthesis. The
sufficiency of the real part, magnitude, or angle as specifications of a
network function are taken up and procedures for determining a function
from any of its parts are developed. These include algebraic methods as
well as integral relationships given by the Bode formulas. Integral
formulas relating the real and imaginary parts of a network function to
the impulse or step response are also developed. For passive networks the
energy functions provide a basis for establishing analytic properties of
network functions. Positive real functions are introduced and the proper
ties of reactance functions and RC impedance and admittance functions
are derived from them in depth. Synthesis procedures discussed for these
and other network functions include the Darlington procedure and active
RC synthesis techniques.
Chapter 8 presents a thorough treatment of scattering parameters and
the description of multiport networks by scattering matrices. Both real
PREFACE
vii
and complex normalization are treated, the latter including singlefrequency and frequency-independent normalization. Reflection and
transmission properties of multiport networks, both active and passive,
reciprocal and non-reciprocal, are developed in terms of scattering
parameters. Applications to filter design and negative resistance amplifiers
are discussed.
Concepts of feedback and stability are discussed in Chapter 9. Here the
signal flow-graph is introduced as a tool. The Routh-Hurwitz, LinardChipart, and Nyquist criteria are presented.
The final chapter is devoted to time-varying and nonlinear networks.
Emphasis is on general properties of both types of network as developed
through their state equations. Questions of existence and uniqueness of
solutions are discussed, as are numerical methods for obtaining a solution.
Attention is also devoted to Liapunov stability theory.
A rich variety of problems has been presented at the end of each chapter.
There is a total of 460, some of which are routine applications of results
derived in the text. Many, however, require considerable extension of
the text material or proof of collateral results which, but for the lack of
space, could easily have been included in the text. In a number of the
chapters a specific class of problems has been included. Each of these
problems, denoted by an asterisk, requires the preparation of a computer
program for some specific problem treated in the text. Even though writing
computer programs has not been covered and only a minimal discussion
of numerical procedures is included, we feel that readers of this book may
have sufficient background to permit completion of those problems.
A bibliography is presented which serves the purpose of listing some
authors to whom we are indebted for some of our ideas. Furthermore,
it provides additional references which may be consulted for specific
topics.
We have benefited from the comments and criticisms of many colleagues
and students who suggested improvements for which we express our
thanks.
Syracuse, New York
November, 1968
N.
BALABANIAN
T. A.
BlCKART
CONTENTS
1. F U N D A M E N T A L
CONCEPTS
1.1
INTRODUCTION
1.2
Basic Operations
Types of Matrices
Determinants
11
15
Pivotal Condensation
17
Linear Equations
20
Characteristic Equation
26
Similarity
28
Sylvester's Inequality
30
Norm of a Vector
31
1.3
34
1.4
N E T W O R K CLASSIFICATION
36
1.5
Linearity
36
Time-Invariance
37
Passivity
37
Reciprocity
38
N E T W O R K COMPONENTS
39
The Transformer
42
The Gyrator
45
Independent Sources
47
ix
CONTENTS
48
50
PROBLEMS
51
2.
58
2.1
2.2
2.3
2.4
INTRODUCTORY CONCEPTS
58
Kirchhoff's Laws
58
Loop Equations
61
Node Equations
62
63
Solutions of Equations
66
LINEAR GRAPHS
69
Introductory Definitions
70
73
77
81
83
Planar Graphs
88
B A S I C L A W S OF ELECTRIC N E T W O R K S
90
90
95
99
LOOP, N O D E , A N D N O D E - P A I R EQUATIONS
104
Loop Equations
105
Node Equations
110
Node-pair Equations
115
2.5
DUALITY
118
2.6
NONRECIPROCAL A N D ACTIVE N E T W O R K S
122
2.7
M I X E D - V A R I A B L E EQUATIONS
131
PROBLEMS
140
CONTENTS
3.
N E T W O R K FUNCTIONS
3.1
xi
150
150
Driving-Point Functions
153
Transfer Functions
156
3.2
MULTITERMINAL N E T W O R K S
158
3.3
TWO-PORT NETWORKS
160
3.4
161
Hybrid Parameters
163
Chain Parameters
165
Transmission Zeros
165
INTERCONNECTION OF T W O - P O R T N E T W O R K S
169
Cascade Connection
169
171
Permissibility of Interconnection
174
3.5
MULTIPORT N E T W O R K S
176
3.6
T H E I N D E F I N I T E ADMITTANCE M A T R I X
178
182
Suppressing Terminals
183
Networks in Parallel
183
184
3.7
187
3.8
191
191
193
197
200
Two-port Parameters
203
PROBLEMS
208
4.
229
STATE EQUATIONS
4.1
O R D E R OF COMPLEXITY OF A N E T W O R K
230
4.2
234
xii
CONTENTS
4.3
4.4
4.5
247
250
Matrix Exponential
257
FUNCTIONS OF A M A T R I X
258
260
Distinct Eigenvalues
262
Multiple Eigenvalues
265
Constituent Matrices
268
269
271
Resolving Polynomials
276
4.6
245
280
Topological Considerations
282
285
Time-invariant Networks
291
RLC Networks
292
294
302
Output Equations
PROBLEMS
306
314
320
5. I N T E G R A L SOLUTIONS
336
5.1
CONVOLUTION THEOREM
337
5.2
IMPULSE RESPONSE
341
5.3
5.4
345
347
STEP RESPONSE
SUPERPOSITION P R I N C I P L E
351
357
357
360
CONTENTS
5.5
N U M E R I C A L SOLUTION
362
365
State Response
367
Propagating Errors
5.6
370
N U M E R I C A L E V A L U A T I O N OF e
A T
373
Computational Errors
377
377
379
PROBLEMS
382
6. R E P R E S E N T A T I O N S O F N E T W O R K F U N C T I O N S
6.1
6.2
6.3
6.4
6.5
392
392
Locations of Poles
394
396
398
399
MINIMUM-PHASE FUNCTIONS
399
402
404
Hurwitz Polynomials
404
406
Ladder Networks
406
Constant-Resistance Networks
407
D E T E R M I N I N G A N E T W O R K F U N C T I O N FROM ITS M A G N I T U D E
414
416
Chebyshev Response
422
CALCULATION OF A N E T W O R K F U N C T I O N FROM A G I V E N
ANGLE
6.6
xiii
CALCULATION OF
423
NETWORK
REAL PART
FUNCTION
FROM
GIVEN
427
428
429
431
xiv
6.7
CONTENTS
INTEGRAL RELATIONSHIPS B E T W E E N R E A L A N D IMAGINARY
PARTS
6.8
439
441
444
447
F R E Q U E N C Y A N D T I M E - R E S P O N S E RELATIONSHIPS
451
Impulse Response
455
7. F U N D A M E N T A L S O F N E T W O R K S Y N T H E S I S
7.2
7.3
7.4
451
Step Response
PROBLEMS
7.1
433
TRANSFORMATION OF MATRICES
459
467
468
Elementary Transformations
468
Equivalent Matrices
470
Similarity Transformation
472
Congruent Transformation
472
QUADRATIC A N D H E R M I T I A N FORMS
474
Definitions
474
476
478
Hermitian Forms
481
E N E R G Y FUNCTIONS
481
484
488
Condition on Angle
490
POSITIVE R E A L FUNCTIONS
492
497
500
501
503
CONTENTS
7.5
7.6
7.7
R E A C T A N C E FUNCTIONS
xv
504
509
Ladder-Form of Network
512
514
IMPEDANCES A N D ADMITTANCES OF RC
NETWORKS
517
Ladder-Network Realization
523
Resistance-Inductance Networks
525
T W O - P O R T PARAMETERS
525
Resistance-Capacitance Two-Ports
529
7.8
LOSSLESS T W O - P O R T T E R M I N A T E D I N A R E S I S T A N C E
531
7.9
P A S S I V E A N D ACTIVE RC
540
TWO-PORTS
Cascade Connection
541
543
Parallel Connection
546
The RC-Amplifier
Configuration
PROBLEMS
8.
8.1
8.2
8.3
549
553
T H E SCATTERING PARAMETERS
571
The
572
SCATTERING RELATIONS OF A O N E - P O R T
575
Augmented Network
576
578
Power Relations
579
580
583
585
586
T H E SCATTERING M A T R I X A N D P O W E R T R A N S F E R
588
589
xvi
CONTENTS
8.4
8.5
P R O P E R T I E S OF THE SCATTERING M A T R I X
594
596
An ApplicationFiltering or Equalizing
598
601
COMPLEX NORMALIZATION
605
Frequency-Independent Normalization
609
Negative-Resistance Amplifier
617
PROBLEMS
9. S I G N A L - F L O W G R A P H S A N D F E E D B A C K
622
636
9.1
A N OPERATIONAL DIAGRAM
637
9.2
S I G N A L - F L O W GRAPHS
642
9.3
9.4
9.5
Graph Properties
644
Inverting a Graph
646
Reduction of a Graph
647
654
Graph-Gain Formula
655
659
FEEDBACK
664
664
Sensitivity
668
STABILITY
669
Routh Criterion
673
Hurwitz Criterion
674
675
T H E N Y Q U I S T CRITERION
677
Discussion of Assumptions
681
Nyquist Theorem
683
PROBLEMS
690
10. L I N E A R TIME-VARYING A N D N O N L I N E A R
NETWORKS
10.1
705
S T A T E E Q U A T I O N FORMULATION FOR T I M E - V A R Y I N G
NETWORKS
706
CONTENTS
10.2
xvii
706
709
S T A T E - E Q U A T I O N SOLUTION FOR T I M E - V A R Y I N G
NETWORKS
712
714
718
10.3
Uniqueness
721
Periodic Networks
723
P R O P E R T I E S OF THE S T A T E - E Q U A T I O N SOLUTION
727
727
729
734
10.5
10.6
739
744
Topological Formulation
744
Output Equation
754
756
757
761
N U M E R I C A L SOLUTION
768
768
Open Formulas
772
Closed Formulas
774
Euler's Method
776
777
778
780
xviii
10.7
CONTENTS
Milne Method
781
Predictor-Corrector Methods
781
Runge-Kutta Method
782
Errors
783
L I A P U N O V STABILITY
783
Stability Definitions
784
Stability Theorems
787
Instability Theorem
793
795
PROBLEMS
Appendix 1
803
Generalized Functions
829
Al.l
831
A1.2
833
835
836
838
840
842
846
A1.5
Integrodifferential Equations
847
A1.6
850
853
A2.1
Analytic Functions
853
A2.2
Mapping
857
A2.3
Integration
862
863
866
867
A1.3
A1.4
Appendix 2
CONTENTS
A2.4
A2.5
A2.6
xix
Infinite Series
869
Taylor Series
871
Laurent Series
873
875
Multivalued Functions
877
877
879
882
883
886
Jordan's Lemma
888
891
A2.7
Partial-Fraction Expansions
892
A2.8
Analytic Continuation
895
898
Appendix 3
A3.1
A3.2
A3.3
A3.4
898
903
907
907
909
910
Shifting
912
913
Bibliography
917
INDEX
923
.1.
FUNDAMENTAL CONCEPTS
1.1
INTRODUCTION
FUNDAMENTAL CONCEPTS
[Ch. 1
In the case of electric network theory, the model has had great success
in predicting experimental results. As a matter of fact, the model has
become so real that it is difficult for students to distinguish between the
model and the physical world.
The first step in establishing a model is to make detailed observa
tions of the physical world. Experiments are performed in an attempt to
establish universal relationships among the measurable quantities. From
these experiments general conclusions are drawn concerning the behavior
of the quantities involved. These conclusions are regarded as " l a w s , "
and are usually stated in terms of the variables of the mathematical
model.
Needless to say, we shall not be concerned with this step in the process.
The model has by now been well established. We shall, instead, intro
duce the elements of the model without justification or empirical veri
fication. The process of abstracting an appropriate interconnection of the
hypothetical elements of the model in order to describe adequately a
given physical situation is an important consideration, but outside the
scope of this book.
This book is concerned with the theory of linear electric networks. B y
an electric network is meant an interconnection of electrical devices form
ing a structure with accessible points at which signals can be observed.
It is assumed that the electrical devices making up the network are
represented by models, or hypothetical elements whose voltage-current
equations are linear equationsalgebraic equations, difference equations,
ordinary differential equations, or partial differential equations. In this
book we shall be concerned only with lumped networks; hence, we shall
not deal with partial differential equations or difference equations.
The properties of networks can be classified under two general headings.
First, there are those properties of a network that are consequences of its
structurethe topological properties. These properties do not depend on
the specific elements that constitute the branches of the network but
only on how the branches are interconnected; for example, it may be
deduced that the transfer function zeros of a ladder network (a specific
topological structure) lie in the left half-plane regardless of what passive
elements constitute the branches. Second, there are the properties of net
works as signal processors. Signals are applied at the accessible points of
the network, and these signals are modified or processed in certain ways
b y the network. These signal-processing properties depend on the elements
of which the network is composed and also on the topological structure
of the network. Thus if the network elements are lossless, signals are
modified in certain ways no matter what the structure of the network;
further limitations are imposed on these properties by the structure. The
Sec. 1.1]
INTRODUCTION
1.2
Square brackets are placed around the entries to enclose the whole
matrix. It is not necessary to write the whole matrix in order to refer to it.
It is possible to give it a " n a m e " b y assigning it a single symbol, such as
M or V in the above examples. We shall consistently use boldface letters,
either capital or lower case, to represent matrices.
The order of a matrix is an ordered pair of numbers specifying the
[Ch. 1
FUNDAMENTAL CONCEPTS
If the order (m, n) is not of interest, it need not be shown in this expres
sion. The " typical element" is ay. The above simple expression stands
for the same thing as
BASIC OPERATIONS
ij
ij
ij
ij
ij
Sec. 1.2]
A = [a ] and B = [b ],
ij
ij
then
(2)
(3)
Multiplication of Matrices. If A = [a ]
product of A and B is defined as
ij m,n
and B = [b ] ,
ij n,p
then the
(4)
(5)
FUNDAMENTAL CONCEPTS
[Ch. 1
Then
(6)
where
Sec. 1.2]
or
where
1 1
FUNDAMENTAL CONCEPTS
[Ch. 1
(8)
Thus the matrix dA(x)/dx is obtained by replacing each element a (x)
of A(x) with its derivative da (x)/dx.
Now it is easy, and left to you
as an exercise, to show that
ij
ij
(9)
(10)
and
(11)
We see that the familiar rules for differentiation of combinations of
functions apply t o the differentiation of matrices; the one caution is that
the sequence of matrix products must be preserved in (10).
Integration. Let the order of A be n m. Then, for any interval on
Sec. 1.2]
which
a (y)
ij
JXl
A(y) dy
JXi
is defined as
(12)
Thus the (i, j)th element of the integral of A is the integral of the (i, j)th
element of A.
Trace. The trace of a square matrix A is a number denoted by tr A
and defined as
where n is the order of A. Note that tr A is simply the sum of the main
diagonal elements of A.
Transpose. The operation of interchanging the rows and columns of a
matrix is called transposing. The result of this operation on a matrix A
is called the transpose of A and is designated A'. If A = [a ] ,
then
A' = [b ] ,
, where b = a . The transpose of a column matrix is a row
matrix, and vice versa. If, as often happens in analysis, it is necessary to
find the transpose of the product of two matrices, it is important to know
that
(13)
ij mjn
ij n
ij
ji
that is, the transpose of a product equals the product of transposes, but
in the opposite order. This result can be established simply by writing
the typical element of the transpose of the product and showing that it
is the same as the typical element of the product of the transposes.
Conjugate. If each of the elements of a matrix A is replaced by its
complex conjugate, the resulting matrix is said to be the conjugate of A
and is denoted by A. Thus, if A = [a ] ,
then A = [b ] ,
where
b = a and a denotes the complex-conjugate of a .
ij ntm
ij
ij
ij
ij njm
ij
There are two special matrices that have the properties of the scalars
0 and 1. The matrix 0 = [0] which has 0 for each entry is called the zero,
10
FUNDAMENTAL CONCEPTS
[Ch. 1
All elements both above the main diagonal and below the main diagonal
are zero. A diagonal matrix is its own transpose.
If the elements only below the main diagonal or only above the main
diagonal of a square matrix are zero, as in the following,
Sec. 1.2]
11
ij
ji
ji
Then
(16)
Hermitian and Skew-Hermitian Matrices. A square matrix is said to
be Hermitian if it equals its conjugate transpose; that is, A is Hermitian
if A = A* or, equivalently, a = a for all i and j . As another special case,
if a matrix equals the negative of its conjugate transpose, it is called
skew-Hermitian. Thus A is skew-Hermitian if A = A* or, equivalently,
aij = a for all i and j . Observe that a Hermitian matrix having only
real elements is symmetric, and a skew-Hermitian matrix having only
real elements is skew-symmetric.
ij
ji
ji
DETERMINANTS
12
FUNDAMENTAL CONCEPTS
[Ch. 1
equal determinants, this does not imply that the matrices are equal; the
two may even be of different orders.
The determinant of the n n matrix A is defined as
(17)
The product a a
was multiplied by = + 1 because v , v = 1, 2 is an
even permutation of 1, 2; the product a a
was multiplied by = 1
because v , v = 2, 1 is an odd permutation of 1, 2. Determinant evalua
tion for large n by applying the above definition is difficult and not always
necessary. Very often the amount of time consumed in performing the
arithmetic operations needed to evaluate a determinant can be reduced
by applying some of the properties of determinants. A summary of some
of the major properties follows:
11
22
12
21
Sec. 1.2]
13
(19)
It is said that is the cofactor of element a . If i =j, the minor and
cofactor are called principal minor and principal cofactor. More specifically,
a principal minor (cofactor) of A is one whose diagonal elements are also
diagonal elements of A.* The value of a determinant can be obtained by
multiplying each element of a row or column by its corresponding cofactor
and adding the results. Thus
i j
ij
(20a)
(20b)
These expressions are called the cofactor expansions along a row or
column and are established by collecting the terms of (17) or (18) into
groups, each corresponding to an element times its cofactor.
What would happen if the elements of a row or column were multiplied
by the corresponding cofactors of another row or column? It is left to
you as a problem to show that the result would be zero; that is,
(21a)
(21b)
The Kronecker delta is a function denoted by
=1
if
i= j
= 0
if
ij
i j
i j
and is defined as
where i and j are integers. Using the Kronecker delta, we can consolidate
(20a) and (21a) and write
(22)
* This definition does not limit the number of rows and columns deleted from to
form the minor or cofactor. If one row and column are deleted, we should more properly
refer to the first principal cofactor. In general, if n rows and columns are deleted, we
would refer to the result as the nth. principal cofactor.
14
FUNDAMENTAL CONCEPTS
[Ch. 1
and from the cofactor expansion for | A| in (22) or (23) that d | A|/da
ij
= .
i j
Sec. 1.2]
15
The determinant of this matrix is easily seen to be 18. Now let us apply
the Binet-Cauchy theorem. We see that there are three determinants of
order two to be considered. Applying (26), we get
This agrees with the value calculated by direct evaluation of the deter
minant.
THE I N V E R S E OF A MATRIX
U.
- 1
- 1
ji
16
FUNDAMENTAL CONCEPTS
Ch. 1
(27)
Note that the elements in the ith row of adj A are the cofactors of the
elements of the ith column of A. The inverse of A can now be written as
(28)
(29)
Each side of this expression is a matrix, the left side being the product of
two matrices and the right side being a diagonal matrix whose diagonal
elements each equal det A. Taking the determinant of both sides yields
n
n - 1
(30)
In some of the work that follows in later chapters, the product of two
matrices is often encountered. It is desirable, therefore, to evaluate the
result of finding the inverse and adjoint of the product of two matrices
A and B. The results are
(AB)
- 1
= B
- 1
- 1
(31)
(32)
Sec. 1.2]
17
- 1
- 1
(AB)(B A ) = A(BB )A
- 1
- 1
= AA
- 1
= U.
_ 1
Hence A B is the inverse of B A , whence the result. For the second one,
we can form the products (AB)(adj AB) and (AB)(adj B)(adj A) and
show by repeated use of the relationship M(adj M) = U (det M) that both
products equal U (det AB). The result follows.
PIVOTAL C O N D E N S A T I O N
(33)
-1
11
11
and
18
FUNDAMENTAL CONCEPTS
[Ch. 1
(35)
The determinants of the matrices on the right are simply det A and
d e t ( A A A A ) , respectively.
Since the determinant of a product of matrices equals the product of
the determinants, then taking the determinant of both sides of (34) and
using (35) leads to
11
- 1
22
2 1
1 1
1 2
(36)
If A is the scalar a
equation reduces to
1 1
11
1 1
11
(37)
Sec. 1.2]
19
(inter
change of
columns
1 and 3)
20
FUNDAMENTAL CONCEPTS
[Ch. 1
Many of the steps included here for completeness are ordinarily eliminated
by someone who has become facile in using pivotal condensation to
evaluate a determinant.
L I N E A R EQUATIONS
(39)
This fact may be verified by carrying out the multiplication on the left.
In fact, the definition of a matrix product, which may have seemed strange
when it was introduced earlier, was so contrived precisely in order to
permit the writing of a set of linear equations in matrix form.
The expression can be simplified even further by using the matrix
symbols A, x, and y, with obvious definitions, to yield
Ax = y,
(40)
This single matrix equation can represent any set of any number of linear
equations having any number of variables. The great economy of thought
and of expression in the use of matrices should now be evident. The re
maining problem is that of solving this matrix equation, or the corre
sponding set of scalar equations, by which we mean finding a set of
Sec. 1.2]
21
values for the x's that satisfies the equations simultaneously. If a solution
exists, we say the equations are consistent.
Each column (or row) of a matrix is identified by its elements. It can
be thought of as a vector, with the elements playing the role of components
of the vector. Although vectors having more than three dimensions cannot
be visualized geometrically, nevertheless the terminology of space vectors
is useful in the present context and can be extended to
n-dimensional
space. Thus, in (40), x, y, and each column and each row of A are vectors.
If the vector consists of a column of elements, then it is more precisely
called a column vector. Row vector is the complete name for a vector that
is a row of elements. The modifiers " c o l u m n " and " r o w " are used only
if confusion is otherwise likely. Further, when the word " vector " is used
alone, it would most often be interpreted as " column vector."
Now, given a set of vectors, the question arises as to whether there is
some relationship among them or whether they are independent. In
ordinary two-dimensional space, we know that any two vectors are independent of each other, unless they happen to be collinear. Furthermore,
any other vector in the plane can be obtained as some linear combination
of these two, and so three vectors cannot be independent in two-dimensional space.
In the more general case, we will say that a set of m vectors, labeled
x (i = 1 to m), is linearly dependent if a set of constants k can be found
such that
(41)
22
[Ch. 1
FUNDAMENTAL CONCEPTS
rank A = rank [A
This is called the consistency
y].
(42)
condition.
Example
Suppose A is the following matrix of order 3 4:
Sec. 1.2]
23
obtained by deleting the third row and third and fourth columns is nonsingular. Thus rank A = 2. This also tells us that the column vectors
or, since
then
24
FUNDAMENTAL CONCEPTS
G E N E R A L SOLUTION OF y =
[Ch. 1
Ax
(43)
This is done by first determining the rank r b y finding the highest order
submatrix whose determinant is nonzero. The equations are then re
arranged (and the subscripts on the x's and y's modified), so that the first
r rows and columns have a nonzero determinant; that is, A is nonsingular. The equation can now be rewritten as
11
(44a)
(44b)
The second of these is simply disregarded, because each of the equations
in (44b) is a linear combination of the equations in (44a). You can show
that this is a result of assuming that the consistency condition is satisfied.
In (44a), the second term is transposed to the right and the equation is
multiplied through by A
which exists since A is nonsingular. The
result will be
-1
11
11
(45)
This constitutes the solution. The vector x contains r of the elements of
the original vector x; they are here expressed in terms of the elements of
y and the remaining m r elements of x.
Observe that the solution (45) is not unique if n > r. In fact, there are
exactly q = n r variables, the elements of x , which may be selected
arbitrarily. This number q is an attribute of the matrix A and is called the
nullity, or degeneracy, of A.
1
Sec. 1.2]
25
For the special case of homogeneous equations, namely the case where
y = 0 , it should be observed from (45) that a nontrivial solution exists
only if the nullity is nonzero. For the further special case of m = n (i.e.,
when A is a square matrix), the nullity is nonzero and a nontrivial solu
tion exists only if A is singular.
To illustrate the preceding, consider the following set of equations:
W e observe that the first four rows and columns 2, 4, 6, and 8 of A form
a unit matrix (which is nonsingular) and so rank A > 4. In addition, the
fifth row is equal to the negative of the sum of the first four rows. Thus
the rows of A are not linearly independent and rank A < 5. Since 4
rank A < 5, we have established that rank A = 4. For precisely the same
reasons it is found that rank [A
y] = 4. Thus the consistency condi
tion is satisfied. Now the columns can be rearranged and the matrices
partitioned as follows:
26
FUNDAMENTAL CONCEPTS
[Ch. 1
This has been partitioned in the form of (43) with A = U, a unit matrix.
The bottom row of the partitioning is discarded, and the remainder is
rewritten. Thus
11
For each set of values for x , x , x , and x there wili be a set of values for
x , x , x , and x . In a physical problem the former set of variables may
not be arbitrary (though they are, as far as the mathematics is concerned);
they must often be chosen to satisfy other conditions of the problem.
1
CHARACTERISTIC EQUATION
(46)
(47)
(48)
Sec. 1.2]
27
The result is
from which
28
[Ch. 1
FUNDAMENTAL CONCEPTS
SIMILARITY
Two square matrices A and B of the same order are said to be similar
if a nonsingular matrix S exists such that
- 1
S A S = B.
(49)
-1
- 1
- 1
-1
|U - B| = |U - S A S | = | S ( U - A)S|
- 1
- 1
= | S | |U - A| |S| = | S S | |U - A|
=
|U-A|.
Since the eigenvalues of a matrix are the zeros of its characteristic poly
nomial, and since A and B have the same characteristic polynomials, their
eigenvalues must be equal.
An important, special similarity relation is the similarity of A to a
diagonal matrix
Now, if A and are similar, then the diagonal elements of are the
eigenvalues of A. This follows from the fact that A and have the same
Sec. 1.2]
29
S = AS.
(50)
N o w partition S by columns; that is, set S = [Si S ... S ] , where the S are
the column vectors of S. Equating the jth column of AS to the jth column
of S, in accordance with (50), we get
2
S = AS .
j
(51)
- 1
- 1
Example
As an illustration take the previously considered matrix
Earlier we found that i = 3 and = 4 are the eigenvalues and that, for
arbitrary, nonzero sn and s ,
2
12
and
12
= 1; then
30
FUNDAMENTAL CONCEPTS
and
therefore
[Ch. 1
Then, of course,
distinct.
2. A is either symmetric
Hermitian.*
SYLVESTER'S
or
INEQUALITY
PQ
PQ
r + r nr min
P
PQ
{r ,
P
r }.
(52)
Note that n is the number of columns of the first matrix in the product or
the number of rows of the second one.
As a special case, suppose P and Q are nonsingular square matrices of
order n. Then r = r = n, and, by Sylvester's inequality,
nr nor
r
= n. This we also know to be true by the fact that | Q | = | P | | Q | 0,
since | P | 0 and | Q | 0. As another special case, suppose P Q = 0 . Then
r
is obviously zero, and, by Sylvester's inequality, r + r < n.
P
PQ
pQ
PQ
Sec. 1.2]
31
NORM OF A VECTOR
One of the properties of a space vector is its length. For an n-vector the
notion of length no longer has a geometrical interpretation. Nevertheless,
it is a useful concept, which we shall now discuss.
Define the norm of an n-vector x as a non negative number ||x|| that
possesses the following properties:
1. ||x|| = 0 if and only if x = 0 .
2. ||x|| = || ||x||, where is a real or complex number.
3.
+ x ||
||xi
||xi||
xi
and
are two
n-vectors.
(53)
This is the square root of the sum of the squares of the components of the
vector. The Euclidean norm is the one we are most likely to think about
when reference is made to the length of a vector; however, there are
other norms that are easier to work with in numerical calculations. One
such norm is
(54)
that is, the sum of the magnitudes of the vector components. For want of
a better name, we shall call it the sum-magnitude norm. Another such
norm is
(55)
that is, the magnitude of the component having the largest magnitude.
We shall call it the max-magnitude norm. It is a simple matter to show that
||x ||, ||x|| , and ||x|| each satisfy the stated properties of a norm.
That each of these norms is a satisfactory measure of vector length can
be established by several observations. If any one of these norms is non
zero, the other two are nonzero. If any one of them tends toward zero as
a limit, the other two must do likewise.
A matrix is often thought of as a transformation. If A is a matrix of
order m n and x is an n-vector, then we think of A as a matrix that
2
32
FUNDAMENTAL CONCEPTS
[Ch. 1
for all x. The greatest lower bound of all such K is called t h e norm of A
and is denoted by ||A||. It is easy to show that t h e matrix norm has t h e
usual properties of a norm; that is,
1. ||A|| = 0 if and only if A = 0;
2. ||A|| = ||||A||, where is a real or complex number; and
3. | | A + A | | < | | A | | + ||A ||.
1
(57)
(58)
The first step and t h e last step follow from t h e definition of t h e summagnitude norm. The second step is a result of the triangle inequality
for complex numbers. Suppose the sum of magnitudes of a is t h e largest
ij
1 7 1
max
y
J
=l
1 7 1
|a | =
ij
i=l
(59)
Sec. 1.2]
33
(60)
The pattern of steps here is the same as in the preceding norm except that
the max-magnitude norm is used. Again, suppose the sum of magnitudes
m
of a is largest for the kth row; that is, suppose ^ ] j = 1 l 'l =i \ w\.
Then (60) is satisfied as an equality when x = sgn (aA; ). (The function
s g n y equals + 1 when y is positive and 1 when y is negative.)
Therefore
ij
(61)
Thus the max-magnitude norm of a matrix A is the sum-magnitude
norm of that row vector of A which has the largest sum-magnitude norm.
Finally, for the Euclidean norm, although we shall not prove it here,
it can be shown* that
(62)
* Tools for showing this will be provided in Chapter 7.
||A|| = | | / 2 .
2
Example
As an illustration, suppose y = Ax, or
(63)
34
FUNDAMENTAL CONCEPTS
[Ch. 1
Hence
= 26.64 and
We also know, b y substituting the above matrix norms into (57), that
and
1.3
Sec. 1.3]
35
Fig. 1.
indicates the reference for t h e current in a branch. This arrow does not
mean that t h e current is always in the arrow direction. It means that,
whenever the current is in the arrow direction, i(t) will be positive.
Similarly, the plus and minus signs at t h e ends of a branch are the
voltage reference for the branch. Whenever the voltage polarity is actu
ally in the sense indicated by the reference, v(t) will be positive. Actually,
the symbol for voltage reference has some redundancy, since showing
only the plus sign will imply the minus sign also. Whenever there is no
possibility of confusion, the minus sign can be omitted from the reference.
For a given branch the direction chosen as the current reference and the
polarity chosen as the voltage reference are arbitrary. Either of the two
possibilities can be chosen as the current reference, and either of the
two possibilities can be chosen as the voltage reference. Furthermore, t h e
reference for current is independent of the reference for voltage. However,
36
FUNDAMENTAL
CONCEPTS
[Ch. 1
Fig. 2 .
Standard references.
side the branch, if the voltage-reference plus is at the tail of the current
reference, the result is called the standard reference. If it is stated that the
standard reference is being used, then only one of the two need be shown;
the other will be implied. It must be emphasized that there is no requirement for choosing standard references, only convenience.
1.4
NETWORK
CLASSIFICATION
Let the excitation applied to a network that has no initial energy storage
be labeled e(t) and the response resulting therefrom w(t). A linear network
is one in which the response is proportional to the excitation and the
principle of superposition applies. More precisely, if the response to an
excitation ei(t) is wi(t) and the response to an excitation e (t) is w (t),
then the network is linear if the response to the excitation k e (t) +
k e (t) S kiwi(t) +
k w (t).
This scalar definition can be extended to matrix form for multiple
2
Sec. 1.4]
NETWORK CLASSIFICATION
37
and
and
where e ,e ,
etc., are excitations at positions a, b, etc.; and
w ,w ,
etc., are the corresponding responses. Then a network is linear if the
excitation vector kie (t) + k e (t) gives rise to a response vector k w (t)
+
k w (t),
where w is the response vector to the excitation vector e .
a
TIME
INVARIANCE
PASSIVITY
(64)
or
This must be true for any voltage and its resulting current for all t.
38
FUNDAMENTAL CONCEPTS
[Ch. 1
Any network that does not satisfy this condition is called an active
c
network; that is,
v(x) i(x) dx < 0 for some time t.
l
^ -
00
If the network has more than one pair of terminals through which
energy can be supplied from the outside, let the terminal voltage and
current matrices be
(65)
The network will be passive if, for all t,
(66)
RECIPROCITY
Some networks have the property that the response produced at one
point of the network by an excitation at another point is invariant if the
positions of excitation and response are interchanged (excitation and
response being properly interpreted). Specifically, in Fig. 3a the network
Network
Network
(a)
(b)
Fig. 3.
Reciprocity condition.
Sec. 1.4]
NETWORK COMPONENTS
39
1.5
N E T W O R K COMPONENTS
40
FUNDAMENTAL CONCEPTS
[Ch. 1
Parameter
Resistor
Resistance R
Conductance G
Inductor
Inductance L
Inverse Inductance
Capacitor
Capacitance C
Elastance D
Direct
Inverse
Symbol
(68)
(69)
Sec. 1.5]
NETWORK COMPONENTS
41
initial values i(0) or v(0), and in this sense is misleading. Normally one
thinks of the voltage v(t) and the current i(t) as being expressed as
explicit functions such as , sin t, etc., and the antiderivative as being
something unique: ( l / ) , (1/) cos t, etc., which is certainly not
true in general. Also, in many cases the voltage or current may not be
expressible in such a simple fashion for all t; the analytic expression for
v(t) or i(t) may depend on the particular interval of the axis on which the
point t falls. Some such wave shapes are shown in Fig. 4.
- t
- t
(a)
(b)
Fig. 4.
Signal waveshapes.
v(0).
Initially
relaxed
Fig. 5.
v x
()
dx + i(0),
Initially
relaxed
where i(0) is the initial value of the current. This can be considered as a
d-c current source in parallel with an initially relaxed inductor, as shown
in Fig. 5. If these sources are shown explicitly, they will account for all
42
FUNDAMENTAL CONCEPTS
[Ch. 1
Ideal
Ideal
(a)
(b)
Fig. 6.
An ideal transformer.
(71)
Sec. 1.5]
NETWORK COMPONENTS
43
Fig. 7 .
A transformer.
The diagram is almost the same except that the diagram of the ideal
transformer shows the turns ratio directly on it. The transformer is
characterized by the following v-i relationships for the references shown
in Fig. 7:
(73a)
and
(73b)
Thus it is characterized by three parameters: the two self-inductances L\
and L , and the mutual inductance M.
The total energy delivered to the transformer from external sources is
2
(74)
44
FUNDAMENTAL CONCEPTS
[Ch. 1
(75)
Since physical considerations require the transformer to be passive, this
condition must apply. The quantity k is called the coefficient of coupling.
Its maximum value is unity.
A transformer for which the coupling coefficient takes on its maximum
value k = 1 is called a perfect, or perfectly coupled, transformer. A perfect
transformer is not the same thing as an ideal transformer. To find the
difference, turn to the transformer equations (73) and insert the perfecttransformer condition M=L L
; then take the ratio v /v . The result
will be
1
(76)
(77)
N e x t let us consider the current ratio. Since (73) involve the derivatives
of the currents, it will be necessary to integrate. The result of inserting
the perfect-transformer condition M = L1L
and the value n =
L 1 / L , and integrating (73a) from 0 to t will yield, after rearranging,
2
(78)
* A simple approach is to observe (with Li, L , and M all non-negative) that the
only way L\ii + 2 M i i + L i can become negative is for ii and i to be of opposite
sign. So set i = xiu with x any real positive number, and the quantity of interest
becomes Li 2Mx - f L * . If the minimum value of this quadratic in x is non-negative,
then the quantity will be non-negative for any value of x. Differentiate the quadratic
with respect to x and find the minimum value; it will be Li M / L , from which the
result follows.
Since, for actual coils of wire, the inductance is approximately proportional
to the square of the number of turns in the coil, the expression V L1/L2 equals the ratio of
the turns in the primary and secondary of a physical transformer. This is the origin of
the name "turns ratio" for n.
2
Sec. 1.5]
NETWORK COMPONENTS
45
Perfect transformer
Fig. 8.
THE GYRATOR
(b)
(a)
Fig. 9.
A gyrator.
For Fig. 9b
(79b)
46
FUNDAMENTAL CONCEPTS
[Ch. 1
(80)
(81)
2
Thus the equivalent resistance at the input terminals equals r times the
conductance terminating the output terminals. The gyrator thus has the
property of inverting.
The inverting property brings about more unusual results when the
NETWORK COMPONENTS
Sec. 1.5]
47
(82)
I N D E P E N D E N T SOURCES
All the devices introduced so far have been passive. Other network
components are needed to account for the ability to generate voltage,
current, or power.
Two types of sources are defined as follows:
1. A voltage source is a two-terminal device whose voltage at any instant
of time is independent of the current through its terminals. No matter
what network may be connected at the terminals of a voltage source, its
voltage will maintain its magnitude and waveform. (It makes no sense to
short-circuit the terminals of a voltage source, because this imposes two
idealized conflicting requirements at the terminals.) The current in the
source, on the other hand, will be determined by this network. The
diagram is shown in Fig. 12a.
2. A current source is a two-terminal device whose current at any instant
of time is independent of the voltage across its terminals. No matter what
network may be connected at the terminals of a current source, the cur
rent will maintain its magnitude and waveform. (It makes no sense to
48
FUNDAMENTAL CONCEPTS
[Ch. 1
Everyone is familiar enough with the dimming of the house lights when
a large electrical appliance is switched on the line to know that the voltage
of a physical source varies under load. Also, in an actual physical source
the current or voltage generated may depend on some nonelectrical
quantity, such as the speed of a rotating machine, or the concentration
of acid in a battery, or the intensity of light incident on a photoelectric
cell. These relationships are of no interest to us in network analysis,
since we are not concerned with the internal operation of sources, but
only with their terminal behavior. Thus our idealized sources take no cog
nizance of the dependence of voltage or current on nonelectrical quan
tities; they are called independent sources.
CONTROLLED OR D E P E N D E N T SOURCES
The independent sources just introduced cannot account for our ability
to amplify signals. Another class of devices is now introduced; these are
called controlled, or dependent, sources. A controlled voltage source is a
source whose terminal voltage is a function of some other voltage or
current. A controlled current source is defined analogously. The four possi
bilities are shown in Table 2. These devices have two pairs of terminals
one pair designating the controlling quantity; the other, the controlled
quantity. In each of the components in Table 2, the controlled voltage or
current is directly proportional to the controlling quantity, voltage or
current. This is the simplest type of dependence; it would be possible to
introduce a dependent source whose voltage or current is proportional to
the derivative of some other voltage of current, for example. However,
detailed consideration will not be given to any other type of dependence.
Sec. 1.5]
NETWORK COMPONENTS
49
Device
Voltage-controlled
voltage source
(hybrid g)
Current-controlled
voltage source
(impedance)
Voltage-controlled
current source
(admittance)
Current-controlled
current source
(hybrid h)
of interconnected dependent sources and other network elements. Figure
13 shows two such models. These models are not valid representations
of the physical devices under all conditions of operation; for example, at
high enough frequency, the interelectrode capacitances of the tube would
have to be included in the model.
50
FUNDAMENTAL CONCEPTS
[Ch. 1
The last component we shall introduce is the negative converter (NC for
short). It is a device with two pairs of terminals and is defined by the
following v-i equations:
(83a)
or
(83b)
PROBLEMS
51
(a)
(b)
Fig. 14. Negative converters: (a) current-inverting variety: vi = kv and i = kii;
(6) voltage-inverting variety: vi = kv and i = kii.
2
diz/dt.
(84)
PROBLEMS
1 . Is (A + B ) = A + 2AB + B in matrix algebra? if not, give the correct
formula.
2
2.
Let
52
4.
FUNDAMENTAL CONCEPTS
[Ch. 1
Let
Compute AB. What theorem of ordinary algebra is not true for matrices?
5 . Let A and B be conformable and let the submatrices Ay and B be
conformable for all i and k. Verify the statement that the (i, k) submatrix of the product AB is Ay R .
j k
jk
6 . Show that
7.
Prove that
(AB) =
A B
and
(AB)* =
B*A*.
8 . Verify the statement that any square matrix A can be expressed as the
sum of a Hermitian matrix A and a skew-Hermitian matrix ASH. Find
A
and A .
H
( A )
PROBLEMS
53
16. Prove that adj (AB) = (adj B) (adj A) when A and B are nonsingular
square matrices.
17. Prove that
18. Prove that
19. Prove that
20. Show that
21. If A and B are nonsquare matrices, is it true that AB can never equal
BA? Explain.
22. A is of order n and rank n 1. Prove that adj A is of rank 1.
23. Let D be a diagonal matrix with diagonal elements da, and let A = [y]
be a square matrix of the same order. Show that (a) When D
premultiplies A, the elements of the ith row of A are multiplied by
da; and (b) When D postmultiplies A, the elements of the ith column
of A are multiplied by dii.
24. Prove that (a) (AB)' = B'A' and (b) (A + B)' = A' + B'.
25. Let A and B be symmetric and of order n. Prove that (a) the product AB
is symmetric if A and B commute, and (b) A and B commute if the
product AB is symmetric.
26. In the matrix product A = BC, A and C are square nonsingular matrices.
Prove that B is nonsingular.
27. Use pivotal condensation to evaluate the determinant of the following
matrices:
54
FUNDAMENTAL CONCEPTS
[Ch. 1
28. For a set of homogeneous equations the solution as given in (45) will
become xi = A{^ Ai x . If m = n and the matrix A is of rank
r = n 1, determine an expression for each of the x% variables in xi in
terms of cofactors of A.
29. Prove the statement: A determinant is zero if and only if the rows and
columns are linearly dependent.
30. For the following set of equations verify (42).
33. Show the maximum number of linearly independent n-vectors from the
set of all n-vectors x that satisfy 0 = Ax is equal to the nullity of A.
34. If qp, qQ, and qpQ denote the nullity of P, Q, and PQ, respectively,
show that
qQ < qpQ qp
+ qQ-
35. Show that the determinant of a triangular matrix equals the product of
the main diagonal elements.
PROBLEMS
55
36. Let
where A n and A
2 2
(b)
(a)
(c)
(d)
(a)
(b)
(c)
(a)
(b)
(c)
40. A network has the excitation and response pair shown in Fig. P-40. A
second excitation is also shown. If the network is linear and timeinvariant, sketch the response for this second excitation.
56
[Ch. 1
FUNDAMENTAL CONCEPTS
Fig. P-40
(a)
(b)
Fig. P-41
42. Show that the controlled-voltage source shown in the circuit of Fig. P-42
is not a passive device. Comment on the activity or passivity of both
independent and dependent sources.
Fig. P-42
PROBLEMS
44. Establish the terminal equations for the networks shown in Fig. P-44.
(b)
(a)
Fig. P - 4 4
57
.2.
GRAPH THEORY
AND NETWORK EQUATIONS
2.1
INTRODUCTORY CONCEPTS
When two or more of the components defined in the previous chapter are
interconnected, the result is an electric network. (A more abstract definition
is given in Section 2.3.) Such networks store energy, dissipate energy, and
transmit signals from one point to another. A component part of a
network lying between two terminals to which connections can be made
is called a branch. The position where two or more branches are connected
together is called a node, or junction. A simple closed path in a network is
called a loop.
In the first section of this chapter, we shall briefly discuss a number of
ideas with which y o u are no doubt familiar to a greater or lesser degree.
Most of these will be amplified subsequently, but an early introduction in
rather a simple form will serve to focus the discussion of these concepts
before a fuller treatment is given.
KIRCHHOFF'S LAWS
Sec. 2.1]
INTRODUCTORY CONCEPTS
59
sum of all currents leaving any node equals zero at any instant of time. When
this law is applied at a node in a network, an equation relating the branch
currents will result. Proper attention must, of course, be given to the cur
rent references. Thus, in Fig. 1, KCL applied at node A leads to the
following equation:
(1)
Fig. 1.
Since the reference of ii is oriented toward the node, the current " l e a v i n g "
the node through branch 1 is i1; similarly, the current leaving the node
through branch 3 is i .
Kirchhoff's voltage law (KVL) states that in any electric network, the
sum of voltages of all branches forming any loop equals zero at any instant of
time. Application of this law to a loop in an electric network leads to an
equation relating the branch voltages on the loop. In stating KCL, it was
arbitrarily chosen that currents "leaving the n o d e " were to be summed.
It could, alternatively, have been decided to sum the currents "entering
the node." Similarly, in applying K V L , the summing of voltages can be
performed in either of the two ways one can traverse a loop. Thus, going
clockwise in the loop formed by branches 1, 2, 5, and 6 in Fig. 1 leads to
the equation
3
(2)
60
[Ch. 2
many KCL equations as there are nodes in the network and as many K V L
equations as there are loops. Later we shall prove that these equations are
not all independent; if the number of nodes is n + 1 and the number of
branches is b, then we shall prove that the number of independent KCL
equations is n and the number of independent K V L equations is b n.
We shall not dwell on these matters now, but simply observe them and
note that together there are n + (b n) = b independent KCL and K V L
equations.
Now each branch in the network also contributes a relationship between
its voltage and current. This may be an algebraic relationship like v = Ri
or a dynamic relationship like v = Ldi/dt. In any case there will be as
many such equations as branches, or b equations. Altogether there will
be b + b = 2b equations relating b currents and b voltages, or 2b variables.
(As discussed in a later section, independent sources will not be counted
as branches in this context.) Hence the three sets of relations together
namely, KCL, KVL, and the branch v-i relationshipsprovide an ade
quate set of equations to permit a solution for all voltages and currents.
Now 2b is a relatively large number, and a prudent person will try to
avoid having to solve that many equations simultaneously. There are a
number of systematic ways of combining the three basic sets of equations,
each leading to a different formulation requiring the solution of less than
2b simultaneous equations. In this introductory section, we shall briefly
discuss three such procedures and illustrate each one. The bridge in Fig. 2
Fig. 2 .
Sec. 2.1]
INTRODUCTORY CONCEPTS
61
LOOP EQUATIONS
In this network there are six branches (b = 6) and four nodes (n = 3).
Hence there will be b n = 3 independent K V L equations and n = 3
independent KCL equations. In Fig. 2, the circular arrows indicate the
orientations of the loops we have chosen for writing KVL equations.
They do not carry any implication of current (as yet). But suppose we
conceive of fictitious circulating loop currents, with references given by
the loop orientations. Examination of the figure shows that these loop
currents are identical with the branch currents ii, i , and i .
The following equations result if KVL is applied to the loops shown:
2
(3)
These are three equations in six unknowns, and they are observed to be
independent. Into these equations are next substituted the v-i relation
ships of the branches, leading to
(4)
These are still three equations in six unknowns, this time the branch
current unknowns.
There remains KCL. Application of KCL at the nodes labeled A, C,
and D results in
(5)
62
[Ch. 2
in terms of the three currents i1, i , and i , which are identical with the
loop currents. When these equations are substituted into (4) and terms
are collected, the result becomes
2
Suppose now that the order in which the steps were taken is modified.
Suppose we first write KCL equations as in (5); then we insert the i-v rela
tionships. The result (when the terms are written on the same side) becomes
(6)
Sec. 2.1]
INTRODUCTORY CONCEPTS
63
(7)
When these expressions are inserted into (5), and the terms are collected,
the result becomes
(8)
These equations are called the node equations. Like the loop equations,
they are integrodifferential equations. Once these equations are solved
for the node voltages VAB, VCB, and VDB, all the branch voltages will be
known from (7).
To review, the first step in writing node equations is to write KCL
equations at all nodes of a network but one. This particular node is
chosen as a datum, and node voltages are defined as the voltages of the
other nodes relative to this datum. The i-v relationships are inserted into
the KCL equations, changing the variables to branch voltages. The
branch voltages are then expressed in terms of the node voltages. Thus
the order in which KCL, K V L , and the v-i relationships are used for
writing node equations is the reverse of the order for writing loop equa
tions.
STATE E Q U A T I O N S A M I X E D SET
64
[Ch. 2
appear. But this will increase the order of the equations. It would be
better to avoid the appearance of the integrals in the first place.
In the present example it may be seen that an integral appears in the
loop equations when the voltage of a capacitor is eliminated in a KVL
equation by substituting its v-i relationship. Similarly, an integral appears
in the node equations when the current of an inductor is eliminated in a
KCL equation by substituting its i-v relationship. These integrals will
not arise if we leave the capacitor voltages and inductor currents as
variables in the equations.
With this objective in mind, return to the K V L equations in (3).
Eliminate all branch voltages except the capacitor voltage v by using
the branch v-i relations. Since the branch relationship of the capacitor is
not used in this process, add it as another equation to the set. The result
will be
5
(9)
These are four equations in six unknowns, but now one of the unknowns
is a voltage. As before, the KCL equations can be used to eliminate some
of the currents. Substituting them from (5) into (9) and rearranging terms
yields
Sec. 2.1]
INTRODUCTORY CONCEPTS
65
Here we have four equations in four unknowns, and they can now be
solved. But we seem to have complicated matters by increasing the
number of equations that must be solved simultaneously. However, note
that the last two equations in this set are algebraic; they contain no
derivatives or integrals. The first of them can be solved for i , the second
for i , and the results inserted into the previous two equations. The
result of this manipulation will be
1
(10)
where
66
[Ch. 2
(11)
SOLUTIONS OF EQUATIONS
Sec. 2.1]
INTRODUCTORY CONCEPTS
67
(12)
where I and V are the initial values. These can now be solved for
or V (s). For I (s) we get
0
I (s)
2
(13)
and
(14)
* For a detailed discussion of initial conditions, see S. Seshu and N. Balabanian,
Linear Network Analysis, John Wiley & Sons, Inc., New York, 1959, pp. 101-112.
68
[Ch. 2
(15)
Then
(16)
The partial fraction expansion of I (s) puts into evidence all of its
poles. Some of these poles [the second term in (15)] are contributed by
the exciting (source) function, whereas the remainder are contributed by
the network. In the inverse transform we find terms that resemble the
driving function and also other terms that are exponentials. There is an
abundance of terminology relating to these terms that has been accumu
lated from the study of differential equations in mathematics, from the
study of vibrations in mechanics, and from the study of a-c circuit theory,
so that today we have a number of names to choose from. These are the
following:
2
Sec. 2.1]
LINEAR GRAPHS
69
2.2
LINEAR
GRAPHS
70
[Ch. 2
(b)
(a)
Fig. 3 .
what follows, we will use this graph to make some observations from
which generalizations can be drawn. Also, properties to be defined will be
illustrated with this graph as an example.
Branches whose ends fall on a node are said to be incident at the node.
In the example, branches 2, 4, and 5 are incident at node 2.
Each branch of the graph in the example carries an arrow to indicate
its orientation. A graph whose branches are oriented is called an oriented
graph. The elements of a network with which a graph is associated have
both a voltage and a current variable, each with its own reference. In
order to relate the orientation of the branches of a graph to these refer
ences, we shall make the convention that the voltage and current of an
element have the standard referencevoltage-reference " p l u s " at the
tail of the current-reference arrow. The branch orientation of a graph will
be assumed to coincide with the associated current reference. Of course,
Sec. 2.2]
LINEAR GRAPHS
71
72
[Ch. 2
(a)
Fig. 4 .
(b)
Two trees of a given graph.
Sec. 2.2]
LINEAR GRAPHS
73
THE I N C I D E N C E MATRIX
nodes
branches
(17)
74
[Ch.2
links
(18)
LINEAR GRAPHS
Sec. 2.2]
75
(20)
number of trees.
The second line foilows from the fact that a nonsingular submatrix of
A' has the same determinant as the corresponding submatrix of A. Since
each nonzero major equals + 1 , and there are as many nonzero majors as
trees, the last line foilows.
Thus, to find the number of trees of a graph, it is required only to
evaluate det (AA'). For the example of Fig. 3, the incidence matrix was
76
[Ch. 2
Sec. 2.2]
LINEAR GRAPHS
77
Two different people studying network analysis were given this job, and
each came up with a different-looking graph, as shown in Fig. 5. The
apparent problem was that right at the start they placed the nodes on
the paper in a different pattern. However, both graphs have as incidence
matrix the given matrix.
(b)
(a)
Fig. 5.
Isomorphic graphs.
We say two graphs are isomorphic if they have the same incidence
matrix. This requires that they have the same number of nodes and
branches, and that there be a one-to-one correspondence between the
nodes and a one-to-one correspondence between the branches in a
particular waynamely, in a way that leads to the same incidence matrix.
THE LOOP MATRIX
78
Fig. 6.
[Ch. 2
Loop orientations.
(21)
The set of all loops in a graph is quite a large set, as illustrated by this
Sec. 2.2]
LINEAR GRAPHS
79
example. There is a smaller subset of this set of all loops; it has some
interesting properties and will be discussed next.
Given a graph, first select a tree and remove all the links. Then replace
each link in the graph, one at a time. As each link is replaced, it will form
a loop. (If it does not, it must have been a twig.) This loop will be charac
terized b y the fact that all but one of its branches are twigs of the chosen
tree. Loops formed in this way will be called fundamental loops, or f-loops
for short. The orientation of an f-loop will be chosen to coincide with that
of its defining link. There are as many f-loops as there are links; in a graph
having b branches and n + 1 nodes, this number will be b n.
For the example, let the chosen tree be the second one in Fig. 4. The
f-loops formed when the links are replaced one at a time are shown in
Fig. 7. (Note the orientation.) In writing the loop matrix for the f-loops,
(b)
(a)
Fig. 7 .
(c)
Fundamental loops.
let the columns be arranged in the same order as for the reduced incidence
matrix for the same tree; that is, with the twigs first, then the links. Also,
let the order of the loops be the same as the order of the columns of the
corresponding links.
The matrix of f-loops will then be
twigs
links
(22)
The subscript f stands for fundamental. The square matrix formed by the
last three columns corresponding to the links is seen to be a unit matrix;
hence it is nonsingular, and the rank of B / for this example equals the
number of links, b n.
80
[Ch. 2
(23)
The (b n) (b n) square submatrix whose columns correspond to the
links for a particular tree will be a unit matrix from the very way in
which it is formed. Hence the rank of Bf will be b n.
Now the matrix of fundamental loops is a submatrix of the matrix of
all loops. Hence the rank of B is no less than that of Bf; namely, b n.
We shall next show that the rank of B a is no more than b n, and so it
is exactly b n. To do this, we shall use a result that is of great importance
in its own right.
Given a graph, let the columns of the two matrices A a and B a be
arranged in the same order. Then it will be true that
a
(24)
and
(25)
Of course, the second one will be true if the first one is, since B A'a =
(A B )'.
The relationships in (24) and (25) are called the orthogonality
relations and can be proved as follows.
The matrices A and B'a will have the following forms:
a
loops
branches
nodes
Focus attention
rows of Aa; that
is on the loop or
incident at the
branches
Sec. 2.2]
LINEAR GRAPHS
81
element in a column of
there will be a zero element in the row of A ;
so the product will yield zero. If the node is on the loop, then exactly
two of the branches incident at the node will lie on the loop. If these two
branches are similarly oriented relative to the node (either both oriented
away or both toward), they will be oppositely oriented relative to the
loop, and vice versa. In terms of the matrices, if the elements in a row
of A corresponding to two branches are both + 1 or both 1 , the corre
sponding two elements in a column ofB'awill be of opposite sign, and vice
versa. When the product is formed, the result will be zero. The theorem
is thus proved.
With the preceding result it is now possible to determine the rank of
Ra by invoking Sylvester's law of nullity, which was discussed in Chapter
1. According to this law, if the product of two matrices equals zero,
the sum of the ranks of the two matrices is not greater than the number
of columns of the first matrix in the product. In the present case the
number of columns equals the number of branches b of the graph. So,
since the rank of a matrix is the same as the rank of its transpose,
(26)
The rank of A a has already been determined to be n. Hence
(27)
Since it was previously established that the rank of B a is no less than b n
and it is now found that it can be no greater than b n, then the rank of
B is exactly b n.
Observe that the removal of any number of rows from A or any
number of rows from B will not invalidate the result of (26) and (27).
Let B be any submatrix of B having b n rows and of rank b n. (One
possibility is the matrix of f-loops, Bf.) Then the orthogonality relations
can be written as
a
(28)
RELATIONSHIPS B E T W E E N SUBMATRICES OF A A N D B
82
[Ch. 2
(30)
Since A is nonsingular,
t
(31)
Finally matrix B becomes
(32)
N o w let the same procedure be carried out starting at (30), but this
time with the matrix B of the f-loops for some tree, with B partitioned
in the form
f
(33)
the subscript f on B being used to avoid confusion. The details will be
left for you to carry out; the result will be
f t
(34)
B y comparing this with (32), it foilows that
(35)
Since B and B are both of rank b n, then B must be nonsingular. This
foilows from (52) in Chapter 1. The result is thus proved.
The converse is also true; that is, if the loop-matrix B is partitioned
into two matrices as in (29), one of them being square of order b n and
nonsingular, the columns of this matrix will correspond to the links for
some tree. The proof is left for you to carry out. (See Problem 6.)
Since B in (35) is nonsingular, the matrices B and B are row-equivalent
matrices. (For a discussion of equivalent matrices, see Chapter 7.) Hence,
the rows of B are linear combinations of the rows of B , and vice versa.
f
Sec. 2.2]
LINEAR GRAPHS
83
(36)
The last step follows because the transpose of a product equals
of transposes in the reverse order. The preceding step follows
operations of transpose and inverse are commutative for a
matrix.
When this is inserted into the partitioned form of the A
result will be
the product
because the
nonsingular
matrix, the
(37)
This should be compared with (32), which gives the loop matrix in a
similar form.
CUT-SETS A N D THE CUT-SET MATRIX
(a)
(b)
Fig. 8.
(c)
84
[Ch. 2
Fig. 9.
the graph into three parts, of which the isolated node 1 will be one part.
So this set of branches is not a cut-set.
However, the graph shown in Fig. 9 is a peculiar kind of graph, and
node 1 is a peculiar node. We define a hinged graph as a graph in which
there is at least one subgraph which has only one node in common with its
LINEAR GRAPHS
Sec. 2.2]
85
branches
where the first four rows are identical with A a and the last three rows
correspond to the cut-sets {1, 3, 4, 5 } , {2, 3, 5, 6 } , and {1, 2, 4, 6 } , respec
tively.
The cut-set matrix Q a of a graph is seen to have more rows than its inci
dence matrix. The question of the rank of the cut-set matrix then arises. To
answer this question, consider a special kind of cut-set formed as follows.
Given a connected graph, first select a tree and focus on a branch b of
the tree. Removing this branch from the tree unconnects the tree into two
pieces. All the links which go from one part of this unconnected tree to
the other part, together with b , will constitute a cut-set. We call this a
fundamental cut-set, or f-cut-set
for short. For each twig, there will be
k
86
[Ch. 2
links
LINEAR GRAPHS
Sec. 2.2]
87
Q matrix will be of rank n. This still does not tell us everything about
the rank of Qa. But since the matrix of f-cut-sets is just a submatrix of
Qa, the rank of Qa can be no less than that of Q , or rank of Qa > n.
When seeking to find the rank of B a , it became necessary to use the
orthogonality relation A B ^ = 0 . But Qa is a matrix that contains Aa
as a submatrix, and it might be suspected that this same relationship is
satisfied with Aa replaced by Qa.* This is true and can be proved in the
same way as before. It is only necessary to establish that if a cut-set has
any branches in common with a loop, it must have an even number. This
fact can be readily appreciated by reference to Fig. 11, which shows a
f
cut-set separating a graph into two parts. Suppose branch 1 of the cut-set
is in a loop. If we start at its P1 end and traverse the branch to P , it will
be necessary to return to P1 via another branch of the cut-set in order to
form a closed path. The path need not be closed b y only one excursion
between P1 and P , but any single excursion will use up two branches
of the cut-set. If these two branches have the same orientation relative
to the cut-set, they will have the opposite orientation relative to the
loop, and vice versa. Hence, b y the same reasoning used in obtaining (24),
it follows that
2
(39)
The rank of Qa can now be determined. Using Sylvester's law of nullity
and the known rank of B, it will follow that the rank of Qa is no greater
than n. (You may carry out the details.) And since Q is a submatrix of
Qa having a rank of n, the rank of Qa is no less than n. Hence the rank of
Qa is exactly n.
Removal of any number of rows from Qa will not invalidate (39). Let
Q be any n-rowed submatrix of Qa of rank n. (One possibility is the matrix
of f-cut-sets,
Q .) Then
f
(40)
* This statement applies only for a nonhinged graph.
88
[Ch. 2
or
(41)
and, finally,
(42)
Something very interesting follows from this expression. Comparing it
with (37) shows that
(43)
Since A is a nonsingular matrix, the incidence matrix of a graph is
row equivalent to the fundamental cut-set matrix for some tree. Thus
the rows of A are linear combinations of the rows of Q , and vice versa.
t
P L A N A R GRAPHS
Sec. 2.2]
LINEAR GRAPHS
89
branches lie and the infinite region. The infinite region can be looked upon
as the "interior" of this set of branches. It is the complement of the
finite region. Hence this set of branches can also be considered a mesh
and is called the outside mesh. In Fig. 12a the outside mesh is {1, 2, 6,
7, 8, 5 } . However, when the meshes of a graph are enumerated, the outside
mesh is not counted.
(a)
Fig. 12.
(b)
Planar (a) and nonplanar (b) graphs.
(a)
Fig. 13.
(b)
Meshes may or may not be f-loops for some tree.
very similar planar graphs having the same number of nodes, branches,
and meshes. The branches common between meshes are shown darkened.
These must be twigs for a tree if the meshes are to be f-loops. However,
90
[Ch. 2
in the first graph, these branches form a loop, and so the desired result
is not possible; whereas it is possible for the second graph.
This completes the discussion of linear graphs as such.
2.3
(45)
where A is the incidence matrix, i(t) is a column matrix of branch currents,
and I(s) is the column matrix of Laplace transforms of the branch currents.
(46)
Sec. 2.3]
91
(Of course, it is also true that A a i(t) = 0 if all the nodes are included.)
Since the rank of A is n, all of the equations in this set are linearly
independent.
Let the incidence matrix be partitioned in the form A = [A A ] for
some choice of tree, and let the matrix i be similarly partitioned as
t
(47)
or
(48)
since A is a nonsingular matrix.
The message carried by this expression is that, given a tree, there is a
linear relationship by which the twig currents are determined from the
link currents. This means that, if the link currents can be determined
by some other means, the twig currents become known from (48). Of all
the b branch currents, only b n of them need be determined indepen
dently.
Using (48), the matrix of all currents can now be written as
t
(49)
Comparing the matrix on the right with the one in (34) and also using
(35), there follows that
(50)
(50b)
Each of these equations expresses all the branch currents of a network in
terms of the link currents for some tree by means of a transformation
92
[Ch. 2
that is called a loop transformation. The link currents for a tree are seen
to be a basis for the set of all currents. We shall shortly discuss sets of
basis currents other than the link currents for a tree.
Other sets of equations equivalent to KCL in (45) can be obtained.
(Recall that two sets of equations are equivalent if they have the same
solution.) Consider a particular cut-set of a network. It will separate the
network into two parts, P1 and P . Write the KCL equations at all the
nodes in one of the parts, say P i , and consider the columns. If both ends
of a branch are incident at nodes in P1, the corresponding column will
have two nonzero elements, a + 1 and a 1 . But if one end of a branch is
incident at a node in P1 and the other end at a node in P (i.e., if this
branch is in the cut-set), this column will have but a single nonzero
element. N o w suppose these KCL equations are added; only the currents
of the cut-set will have nonzero coefficients in the sum. The result will be
called a cut-set equation. A cut-set equation is, then, a linear combination
of KCL equations. The set of all such cut-set equations will be precisely
Qai(t) = 0 , where Qa is the previously defined cut-set matrix for all
cut-sets. But the rank of Qa is n, which is less than the number of equa
tions. These equations are thus not independent. Let Q be a cut-set
matrix of n cut-sets and of rank n. (One possibility is the matrix off-cut
sets, Q . ) Then
2
(51)
or
(52)
which is the same as (48) in view of (43). This expression can be inserted
into the partitioned form of the matrix of all currents as in the case of
(48) to yield
(53)
Sec. 2.3]
93
(The final step comes from the results of Problem 17.) This is again a
loop transformation identical with (50a). Note that the matrix of the
transformation is the transpose of the matrix of fundamental loops.
We have seen that the link currents for a given tree are basis currents
in terms of which all currents in a network can be expressed. Another set
of basis currents are the loop currents, which are fictitious circulating
currents on the contours of closed loops. This can best be illustrated by
means of an example. Figure 14 is a redrawing of Fig. 13a, with the
branches and nodes appropriately numbered. This graph is planar, but
it is not possible to find a tree for which the meshes are f-loops, as discussed
earlier. Let a loop matrix be written for the loops specified in the figure.
(This set of loops is neither a set of f-loops nor a set of meshes.) The
orientation of the loops is given by the ordering of the nodes; it is also
shown by means of the arrows on the diagram. The B matrix will be
branches
loops
(54)
(This matrix is of rank 4 since the submatrix consisting of the last four
columns is nonsingular.) Now suppose a set of circulating currents, i ,
i , etc., is defined on the contours of the same loops for which the B
matrix is written and having the same orientation. B y inspection of the
m1
m 2
94
[Ch. 2
Fig. 14.
Illustrative example.
(55)
(56)
Sec. 2.3]
95
where b has the same definition as the elements of the loop matrix. In
matrix form, K V L becomes
jk
(58)
(59)
If all the loops in the network are included, the coefficient matrix will be
B . However, the rank of B a is b n, and the equations in this set will
not be independent.
Let B have b n rows and be of rank b n. (One possibility is the
matrix of f-loops.) It can be partitioned in the form B = [B B ] for
some choice of tree. Let v also be partitioned conformally as
a
96
[Ch. 2
from which
(60)
since B is a nonsingular matrix.
The message carried by this expression is that, given a tree of a graph,
there is a linear relationship b y which the link voltages are determined
from the twig voltages. If the twig voltages can be determined by some
other means, the link voltages become known by (60). Of all the b branch
voltages, only n of them, those of the twigs, need be determined inde
pendently.
Now let (60) be inserted into the partitioned form of the branch
voltage matrix v(t). Then
l
(61)
Sec. 2.3]
97
(64)
This follows from the fact that each column of the A matrix pertains to a
specific branch. The nonzero elements in a column specify the nodes on
which that branch is incident, the sign indicating its orientation. Hence
each column of A will specify the corresponding branch voltage in terms
of the node voltages.
The following example will illustrate this result. Figure 15 is the same
Fig. 15.
98
[Ch. 2
and
(65)
n2
n3
n4
(66)
Sec. 2.3]
99
be the same as that in (65) but with the last row replaced b y [1
1
1 1 0 0 0 0]. Whatever tree is now chosen, A will not be a
unit matrix, and (63) will not apply. However (64), in terms of the new
node voltages, will still apply. You are urged to verify this.
t
THE B R A N C H RELATIONS
(a)
(b)
Fig. 16. Illustrating the v-shift and i-shift.
100
[Ch. 2
Fig. 17.
A general branch.
and a current source in parallel with this combination. (In Fig. 16b, for
example, the left-hand current source can be considered to be in parallel
Sec. 2.3]
101
g k
gk
(67)
(68)
(69)
where i and are column matrices of source currents and voltages.
Similarly, the transformations from branch variables to loop currents
or node voltages must be replaced by the following:
g
(70)
(71)
(72)
Since sources can be handled in this way independently of the passive
parts of a branch, we shall henceforth concentrate on the passive compo
nents. When doing so, the sources will be made to vanish, which is done
by replacing the voltage sources by short circuits and the current sources
by open circuits.
N o w we can turn to a consideration of the relationships between
voltage and current of the branches of a graph. At the outset, we shall
make no special conventions regarding the manner in which branches are
selected and numbered. We shall deal with the Laplace-transformed
variables and assume that initial conditions have been represented as
equivalent sources. The impedance and admittance of branch k will be
represented b y lower case z and y , respectively, whereas the corres
ponding matrices will be Z and Y .
The branch relationships can be written as follows:
k
Branch k
Network
(73)
102
[Ch. 2
(b)
(a)
Fig. 18.
Illustrative example.
The nature of these branch matrices can be illustrated with the example
shown in Fig. 18. The two inductive branches 1 and 2 are mutually
coupled. N o w (73) with the branch-impedance matrix Z shown in detail
can be written as follows:
(74)
Sec. 2.3]
103
Note that, because of the way the branches were numberedwith each
element a separate branch and with inductance first, then resistance,
then capacitancethe matrix can be partitioned as shown, with obvious
meanings for the submatrices. The resistance and inverse-capacitance
matrices are diagonal because there is no coupling from one branch to the
other. This is not the case for the h matrix because of the inductive
coupling.
It is clear that these properties of the submatrices of the impedance
matrix are quite general if the above numbering scheme for branches is
used. Such a numbering scheme is useful, so we shall adopt it when
convenient for the purpose of determining properties of the corresponding
matrices. There will be times, however, when we will want greater
flexibility and will number the branches differently.
In the general case, then, if each element is counted as a separate
branch and if the inductive branches are numbered first, then the resistive
branches, and then the capacitive branches, the branch impedance and
admittance matrices can be written as follows:
p
(75)
- 1
-1
(76)
for the branch resistance matrix and similarly for the others. If this is
104
[Ch. 2
(77a)
and
(77b)
2.4
Sec. 2.4]
105
for a tree or the loop currents. Similarly, the branch voltages could all be
determined in terms of a smaller subset of voltages. We shall now consider
a number of procedures for utilizing these results to carry out an analysis
of a network problem. The outcome depends on the order in which the
three basic relationships are used.
LOOP EQUATIONS
or
(80b)
where E is shorthand for B {V Z I } and
g
(81)
This matrix equation represents a set of b n equations, called the
loop equations, in the b n loop-current variables. The coefficient matrix
Z (s) is called the loop-impedance matrix, not to be confused with the
branch-impedance
matrix Z. For a passive reciprocal network, Z is a
symmetric matrix. Hence (see Problem 1.15) Z is also symmetric.
The loop-impedance matrix can be written explicitly in terms of the
m
106
[Ch.2
(82)
where
(83a)
(83b)
(83c)
are the loop-parameter matrices.
To illustrate (80), consider the network of Fig. 18 for which the branchimpedance matrix was given in (74). Its graph is redrawn in Fig. 19 to
Fig. 19.
show the loops chosen. This is a planar graph, and perhaps the simplest
loops would be the meshes. However, for illustrative purposes another
set of loops is chosen. The B matrix for this choice is the following:
107
108
[Ch.2
From the network we observe that the main diagonal elements in this
matrix are simply the total resistance on the contour of the corresponding
loop; the off-diagonal elements are plus or minus the resistance common
to the corresponding loops: plus if the orientations of the two loops are
the same through the common resistance, minus if they are opposite.
The source vectors will be
Sec. 2.4]
109
Then Z I has a nonzero entry only in the seventh row, and its value is
I /sC .
Hence the right-hand side of (80) becomes
g
(84)
110
[Ch. 2
further work. We shall postpone until the next chapter further considera
tion of this subject.
In reviewing the preceding discussion of loop equations it should be
noted that, at least for the example considered, less effort will be required
to write the final loop equations if a straightforward scalar approach is
used. In fact, it is possible to write the loop-impedance matrix Z and
the equivalent source matrix E by no more than inspection of the network,
once a set of loops is chosen. We seem to have introduced a matrix pro
cedure that is more complicated than necessary. Three comments are
appropriate to this point. In the first place, the general approach discussed
here should not be used for writing loop equations for networks with a
B matrix of relatively low rank. The general procedure becomes preferable
when dealing with networks having large B matriceswith tens of rows.
Secondly, the general approach using topological relationships is amenable
to digital computation, which makes it quite valuable. Finally, the
general form constitutes an "existence theorem"; it is a verification
that loop equations can always be written for the networks under con
sideration.
m
N O D E EQUATIONS
Sec. 2.4]
111
(89)
This matrix equation represents a set of n equations, called the node
equations, in the n node-voltage variables. The coefficient matrix Y (s) is
called the node-admittance
matrix.
This time the right-hand side, J, is the equivalent node current-source
vector whose entries are algebraic sums of current sources (including the
Norton equivalents of voltage sources) incident at the corresponding
nodes, with the references chosen so that they enter the node.
The node-admittance matrix can be written explicitly in terms of the
branch-parameter matrices by inserting (77) into (89). The result will be
n
(90)
where
(91a)
(91b)
(91c)
are the node-parameter
matrices.
Once the node equations are available in the form
(92)
the solution is readily obtained by inverting:
(93)
Again, this is essentially a symbolic solution. In the next chapter we shall
consider the details of the solution.
Let us now illustrate the use of node equations with the example of
112
[Ch. 2
Fig. 18, which is redrawn as Fig. 20. Let node 5 be chosen as the datum
node and as the node that is omitted in writing the A matrix.
where = L L
11
22
12
L.
21
Then
Sec. 2.4]
113
114
[Ch. 2
The source matrices V and I are the same as before. Hence we get
g
This agrees with the node-capacitance matrix obtained from the previously
found Y matrix.
N
Sec. 2.4]
115
N O D E - P A I R EQUATIONS
The variables in terms of which the node equations are written are the
voltages of the nodes all related to a datum node. This set of variables is
a basis for all branch variables. It was observed earlier that the twig
voltages for a given tree also constitute a basis for all branch voltages.
Hence we should expect the possibility of another set of equations similar
to the node equations, but with twig voltages for a tree as the variables;
this expectation is fulfilled.
Given a network, the first task is to select a tree and to apply KCL to
the fundamental cut sets, arriving at (69), which is repeated here:
(94)
Here Q is of order n b and of rank n. (The subscript f is omitted for
simplicity.) Into this expression we next insert the branch relations of
(73) getting
(95)
Finally, we express the branch voltages in terms of twig voltages through
the transformation in (71). The result is
(96)
or
(96b)
where J is simply shorthand for Q {I Y(s)V } and
t
(97)
Note that this expression is quite similar to the node equations given
in (88), the difference being that the f-cut-set
matrix Q replaces the
incidence matrix A, and the variables here are not node voltages but
node-pair voltages. We shall call these equations the node-pair
equations.
The coefficient matrix of the node-pair equations Y (s) is called the
node-pair admittance matrix; it can be written explicitly in terms of the
branch-parameter matrices by inserting (77). The result will be
t
(98)
116
[Ch. 2
where
(99a)
(99b)
(99c)
are the node-pair parameter matrices.
The same example will be used (Fig. 18) to illustrate the node-pair
equations as used earlier for the loop and node equations, except that it
will be assumed there is no mutual coupling between branches 1 and 2.
The diagram is repeated here as Fig. 21. The tree consisting of branches
The order of the columns is the same as the original numbering of the
branches, not the order for which Q can be partitioned into [U Q ] . The
reason for this is that the branch-admittance matrix was already written
for that order when the node equations were written. Refer back to the
branch-admittance matrix and note that L / is now replaced by 1/L ,
l
22
Sec. 2.4]
117
and Ln/ by 1/L ; also the off-diagonal elements are zero, since there is
no mutual coupling. The node-pair equations are found to be
2
118
[Ch. 2
2.5
DUALITY
DUALITY
Sec. 2.5]
119
(100)
where B is a loop matrix of N . Clearly, the number of branches of the
two networks must be equal, and the rank of A must equal the rank of
B . Thus
2
(101a)
(1016)
where b and b refer to the number of branches; n + 1 and n + 1 refer
to the number of nodes of the two networks, respectively.
Evidently these relationships constitute conditions on the structure of
the two networks. First, there must be a correspondence between the
branches of the two networks, as defined by the ordering of the columns
in the matrices A and B to satisfy (100). Secondly, there must be a
correspondence between the nodes of N (rows of A ) and the loops of
N (rows of B ) .
Two structures that are related b y (100) are called dual graphs. We
shall not discuss the abstract properties of dual graphs here but shall
state some of the simpler results.*
The basic result is that a network will have a geometrical (structural)
dual if and only if it is planar. If two planar networks can be superimposed
such that each junction but one of N lies inside a mesh of N and the
references of corresponding branches are suitably oriented, then a row
of A and the corresponding row of B will be identical. Such a loop and
node are shown in Fig. 22. The branches and loops of N are primed, and
the branches and nodes of N are unprimed. Node 1 in N corresponds
to loop 1' in N . You should verify that the KCL equation at node 1
has the same coefficients as those of the K V L equation for loop 1'. The
coefficient of i in the KCL equation at node 2 is + 1 . In order to make the
1
* For a more detailed account, see H. Whitney, "Nonseparable and Planar Graphs,"
Trans. Amer. Math. Soc, vol. 34, No. 2, pp. 339-362, 1932, and C. Kuratowski, "Sur
le probleme des courbes gauches en topologie," Fundamenta Mathematicae, vol. 15,
pp. 271-283, 1930.
120
Fig. 22.
[Ch. 2
coefficient of v in the KVL equation for loop 2' the same, namely + 1 ,
loop 2' must be oriented as shown. B y following through the entire graph
in the same manner, it can be seen that all the loops must be oriented in the
same sense (all clockwise if the branch references are as shown in the
figure, or all counterclockwise if the branch references are reversed).
If the meshes of a planar graph are chosen for writing K V L equations,
and all loops are oriented in the same sense, then the off-diagonal terms in
the loop equations will all carry negative signs, just as they do in node
equations.
The second condition of duality has to do with the branch relationships.
Figure 23 shows dual pairs of v-i relationships. For mutual inductance there
is no dual relationship. From the definition, then, only planar networks
without mutual inductance have duals.
Given such a network, say N , construction of the dual network
2
Fig. 23.
Dual branches.
Sec. 2.5]
DUALITY
121
(102)
and
(102b)
Since these matrices are equal, their determinants and cofactors will be
equal.
As an illustration, consider the diagram in Fig. 24a. A node is placed
(a)
(b)
(e)
Fig. 24.
Construction of a dual.
* In this process it is convenient to consider the sources as separate branches for the
purpose of constructing the dual.
122
[Ch. 2
within each mesh, and an additional one is placed outside. In part (b)
of the figure, dashed lines are used to represent dual branches crossing
each of the branches of N . Finally, the dual is shown in part (c). You
should verify (100) and (102) for this example.
1
2.6
The coefficient matrices of the loop equations and the node equations
are, respectively, BZB' and AYA'. The success of carrying out a loop
analysis or node analysis in this general form, then, depends on the
existence of a branch-impedance matrix Z or a branch-admittance
matrix Y. For the passive, reciprocal networks dealt with up to this
point, both of these matrices exist. (This statement must be qualified for
a perfectly coupled transformer, for which Y does not exist.)
Now we shall consider networks containing active and/or nonreciprocal
devices, in addition to passive ones having more than two terminals.
Table 1 shows such components, together with their representations.
There are two points to consider when dealing with these components.
One has to do with how the graphs of such multiterminal components are
to be represented. This will influence the number of KCL and K V L
equations and, hence, matrices A and B. The other point concerns the
existence of a branch representation that can be used in a loop or node
analysis. We shall initially take up the former point and consider the
graphs of the components.
Each of the components shown in Table 1 has four terminals. However,
the terminals are always taken in pairs, so that it is more appropriate to
consider them as having two pairs of terminals. It is possible, of course,
to identify (connect together) one terminal from each pair without
influencing the v-i relationships of the components. Thus each component
can be looked upon as a three-terminal component. The behavior of each
component in the table is specified by two relationships among two pairs
of variablestwo currents and two voltages. This is a special case of a
general condition; namely, that the behavior of an n-terminal component
can be completely specified in terms of n 1 relationships among n 1
pairs of voltage and current variables. (This condition is, in effect, a
postulate and, as such, not susceptible to proof.)
For a component with one pair of terminals, described by one voltage
and one current, the graph is represented by a single branch. The compo
nents in Table 1 have two pairs of terminals, and two voltages and
currents; their graph will be represented by two branches across the
Sec. 2.6]
123
Table 1
Device
Symbol
Equations
Type of
Representation
a
Voltagecontrolled
current-source
Admittance
Currentcontrolled
voltage-source
Impedance
Voltagecontrolled
voltage-source
Hybrid g
Currentcontrolled
current-source
Hybrid h
Gyrator
Impedance
or
admittance
Negative
converter
Hybrid h
or
Hybrid g
Ideal
transformer
Hybrid h
or
Hybrid g
a
Graph
124
[Ch. 2
pairs of terminals, as shown in Fig. 25. If one terminal from each pair is
(b)
(a)
Fig. 25.
(103)
Sec. 2.6]
(b)
(a)
Fig. 26.
125
Fig. 27.
126
[Ch. 2
and
Sec. 2.6]
127
You are urged to verify these. The presence of a gyrator, then, requires
nothing special for writing loop and node equations.
N e x t let us consider the v-controlled i-source. This was already done
in the illustration in Fig. 26. We saw that the input branch of the graph
of this device can be eliminated when the controlling voltage is expressed
in terms of other branch voltages. The device is then represented by a
single branch, and a branch-admittance matrix is easily written, leading
to a successful node analysis.
If a loop analysis is required, however, a problem will arise. Loop
equations require an impedance representation of the components. In
Fig. 26, what is needed is to write an expression for v explicitly in terms
of currents. Since v does not appear explicitly in (103), which is the
pertinent branch equation, we appear to be at an impasse. However, a
remedy can be found if branch 4', which is in parallel with 4 in Fig. 26,
is lumped together with 4 into a single branch. If i is the current of the
combined branch, then the following equation will replace (103):
4
(104)
Now it is possible to solve for v , after each of the other voltages has been
eliminated by inserting its branch relationship. Thus
4
(105)
128
[Ch. 2
(b)
(a)
(c)
Fig. 28.
Sec. 2.6]
129
(106)
Now let us write the branch relations for the transformer, this time
combining them with their accompanying branches. Let the sum of I
and the current in branch 3 be written I , and the sum of V and the
voltage across branch 4 be V . Then
a
(107a)
(107b)
Representation
Admittance
Representation
130
[Ch. 2
(180)
Fig. 29.
Sec. 2.7]
MIXED-VARIABLE EQUATIONS
131
of the device. In later work, we shall find other ways of dealing with the
branches of the graphs of these devices.
The comment should be made that for low-order networks the general
systematic formulation we have carried out requires more effort than
needed to write loop or node equations using a straightforward scalar
approach. Furthermore, the work is increased by the requirement that
the components be represented in a fashion that is not natural to them. In
the next section we shall consider an alternative approach, which utilizes
the hybrid representation directly.
One final point: the presence of nonreciprocal and/or active devices in
a network may mean that a unique solution cannot be found for certain
values of network parameters. This is illustrated b y the network in
Fig. 29, which contains a v-controlled v-source. For this network the
node-admittance matrix will be singular if G = (k 1 ) G . Hence a
solution will not exist in this case. Verify this.
1
2.7
MIXED-VARIABLE EQUATIONS
We observed in the last section that the desire to write loop or node
equations for networks containing multiterminal elements meets with
some difficulty for certain kinds of elements. The success of writing loop
or node equations depends on the existence of an impedance or admittance
representation for the branch relationships. When a multiterminal
device has no such representation, certain preliminary manipulations and
combinations of branches can be performed in order to write the desired
equations.
We shall now discuss a scheme that avoids these preliminary manipula
tions and permits all branch relationships to appear in their natural form.
This desirable result must be purchased at a price, however, as will be
evident as we proceed.
Of all the elements in our arsenal, some (including R, L, C, and gyrators)
have both impedance and admittance representations; one (the vcontrolled i-source) has only an admittance; one (the i-controlled
v-source)
has only an impedance; and some (transformer, NC,
v-controlled
v-source and i-controlled i-source) have only a mixed, or hybrid, repre
sentation. Whatever scheme we come up with must accommodate this
fact. We should also recall that twig voltages form a basis set of voltages
in terms of which all branch voltages can be expressed. Similarly, link
currents form a basis set of currents in terms of which all branch currents
can be expressed.
132
[Ch. 2
(109)
or
(110a)
(110b)
(The orders of the submatrices are as indicated.) The notation for the
coefficient matrices is chosen as an aid to the memory. For uniformity
the first submatrix in the first row should be written H , and the second
one in the second row should be H . But since they have the dimensions
of admittance and impedance, respectively, the simpler and more sugges
tive notation was chosen.
In these equations we express twig currents and link voltages in terms
of a mixed set of variables consisting of twig voltages and link currents.
Here we see the possibility of accommodating hybrid branch representa
tions if the branches having such representation are chosen as links or
twigs appropriately. This point will be amplified shortly.
There remain K V L and KCL to apply. Let us assume that K V L is
applied to the fundamental loops for the given tree, and KCL is applied
at the f-cut-sets.
Using (68) and (69) for K V L and KCL; partitioning the
B and Q matrices in the usual manner, with B = [B U] and Q = [U Q ] ;
and remembering that B = Q ' from (41), we get
1 1
2 2
(111a)
and
(111b)
Sec. 2.7]
MIXED-VARIABLE EQUATIONS
133
(112)
1 2
2 1
134
[Ch. 2
Table 2
Twig or Link
Device
Type of
Representation
Controlling
Branch
Controlled
Branch
Gyrator
Admittance
or
impedance
Twig
or
link
Twig
or
link
Voltage-controlled
current source
Admittance
Twig
Twig
Current-controlled
voltage source
Impedance
Link
Link
Ideal transformer
or NC
Hybrid h
or g
Twig
or
link
Link
or
twig
Voltage-controlled
voltage source
Hybrid g
Twig
Link
Current-controlled
current source
Hybrid h
Link
Twig
(113)
In this case, the twig and link designations in the last two columns of
Sec. 2.7]
MIXED-VARIABLE EQUATIONS
135
Table 2 will have to be reversed. It turns out that the equation corres
ponding to (112) using the G system will be somewhat more complicated.
The details of obtaining this equation will be left as a problem. In any
case, if one of these hybrid representations fails, the other one might
succeed.*
To illustrate the mixed-variable equations, consider again the network
of Fig. 28, which is redrawn here as Fig. 30. A possible tree is shown in
the graph b y heavy lines. For purposes of comparison, the same branch
numbering is made as before, but this is not the natural one for the tree
selected; care must be exercised about the ordering of branches when
writing the equations. The matrix off-cut sets, taken in the order 1, a,
4, 5, is found to be
To write the V-I relations in the form of (109), the twig and link
variables are listed in column matrices in the same order as for Q. Thus
* Matters can be salvaged, even if both fail, by a modification that uses two trees.
It should be clear, however, that failure of both representations can occur only in net
works having a proliferation of controlled sources in peculiar connections. Hence the
need for such a modification is so rare that further discussion is not warranted.
136
Fig. 30.
[Ch. 1
Now the H matrix is filled in. The only branches that will yield offdiagonal terms are those of the transformer. The appropriate equations
are V = Va/n, and I = I /n. The result will be
b
g t
gt
Now all the submatrices that make up (112) are determined. Putting
everything together leads to
Sec. 2.7]
MIXED-VARIABLE EQUATIONS
137
(a)
Fig. 31.
(b)
138
[Ch. 2
As for the sources, assume that the capacitors have initial voltages
v
and v , respectively. These can be represented as voltage sources
v /s and v /s. The source matrices are, therefore,
40
40
60
60
Sec. 2.7]
MIXED-VARIABLE EQUATIONS
139
Hence QI becomes I = [I
0 0 0 0]', and B V becomes Q/V +
V = [v /s
v /s
0 0]'.
When all of the above are inserted into (112), the final equations become
g
g l
60
g t
gt
40
Again the size of the matrix may appear to be excessively large but again
it is quite sparse.
To summarize the features of the mixed-variable equations, after a
tree has been selected, the V-I relationships of the branches are written
as a single matrix equation in which twig currents and link voltages are
expressed in terms of twig voltages and link currents. Next, KCL applied
to the f-cut-sets for the chosen tree and K V L applied to the f-loops are
used to eliminate the twig currents and link voltages from this expression.
The result is rearranged to give a set of equations in the twig-voltage and
link-current variables. The number of equations equals the number of
branches, which is the same as the sum of the number of loop equations
and node equations. This is the major drawback of the approach. The
virtue is the fact that multiterminal elements can be accommodated
quite naturally.
One other observation should be made here. In selecting a tree the
only restrictions involve the branches of multiterminal components. No
special pains are taken in assigning two-terminal elements to the tree or
the cotree, because the equations require no special distinctions among
capacitors, inductors, or resistors. As we shall see in a later chapter, this
is not the case when dealing with the state equations, and there are
reasons for assigning capacitors and inductors uniquely to the tree or
cotree.
140
[Ch. 2
PROBLEMS
1. When writing the state equations for the bridge network in Fig. 2, the
branch relations were inserted into KVL to eliminate all branch
voltages except that of the capacitor. Then KCL was used to eliminate
some of the branch currents.
Instead, start with KCL and use the branch relations to eliminate
appropriate currents. Then use KVL to eliminate some of the branch
voltages. The final result should be the same as (10).
2. Let A be an nth order nonsingular submatrix of the incidence matrix
of a connected linear graph, where n + 1 is the number of nodes. Prove
that the columns of A correspond to twigs for some tree. (This is the
converse of the theorem on p. 74).
t
7. A linear graph has five nodes and seven branches. The reduced incidence
matrix for this graph is given as
PROBLEMS
141
(b) For this tree write the matrix of f-loops, B (again, without drawing
a graph).
(c) For the same tree determine the matrix of f-cut-sets, Q . (No
graphs, please.)
(d) Determine the number of trees in the graph.
(e) Draw the graph and verify the preceding results.
f
8. Repeat Problem 7 for the following incidence matrices and the specified
branches:
branches: {2, 3, 4}
(a)
branches: {1, 3, 5, 6}
(b)
(a)
(b)
(c)
ij
142
[Ch. 2
Fig. P - 1 2
- 1
_ 1
15. (a) Two branches are in parallel in a graph. Determine the relationship
of the columns corresponding to these two branches in the Q matrix,
(b) Repeat if the two branches are in series.
f
PROBLEMS
(a)
(b)
Fig. P - 1 8
19. Prove that a graph contains at least one loop if two or more branches
are incident at each node.
20. For a connected planar graph of b branches and n + 1 nodes show that
the B matrix of the meshes is of rank b n. This will mean that the
mesh currents will be an adequate basis for expressing all branch
currents.
21. Using the concept of duality, show that KVL equations written for the
meshes of a planar graph are linearly independent.
22. Let a branch t of a graph be a twig for some tree. The f-cut-set
determined by t contains a set of links l , l , ... for that tree. Each of
these links defines an f-loop. Show that every one of the f-loops formed
by links l , l , etc. contains twig t.
1
23. For the graph in Fig. P-23, let B be the loop matrix for the meshes.
Fig. P - 2 3
-1
(a) Form B ' ( B ) ' and verify that the branch currents are correctly
given in terms of the link currents for the tree shown.
(b) For the same tree determine B directly and verify that
Bj=B'(Bl )'.
(c) Verify the mesh transformation.
l
-1
143
144
[Ch. 2
2 4 . In the graph of Fig. P-18b, branches {2, 3, 7, 6} form a loop. Verify the
result of Problem 3 for this set of branches.
2 5 . In the graph of Fig. P-18b, branches {2, 4, 5, 6} form a tree. Partition
the A matrix into [A A ] by using this tree. From this, letting
B = [B U] and Q = [U Q ] , determine B and Q and verify that
t
BT=-Q'L.
2 6 . It is possible for two f-loops of a graph for a given tree to have twigs
and nodes in common. Prove that it is possible for two f-loops to have
two nodes in common only if the path in the tree between the nodes is
common to the two loops.
2 7 . In Fig. P-27 the following loops exist:
loop 1: , e, g, b
loop 2: d, c, g,f
loop 3: , d, f, j , h, b
loop 4: e, c, h, j
Fig. P-27
(a) Do the KVL equations for these loops constitute an independent set?
(b) Either (1) find a set of links that form f-loops that are meshes of
the graph or (2) prove that there is no such set.
2 8 . In a linear network let the power in branch j be defined as p (t)
Vj(t) i (t), with standard references for the variables.
j
(a) For a given network assume that KCL and KVL are satisfied. Show
that p (t) = v(t)' i(t) = 0, where the summation is over all branches of
the network. The result v'i = 0 is called Tellegen's theorem.
(b) Next assume that Tellegen's theorem and KVL are both true. Show
that KCL follows as a consequence.
(c) Finally, assume that Tellegen's theorem and KCL are both true.
Show that KVL follows as a consequence.
j
This problem demonstrates that, of the three laws, KCL, KVL and
Tellegen's theorem, any two can be taken as fundamental; the third
will follow as a consequence.
PROBLEMS
145
29. Construct the duals of the networks in Fig. P-29. The numbers shown
are values of R, L, or C.
(a)
(b)
(d)
(c)
Fig. P - 2 9
30. (a) If two branches are parallel or in series in a graph, how are the
corresponding branches in the dual graph related? Verify in terms of
Fig. P-29.
(b) Figure P-30 represents, the network in Fig. P-29b. Within the box
Fig. P - 3 0
146
[Ch. 2
(b) Choose some tree [different from part (a)] and write loop equations
for the f-loops.
32. For the networks of Fig. P-29 choose the lowest node as datum and
write a set of node equations.
33. For the networks of Fig. P-29 choose a tree and write a set of nodepair equations with the twig voltages for this tree as basis variables.
34. For the network in Fig. P-34 (a) write a set of node-pair equations for
the tree shown, (b) Write a set of node equations using a convenient
node for datum.
Fig. P - 3 4
35. For the network in Fig. 29 in the text write a set of node equations
and verify that the node-admittance matrix will be singular when
G =
(k-1)G .
1
36. Fig. P-36 shows a network having a potentially singular nodeadmittance or loop-impedance matrix. Determine conditions among the
parameters that will make it singular
Fig. P - 3 6
PROBLEMS
147
37. For the network in Fig. P-37 set up in matrix form (a) the node
equations; (b) the mixed-variable equations, and (c) for the latter, show
all possible trees of the graph.
38. For the network of Fig. P-38 set up (a) the loop equations, (b) the
mixed-variable equations (use the usual small-signal linear equivalent
for the triode), and (c) for the latter, show all possible trees.
Fig. P-38
39. Find all possible trees for which mixed-variable equations can be
written for the network in Fig. 31 in the text.
40. Show that the loop-impedance matrix Z = B Z B ' f can be written as
Z + Q'lZ Q , where Q = [U Q ] and Z is conformally partitioned.
m
41. Show by doing it that, for RLC networks, the mixed-variable equations
in (112) can be converted into loop equations or node-pair equations.
The next six problems involve the preparation of a computer program to help
in implementing the solution of some problems. In each case prepare a program
148
[Ch. 2
flow chart and a set of program instructions, in some user language like
FORTRAN IV, for a digital computer program to carry out the job specified
in the problem. Include a set of user instructions for the program.
*42. Prepare a program to identify a tree of a connected network when each
branch and its orientation in the graph are specified by a sequence of
triplets of numbers: The first number identifies the branch, the second
number identifies the node at the branch tail, and the third number
identifies the node at the branch head. The program must also renumber
the branches such that twigs are numbered from 1 to n and links are
numbered from n + 1 to b. An example of a typical set of data is given
in Fig. P42 for the network graph shown there.
Fig. P-42
The output data from Problem 43 should be taken as the input data for
this problem.
*45. Prepare a program to determine the number of trees in a network by
(20) with pivotal condensation used to evaluate the determinant.
Specify the input data format.
*46. Prepare a program to determine the (a) node-admittance matrix,
(b) loop-impedance matrix, and (c) node-pair-admittance matrix of an
RLC network by evaluating the (1) node-parameter matrices of (91),
(2) loop-parameter matrices of (83), and (3) node-pair-parameter
PROBLEMS
149
Fourth
Number
Branch
Type
1
2
3
Capacitor
Resistor
Inductor
*47. Combine the programs of Problems 43, 44, and 46 to create a single
program that will determine at the program user's option the nodeadmittance matrix, loop-impedance matrix, and/or node-pair-admittance
matrix of an RLC network.
.3.
NETWORK FUNCTIONS
3.1
Sec. 3.1]
151
(lb)
(lc)
The right-hand sides are the contributions of the sources, including the
initial-condition equivalent sources; for example, J = [ J i ] , where Ji is
the sum of current sources (including Norton equivalents of accompanied
voltage sources) connected at node i, with due regard for the orientations.
The symbolic solution of these equations can be written easily and is
obtained by multiplying each equation by the inverse of the corresponding
coefficient matrix. Thus
(2a)
(2b)
(2c)
Each of these has the same form. For purposes of illustration the second
one will be shown in expanded form. Thus
(3)
(4)
152
[Ch. 3
NETWORK FUNCTIONS
Fig. 1.
When
tuted
linear
for V
(5)
As a general statement, then, it can be said that any response transform
can be written as a linear combination of excitation transforms. The
coefficients of this linear combination are themselves linear combinations
of some functions of s. These functions are ratios of two determinants,
the denominator one being the determinant of Z , Y , or Y , and the
numerator one being some cofactor of these matrices; for example, in
(5) the denominator determinant is det Y , and the coefficient of I
is
the difference between two such ratios of determinants.
Once the functions relating any response transform (whether voltage
or current) to any excitation transform (V or I) are known, then the
m
g2
Sec. 3.1]
153
gl
g2
g3
D R I V I N G - P O I N T FUNCTIONS
(a)
Fig. 2 .
(b)
Driving-point functions.
(c)
(6)
where V\ and 1\ are the transforms of the terminal voltage and current
with the references as shown in Fig. 2. In making this definition nothing
is said about how the terminals are excited or what is connected to them.
The implication is that it makes no difference. (Is this completely intuitive
or does it require proof?) The conditions of no independent sources and
zero initial conditions are essential to the definition. Clearly, if the network
contains independent sources or initial conditions, then Z or Y can take
on different values, depending on what is connected at the terminals;
thus it will not be an invariant characteristic of the network itself.
154
NETWORK FUNCTIONS
[Ch. 3
(7)
(8)
where Y is the driving-point admittance and the notation z means that
is the determinant of the loop impedance matrix.
Expressions 7 and 8 are useful for calculating Z and Y, but it should be
remembered that they apply when there are no controlled sources or
other such devices. They may also apply in some cases when controlled
sources are present, but not always.
As an illustration of a simple case in which (7) and (8) do not apply
when a controlled source is present, consider the grounded-grid amplifier
shown in Fig. 3a. The linear model is shown in Fig. 3b, in which a voltagecontrolled voltage source appears. Since this does not have either an
Sec. 3.1]
(a)
155
(b)
(c)
Fig. 3 .
impedance or an admittance representation, let us express the controlling voltage V in terms of a branch current; from Fig. 3b the proper
expression is V= R I
V . W^^th this expression replacing the control
ling voltage, the controlling branch is not even shown in the graph of
Fig. 3c. The V-I relation of the controlled branch (branch 2 in the graph)
is V = V= R I
V . This gives an impedance representation of
the branch. (V is simply an independent accompanying source and
will appear in the source-voltage matrix.) The tree in solid lines is chosen
so that the exciting source appears in only one loop. The loop matrix B,
the branch impedance matrix Z, and the source-voltage matrix V can
be written as follows:
k
156
NETWORK FUNCTIONS
[Ch. 3
Thus, even though we were careful to choose only one loop through the
exciting source, the source voltage V appears in both loop equations,
and the final result differs from (8). It may be concluded that one should
not rely on special formulas such as (7) and (8) but should go to definitions
such as (6).
1
TRANSFER
FUNCTIONS
(c)
(b)
(a)
Fig. 4 .
Transfer functions.
either VL(S) or Iz,(s) as the response and either V (s) or I (s) as the exci
tation. Again, in making these definitions there is no stipulation as to
1
Sec. 3.1]
157
From these, and from the fact that I L = Y L Vz,, each of the transfer
functions Vz,/V , VL/Ii, IL/V , and IL/I can be obtained. In a similar
manner, by assuming a voltage source is applied, as shown in Fig. 4c,
the loop equations may be written. The resulting expressions from both
the node and loop equations will be
Transfer
impedance:
1
(9a)
Transfer
admittance:
(9b)
Voltage gain, or transfer voltage
ratio:
(9c)
ratio:
(9d)
Let us emphasize t h a t
1. These formulas are valid in the absence of controlled sources and
other nonpassive, nonreciprocal devices.
158
NETWORK FUNCTIONS
[Ch. 3
3.2
MULTITERMINAL NETWORKS
(b)
(a)
Fig. 5.
Six-terminal network.
external connections can be made. Exactly how should the voltage-andcurrent variables be defined? Consider Fig. 5b. Should the voltages he
defined so that each terminal voltage is referred to some arbitrary datum
or ground, such as V ? Should they be defined as the voltages between
pairs of terminals, such as V , or V ? Should the currents be the ter2
12
46
Sec. 3.2]
MULTITERMINAL NETWORKS
159
(a)
Fig. 6.
(b)
Six-terminal network connected (a) as a three-port and (b) as a five-port.
160
[Ch. 3
NETWORK FUNCTIONS
3 . 3 TWO-PORT NETWORKS
At this point it is possible to proceed by treating the general multiport
network and discussing sets of equations relating the port variables. After
this is done, the results can be applied to the special case of a two-port.
An alternative approach is to treat the simplest multiport (namely, the
two-port) first. This might be done because of the importance of the twoport in its own right, and because treating the simplest case first can lead to
insights into the general case that will not be obvious without experience
with the simplest case. We shall take this second approach.
A two-port network is illustrated in Fig. 7. Because of the application
Fig. 7.
Two-port network.
Sec. 3.3]
TWO-PORT NETWORKS
161
(10)
(11)
162
[Ch. 3
NETWORK FUNCTIONS
11
22
21
12
21
12
(12)
This time interpretations are obtained by letting each current be zero
in turn. Then
(13)
2 2
2 1
2 1
s c
o c
(14)
From this it follows that
(15)
Demonstration of this is left as an exercise.
Sec. 3.3]
TWO-PORT NETWORKS
163
12
1 2
2 1
(16)
which means that both Y
works.
s c
and Z
o c
H Y B R I D PARAMETERS
The z and y representations are two of the ways in which the relation
ships among the port variables can be expressed. They express the two
voltages in terms of the two currents, and vice versa. Two other sets
of equations can be obtained b y expressing a current and voltage from
opposite ports in terms of the other voltage and current. Thus
(17)
and
(18)
(19)
164
NETWORK FUNCTIONS
[Ch. 3
Thus we see that the h- and g-parameters are interpreted under a mixed
set of terminal conditions, some of them under open-circuit and some
under short-circuit conditions. They are called the hybrid h- and hybrid
g-parameters. From these interpretations we see that h
and g
are
impedances, whereas h and g
are admittances. They are related to
the z's and y's b y
11
22
22
11
(20)
The transfer g's and h's are dimensionless. The quantity h is the forward
short-circuit current gain, and g is the reverse short-circuit current gain.
The other two are voltage ratios: g is the forward open-circuit voltage
gain, whereas h is the reverse open-circuit voltage gain. We shall use H
and G to represent the corresponding matrices.
B y direct computation we find the following relations among the
transfer parameters:
21
12
21
12
(21a)
(21b)
21
12
o c
21
S C
(22)
TWO-PORT NETWORKS
Sec. 3.3]
165
CHAIN PARAMETERS
The remaining two sets of equations relating the port variables express
the voltage and current at one port in terms of the voltage and current
at the other. These were, in fact, historically the first set usedin the
analysis of transmission lines. One of these equations is
(23)
They are called the chain, or ABCD, parameters. The first name comes
from the fact that they are the natural ones to use in a cascade, or tandem,
or chain connection typical of a transmission system. Note the negative
sign in I , which is a consequence of the choice of reference for I .
Note that we are using the historical symbols for these parameters
rather than using, say, ai for i and j equal 1 and 2, to make the system of
notation uniform for all the parameters. We are also not introducing
further notation to define the inverse parameters obtained by inverting
(23), simply to avoid further proliferation of symbols.
The determinant of the chain matrix can be computed in terms of z's
and y's. It is found to be
2
(24)
Table 1
Open-Circuit
Impedance
Parameters
Short-Circuit
Admittance
Parameters
Chain
Parameters
Hybrid
h-Parameters
Hybrid
g-Parameters
166
NETWORK FUNCTIONS
[Ch.3
Sec. 3.3]
TWO-PORT NETWORKS
167
(25a)
(25b)
Except for possible cancellations, all of these transfer functions will have
the same zeros. We use the generic term transmission zero to refer to a
value of s for which there is a transfer-function zero, without having to
specify which transfer functionwhether current gain, transfer admit
tance, or any other.
Example
As an illustrative example of the computation of two-port parameters,
consider the network shown in Fig. 8, which can be considered as a
Fig. 8.
168
NETWORK FUNCTIONS
Fig. 9.
[Ch. 3
21
22
and
y:
12
We see that y
is different from y , as it should be, because of the
presence of the controlled source.
If the y-parameters
are known, any of the other sets of parameters
can be computed b y using Table 1. Note that even under the conditions
that C and C are zero and R infinite, the y-parameters exist, but the
z-parameters do not (z , z , and z become infinite).
12
21
11
22
21
Sec. 3.4]
3.4
169
we can
170
NETWORK FUNCTIONS
[Ch. 3
(26)
(27)
21
21
21b
1b
Sec. 3.4]
171
1b
Hence
Finally,
(28)
2 1 a
21b
12
172
NETWORK FUNCTIONS
[Ch. 3
(a)
Fig. 13.
(b)
the input and output voltages of the component two-ports are forced to
be the same, whereas the overall port currents equal the sums of the
corresponding component port currents. This statement assumes that
the port relationships of the individual two-ports are not altered when
the connection is made. In this case the overall port relationship can be
written as
Sec. 3.4]
173
(31)
Fig. 14.
Parallel-ladders network.
174
NETWORK FUNCTIONS
[Ch. 3
(d)
(b)
(a)
Fig. 15.
PERMISSIBILITY OF INTERCONNECTION
(a)
Fig. 16.
(b)
Test for parallel-connected two-ports.
Sec. 3.4]
175
connected in parallel, whereas the other ports are individually shortcircuited. The short circuits are employed because the parameters
characterizing the individual two-ports and the overall two-port are the
short-circuit admittance parameters. If the voltage V shown in Fig. 16
is nonzero, then when the second ports are connected there will be a
circulating current, as suggested in the diagram. Hence the condition
that the current leaving one terminal of a port be equal to the current
entering the other terminal of each individual two-port is violated, and
the port relationships of the individual two-ports are altered.
For the case of the series connection, consider Fig. 17. A pair of ports,
(a)
Fig. 17.
(b)
Test for series-connected two-ports.
one from each two-port, is connected in series, whereas the other ports
are left open. The open circuits are employed because the parameters
characterizing the individual two-ports and the overall two-port are the
open-circuit impedance parameters. If the voltage V is nonzero, then
when the second ports are connected in series there will be a circulating
current, as suggested in the diagram. Again, the port relationships of the
individual two-ports will be modified by the connection, and hence the
addition of impedance parameters will not be valid for the overall network.
Obvious modifications of these tests apply to the series-parallel and
parallel-series connections. The preceding discussion of the conditions
under which the overall parameters for interconnected two-ports can be
obtained by adding the component two-port parameters has been in rather
skeletal form. We leave to you the task of supplying details.
When it is discovered that a particular interconnection cannot be
made because circulating currents will be introduced, there is a way of
stopping such currents and thus permitting the connection to be made. The
approach is simply to put an isolating ideal transformer of 1 : 1 turns
ratio at one of the ports, as illustrated in Fig. 18 for the case of the parallel
connection.
176
NETWORK FUNCTIONS
Fig. 18.
3.5
[Ch. 3
MULTIPORT NETWORKS
Fig. 19.
Multiport network.
(33a)
Sec. 3.5]
MULTIPORT NETWORKS
177
or
(33b)
B y direct observation, it is seen that the parameters can be interpreted
as
(34)
(35a)
where
(35b)
178
NETWORK FUNCTIONS
[Ch. 3
where each vector contains both the current and voltage of one particular
port. The former category are like the hybrid h and hybrid g representa
tions of a two-port. The latter has some of the features of the chainmatrix representation. It is clearly not very productive to pursue this
topic of possible representations in the general case.
Just as in the case of two-ports, it is possible to interconnect multiports.
Two multiports are said to be connected in parallel if their ports are
connected in parallel in pairs. It is not, in fact, necessary for the two
multiports to have the same number of ports. The ports are connected in
parallel in pairs until we run out of ports. It does not matter whether we
run out for both networks at the same time or earlier for one network.
Similarly, two multiports are said to be connected in series if their ports
are connected in series in pairs. Again, the two multiports need not have
the same number of ports.
As in the case of two-ports, the overall y-matrix for two n-ports
connected in parallel equals the sum of the y-matrices of the individual
n-ports. Similarly, the overall z-matrix of two n-ports connected in
series equals the sum of the z-matrices of the individual n-ports. This
assumes, of course, that the interconnection does not alter the parameters
of the individual n-ports.
3.6
Sec. 3.6]
Fig. 20.
(a;
Definition of terminal variables.
179
(b)
(36)
180
NETWORK FUNCTIONS
[Ch. 3
The quantity within each pair of parentheses is the sum of the elements
in a column of Y i . The terminal voltages are all independent. Suppose all
terminals but one, say the kth one, are short-circuited. This expression
then reduces to
(38)
Since V 0, this means the sum of elements in each column of the
indefinite admittance matrix equals zero. Thus the columns are not all
independent and Yi is a singular matrix.
What is true of the columns is also true of the rows. This can be shown
as follows. Suppose all but the kth terminal are left open, and to the kth
is applied a voltage source V . Assuming, as we did, that none of the
terminals is isolated, the voltages of all other terminals will also equal
V . All terminal currents will be zeroobviously, ail but I because the
terminals are open, and I because of Kirchhoff's current law. With ail
the voltages equal in (36), the jth current can be written as
k
ac
bc
Sec. 3.6]
(b)
(a)
Fig. 21.
181
bc
ac
(39)
The coefficient matrix of these equations is Y . Notice how it could have been
formed immediately from the original Y - m a t r i x by the process of add
ing a row and column, using the zero-sum property of rows and columns.
The preceding discussion provides a method for taking the Y - m a t r i x
of a common-terminal multiport with one terminal as the common termi
nal and from it easily writing the Y -matrix of the common-terminal
multiport with any other terminal taken as the common ground.
This is especially useful in obtaining, say, the grounded-grid representa
tion of a triode amplifier from the grounded-cathode representation or
the common-base representation of a transistor amplifier from the
common-emitter representation. To illustrate the approach, consider
the common-terminal two-port shown in Fig. 22a. The short-circuit
admittance matrix of this two-port is the following:
i
sc
sc
sc
182
NETWORK FUNCTIONS
(a)
Fig. 22.
[Ch. 3
(b)
(The letters at the top and the end identify the columns and the rows
with specific terminals.) From this the indefinite admittance matrix is
immediately written as
(40)
Sec. 3.6]
183
S U P P R E S S I N G TERMINALS
nn
or
(41)
(42)
184
NETWORK FUNCTIONS
[Ch. 3
terminal of the other and both have a common datum for voltage. It is
not necessary that the two networks have the same number of terminals.
If they do not, then rows and columns of zeros are appended to the Y matrix of the network having the fewer terminals. In particular, a
simple two-terminal branch connected across two terminals of a multi
terminal network can be considered to be connected in parallel with it.
Note that the indefinite admittance matrix of a branch having admittance
Y is
i
where is the (j, k)th cofactor of det Y . Because the elements of the
jth row sum to zero, we can write one of the elements as the negative
sum of all the others. Thus
j k
When this is inserted into the preceding equation and terms are collected,
the result becomes
jk
(44)
This means all cofactors of elements of any row are equal.
The same procedure, starting with an expansion of det Y along a
column, will yield a similar result concerning the equality of cofactors of
elements of any column. Since each row and column has a common
i
Sec. 3.6]
185
element, the cofactor of this element equals ail cofactors of that row and
column. The conclusion is that all (first) cofactors of the indefinite
admittance matrix are equal. This property has led to the name equicofactor matrix for Y .
i
Example
Let U S now illustrate with an example how the indefinite admittance
matrix can be used for certain network calculations. Consider the network
shown in Fig. 23a. It is desired to find the short-circuit admittance matrix
of the common-terminal two-port shown. We shail do this by (1) finding
the indefinite admittance matrix of the four-terminal network in Fig. 23b,
(2) adding to it that of the single branch, (3) suppressing terminal 3, and
finally (4) making terminal 4 the datum.
(c)
(b)
(a)
Fig. 23.
To find Y in Fig. 23b, we shall first treat that network as a commonterminal three-port with terminal 4 as datum. The y-parameters of this
three-port can be found from the definitions; for example, apply a voltage
to the left-hand port and short-circuit the other two, as in Fig. 24. Three
i
Fig. 24.
Calculating Y
sc
of the y-parameters are easily found from the diagram shown there. The
remaining y's are found in a similar way, with the result
186
[Ch. 3
NETWORK FUNCTIONS
Then
Sec. 3.6]
187
Finally, terminal 4 is made the common terminal by deleting the last row
and column. The desired y-matrix of the two-port is
3.7
188
NETWORK FUNCTIONS
[Ch. 3
(a)
Fig. 25.
(b)
Variables for the indefinite impedance matrix.
For the linear networks with which we are concerned, linear relations
expressing these voltage variables in terms of the terminal currents can
be written. However, it is more convenient to define a set of currents
other than the terminal currents, as shown in Fig. 25b. The currents J
are external loop currents between pairs of terminals. There is a simple
linear relationship between the terminal currents and these loop currents;
namely, I = J Jk+i- Hence a linear relationship expressing the volt
ages in terms of these loop currents can be written just as readily as one
expressing them in terms of the terminal currents. Such a relationship will be
k
(45)
Sec. 3.7]
189
This implies shorting all terminals together, which makes (1) the last
voltage also zero, by Kirchhoff's voltage law, and (2) all the external
loop currents equal. The details of these steps are left to you. Then, as for
Y , the indefinite impedance matrix is singular.
Recall that the indefinite admittance matrix was simply related to the
short-circuit admittance matrix of the common terminal (n 1) port
derived by short-circuiting one terminal of the network in Fig. 20 to the
datum. The corresponding situation here is: The indefinite impedance
matrix Z is simply related to the open-circuit impedance matrix Z of
the (n 1) port derived by opening one pair of terminals of the network
of Fig. 25. (The open-circuit impedance matrix thus established will be
called the common-loop open-circuit impedance matrix.) In fact, Z is
obtained by first adding a row to Z , each element of that row being the
negative sum of all elements in the corresponding column, and then adding
a column to the resulting matrix, each element of that column being the
negative sum of all the elements in the corresponding row.
For the Y -matrix the zero-sum property of rows and columns led to the
equicofactor property. In the same way the (first) cofactors of the indefinite
impedance matrix are all equal.
Because common-terminal port relations are often encountered, let us
examine the relation of a common-terminal open-circuit impedance
matrix to a common-loop open-circuit impedance matrix. In Fig. 25,
suppose terminal n is to be the common terminal and the voltage variables
for the port representation are to be the terminal voltages taken with
respect to terminal n as datum. Thus V = 0. Since the sum of the
terminal currents equals zero, by Kirchhoff's current law, one of them is
redundant. Since each terminal current is the difference between two
external loop currents, the terminal currents will not change if all the
external currents are increased or reduced by the same amount. Suppose
J is taken to be zero. This is equivalent to subtracting J from each
external current, so the terminal currents remain unchanged. From
Fig. 25, we can write the following:
i
o c
o c
190
NETWORK FUNCTIONS
[Ch. 3
(47)
(48)
n1
o c
(49)
(Keep in mind that Z (Z) here is the Z of (45) with its last row and column
removed.) This equation relates the port voltages to the port currents.
Hence the common-terminal open-circuit impedance matrix Z is
0C
o c
(50a)
and
(50b)
The last follows from the fact that a triangular matrix with nonzero
diagonal terms is nonsingular. The relationship here is seen to be much
Sec. 3.7]
191
more complicated than the corresponding one relating the commonterminal short-circuit admittance matrix Y to Y . Given Z , (50a)
permits finding Z after first writing Z
. Conversely, given Z for a
common-terminal (n 1) port, (50b) gives Z
. From this, Z is obtained
by adding a row and column, using the zero-sum property of rows and
columns.
s c
o c
o c
o c ( l )
3.8
o c ( l )
Let us pause briefly and review what has been done in this chapter.
In the first section we defined network functions as the ratios of Laplace
transforms of a response to an excitation. These may be driving-point
functions or transfer functions; but for linear, lumped networks they will
all be rational functions of the complex frequency variable s. Expressions
for any of these functions can be found by solving the loop or node
equations. In all cases the network functions can be expressed in terms of
ratios of the determinant of the node admittance matrix and its cofactors
or the loop impedance matrix and its cofactors. The subsequent sections
of the chapter were devoted to a discussion of different ways of describing
the external behavior of networks. These descriptions all entail sets of
network functions defined under various conditions imposed on the
terminals (open-circuit impedance, short-circuit admittance, etc.). Any
of these functions can be evaluated according to the discussion in the
first section. They all boil down to calculating a determinant and its
cofactor.
We shall now turn our attention to the task of finding relatively simple
means for evaluating determinants. The usual methods of evaluating
determinants (e.g., cofactor expansion or pivotal condensation) require
that many terms be multiplied together and then added. In this process
many terms eventually cancel, but only after extensive calculations have
been made. It would be of tremendous value to know which terms will
cancel in the end. The method we shall now discuss achieves exactly this
result.
D E T E R M I N A N T OF THE N O D E ADMITTANCE MATRIX
(51)
in which A is the incidence matrix and Y is the branch admittance
192
NETWORK FUNCTIONS
[Ch. 3
(53)
Sec. 3.8]
Fig. 26.
193
124
125
126
134
135
136
145
156
234
235
236
245
246
346
356
456
Note that, in this example, locating all the trees may not be a difficult
job. Nevertheless, it is worthwhile to tackle the problem systematically.
Since each tree contains n twigs, in a network having n + 1 nodes, one
procedure is to list all combinations of b branches taken n at a time. From
this set are then eliminated all those combinations that form a loop and
hence cannot be a tree.
To return to the example, once the trees are listed, the tree admittance
products are formed. In fact, this can be done in such a way that terms
with like powers of s are written together. The result will be
j k
194
NETWORK FUNCTIONS
[Ch. 3
(54)
(55)
This expression can be used to find ; but it would be even more useful
to relate to the original network N.
N o w N_ has one less node than N; hence, one less twig in a tree. A tree
of N_ cannot be a tree of N, and it cannot contain a loop of N. This
statement is an obvious consequence of the fact that a tree of N_ contains
no closed path in N_ and, thus, cannot contain one in N. Then, since a
tree of N_ has one less twig than a tree of N, it must be contained in a
tree of N. Let T_ denote a tree of N_ and let it be contained in T, a tree
of N. Clearly, T_ is a two-part subgraph of T. (This does not contradict
the fact that it is a one-part subgraph of N_ . Why?) Since node j and
the datum node d are short-circuited in T_ as a subgraph of N_ , there is
no path between them in T_ as a subgraph of T. Hence nodes j and
d are each in a different part of T- as a subgraph of T. Such a structure
is called a two-tree. Specifically, in a network of n + 1 nodes, a two-tree is
a set of n 1 branches that forms no loops and separates some tree of the
network into two connected parts. The product of branch admittances
constituting a two-tree is called a two-tree admittance product and is
labeled T(y). Subscripts are used to indicate the nodes that are required
to be in different parts. Thus T (y)
means a two-tree admittance
product in which nodes j and d are in separate parts.
j j
j j
j,d
For the example in Fig. 26, the network N_i is formed by shortcircuiting nodes 1 and 4 together, as shown in Fig. 27. Branch sets 13,
34, 45, and 24 are four of the trees of this network. In the original network
N these branch sets have the configurations shown in Fig. 28. Each of
these is a two-tree with nodes 1 and 4 in different parts. In some of them
Sec. 3.8]
Fig. 27.
195
node 1 or 4 is isolated; in another, not. Besides these, there are four other
two-trees (12, 15, 23, and 35) in which nodes 1 and 4 are in different parts.
(Verify them.) All of these contribute to 11.
Fig. 28.
Fig. 29.
196
NETWORKS FUNCTION
[Ch.3
(57)
Substituting for and
1 1
(58)
This is truly a powerful result. It permits the evaluation of drivingpoint functions of simple networks and even those of moderate complexity
essentially by inspection, without extensive analysis. Furthermore, it
provides an approach by which the digital computer can be applied in
more extensive networks, first for searching out all the trees and two-trees
and, second, in forming the required products.
Let us apply the formula to the high-pass filter network in Fig. 30
Fig. 30.
for which it is desired to find the driving-point function. There are four
nodes, hence three twigs in a tree and two branches in a two-tree. The
trees and two-trees ( 1 , 0 ) are the following:
Trees
Two-trees ( 1 , 0 )
Sec. 3.8]
197
out of a two-tree.) The admittance products are now formed. They can be
written in any order, of course, but we can note the powers of s to which
each tree or two-tree will lead and can group those with like powers of s,
as we have actually done in the listing. The result will be
(59)
+j
i j
ij
ij
ij
(60)
As before, a nonzero major of A _ Y corresponds to a two-tree in which
nodes i and datum are in separate parts. Similarly, a nonzero major of
A-j (which equals + 1 ) corresponds to a two-tree in which nodes j and
datum are in separate parts. Since, to be of interest, each factor of the
product in (60) must be nonzero, the subnetworks that contribute to
must be two-trees with nodes i and d, as well as j and d, in separate
parts. Since there are only two parts to a two-tree and d is in one of them,
we conclude that both i and j must be in the same part. Thus two-trees
with nodes i and j in one part and the datum node in the other are the
only ones that contribute to . Such two-tree products are designated
T (y).
The only reservation lies in the sign. Since A
and A _ are
different, there is no assurance concerning the signs of corresponding
majors of A - Y and A'_j. However, it turns out* that the sign of the
i
i j
i j
ij>d
* For a proof see S. Seshu and M. B. Reed, Linear Graphs and Electric Networks
Addison-Wesley, 1961.
198
NETWORK FUNCTIONS
[Ch. 3
i+j
Fig. 31.
ratio V (s)/V1(s).
Suppose a current source is applied as shown and
node 0 is taken as datum. The node equations can be solved for node
voltages V1, V , and V . Since V = V V , we get
23
23
(62)
As a specific example, look back at the network in Fig. 30. Let the
output voltage be the voltage across branch 5. The T ( y ) products
were already determined and given as the numerator in (59), so let us
concentrate on the numerator two-tree products in (62). For the two-tree
containing both nodes 1 and 2, branch 4 must be present. Branch 1 con
nects nodes 0 and 1; branch 2 connects nodes 0 and 2. Neither of these
branches can be present in T
0(y). Similarly, in order for nodes 1 and
3 to be in one connected part, branches 4 and 5 must both be present.
Since the two-tree has only two branches, no others are possible. Hence
2
1 , 0
1 2 ,
(63a)
(63b)
The transfer function, therefore, is
(64)
Sec. 3.8]
199
1 ,
1,
12,
13,
ij,
j,
di
j,
(65)
Note that they have some common terms. When these are inserted into
(62), the result becomes
(67)
and so
14
200
NETWORK FUNCTIONS
[Ch.3
From notions of duality one would expect that what has been done for
the node admittance matrix can also be done for the loop impedance
matrix. This is true, with some characteristic differences, as we shall now
discuss. One of these differences is that determinants of nonsingular
submatrices of the loop matrix B do not necessarily equal + 1 . They do,
however, if B is the matrix of fundamental loops. Hence we shall make
the assumption that fundamental loops have been chosen in writing the
B matrix.
The beginning point here is the loop impedance matrix BZB'. Again
the Binet-Cauchy theorem is applied. This time the nonsingular sub
matrices (of B) correspond to cotrees (complements of trees) instead of
trees. We define a cotree impedance product as the product of link impe
dances for some tree. We use the symbol C[T(z)] to indicate this product.
B y following the same kind of proof as for the node admittance matrix
(whose details you should supply) we find that
(68)
that is, to find the determinant , we must locate all the trees, from which
we find all the cotrees, multiply together the link impedances of each
cotree, and then add the resulting products.
The question no doubt has occurred to you as to whether there is any
relationship between the determinants of the loop impedance matrix and
the node admittance matrix. We shall now examine this question. Suppose
a tree admittance product is multiplied by the product of all the branch
impedances of the network. The twig impedances will cancel with the
twig admittances, leaving a cotree impedance product. If we do this
for all the tree admittance products and add, the result will be a sum of
all cotree impedance products; that is to say
(69)
Since the branch impedance matrix Z is diagonal, the product of impe
dances in this expression is simply the determinant of Z. Hence it can be
rewritten as
(70)
Sec. 3.8]
201
This is a very significant result. It states that the loop and node
determinants, although arising from different matrices (which are, in
general, of different orders), are related in a very simple way. In particular,
if we take each R, L, and C to be a branch of the network, the two deter
minants can differ at most by a multiplicative factor ksp. This means the
loop and the node determinants always have the same zeros, except
possibly at s = 0; that is, the nonzero natural frequencies of a network
are independent of whether the loop or the node basis is chosen for
analysis. (In this form, the relationship applies also when mutual induc
tance and transformers are present.)
It must be emphasized that this relationship between the determi
nants applies when they refer to the same network. There is the possibility
of going astray on this point when different sources are applied to a network.
Consider the situation in Fig. 32a. Suppose a voltage source is applied
1
(b)
(a)
Fig. 32.
j j
j j
202
NETWORK FUNCTIONS
[Ch. 3
Fig. 33.
Driving-point function.
Let N be the network when the terminals are open; that is, when loop 1
is opened. Then, with a voltage source applied, the resulting network is
N , since the source is just a short circuit for the evaluation of deter
minants. This means | is evaluated from (68) for network N . Recall
that tree admittance products for N
are two-tree admittance products
for N. Hence cotree impedance products for N
are co-two-tree impedance
products for N.
The driving-point admittance at the terminals of N is given by (8)
as Y(s) = / | , where is the determinant of the network N that
results when loop 1 in network N
is opened. Then, since Z(s) = 1/Y(s),
we find from (68) that
1
1 1
1 1
(71)
The notation appears cumbersome, but the ideas are simple. Thus
C[ T
(z)] is a blueprint for certain operations. It says: Find a two-tree
in which nodes 1 and 0 are in separate parts; take the branches that are
not in the two-tree (they are in the complement of the two-tree, the co-twotree) and multiply together their impedances. The numerator of (71) is
simply the sum of such terms for all two-trees (1, 0).
Note that this result could have been anticipated from the expression
for the impedance in (58) arrived at from the node admittance matrix.
Suppose numerator and denominator of that expression are multiplied
b y det Z. As discussed above, each term, which consists of a product of
admittances of certain branches, is converted to a product of impedances
of the complements of those branches. Hence (71) follows.
FinaUy we turn to the unsymmetrical cofactors of the loop impedance
matrix. Specifically, we look back at Fig. 31 with the change that a
2
1,
Sec. 3.8]
203
23
1 2
1 2
(72)
in which the complements are computed without the load Z L *
Consider, for example, the transfer voltage ratio for Fig. 31. In terms
of the loop equations, this is given by
(73)
An expression for this voltage ratio was previously given in (67). Suppose
the numerator and denominator of that expression are multiplied b y
det Z for the network with loop 1 open. (Why is det Z for this particular
network used?) N o w ZL, is a factor of det Z. Since Y L did not appear in
the numerator, Z L will not cancel in any term of the numerator. Hence it
can be factored out. The result will be
(74)
TWO-PORT P A R A M E T E R S
204
NETWORK FUNCTIONS
[Ch. 3
Fig. 34,
a datum for defining node voltages. Note that the port voltage V is the
difference between node voltages V
and V . Assume that current
sources are applied at the input and output ports. The node equations
will be
2
20
30
(75)
and V
30
:
(16a)
(76b)
(76c)
Observing that V = V
2
20
V,
30
(77)
To find the short-circuit admittance matrix we must invert Z . Let
be the determinant of Z . From (77), this determinant is
o c
o c
o c
(78)
Sec. 3.8]
205
(79)
where
is the cofactor formed by deleting rows i and j , and columns
i and k from . When these identities are used (78) becomes
i i j k
(80)
(81)
We have topological formulas for all the cofactors except those with
four subscripts. Let us define a three-tree of a graph having n + 1 nodes
as a set of three unconnected subgraphs having a total of n 2 branches and
containing no loops. We denote a three-tree with the symbol T together
with subscripts indicating which nodes are in the separate parts. N o w
just as was shown to equal the sum of the admittance products over all
trees of the graph and was shown to equal two-tree (ij, d) admittance
products over all two-trees of the type in which i and j are in one part
and d in the other, in the same way it can be shown that
3
i j
(82)
where T1,
(y) is a three-tree admittance product with nodes 1, 2, and
0 in separate parts, and the other three-trees have similar interpretations.
But
2,
(83)
* See, for example, A. C. Aitken, Determinants and Matrices, 9th ed., Interscience
Publishers, New York, 1956.
206
NETWORK FUNCTIONS
[Ch. 3
(84)
Before inserting the topological formulas for the cofactors into (81),
note also that it is possible to simplify the expression for the first element
in the matrix there. Thus
Hence
(85)
When all the above is collected and inserted into (77) and (81), the
results are
where ~ I\y) is defined in (84). Note that the numerators of the transfer
impedances and of the transfer admittances differ in sign only. This
verifies our earlier observation that these functions have the same zeros
unless there is a cancellation with the denominators.
As an illustration, the network of Fig. 30 is redrawn as a two-port in
Fig. 35. Let us find Y . For this network the two-trees (1,0) were already
listed under (58). We shall repeat them here together with the other
required two-trees:
s c
two-trees
two-trees
two-trees
two-trees
(1, 0):
(12, 30):
(13, 20):
(2, 3):
Sec. 3.8]
Fig. 35.
207
In this example there are four nodes. Hence a three-tree has only one
branch. The three-trees are, therefore, easily determined. They are
three-trees
three-trees
three-trees
three-trees
208
NETWORK FUNCTIONS
[Ch. 3
setting up loop and node equations in the last chapter, requiring exten
sive matrix multiplications to obtain AYA' and BZB' turns out to be of
great value, because the Cauchy-Binet theorem permits some further
mathematical derivations culminating in some simple formulas that
require no matrix operations.
PROBLEMS
1. In the network of Fig. PI solve for the voltage-gain function V (s)/Vi(s).
Do this by (a) using mixed-variable equations; (b) using node equations
after expressing I in terms of appropriate voltages.
2
Fig. PI
PROBLEMS
(b)
(a)
(c)
Fig. P2
(b)
(a)
(c)
Fig. P3
209
210
[Ch. 3
NETWORK FUNCTIONS
Fig. P4
Fig. P5
6. The two-port shown in Fig. P6 is a potential negative converter.
(a) Find the hybrid h-parameters.
(b) Specify the ratio R /R1 in terms of to make h h
= 1.
(c) Comment on the relative values of R and R for = 50. ( = 50 is
an easily realizable current gain.)
(d) Is this a voltage or a current negative converter?
2
12
21
12
21
PROBLEMS
Fig. P6
Fig. P7
Fig. P8
211
212
[Ch. 3
NETWORK FUNCTIONS
(a)
(b)
Fig. P9
()
(a)
Fig. P10
o c
PROBLEMS
213
Fig. P11
21
Fig. P14
Voltage
negative
converter
Fig. P15
214
NETWORK FUNCTIONS
[Ch. 3
Fig. P16
(b)
(a)
Fig. P17
18. Show that the Y matrix of the two-port shown in Fig. P18a, consider
ing each transistor as an ideal current-dependent current source as
shown in Fig. P18b, is
s c
where G = R/R R
and g = G + 1/R , under the assumption that
R e / 1 << R ^ R e . Verify that the two-port in Fig. P18c is equiva
lent to this two-port.
19. The hybrid h-matrix of a two-port device has one of the following
forms:
e1
e2
el
PROBLEMS
(a)
(b)
215
(c)
Fig. P18
20. Find the hybrid h-matrices for each of the networks in Fig. P20.
(Replace each transistor by the simplest possible small-signal equivalent.)
(a)
(b)
Fig. P20
21. In Fig. P21, the two-port is a general converter having the hybrid
h-matrix shown. Find the impedance Z.
22. For the diagram of Fig. P22, show that the voltage transfer ratio is
(Observe that the conductances Gi, G , and G can account for the input,
output, and feedback impedances of an actual amplifier of which the
controlled source is the idealization.)
2
216
NETWORK FUNCTIONS
[Ch. 3
Fig. P21
Fig. P22
23. For each of the networks in Fig. P23 find the short-circuit current-gain
function h . Use the y-parameters of the two-port N.
21
24. Find the voltage transfer ratio for the two networks in Fig. P24 in
terms of the y-parameters of the two-ports N and N , and the amplifier
gain . Verify for Fig. P24b that the limiting value as -> is
V /Vi=
y /y .
a
21b
21a
26. A two-port N with a resistance R across both its input and output ports
is shown in Fig. P26a. The resulting two-part is denoted N , and the
z-parameters of N are z , z , z , and z . The network N ,
after introducing either a series or shunt resistance R at its ports, is to
be cascaded with the feedback amplifier of Problem 25. The two cascade
configurations to be considered are shown in Fig. P26b and c. Show that
c
l l b
1 2 b
2 1 b
2 2 b
PROBLEMS
(a)
(b)
(c)
(d)
Fig. P23
(a)
(b)
Fig. P24
217
218
NETWORK FUNCTIONS
[Ch. 3
Fig. P25
(a)
(c)
(b)
Fig. P26
27. Let a transmission zero of a two-port be defined as a zero of the shortcircuit transfer admittance y (s). Show that the output current or
voltage of the terminated two-port in Fig. P27 will be zero with either
a voltage or current excitation, even if yn or y also have a zero at
this frequency, leading to a cancellation in (25) and causing z , h ,
or g , or all three to be nonzero. Comment on the appropriateness of
the term " transmission zero ".
21
22
21
21
21
PROBLEMS
219
Fig. P27
(a)
(b)
29. Find the chain matrix of the two-ports in Fig. P29. The transformers
are perfect.
(a)
(b)
Fig. P29
30. Treat the bridged-tee network of Fig. P30 first as the parallel connection
of 2 two-ports and then as the series connection of 2 two-ports to find
the overall y-parameters. (The answers should be the same.)
31. Find the y-parameters of the two-ports in Fig. P31 by decomposing
them into suitable parallel-connected two-ports.
220
NETWORK FUNCTIONS
[Ch. 3
Fig. P30
(a)
(b)
Fig. P31
32. The short-circuit admittance matrix of the network in Fig. P32 with
terminal 3 as the common terminal is
Fig. P32
PROBLEMS
221
(a)
(b)
Fig. P33
34. Figure P34a shows a four-terminal network connected as a commonterminal three-port. The short-circuit equations of this three-port are as
shown. It is desired to connect a unit capacitor between terminals 1
and 2, as shown in Fig. P34b. Find the short-circuit admittance matrix
of the network when it is considered as a two-port with the ports
shown in Fig. P34b.
(b)
(a)
Fig. P34
222
NETWORK FUNCTIONS
[Ch. 3
35. The n-terminal network shown in Fig. P35 is linear, lumped, and
time invariant. It is represented by I = Y V, where Y is the indefinite
admittance matrix, and the currents and voltages are defined on the
diagram. It is proposed to retain the first k terminals as terminals and
to connect the remaining ones to ground through impedances, as shown.
Let Z be the diagonal matrix whose diagonal elements are the impe
dances Zj. Find an expression relating the new terminal currents to the
voltages in terms of Z, Y , and submatrices thereof.
i
(a)
(b)
Fig. P35
36. The diagram in Fig. P36 is a linear RLC network with no transformers.
Determine a relationship for the voltage transform V(s) using topologi
cal expressions for the determinant of the node admittance matrix and
its cofactors, taking node 5 as the datum. Simplify as much as
possible.
Fig. P36
PROBLEMS
223
Fig. P37
40. Find the driving-point admittance of each of the networks in Fig. P40
by using topological formulas. Do it twice, once with the node admit
tance matrix and once with the loop impedance matrix.
(a)
(b)
Fig. P40
41. In the networks of Fig. P41 find the driving-point admittance at the
left-hand port by using topological formulas. Do it twice, once with
the node admittance matrix and once with the loop impedance matrix.
224
NETWORK FUNCTIONS
[Ch. 3
(a)
(b)
Fig. P41
42. In the networks of Fig. P41, find the transfer voltage ratio V2/V1 by
using topological formulas.
43. For a common-terminal two-port the topological formulas for Y
Z will simplify to some extent. Find these simplified formulas.
s c
and
o c
44. For the networks in Fig. P44, find the open-circuit impedance matrix
Z using topological formulas.
o c
(a)
(c)
(6)
(e)
(d)
Fig. P44
45. (a)
PROBLEMS
225
inductance will have a pole at s = 0 if and only if there is an allcapacitor cut-set that separates the two terminals.
(b) Prove that the impedance will have a pole at infinity if and only
if there is an all-inductor cut-set separating the terminals as in (a).
4 6 . (a) Prove that the admittance of an RLC network without mutual
inductance will have a pole at s = 0 if and only if there is an allinductor path between the terminals.
(b) Prove that the admittance will have a pole at infinity if and only
if there is an all-capacitor path between the terminals.
47,
Let
22
22
21
ll9
22
21
4 8 . (a) For the network of Fig. P48 find the short circuit admittance
parameters by direct application of the definition. Fialkow's condition
is apparently not satisfied.
(b) Find the parameters again, using topological formulas and compare
the two answers. State a condition that must be assured in order for
Fialkow's condition to be valid.
226
NETWORK FUNCTIONS
[Ch. 3
Fig. P48
49. Let the hybrid h-matrix for a transistor in the common-emitter con
nection be
Find the h-matrix of the transistor in the common-base and commoncollector configurations through the agency of the indefinite admittance
matrix.
50. The diagram in Fig. P50 is a passive, reciprocal network in which the
resistor R is shown explicitly. The driving-point impedance of the
network is Z(s). Suppose the branch containing R is opened and the
terminals so formed constitute the input port of a two-port whose other
port is the original pair of terminals of the network. Let g (s) be the
forward voltage-gain function of this two-port. Show that, if the net
work contains n resistors,
k
21k
Fig. P50
PROBLEMS
227
z and y = y , y , respectively. If we consider bisecting the twoport at its structural line of symmetry, a number of terminals (two
or more) will be created at the junction between the two halves. Assume
that none of the leads from which these terminals are formed are
crossed. Now consider the two cases shown in Fig. P51 in which these
terminals are left open and short-circuited, respectively. The input
impedance and input admittance are designated z
and y
in the two
cases, respectively, where the subscript h stands for "half." Show that
12
11
2 2
12
l l h
l l h
or
Fig. P51
52. The current and voltage variables in the loop, node and node-pair
equations in (1) are Laplace transforms. The solution, say, for one of
the node voltages is as given in (4).
Now suppose the excitations are all exponentials, so that the ith
equivalent current source is I e 0 . I is a complex number, called a
phasor. Assume that s =j is not a natural frequency of the network
and assume the network is initially relaxed. The forced response to the
j
228
NETWORK FUNCTIONS
[Ch. 3
Fig. P53
(b)
(c)
.4.
STATE EQUATIONS
230
STATE EQUATIONS
[Ch. 4
4.1
Sec. 4.1]
231
Fig. 1.
(2b)
* To avoid repetition, we shall use the term "all-capacitor loop" to mean a loop con
taining only capacitors or only capacitors and independent voltage sources. Likewise, we
shall use the term "all-inductor cut-set" to mean a cut-set containing only inductors or
only inductors and independent current-sources.
232
STATE EQUATIONS
[Ch. 4
(3)
Fig. 2.
j j
Sec. 4.1]
233
234
STATE EQUATIONS
[Ch. 4
All-capacitor cut-sets
5, 6,7
6,8
5, 7,8
All-inductor cut-sets
1,2,3
3,4
1,2,4
All-inductor loops
None
Fig. 3.
other two. The same is true of the three all-capacitor cut-sets: only two are
independent. Since there are a total of 11 inductors and capacitors, and
3 linear constraints (one all-capacitor loop and two all-inductor cut-sets),
the order of complexity and the number of natural frequencies is
11 3 = 8. Of these natural frequencies, two are zero, corresponding to
the two independent all-capacitor cut-sets. Thus there are 8 2 = 6
nonzero natural frequencies.
4.2
We are now ready to begin the development of the state equations. The
basic equations at our disposal are still K V L , KCL, and the v-i relation
ships. It is the particular combination and the particular order in which
these are invoked that we must choose. The decision is made on the basis
of a number of considerations:
1. We want the final equations to contain no integrals. Integrals arise
from the substitution of i = j" v dxjL + i(0) for an inductor current
in KCL and the substitution of = J i dx/C + v(0) for a capacitor
voltage in KVL. So, we shall simply not make these eliminations of
inductor currents and capacitor voltages but keep them as variables.
2. We want the final equations to be first-order differential equations.
0
235
236
[Ch. 4
STATE EQUATIONS
current sources. (If the network is not connected, the corresponding term
is "normal forest." For simplicity, we shall later refer to a normal tree,
sometimes reverting to the use of "normal forest" for emphasis.)
If there are no all-capacitor loops, all the capacitors will be twigs of the
normal tree. Also, if there are no all-inductor cut-sets, none of the induc
tors will be twigs of the normal tree; they will all be links. This can be
proved by contradiction. Suppose there is a capacitor link in the absence
of an all-capacitor loop. Both end nodes of the capacitor link lie on the
corresponding normal tree. If this capacitor link is added to the tree, it
will form a loop that, by hypothesis, is not an all-capacitor loop. If from
this loop a noncapacitor branch is removed, the result will be a new tree
that will have one more capacitor than the preceding one. This is not
possible, since the preceding normal tree has the maximum number of
capacitors, by definition. A similar proof applies for the inductors.
Given a network, we first select a normal tree (or a normal forest if the
network is not connected). We then write KVL equations for the f-loops
and KCL equations for the f-cut-sets
of this tree. We use the branch
v-i relationships to eliminate capacitor currents and inductor voltages but
we have not yet discussed how to handle the variables of the resistive
elements, including controlled sources, gyrators, as well as resistors.
Before considering this problem in a general way, let us spend some time
in discussing some examples through which the general approach can
evolve.
As a first illustration consider the network shown in Fig. 4a. It is de
sired t o find the output voltage v (t) when voltages v (t) and v (t) are
the inputs. There are three reactive elements and no degeneracies.
Hence the order of complexity is 3. There are six nodes and nine branches.
(Remember that the two voltage sources are counted as separate
branches.) A normal tree must contain both voltage sources and both
0
gl
g2
(b)
(a)
(c)
Fig. 4.
237
capacitors, but not the inductor. This information is shown in Fig. 4b,
where the solid lines show a partial normal tree and the dashed line re
presents the inductor link. We need one more twig, and this must connect
node c to the rest. There are clearly two possibilities among the resistive
branchesbranch 5 or 6. We have chosen 5 to complete the tree, as
shown in Fig. 4c. Notice that the branches have been numbered with the
twigs first, then the links.
Now let us write the KCL equations for the f-cut-sets
and the K V L
equations for the f-loops.
With the usual partitioning, these can be
written as
(6a)
(6b)
or
(7a)
(7b)
Note that, because the sources are counted as separate branches, the
right-hand sides in (6) are 0 and not Qi and B v , as we have been used to
writing them. In scalar form (7) leads to the following:
g
(8a)
(8b)
(8c)
(8d)
(8e)
(8f)
(8g)
(8h)
(8i)
238
STATE EQUATIONS
[Ch. 4
gl
g2
(9a)
(9b)
(9c)
(9d)
(9e)
The resistor equations are either of the form v = Ri or i = Gv, but we have
no guidelines yet as to how they should be written. According to our
earlier discussion, we want to eliminate capacitor currents and inductor
voltages. Hence we substitute i , i , and v9 from the last three equations
into the appropriate KCL and K V L equations in (8). This step leads to
3
(10a)
(10b)
(10c)
There are two classes of variables on the right-hand sides: (1) capacitor
voltages and inductor currents (v and i9), which we want to keep; and
(2) resistor currents and voltages. There are four of the latter kind of
variable; namely, v , i , i , and i . Note that none of these appears
explicitly solved for in the Kirchhoff equations in (8), but their comple
mentary variables do. To change these complementary variables to the
desired ones, we can combine the appropriate v-i relationships with (8).
3
239
(11b)
(11c)
(11d)
Thus for the twigs we use the form v = Ri; and for the links i = Gv.
When these are inserted into the appropriate four equations in (8), the
result can be rewritten in the following form:
(12a)
(12b)
(12c)
(12d)
These are purely algebraic equations giving resistor voltages or currents
in terms of (1) source voltages, (2) capacitor voltages, and (3) inductor
currents. These algebraic equations can be easily solved (the last two
trivially) to yield
(13a)
(13b)
(13c)
(13d)
Finally, these equations can be substituted into (10) to yield, after re
arrangement,
(14a)
240
STATE EQUATIONS
[Ch. 4
(14b)
(14c)
(15)
Input
vector
The elements of the state vector are state variables. We refer to the matrix
equation as a state equation. Equation (15) can be written in compact
241
matrix notation as
(16)
where the meanings of the matrices A and B are obvious. This is called
the normal form of the state equation. The derivative of the state vector
is given as a linear combination of the state vector itself and the input,
or excitation, vector. (The letter " e " stands for excitation.)
The desired output quantities may be state variables or any other
variables in the network. In the present case we had wanted the output to
be v (t). From the network it is found that vo = v v , which can be
written in matrix form as
0
or more compactly as
(17)
where w is the output vector.
The next step would be to solve the state equations, a project that will
occupy a major part of the remainder of this chapter. However, before
undertaking this major effort, we shall consider another example, which
will introduce some features not present in this past example.
The network for this example is shown in Fig. 5. It has an all-capacitor
loop (capacitors and voltage sources). A normal tree must contain both
voltage sources but clearly cannot include all three capacitors. The ones
numbered 3 and 4 have been placed on the tree. The normal tree can be
completed by any one of the three resistors; the one numbered 5 is chosen
here.
Fig. 5.
Illustrative example of circuit with all-C loop for writing state equations.
242
STATE EQUATIONS
[Ch. 4
and K V L
(18a)
(18b)
(18c)
(18d)
(18e)
(18f)
(18g)
(18h)
If we followed the approach of the preceding example, we would next
write the v-i equations of all the capacitors (the only reactive elements in
this case) in order to eliminate the capacitor currents from (18) and retain
only the capacitor voltages. However, because of the all-capacitor loop,
not all these voltages are dynamically independent. Therefore we write
the v-i equations of only those capacitors that are in the normal tree.
Thus
(19a)
(19b)
These are now inserted into the appropriate KCL equations in (18) to
yield
(20a)
(20b)
Of the three variables on the right sides, the capacitor link current i
and the resistor link currents i and i are treated separately. The K V L
equation for v in (18f) is inserted into the v-i equation for i , yielding
6
243
(21)
which gives i in terms of "desirable" variables only.
This leaves us with i and i to eliminate from (20). These do not
appear explicitly solved for in (18). Again we write the v-i relations for
the resistor links as v = Ri and for the twig (branch 5) as i = Gv. The
appropriate three equations in (18) can then be rewritten as follows:
6
(22a)
(22b)
(22c)
This is simply a set of algebraic equations that can be solved to yield
(23a)
(23b)
(23c)
where K = 1 + R (C + G ). The last two of these equations together with
(21), when inserted into (20), will eliminate the unwanted variables.
After rearrangement, the result in matrix form becomes
7
(24)
244
[Ch. 4
STATE EQUATIONS
(25)
The derivative of
(26a)
(26b)
245
terminology, the two equations together are called the state equations.
The second one is called the output equation. Our next task is to carry out
a solution of these equations.
4.3
In the examples of the last section we found that the input and output
variables are related through equations such as (26). We shall find in later
sections that such equations result for all networks of the class we are
considering. Observe that a somewhat simpler form is obtained b y setting
(27)
in (26), which then becomes
The derivative of the excitation has been eliminated in the first equation
but not in the second. For simplicity we shall simply remove the bar and
write x instead of 5. Furthermore, we shall replace B + AB
b y B,
D +CB
by D, and D by D. The equations we shall treat will, there
fore, have the forms
1
(28a)
(28b)
246
[Ch. 4
STATE EQUATIONS
(29)
in which Y(t) is a square matrix of order n that is assumed to be non
singular for all finite t>t .*
Insert this transformation into (28). The
result after suitable arrangement of terms will be
0
(30)
(31)
(32)
(33)
* The development of this chapter is extravagant in the use of symbols. The need far
outstrips the availability. This forces us to use a symbol in one context when it already
has a well-defined meaning elsewhere. Thus in earlier work Y is an admittance matrix.
Its use here with a different meaning will hopefully cause no confusion.
Sec. 4.3]
247
- 1
(35)
This is a very significant result. It tells us that in order to solve (28) we
first solve (31) with some nonsingular initial condition, such as Y(to) = U.
We then carry out the indicated integration in (35). However, the
integrand requires us first to find the inverse of Y, which is a considerable
chore.
It turns out that this can be avoided, because Y(t) Y ( ) is a matrix
function of t r, which, as will be seen shortly, is easily determined. We
express this relationship symbolically as
- 1
(36)
When this is inserted into (35), the result becomes
(37)
The matrix is called the state-transition matrix. The name derives from
the idea that when e = 0 the transition from the " state " of the network
at time t to the " state " at time t is governed by , as (37) illustrates.
Equation (37) constitutes the time-domain solution of the original
nonhomogeneous differential equation in (28). Its importance cannot be
overemphasized. However, it is really a symbolic solution, because we are
still required to solve the homogeneous equation (31) before the job is
complete. This will be our task now.
0
248
STATE EQUATIONS
[Ch. 4
which may be verified by direct substitution into the equation. Since the
form of the matrix equation (31) is identical with that of the scalar
equation, it is tempting to seek an exponential solution:
(38)
The only trouble is, we do not know the meaning of an exponential with
a matrix in the exponent. You have no doubt encountered a similar
difficulty in defining an exponential with a complex-number exponent.
This is handled by defining a complex exponential in terms of the series
expansion of the real exponential. (See Appendix 2.) We shall do the same
thing here and define
(39)
A t
Then
(40)
At
Sec. 4.3]
249
(41)
that is, the formula for the derivative of a matrix exponential is the same
as it is for a scalar exponential. When this result is used it is found that
Y(t) = e o) given in (38) is the (unique) solution satisfying (31) and the
initial condition Y(to) = U.
Recall that in obtaining (32) it was assumed that Y(t) is nonsingular
for all finite time following t . We must now show that it is, in fact,
nonsingular. This is not difficult. From the series definition of a matrix
exponential, we can write
A(t
(42)
Now let this series be multiplied by the series for the positive exponential
in (39). The result will be
(43)
All other terms cancel. This term-by-term multiplication is permissible
because of the absolute convergence of the two series for all finite t.
The result tells us that we have found a matrix ( e ) , which, when multi
plied by e , gives a unit matrix. B y definition, it is the inverse of e .
Hence Y(t) is nonsingular for t t .
There is only one thing left to do. We must give an explicit expression
for the state-transition matrix (t T) = Y(t) Y ( ) . This is an easy
task. We know that Y(t) = e
and Y ( ) = e
; therefore
A t
A t
A t
A ( t t o )
- 1
A ( t o )
(44)
and is, like Y(t), a matrix exponentialonly the scalar time variable is
different. This relation can now be inserted into (37) to yield
(45)
250
[Ch. 4
STATE EQUATIONS
/ o
A L T E R N A T I V E METHOD OF
SOLUTION
We have just treated the solution of the state equation in the general
case. In a particular situation, suppose there is no excitation (e = 0) or
for a given network B = 0. Then the state equation reduces to the homo
geneous equation
(46)
Comparing this with (31) shows that they are of the same form. There is,
however, one difference: whereas Y (or, equivalently, the state-transition
matrix ) is a square matrix, in the present equation x is a column
vector. Among other things, this means the initial value in the present
case cannot be a unit matrix but must be represented by the initial-value
vector x(t ).
From the general solution in (45) the solution of (46) can be written as
0
(47)
This is, of course, much simpler than the general solution when Be 0.
It would, therefore, be of considerable value if, by some modification of
variables, it would be possible to convert the general nonhomogeneous
state equation into a homogeneous one. This is what we shall pursue in
this section.
Consider the state equation (28a), repeated here for convenience:
(48)
(49)
Sec. 4.3]
251
(50)
(51)
(52)
just as (47) was the solution of (46). (The notation exp(u) stands for e .)
The solution for x(t) is, of course, just the first n-elements of this solution
for [x f ] ' .
There is one major drawback to this method of obtaining a homo
geneous differential equation equivalent to the nonhomogeneous state
equation. Suppose f is an m-vector. Then the matrix exponential in (52)
is of order n + m. In the solution (45) of the original state equation the
order of the matrix exponential is just n. Since m may very easily be large,
the increase in the order of the matrix exponential by m can result in a
substantial computing effort merely to eliminate computing the integral
appearing in (45).
It is possible to use an alternative procedure that will lead to a homo
geneous differential equation without increasing the order of the matrix
exponential to be evaluated. As might be expected, this is achieved at a
price. Let
(53)
or, equivalently,
(54)
252
STATE EQUATIONS
[Ch. 4
Then substitute (49) and (50) into this equation and rearrange terms to
obtain
(56)
If an
can be found that satisfies the following two-sided, linear, alge
braic matrix equation
(57)
then (56) becomes
(58)
which is the same homogeneous differential equation as in (46). Its solu
tion, therefore, is
(59)
in which
(60)
from the definition of y in (53). The solution for x(t) is obtained by sub
stituting y(t) from (59) and f (t) from the solution of (49) into (54).
The solution of the two-sided matrix equation (57) is not a trivial
matter.* Since Sf will be an n m matrix, (57) is equivalent to nm
linear algebraic equations for the nm unknown elements of Sf.
To illustrate these two methods of obtaining an equivalent homogene
ous differential equation, let us start with the state equation
* It turns out that, if the eigenvalues of F are different from those of A, the solution
for S can be expressed in closed form using some of the results of the next section. This
closed-form solution will be given in Problem 17. You will find a proof for that solution
in: J. S. Frame, "Matrix Functions and ApplicationsPart IV," IEEE Spectrum,
Vol. 1, No. 6, June 1964, pp. 123-131. A second closed form solution will be given in
Problem 35. You will find a proof for that solution in: A. Jameson, "Solution of the
Equation A X + X B = C by Inversion of an M M or N N Matrix," SI AM Jour, of
Applied Mathematics, Vol. 16, No. 5, Sept. 1968, pp. 1020-1023.
Sec. 4.3]
253
Observe that
Therefore
The matrices A , B, and F are obvious from the state equation and the
differential equation for f. The vector differential equation corresponding
to (51) is, therefore,
254
STATE EQUATIONS
[Ch. 4
whose solution is
Thus S exists, and the solution for y(t), and then x(t), can be obtained by
the use of this method.
In this example we converted the two-sided matrix equation for S into
an equivalent vector equation for a vector with the same elements as S.
Let us indicate how this is accomplished in general. Let s and k
denote the ith column vectors of S and K, respectively. Then the vector
equation
i
(61)
where
(62)
is equivalent to the two-sided matrix equation (57) in the sense that the
solution of (61) above yields values for all the elements of S.
Sec. 4.3]
255
Table 1
f(t)
f(0)
Since a network excitation vector is very apt to have elements that are
combinations of the elements of standard excitation vectors with different
values of , , and k, it may be necessary to combine the several differen
tial equations for the different standard excitation vectors into a single
differential equation. To make this point clear we shall consider a simple
example. Let
256
STATE EQUATIONS
[Ch. 4
Based on Table 1, there are five differential equations, each for a part of
the f-vector on the right; they are
Sec. 4.3]
257
MATRIX
EXPONENTIAL
or
258
STATE EQUATIONS
[Ch. 4
Finally, we take the inverse transform to get Y(t). Because we have taken
to = 0, Y(t) will also equal e . Hence
A t
(63)
This is very interesting. Let us apply it to the simple matrix considered
earlier in (40). The matrix (sU A), its determinant, and its inverse are
easily obtained as
(64)
4.4
FUNCTIONS OF A MATRIX
At
Sec. 4.4]
FUNCTIONS OF A MATRIX
259
(65)
(66)
260
STATE EQUATIONS
[Ch. 4
equation.
and
then
FUNCTIONS OF A MATRIX
Sec. 4.4]
261
- 1
(69)
(70)
where q(s) is an analytic " q u o t i e n t " function, which is regular at the
zeros of the polynamial a(s), and where g(s) is a " remainder " polynomial
whose order is less than the order of a(s).
Suppose the polynomial a(s) is an annihilating polynomial of matrix A;
that is, a(A) = 0. This means that, with s replaced by A in (70), we get
(71)
where f(A) is a function and g(A) is a polynomial.
This is a very interesting result. Remember that f(s) is an arbitrary
function; thus this result states that any analytic function of a matrix A
can be expressed as a polynomial in A of order no greater than one less than
the order of A.
We still have the job of determining the " r e m a i n d e r " polynomial
g(s). Before doing this, let us look at annihilating polynomials a little
further. The Cayley-Hamilton theorem assures as that a square matrix
has at least one annihilating polynomial. (You should show that this
implies there are an infinite number of annihilating polynomials of the
262
STATE EQUATIONS
[Ch. 4
matrix.) Let the one having the lowest degree and with unity leading
coefficient be labeled m(s) and be called the minimal
polynomial.
One interesting fact about a minimal polynomial is given in the follow
ing theorem:
Theorem.
The minimal polynomial of any square matrix s is a factor
of every annihilating polynomial of A.
This is easy to prove. Given a(s) and m(s), where m(s) is of no higher degree
than a(s), we can divide a(s) by m(s) to obtain a quotient qi(s) and a
remainder ri(s) of degree lower than that of m(s). After multiplying
through by m(s) the result will be
EIGENVALUES
We shall first assume that the eigenvalues of A are distinct and write
d(s) in factored form as
(74)
Now let us evaluate (73) at each of the eigenvalues s i . Since d(si) = 0,
Sec. 4.4]
FUNCTIONS OF A MATRIX
263
we find
(75)
There are n such relationships. When (72) is used for g(s) these n relation
ships become
(76)
The right-hand sides are known quantities, since f(s) is the originally
given function. This is a set of n equations in n unknown g coefficients.
Inversion of this set of equations gives the solution.
Let us illustrate the process with the same simple example considered
before. For the A in (40), the characteristic polynomial was given in (64).
They are repeated here:
i
A t
St
from which
Hence,
264
[Ch. 4
STATE EQUATIONS
or
What we have done in the last step is to rearrange the terms so that it
is not the powers of s on which we focus, but on the values f(s ). Since the
equations of (76) are all so similar, it must be possible to write this result
in a simpler form, and indeed it is. The result was first given by Lagrange
in the context of passing a polynomial of degree n 1 through n points.
It is called the Lagrange interpolation formula and converts the summation
j
Sec. 4.4]
FUNCTIONS OF A MATRIX
265
is easily obtained.
(78)
MULTIPLE
EIGENVALUES
If the eigenvalues of A are not distinct and there are repeated values,
a modification of this procedure is necessary. Let d(s) be written as
(79)
where the multiplicities are obviously r for the ith eigenvalue.
Let us now consider differentiating (73), after which we shall evaluate
the result for s = s . Except for the derivatives of the product (s)d(s),
the derivatives of f and g will be equal. Thus,
i
(80)
(81)
and for derivative orders j = 0, 1, 2, ..., (r 1); that is, for derivatives of
k
266
STATE EQUATIONS
[Ch. 4
Similar sets of equations will result for each distinct eigenvalue. The
entire result in matrix form will be
(82)
FUNCTIONS OF A MATRIX
Sec. 4.4]
267
st
Slt
since df(si)/ds = te
inversion, as
-2t
= te .
268
STATE EQUATIONS
[Ch. 4
by replacing s by A:
The remaining work is just arithmetic, after A is inserted here. The final
result is obtained from (71): f(A) =g(A)
leads to
This completes the example. (You should verify that this equation follows
from the immediately preceding one upon the insertion of A.)
CONSTITUENT
MATRICES
Let us look back at (82). This equation can be solved for the g coeffi
cients which are then inserted into the g(s) polynomial, just as in the case
of (76). Again we rearrange the terms so that the focus is on the elements
of the right-hand vector in (82), rather than on the powers of s. The re
arranged expression can be written as follows:
i
(83)
Sec. 4.4]
FUNCTIONS OF A MATRIX
269
It is only for later convenience that the K 's have been chosen as the
coefficients of the derivatives divided by the factorials rather than just
the derivatives themselves. When the eigenvalues are of single multipli
city this complicated expression reduces to just the first column, which is
simply (77).
The next step is to replace s by A. Recalling that g(A) = f(A),
we
now get
ij
(84)
ij
i j
ij
ij
THE R E S O L V E N T
MATRIX
ij
ij
-1
270
[Ch. 4
STATE EQUATIONS
(85)
where s are specific values of s'. Now substitute (85) in (84) and replace
s' by A inf(s') = (s s') . The result will be (sU A)
in terms of the
constituent matrices. Next, from (67) this can be written
i
-1
-1
(86)
The numerator of the right side is a matrix whose elements are poly
nomials in s since they are cofactors of the matrix (sU A). Since each
element of the numerator is divided by d(s), the whole thing is a matrix
of rational functions. A partial-fraction expansion of the right-hand side
can be carried out and leads to
(87)
In view of (85), this expression is exactly of the form of (84), and our
anticipation in using the same symbols K = K (A)
for the coefficients
of this partial-fraction expansion as for the constituent matrices is
justified. That is, the constituent matrices are the coefficient matrices in
the partial-fraction expansion of (sUA) .
The matrix (sUA)
is
called the resolvent matrix.
Given a matrix A and a function f(s), the determination of f(A) in the
form of (84) is carried out by expanding the resolvent matrix (sU A)
into partial fractions. The coefficients of the expansion (which are the
residues if the eigenvalues are simple) are the constituent matrices.
i j
-1
ij
-1
-1
Sec. 4.4]
FUNCTIONS OF A MATRIX
271
Finally, these constituent matrices are inserted into (84) to yield for the
matrix exponential
ALGORITHM
272
STATE EQUATIONS
[Ch. 4
-1
(91)
* Early reports of this algorithm may be found in: J. M. Souriau, " Une mthode pour
la Decomposition spectrale inversion des matrices," Compt. Rend., Vol. 227, pp.
1010-1011, 1948; D. K. Fadeev and I. S. Sominskii, "Collection of Problems on Higher
Algebra," 2nd ed. (in Russian), Gostekhizdat, Moscow, 1949; J. S. Frame, "A Simple
Recursion Formula for Inverting a Matrix," Bull. Am. Math. Soc, Vol. 55, p. 1045,
1949. H. E. Fettis, "A Method for Obtaining the Characteristic Equation of a Matrix and
Computing the Associated Model Columns," Quart. Appl. Math., Vol. 8, pp. 206-212,
1950.
Sec. 4.4]
FUNCTIONS OF A MATRIX
273
(92)
We shall now show that tr [P(s)] equals the derivative of d(s). Write
d(s) as
(93)
ij
i j
ij
ij
ij
kj
(94)
In the present case the determinant is d(s) = det (sU A). Hence, using
(94), we get
(95)
(96)
Finally, we substitute the expressions for P(s) and d(s) from (88) and (89)
274
STATE EQUATIONS
[Ch. 4
(97)
This set of expressions for the d coefficients, together with (91) for the
P matrices constitute an algorithm, with a finite number of steps, to
compute the resolvent matrix, ( s U A ) . We shall write them again,
side by side, showing the sequence of steps:
k
- 1
(98)
(check).
Sec. 4.4]
FUNCTIONS OF A MATRIX
275
The last equation in this set can be used as a check since all its components
have already been determined in the preceding steps. If the equation is
not satisfied, then an error (or more than one) has been made.
The important point concerning the resolving matrix algorithm is the
fact that all of the steps involve purely numerical operations; the variable
s does not appear. Consequently, although it might appear that there is a
prodigious amount of matrix arithmetic, the algorithm can be easily
programmed for a computer.
A side result of the algorithm is an evaluation of the inverse of A when
zero is not an eigenvalue of A. In that case d = d(0) 0. From (86),
(sU A)- = P(s)/d(s). Setting s = 0 gives A - 1 = P(0)/d(0), or
n
(99)
276
STATE EQUATIONS
[Ch. 4
In the last step the result was again rearranged into a single matrix.
Compare this with the earlier solution. Incidentally, since zero is not an
eigenvalue of A, (99) gives the collateral result that
RESOLVING POLYNOMIALS
- 1
* The quotient-difference algorithm is one of the most widely known of the methods
for determining the zeros of a polynomial. The algorithm is described in: P. Henrici,
Elements of Numerical Analysis, John Wiley, New York, 1964, Chap. 8.
Sec. 4.4]
FUNCTIONS OF A MATRIX
277
(100)
We shall call this the resolving equation. Although the vector elements
are square matrices of order n, these matrices are treated as single quanti
ties when the indicated matrix multiplication is interpreted. Thus the
matrix multiplication, when performed, gives terms in which a matrix is
multiplied by a scalara perfectly legitimate operation; for example, in
the first row of the product.
(101)
(102)
Rather than writing the general expression for this case, suppose, for
278
STATE EQUATIONS
[Ch. 4
2
Then
(103)
It is clear that the elements of the right-hand vector are easy to determine
in this case but that inverting the matrix will require considerable effort,
especially if n is much larger.
As a more explicit illustration let us consider the example given earlier
in which
(104)
With the resolving polynomials chosen according to (102), we get for (100)
(105)
Sec. 4.4]
FUNCTIONS OF A MATRIX
279
where the s 's are the eigenvalues. In this case the evaluation of f (A)
will require a large effort, but the matrix in (100) will be easy to invert.
Again we shall consider the particular case in which d(s) = (s s1)
(s s ) . Then
i
(107)
280
STATE EQUATIONS
[Ch. 4
The coefficient matrix is seen to be upper triangular in this case and can be
easily inverted. This is true in general for this selection of resolving
polynomials.
For the specific example treated before, given in (104), the resolving
equation becomes
On comparing the last step with (105) we find they are identical, as they
should be.
In this section we have treated a number of methods for evaluating a
function of a matrix. Each method has certain advantages and dis
advantages. Some are more readily applied to low-order matrices; others
lend themselves to numerical evaluation by computer. Our basic interest
is in determining equivalent closed-form expressions for the function e ,
which constitutes the solution to a homogeneous state equation.
A t
4.5
Let us briefly review what has been done in this chapter. We started
by considering the order of complexity of a network. We discovered
that the number of dynamically independent variables for RLC networks
equals the number of reactive elements, minus the number of all-capacitor
loops and the number of all-inductor cut-sets. For a network containing
multiterminal components (controlled sources, etc.), additional algebraic
constraints among capacitor voltages and inductor currents may be intro
duced, thus further reducing the order of complexity. We shall here
assume that in all cases the order of complexity is the same as would be
computed for an RLC network. If it turns out that the assumption is
false in a particular network, then it will be impossible to obtain the
equations in the desired form by the process we shall describe. An illustra
tion will be provided shortly.
Sec. 4.5]
281
(108b)
However, by the transformation x -> x + B e,
2
(109b)
where
Equation (109a) is the normal form for the state equation; x is the state
vector, and its elements are the state variables. Actually the state vector
in the last pair of equations is a linear combination of the original " s t a t e
v e c t o r " (with capacitor twig voltages and inductor link currents as
variables) and the source vector e. Even with this transformation the
second equation of the pairthe output equationmay still contain the
de/dt term. We shall shortly clarify the conditions under which this will
occur. For purposes of accurate reference we shall refer to the first equa
tion of either pair as the state equation and the second equation of either
pair as the output equation. The two equations together will be referred
to as the state equations.
Our next task was to solve the state equation, and this was done by
finding a solution first for the homogeneous equation (with e = 0).
Symbolically, this solution involves the matrix exponential e , so we
devoted some effort to determine methods for evaluating such functions
of a matrix. Once the matrix exponential is evaluated, the state vector x
is found from (45). We shall defer further consideration of the evaluation
of this integral and its ramifications to the next chapter.
A t
282
[Ch. 4
STATE EQUATIONS
We must now formalize the writing of the state equations (109) for a
given network and show that this is the general form. Let us observe at
the outset that it is possible to choose some variables other than the
capacitor voltages and inductor currents as state variables. In Fig. 6,
Fig. 6.
TOPOLOGICAL
CONSIDERATIONS
The first step is the selection of a normal tree (or normal forest).
Generally this is not unique. If there are no degeneracies (no all-capacitor
loops or all-inductor cut-sets), at least the reactive elements will be
uniquely assignable to the normal tree and cotreebut not the resistive
elements. However, when there are degeneracies, there will be a choice
even among the reactive elements.
According to our usual convention, in writing a loop or cut-set matrix
we first number the twigs and then the links. We shall here make a more
detailed convention of branch numbering and adopt the following order
within the twig and link categories:
Twigs
1.
2.
3.
4.
Voltage-source twigs
Capacitor twigs
Resistor twigs
Inductor twigs
Links
5.
6.
7.
8.
Capacitor links
Resistor links
Inductor links
Current-source links
Note that the terms " resistor twig " and " resistor link " include branches
of multiterminal devices, such as gyrators and controlled sources, whose
v-i relationships are algebraic like that of a resistor. We are imposing no
specific order in the numbering of such branches but are including them
Sec. 4.5]
283
(110a)
(110b)
R l
(111a)
(111b)
where the usual partitioning is Q = [U Q ], B = [B
U ] . The last step
follows from B = Q'l. If the current-and-voltage vectors partitioned
according to (110) are to be inserted here, we must also partition the Q
matrix conformally; that is, into four rows and columns. Now each row
of Q corresponds to an f-cut-set defined by a twig for the normal tree.
The columns correspond to links. If we arrange the columns and rows in
the conventional order decided upon, Q must take the form
l
284
STATE EQUATIONS
[Ch. 4
(112)
When this is inserted into the Kirchhoff equations in (111) the result can
be expanded into
(113a)
(113b)
(113c)
(113d)
(113e)
(113f)
(113g)
(113h)
To illustrate this partitioning return to the examples considered earlier
Sec. 4.5]
285
Since there are no inductor twigs, capacitor links, or current sources in the
network, the Q matrix is less than full.
l
ELIMINATING
UNWANTED
VARIABLES
Up to this point the discussion has been topological. We must now bring
in the v-i relationships. First, let us write them for the reactive elements;
thus
(114a)
and
(114b)
tl
t t
l l
l t
C t
l t
tl
286
STATE EQUATIONS
[Ch. 4
C l
, and i .
C l
(115)
Into the left-hand side we insert the capacitor v-i relationship from (114).
This side then becomes
(116)
C l
from (113e). To
(117)
(118)
which is the zero matrix when there are no loops containing just capacitors
and independent voltage sources. Then, with the last two equations in
serted into (115), we get
(119)
Sec. 4.5]
287
(120)
Into this we next insert the inductor v-i relationship from (114). The lefthand side becomes
(121)
L t
from (113). To
(122)
which equals L
l l
(123)
which is the zero matrix when there are no cut-sets containing just
288
[Ch. 4
STATE EQUATIONS
inductors and independent current sources. With the last two equations
inserted into (120) there results
(124)
R l
R l
R t
(125a)
(125b)
This is one of the hybrid-parameter forms.* It is the same form used in
Chapter 2 for the mixed-variable equations. For simple RLC networks
the G and G matrices will be diagonal, and G and G will be zero
matrices. More generally, none of the matrices need be diagonal, and they
may all be nonzero. There is no assurance at the outset that equations of
the form of (125) exist for a given network; but unless they exist, the
method we are developing will not work. This does not mean that state
equations do not exist, but only that our method will fail.
As a simple example, consider the network of Fig. 7. The normal tree
includes only the two capacitors. The resistor branches of the graph are
all links, but their v-i relationships have the form
ll
tt
lt
tl
Sec. 4.5]
(a)
289
(b)
R t
(126a)
(126b)
These are a pair of vector algebraic equations in the two variables i and
v . They will have a solution if the following matrices have inverses:
R l
R t
(127a)
and
(127b)
or
(128a)
290
[Ch. 4
STATE EQUATIONS
and
(128b)
R t
Ct
C t
Ll
L l
(129a)
(129b)
The symbols we have used for the matrices take into account the dimen
sions. Thus y and y relate a current vector to a voltage vector and so
have the dimensions of admittance. The H and G are dimensionless; they
correspond to the hybrid h- and hybrid g-matrices. In this form the equa
tions apply to time-varying as well as time-invariant networks. Also,
they can more readily be generalized to nonlinear networks. Look over
the preceding development and note that, in arriving at these equations,
we have used all the v-i relationships and all the topological (Kirchhoff)
relationships in (113) except the fitst and last, relating to the voltagesource currents and the current-source voltages. These two will be used in
the determination of output variables, assuming that elements of i
and v are output variables.
In fact, we should establish that, once the state equation and its
solution is available, all other variables can be expressed in terms of
the state variables v
and i , the source quantities v and i , and the
derivatives of the latter, as in (109). It is a matter only of looking over the
previously developed equations to verify that this is the case. It will be
left as an exercise for you to do at this point.
E
C t
L l
Sec. 4.5]
291
One point should be clear after contemplating (129) and the way in
which any output variable is expressed in terms of the state variables.
This is that source-voltage derivatives will appear only when there is
an all-capacitor loopand even then only when this loop includes a
voltage source, making C = Q C Q
nonzero. Similarly, sourcecurrent derivatives will appear only when there is an all-inductor cut
setand only when this cut-set includes a current source, making L =
Q _ L T TQ L J + L Q
nonzero. It is only in these cases that deriv
atives of source quantities can appear in the state equations.
C C
TIME-INVARIANT
L T
E C
L J
NETWORKS
(130)
(132)
(133)
(134)
292
[Ch. 4
STATE EQUATIONS
This is the desired result. What has been done here is to present a
procedure for arriving at a first-order vector differential equation for a
given network in the form of (131). However, we have not derived formu
las for the A, B1, and B matrices directly in terms of branch-parameter
matrices and submatrices of Q , because such formulas would be extremely
complicated and impossible to use. The result depends crucially on the
existence of the inverse of the matrices in (127) or (128) and of the matrices
C and L . Unfortunately, there are no simple necessary and sufficient
conditions to tell us when these inverses exist and when this procedure
will work.
2
RLC NETWORKS
There is, however, one class of networks for which the above-mentioned
procedure will always work. This is the class of time-invariant RLC
networks. It is of interest to carry through the development for this class,
because the results can be written out explicitly and provide insight into
the more general case.
The first simplification comes in the v-i relations of the resistor branches
in (125). There will be no coupling terms in the parameter matrices.
Hence G and G are both zero matrices, and the matrices G and G are
diagonal and, hence, nonsingular. Let us rename these matrices according
to the dimensions of their elements; G is dimensionally conductance, and
G is resistance. Set
tl
lt
ll
tt
ll
tt
from which
(135a)
(135b)
Equations (126) reduces to
(136a)
(136b)
The conditions for the existence of a solution reduce to the existence of
Sec. 4.5]
the inverse of K
or K
293
(137a)
(137b)
Define
(138a)
(138b)
so that K = R G and K = G R. Thus K and K will be nonsingular
if G and R are nonsingular. We shall shortly show that R and G can be
interpreted as loop-and-node parameter matrices and are consequently
nonsingular. Accepting this fact here we conclude that a solution for
(136) always exists. We now solve this equation for i
and v
and
substitute into (119) and (124). The details of the process are tedious and
will not be given here. The result will be
1
R l
R t
(139)
where
(140)
and
(141)
Note that the matrix G in the case of the reciprocal networks under
consideration is the negative transpose of H, which is something we
would expect. The form of (139) is the same as that of (129) for the general
network. The difference in the present case is that we have explicit
expressions for the coefficient matrices of the state and source vectors.
294
STATE EQUATIONS
[Ch. 4
(142)
State Equation
Resistor
Multiport
Matrices
Parameter
Matrices
RLC Networks
Voltage-Current
Relationships
Topological
Relationships
Sec. 4.5]
Table 2
SYSTEMATIC FORMULATION OF THE STATE EQUATIONS
295
296
[Ch. 4
STATE EQUATIONS
in mind that the row and column corresponding to sources are now absent.
Now, because of the order in which the elements are arranged in Z,
B U but is a rearrangement of the columns of U. Thus the partitioning
of B becomes
l
R R
l l
R R
t t
L L
C C
R R
C C
R R
Sec. 4.5]
297
formed by branches
(b)
(a)
298
STATE EQUATIONS
[Ch. 4
normal tree. There are a total of six reactive elements, but there is an
all-C loop (including a voltage source) and an all-L cut-set. Hence the
order of complexity will be 4, so there will be four state variables. One of
the capacitors must be a link; and one of the inductors, a twig. A possible
normal tree is shown in heavy lines in Fig. 8b. The branches are numbered
according to the scheme: twig v-source, C, R, and L; then link C, R, L,
and i-source. For purposes of simplification we will assume that the branch
numbers are also the numerical values of the elements: resistance in
ohms, inductance in henries, and capacitance in farads. With this choice
of normal tree, the branch parameters are
The various submatrices are evident from the partitioning. The parameter
matrices are now computed, as follows:
Sec. 4.5]
matrices:
299
300
STATE EQUATIONS
[Ch. 4
When all of this is inserted into (139) and the resulting equation is premultiplied by
we obtain
(143)
This is the state equation. Observe that it is not in normal form. If we set
Sec. 4.5]
301
and substitute it into the above state equation, we obtain the following
normal-form state equation:
(144)
You should verify this result. The new state variables x\ and x are linear
combinations of a capacitor voltage and the voltage of the source. They
cannot be identified on the network diagram as measurable voltages.
In looking over the effort just completed, you may despair at the large
amount of work involved. But observe the kinds of mathematical
operations that occur. They are largely matrix multiplications and addi
tions. Such operations are easily programmable for a computer, and so the
work reduces to writing a convenient program.
Note, in this case, that the parameter matrices could have been written
by inspection; for example, L is the inductance submatrix of f-loops
defined by inductor links 9 and 10. Each loop also contains inductor 6
whose orientation coincides with that of the first loop but is opposite to
that of the second. Hence
2
302
[Ch. 4
STATE EQUATIONS
the formulas, for any given problem we could also proceed by actually
retracing the steps of the derivation. This may sometimes require less
effort than inserting into formulas. You might carry out the solution for
this example that way and compare the amount of work.
CONSIDERATIONS
I N H A N D L I N G CONTROLLED
SOURCES
R l
Sec. 4.5]
(a)
(6)
Fig. 9.
303
304
STATE EQUATIONS
[Ch. 4
Using these, the resistor branch v-i relations according to (125) become
We now have all the submatrices to insert into (126). These equations
become
R t
R l
R l
R t
Sec. 4.5]
305
these steps is
is
and hence
- 1
306
STATE EQUATIONS
[Ch. 4
(145)
In looking over this example you will notice that many of the sub
matrices are sparse (meaning that many elements are zero). This results
in the need for many operations whose result is zero. It is possible to
carry out the same steps with the equations written in scalar form. This
obviates the need for writing large numbers of zeros, but sacrifices
compactness. You might parallel the steps of the solution with the equa
tions in scalar form to observe the difference.
It should also be noted in this example that G H' because of the
presence of g ; but if g = 0, then G will equal H'.
m
4.6
Sec. 4.6]
307
Similarly, from (120) and (121), together with the definition of L and
in (122) and (123), respectively, we get
These two can be combined into a single matrix equation. Then the deriva
tives can be eliminated by substituting the state equation of (129),
which applies to the general network. The result will be
(146)
Current
sources
Capacitor
subnetwork
Subnetwork of
resistor
branches
Inductor
subnetwork
Voltage sources
Fig. 10.
308
STATE EQUATIONS
[Ch. 4
Resistor
multi-port
Fig. 11.
Multiport interpretation.
L t
(147)
Sec. 4.6]
309
L t
(148)
We conclude that Z is the open-circuit impedance matrix of the resistive
multiport whose ports are the terminals of the inductor links, when all
other reactive elements are shorted and all sources are removed.
Matrices H and G can be found by similar means. We shall state the
results and ask you to supply the details.
(149a)
(149b)
Thus H is the current-transfer matrix of the resistive multiport, the
input ports being the terminals of the inductor links that are replaced by
current sources, and the output ports the shorted terminals of the capacitor
twigs when the capacitor links are open-circuited and all independent
sources are removed. The inductor twigs will be replaced by current
sources as dictated by KCL at the all-inductor cut-sets.
Finally, G is the voltage-transfer matrix of the resistive multiport,
the input ports being the terminals of the capacitor twigs that are re
placed by voltage sources, and the output ports the
open-circuited
terminals of the inductor links, when the inductor twigs are shorted and
all independent sources are removed.
To illustrate these interpretations return to the example considered in
Fig. 9. We shall carry through the calculations of each of the matrices
in turn. To find y, we are to open-circuit the two inductors, remove the
independent voltage source by shorting it, replace the capacitor with a
voltage source v having the same polarity as the capacitor voltage, and
then find the current i in this voltage source. The appropriate diagram
2
310
STATE EQUATIONS
[Ch. 4
Calculation of G
Fig. 12.
Calculation of y
Computation of y and G.
Since there are no inductor twigs, the diagram for the calculation of G
is the same as the one in Fig. 12. However, now the desired outputs are
the voltages across the open-circuited inductors. The computation is also
shown in Fig. 12.
Sec. 4.61
311
1 0
10
1 0
10
10
1 0
Calculation of Z
Fig. 13.
STATE EQUATIONS
Computation of Z and H.
Calculation of H
312
[Ch. 4
Sec. 4.6]
313
Fig. 14.
Computation of y and G.
mines the current of the controlled source, and KCL then gives i . The
voltages vg and v are also trivially obtained. The details are shown in
Fig. 14, with the result
2
10
314
STATE EQUATIONS
[Ch. 4
OUTPUT
E C
L J
EQUATIONS
L l
C l
C t
L t
L l
Consider the first set. Into the v-i relation i = C d /dt insert the K V L
equation in (113e). Remembering the assumption that Q
= 0, we
obtain i = C Q dv /dt.
But dv /dt
is expressed in terms of state
variables and sources by the state equation. When this is substituted
from (129) there results
C l
Cl
E C
C l
C C
Ct
Ct
(152a)
Sec. 4.6]
L t
315
gives
(152b)
Thus both capacitor link voltages and inductor twig currents can be
expressed as output variables in terms of state variables and source
variables. Aside from the matrices already found when forming the state
equations (i.e., L, C, y, and H), we require a knowledge of the topological
matrices Q
and Q , and the parameter matrices C , L , and L .
As for i and v , which are the variables complementary to the state
variables, we already have (146). This, however, is not exactly in the
desired form for output equations because of the term containing i and
v . However, (152) can be substituted for these, so that (146) can be
rewritten as
C C
C t
L L
t t
t l
L l
C l
L t
(153a)
(153b)
We observe that any reactive-component voltage or current variable can
be written as an output in the standard form in terms of state variables
and sources. Except for the matrices C and L L Q , this is done in
terms of matrices already found in writing the state equations.
This leaves the output variables v and i . Recall that the topological
equations expressing these variables in (113a) and (113h) had not been
used in arriving at the state equations. When the solutions for v
and
i of (126) are inserted into these equations, the result will be in terms of
state variables and sources. It will have the following form:
t
l l
l t
L L
R t
R l
(154)
316
STATE EQUATIONS
[Ch. 4
C t
Table 3
Multiport
interpretation
Output equations
To illustrate, consider the network shown in Fig. 15a. With the usual
equivalent circuit for the triode, the result is redrawn in (15b). Let the
desired output be the voltage v and the current i . Hence we insert a
current source across the right-hand terminals, with i = 0. In the graph
this current source is shown by branch 11. Branch 8 is the controlling
3
Sec. 4.6]
(a)
317
(b)
(c)
Fig. 15.
RC oscillator network.
Since there are no inductors and voltage sources, (146) and (154) reduce to
(155a)
(155b)
Thus to find y and G we open-circuit the capacitor link 6 and the current
source. (In this case the latter step is not needed since the only current
source has zero value.) We then replace the capacitor twigs by voltage
sources. The resulting network is shown in Fig. 16a. It is a matter of
computing the currents i1 i , and i in this resistive network by any
2
318
STATE EQUATIONS
[Ch. 4
(a)
Fig. 16.
(b)
so G is also easily
Sec. 4.6]
319
For this example, since the only source is a zero-value current source,
there was no need of finding H and Z; they will be multiplied b y zero in
the final equations anyway. We went to the trouble of finding them here
only for the purpose of illustration.
We can now easily write the state equation and the output equation.
For simplicity, let us use the following numerical values: C = J; r = R =
10; = 6. Then
C t
- 1
C t
C t
320
[Ch. 4
STATE EQUATIONS
PROBLEMS
1.
(a)
(b)
(c)
(d)
Fig. PI
2. For each of the networks in Fig. P1, draw at least one normal tree.
3. Show at least three normal trees for the network in Fig. 3 in the text.
4. How many normal trees are there in Fig. 5 in the text?
5. Equation 24 relating to Fig. 5 contains the derivative of the source
voltages. Make a transformation of variables so that the corresponding
equation in the new variables does not contain derivatives of source
voltages. Can you interpret the new variables in terms of the network?
6. Assume A and B are square matrices of the same order. Under what
condition is the following relation valid:
PROBLEMS
321
(a)
(b)
(c)
(d)
(e)
(f)
(a)
(b)
(c)
10. For a given matrix A, observe that d(s) equals the minimal polynomial
m(s) if the zeros of d(s) are simple. For each of the following matrices
322
STATE EQUATIONS
[Ch. 4
A t
(a)
(d)
(b)
(e)
(c)
(f)
11. For each of the matrices of Problem 10, set up equations (76) or (82) if
the eigenvalues are not simple, and solve them for the g coefficients
using f(s) = e . From this determine e and compare with the previous
result.
i
5t
A t
12. For each of the matrices of Problem 10, use the resolving matrix
algorithm to evaluate [sU A] .
Then make a partial-fraction expansion
of [ s U A ] to determine the constituent matrices of A.
-1
- 1
13. For each of the matrices of Problem 10, determine the constituent
matrices by the method of resolving polynomials, using f (s) = s .
i - 1
14. For each of the matrices of Problem 10, determine the constituent
matrices by the method of resolving polynomials using the set of poly
nomials in (106).
15. Evaluate the following matrix functions:
(a)
(b)
PROBLEMS
323
(c)
16. Solve the following sets of state equations with the state vector evaluated
using first (45), then (52), and finally (54) with (49) and (59). In each
case evaluate e by first finding the constituent matrices of A and then
applying (84).
A t
(a)
(b)
(c)
324
STATE EQUATIONS
(d)
(e)
(f)
[Ch. 4
(b)
(a)
(d)
(c)
(e)
(f)
(g)
(h)
(i)
(J)
(k)
Fig. P18
326
STATE EQUATIONS
[Ch. 4
where the K are the constituent matrices and the s are the eigenvalues
of A. We let r denote the multiplicity of s and k the number of
distinct eigenvalues. In each of the parts of Problem 16 solve for S
by using the above formula, when the eigenvalues of A and F are
different.
ij
18. For each of the networks in Fig. P18, derive the state equations by
(a) the formal matrix approach and (b) the method of resistive multiport
parameter evaluation.
19. Derive the state equations for the network in Fig. P19 by both the
matrix method and the method of resistive multiport parameter evalu
ation. For the pentode use the equivalent circuit shown. The response
variables are the volages across all the inductors.
(b)
(a)
Fig. P19
22. Derive the state equations for the single-stage amplifier shown in Fig.
P22a. Use the hybrid-pi-equivalent circuit for the transistor, as shown in
Fig. P22b. The voltage v is the network response.
2
PROBLEMS
(b)
(a)
(c)
Fig. P20
(b)
(a)
Fig. P21
(a)
(b)
Fig. P22
327
328
STATE EQUATIONS
[Ch. 4
23. Derive the state equations for the network of Fig. P23. Solve for the
state-transition matrix. Set up the equation for the solution of the
response vector w = [v v ]'. The capacitors are uncharged at time
to = 0. Carry out the solution.
2
Fig. P23
(b)
(a)
Fig. P24
PROBLEMS
(a)
(b)
(o)
(d)
Fig. P25
(a)
(b)
Fig. P26
329
330
[Ch. 4
STATE EQUATIONS
27. Derive the state equations for the three-stage amplifier network shown
in Fig. P27a. Use the triode model shown in Fig. P27b.
(a)
(b)
Fig. P27
R l
R t
R t
(a)
(d)
(b)
(e)
(c)
29. In the networks in Figs. 4 and 5 in the text, solve for i and v
using the method based on (125) discussed in the text. Then solve for
i and v using one of the other representations in Problem 28.
Do some representations require less computational effort? Discuss.
R l
R t
R t
R t
PROBLEMS
331
30. Derive the state equations for the network shown in Fig. P30. Use the
transistor model shown in Fig. P26b.
Fig. P30
31. Derive the state equations for the differential amplifier network shown
in Fig. P31. Use the transistor model shown in Fig. P26b.
Fig. P31
32. Derive the state equations for the network shown in Fig. P32 using the
transistor model shown in Fig. P26b.
Fig. P32
332
STATE EQUATIONS
[Ch. 4
33. Determine the state equation for each of the oscillator networks shown
in Figs. P33a through c by using the transistor model shown in Fig. P33d.
(a)
(b)
(c)
(d)
Fig. P33
Fig. P34
n
35. Suppose A and F in (57) have no eigenvalues in common. Let
i=0
m
and
m - i
respectively.
F,
n - i
PROBLEMS
333
Let
In each of the parts of Problem 16, solve for S using each of the above
formulas, when the eigenvalues of A and F are different.
36. (a) For the network shown in Fig. P36a, specify the number of natural
frequencies and the number of nonzero natural frequencies.
(b) Repeat for the network in Fig. P36b. State whether the values of the
natural frequencies (not their number) are the same or different in the
two cases. Explain.
(b)
(a)
Fig. P36
by the resolvent-matrix
334
[Ch. 4
STATE EQUATIONS
by (99).
Type of Branch
Independent voltage source
Independent current source
Capacitor
Resistor
Inductor
Fig. P39
(a)
PROBLEMS
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
335
.5.
INTEGRAL SOLUTIONS
We have now reached a level of ability where given any linear, timeinvariant network and an arbitrary excitation the complete response can
be obtained. The complex-frequency-domain methods of Chapters 2 and 3
are very useful in determining an analytic expression for the response.
In particular, when the network is initially relaxed, we have seen that it
can be characterized by its transfer function. Hence it is not even neces
sary that the network be given, as long as its transfer function is known.
The time-domain methods of Chapter 4 are also useful in establishing an
analytic expression for the response, but they are particularly suited to
numerical evaluation of the response in the time domain. As with the
complex-frequency-domain methods, if the network is initially relaxed,
it need not be given; it is sufficient to know the integral relation giving the
response in terms of the network excitation.
In this chapter we shall be concerned first with the problem of determin
ing the response of a network to an arbitrary excitationnot when the
network is given, but when its response to some standard excitations is
given. Step and impulse functions will be used in defining these standard
excitations. We shall establish analytic results by using both time-domain
and complex-frequency-domain methods. Furthermore, we shall treat the
problem of obtaining numerical results in the time domain.
To start with, let us relate the network response in the complexfrequency domain to the response in the time domain. To achieve this
goal we shall need a result from the theory of Laplace transforms. How
ever, this result is probably less familiar than such standard things as
partial-fraction expansions. Hence we shall spend some time discussing it.
336
Sec. 5.1]
5.1.
CONVOLUTION THEOREM
337
CONVOLUTION THEOREM
Fig.l
voltages v and v , and the current i . The transforms of the excitation
and response vectors for this network are as follows:
e
(i)
In this example H is of order (3, 2), but the relationship is quite general.
In the general case E is a p-vector; and W, an r-vector; hence H is of order
(r,p).
Now, with W(s) known, w(t) is most often found b y first making a
partial-fraction expansion of W(s) and then inverting each term in the
338
INTEGRAL SOLUTIONS
[Ch. 5
(2)
then, from the definition of the Laplace transform, we can conclude that
whatever is in the parentheses is the desired response vector. What we
plan to do does not depend on the interpretations of H(s) as a transfer
matrix and E(s) as an excitation vector. Hence we shall use a more general
notation in developing this result.
Let F1(s) and F (s) be the Laplace transforms of the matrix functions
F1(t) = [f (t)]
and F (t) = [f (t)],
respectively; that is,
2
lij
2ij
(3a)
(3b)
(4)
The last step is clearly justifiable, since each integral in the second line
is a constant with respect to the other variable of integration. The product
of integrals in the second line can be interpreted as a double integral
over an area whose coordinate axes are u and v. The integration is to be
performed over the entire first quadrant, as indicated in Fig. 2a.
CONVOLUTION THEOREM
Sec. 5.1]
339
(b)
(a)
Fig. 2.
Region of integration.
(5)
(6)
Computing the partial derivatives from (5) and substituting here leads
to the result that d dt = du dv.
To complete the change of variables we must determine the new limits
of integration. Note that, since t=u + v = + v, and since v takes on
only positive values, t can be no less than T . The line t = in the T - t
plane bisects the first quadrant; thus the desired area of integration is the
the area lying between this line and the t-axis, as shown in Fig. 2b.
In order to cover this area we first integrate with respect to from = 0
to T = t; then we integrate with respect to t from 0 to infinity.
With the change of variables given in (5) and with the limits changed
as discussed, (4) yields
(7)
* See Wilfred Kaplan, Advanced Calculus, Addison-Wesley, Cambridge, Mass., 1953,
p. 200.
340
INTEGRAL SOLUTIONS
[Ch.
This is exactly the form of (2), so that we can identify the quantity in
the parentheses as F(t) = { F ( s ) } . It should be clear that, if in (3) we
write F1(s) in terms of the dummy variable and F (s) in terms of u, then
in the result given by (7) the arguments of F1 and F will be interchanged.
The final result can therefore be written in the following two alternative
forms:
-1
(8)
(9)
The operation performed on the two matrices Fi(t) and F (t) repres
ented by these expressions is called convolution. The two matrices are
said to be convolved. The convolution of two matrices is often denoted by
the short-hand notation F1_*F . We can state the above result in the form
of a theorem, as follows.
2
Convolution
Theorem.
Let the two matrices F1(t) and F (t) be Laplace
transformable and have the transforms F1(s) and F ( s ) , respectively.
The
product of F 1 ( S ) with F ( s ) , if they are conformable, is the Laplace transform
of the convolution F1(t) with
F (t).
2
(10)
where
(11)
While we are still in this general notation let us state another useful
result concerning the derivative of the convolution of two matrices.
If the matrices F1(t) and F (t), in addition to being Laplace transformable,
are also differentiable for t > 0 (they need be continuous only at t = 0),
then their convolution will also be differentiable for t > 0. The derivative
will be
2
(12)
Sec. 5.2]
IMPULSE RESPONSE
341
or
(13)
5.2
IMPULSE RESPONSE
(14)
The reason for this choice of notation will be clear shortly. Now the con
volution theorem applied to (1) provides us with the result
(15)
342
INTEGRAL SOLUTIONS
[Ch.
* The impulse function (t) is not an ordinary point function; rather, it is a generalized
function. Symbolically, we may manipulate mathematical relations involving the impulse
function and its derivatives as we would relations involving only ordinary point
functions. On the other hand, mathematical precision requires that we view each function
as a generalized function, and each operation as defined on the space of generalized
functions. A short treatment of the theory of generalized functions is given in Appendix 1.
Here it is enough to observe that the impulse function satisfies the following relations.
With
a<<b,
For other values of T outside the range [a, b], each of the above integrals yields zero.
Sec. 5.2]
IMPULSE RESPONSE
343
Similarly, let w (t) be the response vector resulting from e (t); that is,
w (t) is the collection of all the scalar responses when there is a single
excitation and this excitation is an impulse. Let these w vectors be
arranged as the columns of an r p matrix designated W (t) and called
the impulse response of the network. Note carefully that W (t) is a set of
response vectors, one column for each column of E (t). It is thus not an
observable response in the same sense that each of its columns is an ob
servable response. On the other hand, the sum of elements in each row of
W (t) is an observable (scalar) responsethe response to the sum of all
excitations, these excitations all being impulses. Let us illustrate with the
example of Fig. 1.
j
ol
(16)
Equation (1) relates corresponding columns of the transforms of E (t)
and W (t). Hence, b y using (16), we obtain
(17)
344
INTEGRAL SOLUTIONS
[Ch.
or, equivalently,
(18)
In words, the last equation states that the inverse transform of the network
transfer function is equal to the impulse response of the network. We
anticipated this result in using the notation in (14).
Let us return now to (15). We see that this equation expresses the fact
that, once the impulse response of an initially relaxed network is known,
the response to any other excitation e(t) is determined. What we must do
is premultiply the excitation at each point T b y the impulse responsenot
at the same point, but at a point (t )and then integrate. Another
viewpoint is that the input vector is " weighted " by the impulse response.
This leads to the name of " weighting matrix " used by some authors for
the impulse response.*
Let us elaborate a little on the concept of weighting. Perhaps a simple
example would be more satisfying as a means of communicating this
point. Therefore let us consider a single-input, single-output network with
the transfer function
Fig. 3.
Sec. 5.2]
IMPULSE RESPONSE
Fig. 4.
345
T R A N S F E R F U N C T I O N NONZERO AT
INFINITY
346
[Ch.
INTEGRAL SOLUTIONS
(a)
(b)
Fig. 5.
Illustration of convolution
(19)
where H(s) has a zero at infinity. The impulse response will then be
(20)
Sec. 5.2]
IMPULSE RESPONSE
347
this expression in the second form of (15) to find the response of the net
work to an excitation e(t). The result will be
(21)
The last step follows from the properties of impulse functions and their
derivatives, as given in Appendix 1. This is the general form of the con
volution integral.
A L T E R N A T I V E DERIVATION OF CONVOLUTION
INTEGRAL
(22a)
Recall also that the solution of the differential equation for the state
vector and, from (22b), the solution for the output vector, are
(23a)
(23b)
Observe that part of the response is a consequence of a nonzero initial
state. We shall let
(24)
348
INTEGRAL SOLUTIONS
[Ch. 5
f
denote this part. The superscript f was chosen because w (t) is known as
the free response of the network; this is a sensible name, since w (t) is
independent of the controlling influence of the network excitation. The
remaining part of the response stems from nonzero excitation of the net
work. We shall let
f
(25)
denote this part. The superscript c was chosen because we may think of
w (t) as the controlled response, since w (t) is controlled b y the network
excitation. As a consequence of these definitions we have
c
(26)
f
(27)
Observe that (21) and (27) are identical after the obvious identification
of W (t) with Ce B, K with D, and K with D; the theorem is thus
proved.
One further comment on the need for generalized functions is in order.
Suppose we restrict e(t) and w(t) to the set of ordinary vector-valued
point functions. As (27) then makes evident, if D 0, we must also require
that e(t) have a derivative for t > 0. A restriction such as this is not desir
able, since it is often of value to examine w(t) when elements of e(t) are
discontinuous functions or functions with slope discontinuities. Two very
typical functions of this type are the unit step function
At
IMPULSE RESPONSE
Sec. 5.2]
349
Fig. 6.
Example.
(You may go through the details of finding this; try it by finding the state
equations.) We see that the first element of H(s) has a zero at infinity,
whereas the second one is nonzero at infinity. The impulse response,
therefore, is
350
INTEGRAL SOLUTIONS
[Ch. 5
Sec. 5.3]
STEP RESPONSE
351
Y o u should carry out the details of these calculations and verify that
both expressions give the same value of w at t = 1.
5.3
STEP RESPONSE
Similarly, let w (t) denote the response vector resulting from e (t);
that is, w j(t) is the collection of all the scalar responses when there is a
single excitation and this excitation is a unit step at the jth input.
Suppose these w vectors are arranged as the columns of an r p matrix
designated W (t) and called the step response of the network.
nj
u j
nj
352
[Ch. 5
INTEGRAL SOLUTIONS
(29)
(30)
(31)
The initial value of the step response is readily found from (29), by using
the initial-value theorem. Provided H ( ) is finite, it will be
(32)
where is a real scalar. We conclude that the initial value of the step
response of a network will be zero if the transfer function has a zero at infinity.
Further, if H ( ) is nonzero but finite, the initial value of the step response
will be nonzero and, in addition, the impulse response will itself contain an
impulse. If, on the other hand, H ( ) is nonzero but infiniteH(s) has a
simple pole at infinitythen the step response has an impulse at t = 0,
and the impulse response contains the derivative of an impulse at t = 0.
Note that this does violence to our ordinary concepts of calculus, as em
bodied in (30) and (31). If W (t) is an integrable matrix, then (30) tells
us that W ( 0 ) should be zero (simply by putting zero as the upper limit);
however, if we admit impulses, then our constitution must be strong to
withstand the consequences. Note, however, that if W (t) contains a
first-order impulse, W (t) is not impulsive; further, if W (t) contains the
derivative of an impulse, W (t) has an impulse but not the derivative of an
impulse. Hence W (t) is always better behaved than W (t).
Let us now return to our original task and assume that an arbitrary
Sec. 5 . 3 ]
STEP RESPONSE
353
(33)
(34)
(35)
In each case we have used (29) to obtain the far-right side. To find w(t)
we shall now use the convolution theorem. Focus attention on (33). This
can be written
(36)
where
(37)
B y using the convolution theorem, we can write
(38)
(39)
354
INTEGRAL SOLUTIONS
[Ch. 5
(40)
(41)
This will require that e(t) or W ( t ) , as the case may be, be differentiable
and that, correspondingly, e(0) or W ( 0 ) be finite.
These same expressions can be obtained in an alternative manner,
starting from (34) and (35). To use (34) let us first write
u
(42)
We can now use the convolution theorem on (34). The result will be
(43)
which is the same as (41). In a similar manner, (40) can be obtained starting
from (35). The details are left to you.
For future reference we shall collect all of the forms of these expressions
that have been derived. They are as follows:
(44)
STEP RESPONSE
Sec. 5.3]
355
(45)
(46)
Note that if we differentiate W (t) instead, then, b y using (31) for W and
inserting it into (45), we get back the convolution integral.
Let us use the second form of (44), which should be simpler than the
first form, since V1(t ) = 2[u(t ) u(t T 1)], which is simply the
u
356
INTEGRAL SOLUTIONS
Fig. 7.
[Ch. 5
Example.
SUPERPOSITION PRINCIPLE
Sec. 5.4]
5.4
357
SUPERPOSITION PRINCIPLE
S U P E R P O S I T I O N I N TERMS OF I M P U L S E S
Consider the excitation function sketched in Fig. 8a. The positive time
axis is divided into a sequence of equal-length intervals, the length being
T . It is not necessary that the intervals be equal, but the task ahead is
easier to formulate if they are.
N o w consider the sequence of impulses labeled f(t) shown in Fig. 8b.
The impulse at the point k has a strength e(k), which is the area of
a rectangle formed by the base T and the height of the curve of Fig. 8a
at the point k . The rectangle is shown shaded. The heights of the arrows
in the figure have been drawn proportional to this strength. However,
remember that the impulses are all of infinite height. Hence, for any
finite T , no matter how small, the string of impulses is not a good pointwise representation of the excitation function, which is everywhere
finite. Nevertheless, let us compute the response of the network to this
sequence of impulses. Now f(t) is not a point function; it is a generalized
function that can be expressed as
(47)
358
INTEGRAL SOLUTIONS
[Ch. 5
(a)
(b)
Fig. 8.
w will be equal to the strength of the impulse times the impulse response,
suitably displaced. Thus
k
(48)
is the response at time t to the impulse at time k .
Let us now concentrate on a particular point on the axis, which we can
call T . For a given value of T , this point will be k . If we let get
smaller, we shall have to increase k proportionately so that the value
T = kT will stay the same, since it refers to a fixed point on the axis.
Hence, (47) and (48) can be rewritten as follows:
(49)
(50)
Sec. 5.4]
SUPERPOSITION PRINCIPLE
359
(51)
(52)
where
(53)
360
INTEGRAL SOLUTIONS
[Ch. 5
(55)
N o w w (t) is the response when the jth input is excited by a unit impulse
at time zero and all other inputs have zero excitation. Thus, in the multiinput, multi-output case, an appropriate interpretation of (52) is the
following: The response to the excitation e(t) is the superposition of a set of
responses, each of which is the response of the network to excitation at a
single input, the excitation consisting of a sequence of impulses.
j
S U P E R P O S I T I O N I N TERMS OF STEPS
Fig. 9.
SUPERPOSITION PRINCIPLE
Sec. 5.4]
361
(56)
where the dot indicates differentiation. The factor in brackets is the value
of the step, whereas w (t K ) is the response to a displaced step func
tion.
The total response will be the sum of the contributions for each step.
Again, if we focus attention on the point = K and take the limit as
T approaches zero, we shall get
u
(57)
(58)
(59)
This expression is identical with the first one in (44). We have now demon
strated that the response to an excitation e(t) can be regarded as the
superposition of the responses to a series of step functions that represent
the excitation.
362
INTEGRAL SOLUTIONS
[Ch. 5
(60)
which may be rewritten as
(61)
where ( ) denotes the jth element of and w ( t ) denotes the jth column
vector of W ( t ) . Now w (t) is the response when the jth input is excited
by a unit step at time zero and all other inputs have zero excitation.
Therefore the following interpretation of (61) can be given: The response
to an excitation is the superposition of a set of responses, each of which
is the response of the network to an excitation at a single input, the excita
tion consisting of a sequence of step functions.
i
5.5
uj
uj
NUMERICAL SOLUTIONS
NUMERICAL SOLUTIONS
Sec. 5.5]
363
(a)
(b)
Fig. 10.
364
INTEGRAL SOLUTIONS
[Ch.5
(62)
(63)
where
Let us see the implication of this equation. We notice that the sum on the
right is simply a sum of real numbers, not functions. Thus we can get an
approximate idea of the response simply by adding these numbers,
without integrating functions.
* Since n is the only variable in this equation, we can write this in more conventional
form as
and observe that it is the Cauchy product of two time sequences, for e and w .
NUMERICAL SOLUTIONS
Sec. 5,5]
365
To make this point a little clearer, let us find the approximate response
at t/T= 2, 4, 6, 8, and 10 for the example of Fig. 10, by using the values
given in Table 1. (The invervals chosen are too large for any accuracy,
but the example suffices as an illustration.) Lines 1 and 2 of this table are
found by reading the values of e(kT) and w (kT) from the graphs. The
remaining odd-numbered lines, the lines associated with w [(n k) T],
n = 2, 4 , 6 , 8, and 10, are obtained by copying line 1 backwards, starting at
the column corresponding to k = n 1. The elements in each of these lines,
when multiplied by their corresponding entries in line 2, give the entries
in the even-numbered lines, those associated with w [(n
k)T]e(kT).
The sum of the entries from k = 1 to k = n in each of the evennumbered lines is w (nT)/T.
Thus
This computation has given a numerical value for the response at a few
selected points. The final tabulated values are a time sequence. With
these values, the time series can be written as
(66)
This method of representing everything by time series leads to the socalled z-transform method of analysis used in sampled-data systems.
The same concept of time series is also used in time-domain synthesis.
Quite often the synthesis problem is specified by means of a curve for the
excitation e(t) and a curve for the response w(t) that is desired. Then one
of the procedures in use is to represent e(t) and w(t) as time series and to
use simultaneous equations derived from (66) to find the time series for
w (t). The synthesis then proceeds by finding H(s). The mathematical
problems that arise are too numerous for us to consider this question any
further, and so we leave this application to advanced courses on synthesis.
M U L T I - I N P U T , MULTI-OUTPUT
NETWORKS
3
4
5
6
7
8
9
10
11
12
1
2
Line
No.
Table 1
w [(2-k)T]
w [(2-k)T] e(kT)
t<; [(4-k)T]
w [(4-k)T] e ( k T )
w [(6-k)T]
w [(6-k)T] e(kT)
w [(S-k)T]
w [(S-k)T] e(kT)
w [(lO-k)T]
w [(\0-k)T] e(kT)
w (kT)
e(kT)
k
0.0
0.5
1.0
7.0
14.0
8.5
17.0
2.5
5.0
-1.7
3.4
0.5
2.0
0.0
0.0
3.0
8.1
9.0
24.6
6.0
16.2
-0.3
-0.81
3.0
2.7
0.5
2.0
7.0
28.0
8.5
34.0
2.5
10.0
7.0
4,0
0.0
0.0
3.0
24.0
9.0
72.0
6.0
48.0
9.0
8.0
0.5
6.0
7.0
84.0
8.5
102.0
8.5
12.0
0.0
0.0
3.0
33.0
9.0
99.0
6.0
11.0
0.5
3.0
7.0
42.0
2.5
6.0
0.0
0.0
3.0
7.5
-0.3
2.5
0.5
0.25
-1.7
0.5
0.0
0.0
0.0
10
366
INTEGRAL SOLUTIONS
[Ch. 5
Sec. 5.5]
NUMERICAL SOLUTIONS
367
(68b)
STATE
RESPONSE
Thus far we have sought to solve for the network response directly
without consideration of the network state. Since the network response
could be, without loss of generality, the network state, all that we have
concluded till now concerning the response vector applies equally well
to the state vector. Because, in the method about to be developed, no
significant simplicity is achieved by assuming that the network is initially
relaxed, we shall no longer make this assumption.
According to (23) the state vector is the sum of the free-state response
(70)
and the controlled-state
response
(71)
368
INTEGRAL SOLUTIONS
[Ch. 5
that is,
(72)
(73a)
(73b)
(75)
Sec. 5.5]
NUMERICAL SOLUTIONS
369
A T
have knowledge of e
at only one point in time; namely, t = T. This fact
largely alleviates the computer storage-space problem referred to earlier.
The solution for the network response now proceeds as follows: We
first let x(0) equal the initial state x(0); then (75) is used to solve for the
state vector x(nT) as n successively assumes the values 1, 2, 3 , . . . .
For each value of n, the network response at t = nT, according to (22b),
is given by
(76)
This gives the time sequence of the network at the sampling points nT.
The time-series approximation to the network response then follows
directly.
Example
Let us now illustrate these results with a simple example. Suppose
0.2; then
370
INTEGRAL SOLUTIONS
[Ch. 5
Let the network be initially relaxed and let the excitation be the function
e(t) given in Fig. 10. The values for e(nT) are given in the top line of
Table 2. Using the recursion formula and numerical values above, computations for n = 1, . . . , 5 yield the results given in Table 2.
Table 2
PROPAGATING ERRORS
A T
A T
( n n 0 )
(78)
Sec. 5.5]
NUMERICAL SOLUTIONS
371
A T
A T
A T
( n - n i )
(79)
be the propagated error in x ( n T ) , assuming n > m. A question that is of
great significance is: How does the " size " of the error behave as n is increased? Here we are dealing with an error vector; we specify the " size "
of a vector in terms of its norm.
For a refresher about norms of vectors and matrices, see Chapter 1.
There, the norm of a vector was written as |||| and defined as a nonnegative number having the following properties:
1.
2.
|||| =
3.
||
| | ||||,
||
triangle
where
|| || +
1
|| ||,
2
and
inequality.)
(80a)
(80b)
(80c)
Of these, the most common and familiar one is the Euclidean norm, which
372
[Ch.
INTEGRAL SOLUTIONS
(83a)
where
is the eigenvalue of K*K with the
largest magnitude;
(83b)
(83c)
Our interest is in determining the norm of the error in the state response.
In particular, we want to know in (79) if || || < || || when n > m. In
this equation the two errors are related through the matrix ( e )
K
Taking the norm of both sides of (79), and using (81) and (82) leads to
n
A T
( n m
(84)
A T
Sec. 5.6]
NUMERICAL EVALUATION OF e
A T
373
we find
For this example, any of the norms shows that the norm of the error vector
decreases as n increases.
5.6
NUMERICAL EVALUATION OF e
A T
0 . 2 A
- 0 . 2
- 0 . 9
0 . 2 A
A T
A T
A T
(85)
374
[Ch. 5
INTEGRAL SOLUTIONS
(87)
A T
and
(88)
A T
A T
; that is,
(89)
A T
Sec. 5.6]
NUMERICAL EVALUATION OF e
A T
375
(90)
- 1
(91)
A T
376
INTEGRAL SOLUTIONS
[Ch. 5
The last line follows from the fact that (K + 1)!(K + 2)'/(K + 1 + l)! 1
for all I 0. Now, let K be the least non-negative value of K such that
K + 2 > | | A | | T ; then for all KKo,
we have ||A||T/(K + 2) < 1 and
0
K,
0
(92)
- A T
An upper bound on | |
- A T
A T
(94)
(95)
Sec. 6.5]
NUMERICAL EVALUATION OF e
A T
377
and hence < . Thus the values of K K that satisfy (95) also satisfy
the previously stated criterion for selecting K. Of all the satisfactory
values of K, it is suggested that K should be set equal to the least value
that satisfies (95), since the number of arithmetic operations then needed
to evaluate the truncated power series (87) will be minimized.
0
COMPUTATIONAL
ERRORS
In the previous section we decomposed the state response x(t) into the
sum of the free response x (t) and the controlled response x (t). As shown
in (74), we found that
f
(96)
(97)
ERRORS I N F R E E - S T A T E R E S P O N S E
f
A T
378
INTEGRAL SOLUTIONS
[Ch. 5
will be applied to relations other than just (99). In those other cases n will
not be the exponent.
B y the properties of matrix norms, we show that
(100)
The second line follows from the binomial theorem. The next line results
from the triangle inequality and from (82). Finally, the last line is a result
of the binomial theorem applied backwards. If we require that K
satisfy the inequality relation (95), we know from that equation that
||R|| < | | e - | | . Furthermore, it is easily shown (Problem 15) that
A T
- 1
a n d , therefore, that ||R|| < \\e^ \\. With the added observation that
| | e | | ||A|| + ||R|| and the reasonable assumption that < 1, we get
A T
(101)
(102)
A T
Sec. 5.6]
NUMERICAL EVALUATION OF e
A T
379
(103)
The second line is just the first line with the terms grouped differently.
The last line follows from the fact that [1 (1 ) ] is a bounded, increas
ing function of n that never exceeds n. It is an elementary calculus
problem to find the maximum of the right-hand side of (103) with respect
to n. In doing so, we obtain
n
(104)
ERRORS I N CONTROLLED-STATE R E S P O N S E
c
A T
n -
= 1
(105)
380
INTEGRAL SOLUTIONS
[Ch. 5
After taking the norm of both sides of this equation and applying the
inequality in (102), we establish that
(106)
(107)
n - ;
The equality at the right follows from the fact that a series of the type
m=o < is equal to /(l ) for < 1. We see b y this inequality that
| | ( n T ) | | is bounded and tends to zero with . However, for reasons dis
cussed previously, when computing a numerical value for x ( n T ) , the
accuracy limitations of a computer preclude the error in numerical
evaluation of x ( n T ) being bounded in accord with (108) if is too small.
_
Example
Let us illustrate the many ideas we have developed in this section with
an example. Let us use the same A and T as in the last example; thus,
and T=
0.2. For the norm of A we shall arbitrarily select the least of the
Sec. 5.6]
NUMERICAL EVALUATION OF e
A T
381
two norms (83a) and (83c). We have ||A||1 = 3 and ||A|| = 2; therefore
we shall use the latter norm. To simplify the example, certainly not for
accuracy, we shall require only 0.001.
Recall that K is the least non-negative (integer) value of K such that
K + 2 > | | A | | T . Since ||A||T=0.4,
we find that K = 0. Thus, by (95),
we must find a K 0 such that
and
Observe that
as
||A||oo
is bounded
382
[Ch.
INTEGRAL SOLUTIONS
e ( t ) = = 0.
PROBLEMS
1. In the text, the concept of the convolution of two matrix functions is
introduced; extend the concept to more than two matrix functions.
2. Prove that the convolution of scalar functions shares the following
algebraic properties with ordinary multiplication: Iffi,f , andj are
integrable functions (so thatf|*f = J fi(x)f (t x) dx is defined, as is
/ * / and fffi),
then
2
(a)
fi*f
=f *fi
(commutative law)
(*>)
fi*(f2*f3)
= (fi*f )*f3
(c)
u*f=f*u
(d)
/i*(/2+/8)=/i*/a+/i*/8
(associative law)
(distributivelaw)
PROBLEMS
383
(b)
(a)
(d)
(c)
(f)
(e)
Fig. P4
The unknown matrix function is F(t), and Gi(t) and G (t) are square
matrices and are known (integrable) matrix functions.
2
(a)
384
INTEGRAL SOLUTIONS
[Ch. 5
(b)
(c)
(a)
(b)
(d)
(c)
(e)
Fig. P5
Fig. P6
PROBLEMS
385
()
(b)
11. The triangular voltage pulse shown in Fig. P11a is applied to the network
of Fig. P11b. Find the output-voltage response for all time by using
the convolution theorem.
12. Use the convolution-integral theorem to prove the translation (shifting)
theorem of Laplace-transform theory.
386
INTEGRAL SOLUTIONS
[Ch. 5
(a)
(b)
Fig. P11
13. The trapezoidal pulse shown in Fig. P13a is applied to the network of
Fig. P13b. Find the output-voltage response for all time by using the
convolution theorem.
(a)
(b)
Fig. P13
14. The network of Fig. P14a is excited by two voltage sources. The
voltages as functions of time are shown in Fig. P14b. Using the convolu
tion theorem, compute the indicated network responses v, ii, and i
for all time.
2
AT
||e
- A T
- 1
|| .
Fig. P14
(b)
Find the approximate response of the network for 0 < t < 2, using timeseries representations. Estimate the maximum error in the solution for
the chosen interval.
(a)
PROBLEMS
387
388
INTEGRAL SOLUTIONS
[Ch. 5
Use the staircase approximation and the step response. Estimate the
error.
1 8 . The network of Fig. P18a is excited hy the function of Fig. P18b. Find
the approximate response v (t) for 0 < t < 5.
2
(a)
(b)
Fig. P18
1 9 . For each of the networks of Fig. P4, compute the impulse-response time
sequence with T = 0.1 and for n = 0, 1, . . . , 15. Then compute the
network-response time sequence for n = 0, 1 , . . . , 15, when the network
excitations depicted in Fig. P5 are approximated by their time sequences.
20. The measured impulse response W(t) of a single-input, single-output
network is shown in Fig. P20a. Using the time sequence for the impulse
response, with T= 0.1, compute the network-response time sequence
for n = 0, 1, 2, . . . , 10, when the network excitations e(t) depicted in
Fig. P20b are approximated by their time sequences.
PROBLEMS
389
(a)
(b)
Fig. P20
21. Consider a network for which
390
INTEGRAL SOLUTIONS
[Ch. 5
(a)
(b)
(c)
x(0) = [l
-1
1]',
fi4
= [0.001
0.020
-0.001]'.
2 5 . Repeat Problem 21 with e(t) and x(0) replaced by each of the following:
(a)
(b)
(c)
(d)
(e)
2 6 . Repeat the calculations of the example in Section 5.6 using the other
matrix norms in (83).
2 7 . For each of the following A matrices
(a)
(b)
(c)
Determine K in the truncation (87) and the error bounds for (104)
and (108) when T= 0.1 and = 0.001. Use the three different matrix
norms in (83).
0 . 2 A
PROBLEMS
391
30. If r(t) and e(t) are the response and excitation of a linear time-invariant
network N, then, as illustrated in Fig. P30, r (t) is the response to
(k)
Fig. P30
(k)
A T
.6
REPRESENTATIONS OF
NETWORK FUNCTIONS
6.1
Sec. 6.1]
393
(la)
(lb)
where x, e, and w are the state, excitation, and output vectors, respec
tively. The last term in the second equation can appear only when an
element of the output vector is a capacitor current in a capacitor and
independent voltage-source loop or an inductor voltage in an inductor
and independent current-source cut-set.
Assuming initially relaxed conditions, let us take the Laplace trans
forms of these equations, solve the first one for X(s) and substitute into
the second. The result will be*
(2a)
(2b)
The quantity in braces is the transfer matrix H(s), each of whose elements
is a network function. Thus
(3)
Examine this expression carefully. The last two terms indicate a direct
relationship between excitation and response without the mediation of the
state vector. These terms control the behavior of the response as s
approaches infinity. In fact, as observed in the last chapter, D is the matrix
of residues of H(s) at infinity.
In the first term of (3), C and B are matrices of real numbers. The
* Note that e(0) does not appear in (26) even though the derivative of e appears
in (16). This is dictated by the requirement that the network be initially relaxed. The
initial value of e could appear only if the derivative of the excitation term is present in the
equation. Since it will be present only when there are all-capacitor loops or all-inductor
cut-sets, setting initial capacitor voltages and initial inductor currents to zero will require
initial excitation values also to be zero.
394
[Ch. 6
-1
(4)
N o w the elements of adj (sU A) are simply cofactors of det (sU A)
and hence are polynomials. This fact is not modified when adj (sU A)
is premultiplied by C and postmultiplied b y B. Hence the whole term is
a matrix whose elements are polynomials divided by d(s). We have thus
verified that network functions are rational functions of s.
Something more can be established from the preceding. In preceding
chapters reference has been made to the natural frequencies of a network.
In Chapter 3 we considered these to be the zeros of the determinant of the
loop impedance matrix or the node admittance matrix. There we showed
that these two determinants could differ at most by a multiplier KsP,
and hence their nonzero zeros were the same. However, in Chapter 4 we
treated the natural frequencies as the eigenvalues of the matrix A;
namely, the zeros of d(s). We now see that the zeros of d(s) are the same as
those of the loop impedance determinant and the node admittance
determinant. This follows from the fact that W(s) refers to any output.
Thus, if we choose all the node voltages, and only these, as the outputs,
W(s) is the matrix of node-voltage transforms. Since network solutions
are unique, (2b) must give the same results as the solution of the node
equations. In the latter case the denominator of the solution will be .
Hence and d(s) have the same nonzero zeros. A similar conclusion
follows concerning . We shall state this result as a theorem for ease of
reference.
w
Theorem 1 . The nonzero zeros of det (sU A) are the same as the nonzero
zeros of det (AYA') and det (BZB').
LOCATIONS OF POLES
(5b)
Sec. 6.1]
395
(6)
that is, network functions take on conjugate values at conjugate points
in the complex plane.
Now look at the second form in (5) in which the poles s and the zeros
s
are placed in evidence. Aside from a scale factor K, the network
function is completely specified in terms of its poles and zeros, which
determine its analytic properties. In fact, the poles and zeros provide a
representation of a network function, as illustrated in Fig. 1. The zeros
pk
0k
Fig. 1.
Pole-zero pattern.
are shown by circles; and the poles, by crosses. We refer to such diagrams
as pole-zero patterns or configurations. Because of the reflection property
(6), the poles and zeros of a network function are either real or occur in
complex-conjugate pairs.
Another simple property possessed by network functions follows from
a consideration of stability. We know that the free response is governed
* This statement should be qualified in a trivial way, since it is possible to multiply
every coefficient in the numerator and denominator by an arbitrary complex number
without changing the function. This difficulty is overcome by fixing, say, the coefficient
of the highest power in the denominator to be 1.
396
[Ch. 6
(7)
E V E N A N D O D D PARTS OF A
FUNCTION
Generally speaking, F(s) will have both even and odd powers of s;
it will be neither an even function nor an odd function of s. Hence we can
write
(8)
where E v F(s) means " e v e n part of F(s)," and Od F(s) means " o d d
part of F(s)." Now an even function g(s) is characterized by the property
g(s) =g(s); and an odd function, by the property g(s) = g(s). Using
these properties together with (8), we can express the even and odd parts
of a function as follows:
(9a)
(9b)
Alternative forms can be obtained if the even and odd powers of s are
grouped in both the numerator and denominator of F(s). Thus, write
(10)
Sec. 6.1]
397
where m1 and m are even polynomials, and n1 and n are odd polynomials.
Then, using this in (9), we get
2
(11a)
(11b)
Note that the denominator is the same for both the even and the odd
part of F ( s ) , and it is an even polynomial. The numerator of E v F(s) is
even, and that of Od F(s) is odd, as they should be.
It is of interest to observe where the poles of E v F ( s ) [and Od F(s)
also] lie. From (9) it is clear that E v F(s) has poles where F(s) has poles
and also where F(s) has poles. But the poles of F(s) are the mirror
images about the imaginary axis of the poles of F(s). This can be illustrated
by the following F ( s ) and
F(s):
F(s) has a real negative pole and a complex pair in the left half-plane.
The poles of F(s) are the mirror images of these, as shown in Fig. 2.
Now E v F(s) has all the poles in Fig. 2, both left half-plane (lhp) and
right half-plane (rhp).
Fig. 2.
Poles of Ev F(s).
398
[Ch. 6
FUNCTION
Hence
(14a)
(14b)
Sec. 6.2]
MINIMUM-PHASE FUNCTIONS
399
(15)
FUNCTION
s
6.2
MINIMUM-PHASE FUNCTIONS
400
[Ch. 6
(b)
(a)
Fig. 3.
tude and angle of a transfer function, consider Fig. 3a. This shows a pair
of conjugate zeros in the right half-plane and the left half-plane image of
this pair. Let P {s) and P (s) be quadratics that have the right half-plane
pair of factors and the left half-plane pair of factors, respectively; that is,
r
(17a)
(17b)
It is clear that P (s) = P ( s ) . The geometrical construction in the figure
indicates that the magnitudes of P and P are the same when
s=j.
As for the angles, we find
r
(18a)
(18b)
Sec. 6.2]
MINIMUM-PHASE FUNCTIONS
401
Fig. 4.
(a)
Fig. 5.
(b)
402
[Ch. 6
ALL-PASS A N D M I N I M U M - P H A S E FUNCTIONS
(20)
where
(21)
Sec. 6.2]
MINIMUM-PHASE FUNCTIONS
403
(22)
For positive frequencies this is a negative angle. Thus the angle of an
all-pass function is negative for all positive frequencies.
Using this equation and (20), we can now write
(23)
This result tells us that, at all positive frequencies the angle of a function
having right-half-plane zeros is less than that of the function obtained
when a pair of these zeros is replaced by its left half-plane image.
This procedure of expressing a transfer function as the product of two
others may be repeated. At each step a pair of complex zeros or a real
zero from the right half-plane may be replaced by their left-half-plane
images. A sequence of functions, of which F1 and F2 are the first two,
will be obtained. Each member of the sequence will have fewer right
half-plane zeros than the preceding one. The last member in this sequence
will have no right half-plane zeros. Let us label it F (s). B y definition,
F (s) is a minimum-phase function (as the subscript is meant to imply).
Using (23), and similar results for the other functions, we can write
m
(24)
Each of the functions in this sequence will have the same j-axis magnitude,
but the angles get progressively larger. Paradoxically, the minimumphase function will have the largest angle of all (algebraically, but not
necessarily in magnitude). The reason for this apparent inconsistency is
the following. We have defined transfer functions as ratios of output
transform to i n p u t transform. When the minimum-phase concept was
first introduced by Bode, he defined transfer functions in the opposite
way. With such a definition the inequalities in (24) will be reversed, and
the minimum-phase function will have the smallest angle algebraically.
At each step in the above procedure a second-order or first-order allpass function is obtained. The product of any number of all-pass functions
is again an all-pass function. It follows that any
non-minimum-phase
transfer function can be written as the product of a
minimum-phase
function and an all-pass function; that is,
(25)
where F
404
N E T CHANGE I N
[Ch. 6
ANGLE
POLYNOMIALS
Let us now consider another aspect of minimum-phase and nonminimum-phase functions; namely, some relationships between the coef
ficients of a polynomial and the locations of its zeros. Polynomials with
no zeros in the open right half-plane are called Hurwitz
polynomials.
If, in addition, there are no zeros on the j-axis, the polynomial is called
strictly Hurwitz, for emphasis. Thus the numerator polynomial of a mini
mum-phase function is Hurwitz.
Sec. 6.2]
MINIMUM-PHASE FUNCTIONS
405
Fig. 6.
nonzero coefficients of a polynomial are the first and last (i.e., if P(s) =
s + a ), then there will be a zero on the boundary |arg s| = /n. Note
that the converse of the theorem is not generally true; that is, if the
zeros of a polynomial are excluded from the sector |arg s| < / n , the
coefficients need not all have the same sign. Thus the polynomials
n
all have a real negative zero and the same right-half-plane factor
s 0.8s + 1 whose zeros do not lie in the sector |arg s\ < / 3 ; yet two
of the polynomials have coefficients with different signs.
2
* For a proof using the principle of the argument see Norman Balabanian, Network
Synthesis, Prentice-Hall, Englewood Cliffs, N.J., 1958.
406
6.3
[Ch. 6
Up to this point the discussion has been carried out in terms of functions.
We shall now switch to a consideration of the networks of which these
functions are transfer functions. The locations of transmission zeros of a
network depend both on the types of elements contained in the network
and on the structure of the network. Nothing definitive can be stated as to
restrictions on the locations of transmission zeros due to element types.
Thus RC networks, which have only one type of reactive component,
can have complex transmission zeros, as well as real ones, and can even
have zeros in the right half-plane. But structure alone does place restric
tions.
LADDER
NETWORKS
Fig. 7.
and shunt branches need not be single elements but can be arbitrary
two-terminal networks with no coupling between branches. The first
and last shunt branches may or may not be present. Using topological
formulas for network functions from Chapter 3, the open-circuit transfer
impedance can be written as
(26)
The right-hand side follows because no shunt branch can appear in a
two-tree that includes both nodes 1 and 2 but excludes node 0. The zeros
Sec. 6.3]
407
of z (s) will occur where the numerator has zeros and where the denom
inator has poles that do not cancel with poles of the numerator. Every
tree must contain node 0 and, hence, every tree product must contain at
least one of the shunt-branch admittances, Y1, Y ... Y +1.
Hence the
poles of these admittances must be the poles of the denominator of z (s).
Some of the series branches may also be in a tree, but the poles of the ad
mittances Y , Y4, etc., in these tree products cancel with the poles of the
numerator of z . The conclusion is that the zeros z (s) occur at the
zeros of series-branch admittances Y , Y4, etc., and at the poles of shuntbranch admittances
Yi, Y , etc. But the poles and the zeros of the
admittance of a passive, reciprocal network cannot lie in the right
half-plane. Hence, the result follows.
Although the development was carried out for the z function, the
result is true for other transfer functions also, as discussed in Problem
48 of Chapter 3.
It was shown above that the transmission zeros of a ladder network
occur when a shunt-branch admittance has a pole (the branch is a short
circuit) or a series-branch admittance has a zero (the branch is an open
circuit). It is not true, however, that a transmission zero must occur at
such pointsonly that these are the only points at which it can occur.
Examples where a series-branch admittance zero and a shunt-branch
admittance pole are not zeros of transmission are given in Problem 6.
The foregoing becomes very useful in the synthesis of ladder networks.
We shall not pursue it here, but the realization of a driving-point function
as the open-circuit impedance or short-circuit admittance of a ladder
network with certain prescribed transmission zeros utilizes this knowledge.
However, if the job is to design a filter, say, with transmission zeros in the
right half-plane, we at least know that a ladder network cannot realize
such zeros; thus other structures must be sought.
The simplest structures whose transfer functions can be non-minimumphase are the bridged tee, the twin tee, and the lattice, shown in Fig. 8.
Whether they are actually non-minimum-phase will depend on the types
of elements present and on their numerical values; for example, the twintee network in Fig. 9 will be minimum phase for some values of the resistors
and non-minimum-phase for other values, as shown by two particular
cases in the figure.
21
2m
21
21
21
21
CONSTANT-RESISTANCE
NETWORKS
408
(b)
(a)
[Ch. 6
(c)
Non-minimum-phase
network
Transmission zeros at
Fig. 9.
Minimum-phase
network
Transmission zeros at
Minimumphase
Fig. 10.
All-pass
Cascaded two-ports.
Sec. 6.3]
409
Constant
resistance
Fig. 11.
Constant
resistance
21
(27a)
(27b)
410
(b)
(a)
[Ch. 6
(d)
(c)
2
(28a)
(28b)
When these expressions are compared with (27a), we see that the forms
are the same. Hence we can identify Za directly and then find Z from
the relation Za Z = R . Thus for the first-order lattice we get
b
* Notice that the numerator factor in the first-order case is written as a s rather than
s a. This amounts to changing the sign of the transfer function or inverting the polarity
of the output voltage. Doing this will avoid the discontinuity of radians in the angle.
Sec. 6.3]
411
(29)
Thus the two-ports that realize first- and second-order all-pass functions
take the form shown in Fig. 13.
(a)
Fig. 13.
(b;
412
[Ch. 6
21
W i t h Z a Z = R and R = 1, we get Z = Y .
b
Sec. 6.3]
413
Fig. 14.
Realization of G21
21
(31)
414
[Ch. 6
Sec. 6.4]
415
function, in which case the zeros must lie in the left half-plane. For trans
fer functions, we need not assign to F(s) all the left-half-plane zeros of
G(s). Thus the zeros of F(s) are not uniquely determined from a given
G(s), unless the transfer function is specified to be minimum-phase. In
this case the zeros of F(s), as well as the poles, must lie in the left halfplane.
Let us now consider some examples that illustrate this procedure and
are of practical interest. The requirements on most common electrical
filters involve transfer functions whose j-axis magnitudes are ideally
constant over a given frequency interval, which is referred to as the pass
band, and are ideally zero over the rest of the j-axis, which is referred to
as the stop band. It is not possible for the j-axis magnitude of a rational
function to behave in this ideal manner. (Why?) However, it is possible
to find transfer functions whose j-axis magnitudes approximate the desired
magnitude in some fashion or other.
Consider the ideal low-pass filter function shown in Fig. 15a. Two
(a)
Fig. 15.
(c)
(b)
flat)
(32)
416
[Ch. 6
and
(equal ripple),
(33)
(34)
which reduces on substituting s = j to
(35)
Our problem now is to find the transfer function F(s) when its
squared magnitude is known
MAXIMALLY FLAT
j-axis
RESPONSE
(36)
This function has no finite zeros, so we need only factor the denominator.
In the present case this is a relatively simple task. The zeros of the de
nominator are found b y writing
(37a)
which is simply
(37b)
where the minus sign applies for n even. Taking the 2nth root in (37a),
we find the poles of G(s) to be
(38)
* The use of the letter T for the Chebyshev polynomial is a legacy of the past. Some
of Chebyshev's work was first published in French, leading to the use of the French
spelling "Tschebyscheff," or its variation "Tchebycheff." This spelling of the name has
now been discarded in the American literature.
Sec. 6.4]
417
Thus, there are 2n poles, each of which has unit magnitude. The poles are
uniformly distributed on the unit circle, as shown in Fig. 16 for the case
Fig. 16.
n = 4. Notice that the imaginary parts of the poles lie in the pass band
in the range < 1.
To form F(s) we simply take the n left half-plane poles of C(s). These
are the ones given by values of k from 1 to n. For n = 4 these will be
2
(39)
(40)
2.0000
3.4142
5.2361
7.4641
10.0978
13.1371
16.5817
20.4317
2.6131
5.2361
9.1416
14.5920
21.8462
31.1634
42.8021
1.4142
2.0000
2.6131
3.2361
3.8637
4.4940
5.1528
5.7588
6.3925
2
3
4
5
6
7
8
9
10
b
4
3.2361
7.4641
14.5920
25.6884
41.9864
64.8824
Order
n
Table 1.
3.8637
10.0978
21.8462
41.9864
74.2334
4.4940
13.1371
31.1634
64.8824
5.1258
16.5817
42.8021
5.7588
20.4317
6.3925
418
REPRESENTATIONS OF NETWORK FUNCTIONS
[Ch. 6
1
2
3
4
5
6
7
8
9
10
Order
n
Table 2.
s2+s +
(s +
Factors
s+ l
s + 1.4142s + 1
l)(
l)
(s + 0.7654s + l ) ( s + 1.8478s + 1)
(s + l ) ( s + 0.6180S + l)(s + 1.6180s + 1)
(s + 0.5176s + l)(s + 1.4142s + l)(s + 1.9319s + 1)
(s + l)(s + 0.4450s + l)(s + 1.2470s + l)(s + 1.8019s + 1)
(s + 0.3002s + l)(s + 1.1111s + l)(s + 1.1663s + l)(s + 1.9616s + 1)
(s + l)(s + 0.3473s + l)(s + s + l)(s + 1.5321s + l)(s + 1.8794s + 1)
(s + 0.3129s + l)(s + 0.9080s + l)(s + 1.4142s + l)(s + 1.7820s + l)(s + 1.9754s + 1)
Sec. 6.4]
DETERMINING A NETWORK FUNCTION
419
420
[Ch. 6
2 n
Since the Taylor series for the error starts with the power, this means
the first n 1 derivatives with respect to are 0 at = 0. Hence the
name maximally flat.
The Butterworth function just illustrated is particularly simple, since
all of its zeros are at infinity. It is possible to introduce a modified maxi
mally flat function that will have some finite zeros. A glance back at the
magnitude-squared functions in (32) and (33) shows that they are both of
the form
2
(41)
2
(42)
where
(44)
In the last step a power series is obtained by long division. Again the
series starts with the power, and so the first n 1 derivatives of the
error with respect to are 0 at = 0. The magnitude-square function in
(43) is therefore also maximally flat. In contrast with the Butterworth
function, it has some finite zeros.
As an illustration consider the following magnitude-square function:
2 n
Sec. 6.4]
421
or
Note that the double zeros of F(s) F(s) are assigned equally to F(s)
and to F(s). The locations of the poles and zeros of F(s) are shown in
Fig. 17 and are compared with the poles of the fourth-order Butterworth
function.
Fig. 17.
422
CHEBYSHEV
[Ch. 6
RESPONSE
(45)
(46b)
If we now expand cosh nw in the last equation and set reals and imagin
aries equal on both sides of the equation, we will find the values of x and
y that satisfy the equation. When these values are substituted in (46a)
we find the corresponding values of s. These are the pole locations. If
we designate them b y s = +j ,
the result of the indicated operations
will be
k
(47a)
(47b)
(48)
Sec. 6.4]
423
This is the equation of an ellipse in the s-plane. The major axis of the ellipse
will lie along the j-axis, since the hyperbolic cosine of a real variable is
always greater than the hyperbolic sine. The pole locations for n = 4 are
shown in Fig. 18.
Fig. 18.
Finally, the left half-plane poles of G(s) are alloted to -F(s), and the task
is again complete.
For a typical case, if the permissible ripple is given to be = 0.1 and
the order n = 4, the pole locations are found from (47), and we get the
transfer function
424
[Ch. 6
(49b)
The last step follows from (49b) and (50). If we now invert this last
equation and solve for A(j), we get
(52)
Let us now inquire into the conditions that the tangent function must
satisfy if it is to be a realizable function. Note that
(53)
where R and X are the real and imaginary parts of the network function.
We know that these are, respectively, even and odd functions of .
Sec. 6.5]
425
(54)
(55)
Our problem can now be restated as the problem of finding P(s) and
P (s) when the function on the right side of the last equation is known.
Note that A(s) will always have zeros in the right half-plane, and it will
usually have poles in the right half-plane also. It differs from an all-pass
function in that it may have poles in the right half-plane as well as zeros.
On the other hand, it is similar to an all-pass function in that each zero
is the negative of a pole. As a matter of fact, it can be expressed as the
ratio of two all-pass functions, but this has no utility for our present
purpose. It can have neither zeros nor poles on the j-axis, since, if P1(s)
has a pair of such zeros, so also will P1(s), so that they will cancel in the
ratio; similarly if P (s) has j-axis zeros.
Let us now consider assigning the poles of A(s) to P1(s) or P (s).
If A(s) has any right half-plane poles these must belong to P1(s),
since P (s) cannot have right half-plane zeros. On the other hand, the
left half-plane poles cannot uniquely be assigned to either P (s) or P\( s).
If we assign one of the left half-plane poles of A(s) to P1(s), then P1(s)
will have the corresponding right half-plane factor, indicating that the
transfer function is a non-minimum-phase one. Of course, the distribution
of poles and zeros will be dictated by the permissible degrees of numerator
and denominator of F(s).
Once P (s) and P1(s) have been established from the denominator of
A(s), it is not necessary to examine the numerator, since the transfer
function will now be known; it is only necessary to replace s b y s in
P1(s) to get P1(s).
2
426
[Ch. 6
(56)
The first step is to substitute this into (52) to obtain A(j). The result is
We find that all the poles of A(s) are in the left half-plane, whereas all the
zeros are in the right. Hence there is no unique way to assign the zeros and
poles of F(s). Any one of the following functions will be suitable:
(57a)
(57b)
(57c)
Notice that the last two have right half-plane zeros. Each of these
functions will have the same angle for all values of , but their magnitudes
will be quite different. If F(s) is required to be minimum-phase, the answer
is once again uniquein this case the first function of (57).*
In our computations so far we have assumed that () is specified to be
a continuous function of . If, however, a function F(s) has either poles or
zeros on the j-axis, the corresponding phase function () will have
discontinuities of + at each pole and zero. In such cases we consider the
* Even this uniqueness is only to within a constant multiplier. The angle is obviously
independent of a real positive gain constant.
Sec. 6.6]
427
where () is a continuous function. The index j runs over all the zeros
and poles on the j-axis, and the minus sign applies to the poles.
We now have to identify the step discontinuities. For this we examine
a typical factor in F(s) (pole or zero factor):
c
428
(a)
Fig. 19.
[Ch. 6
(b)
Its j-axis
real part is
that is, the real parts of both admittances are the same, yet the admit
tances themselves are different. The function Y1(s) differs from Y(s) by
having a pair of poles on the j-axis. If the real part is given, we cannot
tell whether to choose Y(s) or Y1(s) corresponding to this real part. As a
matter of fact, an infinite number of functions that differ from Y(s) by
having additional poles on the j-axis will have the same real part on the
j-axis. What we can hope to do from a given real part, then, is to find
the particular function that has no poles on the
j-axis.*
THE B O D E METHOD
Let us turn back to Section 1 and look at the discussion of the real
part of a function starting at (8) and ending at (13). If an even rational
function of with no finite or infinite poles for real is specified to be the
real part of a network function, we replace j by s, and the result will be
the even part of F(s). Thus
(59)
* Such a function is a minimum-susceptance function if it is an admittance, and a
minimum-reactance function if it is an impedance. This condition on driving-point
functions is the analogue of the minimum-phase condition on the transfer function.
Sec. 6.6]
429
The question is: How can we find F(s) from its even part? As discussed
in Section 6.1, the poles of E v F(s) have quadrantal symmetry. Its left
half-plane poles belong to F(s); and its right half-plane poles, to F(s).
If F(s) has a nonzero value at infinity, then F(s) will have this same
value. It is, therefore, clear how to find F(s) from E v F(s): Expand
E v F(s) in partial fractions and group ail the terms contributed by poles
in the left half-plane; if there is a constant term in the expansion, we add
half of this to the group; finally, we multiply b y 2; the result is F(s).
To illustrate, let
(60)
(61)
(62)
430
[Ch. 6
where the m's refer to the even parts of the numerator and denominator
and the n's refer to the odd parts. The even part of F(s) can now be
written as in (11a). Thus
(63)
where the right side has been written in expanded form. When a real-part
function is specified as an even rational function in , the right side of
(63) is obtained when is replaced by s . We first go to work on the
denominator. Since the poles of E v F(s) are those of both F(s) and F(s),
the ones belonging to F(s) are those that lie in the left half-plane. Hence,
when we factor the denominator of (63), we assign all the left half-plane
factors to F(s). In this manner the denominator of F(s) in (62) becomes
known.
Turn now to the numerator. Suppose we write F(s) as a rational function
as in (62) with unknown literal coefficients in the numerator but with
known denominator coefficients. We then form the expression m1m
n1n and set it equal to the numerator of the given function in (63).
Equating coefficients of like powers of s on the two sides of this equation
will permit us to solve for the unknowns. Note that three sets of coeffi
cients are involved: the small a's the capital A's, and the small b's. Of
these, the last two sets are known at this point; only the small a's are
unknown.
Let us carry out the process just indicated. Identifying m1, m , n 1 ,
and n from (62), we can write
2
(64)
(65)
Sec. 6.6]
431
To find the unknown a's, we must solve this set of linear equations
simultaneously.
We shall now illustrate, using the function in (60) already treated by
Bode's method. The left-half-plane factors in the denominator of that
expression are
Since the given R() is zero at infinity, so also must F(s) be zero at infinity.
(Why?) Hence the numerator of F(s) must be of the form
B y inserting the last two equations into (65) and utilizing the fact that
all the capital "A" coefficients are zero except A , which is unity, we get
0
These equations are then solved to yield a = 1, a1 = 4/3 and a = 2/3. The
network function thus obtained verifies the one previously found in (61).
0
(67)
432
[Ch. 6
(68a)
(68b)
(69a)
and
(69b)
The even part of the polynomial q(s) is simply the sum of all its even
powers, if any. If q(s) has any even powers, then the right side of the
last equation will become infinite as s approaches infinity, whereas we
know, by (68b), that the left side does not. The conclusion is that q(s) is an
odd polynomial and has no even powers, so that E v F = E v F and, hence,
E v F = E v F from (68b). Furthermore, this remainder function has the
same poles as the specified function; consequently, it is the desired
functionnamely, F (s) = F(s).
In summary, we may state that when an even rational function
(m1m nin )/(m
n,2 ) is specified, a network function -F (s) whose
even part is l / ( m n ) is determined. This function is then multiplied
b y (m1m ni n ), following which a long division is carried out, yielding
a remainder function with no pole at infinity. This is the desired function
whose even part is the specified function.
To illustrate, let
r
Sec. 6.7]
INTEGRAL RELATIONSHIPS
433
Then
But this is the same function as previously considered in (60) and (61).
Thus
and
Hence
434
[Ch. 6
(70)
In this expression C is a closed contour within and on which F(s) is regular;
z represents points on the contour, whereas s is any point inside. If we let
the contour be a circle and express both z and s in polar coordinates, we
shall be able to express the real and imaginary parts of F(s) in terms of
either its real or its imaginary part on the circle. Finally, by means of a
transformation, the circle is mapped into the imaginary axis. The result
ing expressions relating the real and imaginary parts are referred to as
Hilbert transforms.
An alternative approach, which we shall adopt, is to start with Cauchy's
integral theorem. (See Appendix 2.) This theorem states that the contour
integral of a function around a path within and on which the function is
regular will vanish. In order to apply this theorem, it is necessary to know
(1) the integration contour and (2) the function to be integrated. In the
present problem the contour of integration should include the j-axis,
since we want the final result to involve the j-axis real and imaginary
parts of a network function. Consequently, since the functions we are
dealing with are regular in the entire right half-plane, the contour of
integration we shall choose will consist of the j-axis and an infinite
semicircular arc in the right half-plane. B y Cauchy's theorem, the complete
contour integral will be zero. Hence it remains only to calculate the con
tributions of each part of the contour.
Let F(s) be a network function of either the driving-point or the transfer
type; in the usual way write
(71a)
(71b)
Sec. 6.7]
INTEGRAL RELATIONSHIPS
435
(b)
(a)
Fig. 20.
Path of integration.
436
[Ch. 6
(72)
(73)
Sec. 6.7]
INTEGRAL RELATIONSHIPS
437
(74)
Note that the integration along the imaginary axis must avoid the pole at
z = j in a symmetrical manner. This will yield the principal value of
the integral on the right. In all the subsequent integrals we must keep
this point in mind. N o w collecting all these results and substituting into
(72) we can write
(75)
If we next write F(j) and F(jy) in terms of real and imaginary parts, and
equate reals and imaginaries, we get, finally,
(76a)
(76b)
We are leaving the algebraic details of these steps for you to work out.
The message carried by these two expressions is very important. The
second one states that when a function is specified to be the real part of a
network function over all frequencies, the imaginary part of the function is
completely determined, assuming the network function has no poles on the
j-axis. Similarly, if the imaginary part is specified over ail frequencies,
the real part is completely determined to within an additive constant.
Remember that the same results apply if F(s) is replaced by its logar
ithm. However, now we must require that F(s) be minimum-phase (if
it represents a transfer function). On the other hand, we can relax the
requirement of regularity of F(s) on the j-axis. A simple pole of F(s)
on the j-axis becomes a logarithmic singularity of ln F(s), and such a
singularity will contribute nothing to the integral, as mentioned earlier.
Thus, for minimum-phase transfer functions, (76) with R and X replaced
b y and , relate the gain and phase functions over all frequencies.
438
[Ch. 6
Let us now obtain alternative forms for the two basic expressions in
(76) that will throw additional light on the relationships and will bring
out points that are not at once apparent from these expressions. Remem
ber that the real and imaginary parts are even and odd functions of
frequency, respectively. Let us use this fact and write (76b) as follows:
(77)
In the first of these integrals, replace y and y and change the limits
accordingly. The result is
(78)
The last step follows from the fact that R(y) = R(y).
into (77), we get
Substituting this
(79)
In a completely similar way, starting with (76a) we get
(80)
In the last two expressions it still appears that the integrand goes to
infinity on the path of integration at the point y = . This is really
illusory, since we must understand the integral as the principal value.
E v e n this illusory difficulty can be removed if we note by direct integra
tion that
(81)
again using the principal value of the integral. Hence we can subtract
R()/(y ) from the integrand in (79) and X()/( ) from the
integrand in (80) without changing the values of these integrals. The
results of these steps will be
2
(82a)
(82b)
Sec. 6.7]
INTEGRAL RELATIONSHIPS
439
(83)
Now let us make the change of variable y = 1/p; then dy/y = dp. After
appropriately modifying the limits of integration as well, this equation
becomes
(84)
Note that the integral in (83) or (84) is not a function of and that, for a
given value of the band edge , it will be simply a constant. Thus the
angle will be approximately a linear function of within the pass band.*
Of course, the approximation will get progressively worse as we approach
the band edge, since then can no longer be neglected in comparison to y
in the integrand.
0
The two pairs of expressions obtained so far in (76) and (82) relate the
imaginary part at any frequency to the real part at all frequencies;
or the real part at any frequency to the imaginary part at all frequencies.
* Such a linear phase characteristic corresponds to a constant time delay in the trans
mission of sinusoidal functions over this range of frequencies. Therefore for signals that
have essentially only this frequency range we get a distortionless transmission. For this
reason a linear phase characteristic is desirable.
440
[Ch. 6
We should be able to find limiting forms for these expressions when fre
quency approaches zero or infinity.
First consider (82a) when approaches zero. This leads immediately
to the result
(85)
This expression is referred to as the reactance-integral theorem. It states
that the integral of the imaginary part over all frequencies, weighted by
the reciprocal of frequency, is proportional to the difference of the real
part at the two extreme frequencies. It is also called the phase-area theorem,
since the result remains valid when F(s) is replaced by its logarithm, R by
, and X by .
A more convenient expression is obtained if a change to logarithmic
frequency is made. Define
(86)
where is some arbitrary reference frequency. Then dy/y becomes du,
and (85) can be written as follows:
(87)
Note the change in the lower limit, since u = when y = 0. The argu
ment of X(y) has been retained a s y for simplicity, although the integrand
should more accurately be written as X ( e ) . Alternatively, a new func
tion X\(u) = X(e ) can be defined. However, this introduces additional
new notation to complicate matters. In subsequent equations we shall
retain y as the argument of the integrands and write X(y) or R(y), as the
case may be, with the understanding that we mean to convert to a func
tion of u by the substitution y = e before performing any operations.
Thus we see that the area under the curve of the imaginary part, when plotted
against logarithmic frequency, is proportional to the net change in the real
part between zero and infinite frequency.
N e x t let us multiply both sides of (82b) by and then take the limit as
approaches infinity. Remember that the upper limit on the integral
means that we integrate up to R and then let R approach infinity.
Thus (82b) becomes
u
(88)
Sec. 6.7]
INTEGRAL RELATIONSHIPS
441
(89)
The result expressed by this equation is referred to as the resistanceintegral theorem. (It is also called the attenuation-integral
theorem, since
the result remains valid if F(s) is replaced by its logarithm.) If the asymp
totic behavior of the imaginary part of a network function is specified,
thenno matter how the j-axis real part behaves with frequencythe
area under the curve of the real part, with the horizontal axis shifted
upward by an amount R(), must remain constant. Looking at it from
the opposite viewpoint, when the integral of the real part of a function
over all frequencies is specified, then the infinite-frequency behavior of
the imaginary part is fixed.
Consider the special case in which F(s) has a simple zero at infinity;
then F() = R() = 0. Hence
(90)
However, according to the initial-value theorem, the limit on the righthand side is simply the initial value of the impulse response of the network
represented by F(s). In this case, then, (89) becomes
(91)
-1
NETWORKS
What has just been developed can be used to determine some basic
limitations on the behavior of networks, when allowance is made for
certain inevitable parasitic effects. Consider the situation depicted in
Fig. 21a. The capacitance C accounts for parasitic effects that almost
inevitably occur, such as junction capacitances in a transistor or just
442
(a)
Fig. 21.
[Ch. 6
(b)
(92)
(93)
We see that the shunt capacitance imposes an effective limit on the area
under the curve of the real part. Although this resistance integral evolved
as the limiting value of the general expression relating the real and imag
inary parts of a network function, it appears to provide a figure of merit
of some sort for network capability.
Sec. 6.7]
INTEGRAL RELATIONSHIPS
443
Fig. 22.
Resistance-terminated two-port.
(95a)
power to load =
(95b)
Clearly, the load power cannot exceed the power from the source for a
passive two-port. Hence the second expression can be no greater than the
first; so
(96)
444
[Ch. 6
The equality is valid when the two-port is lossless. Thus the squared
magnitude of the current gain of a lossless two-port is proportional to the
real part of the impedance at the input terminals of the two-port when the
output is terminated in R . Thus, with (96) inserted into (94), and with
R() interpreted as Re Z(j), there results
2
(97)
Suppose the two-port in Fig. 22 is to be a filter with constant power
gain over a given band of frequency and zero outside this band. Then the
integral in (97) will simply equal the constant-power gain times the band
width. In the more general case, even though the transfer function may
not be an ideal-filter function, the area under the curve represented by
this integral is dimensionally power gain times bandwidth. For this
reason the integral in (97) is generally called the gain-bandwidth integral.
Thus we find a basic limitation on the gain-bandwidth product introduced
by the presence of the shunt capacitor C.
A L T E R N A T I V E FORM OF
RELATIONSHIPS
(98)
Sec. 6.7]
INTEGRAL RELATIONSHIPS
445
with
(99)
Note that coth u/2 is an odd function of u, being strictly positive when u
is positive and strictly negative when u is negative. Hence its logarithm
for negative u will be complex, the imaginary part being simply . For
negative u it can be written
(100)
When u = + , ln coth u/2 0; and when u = , ln coth u/2 = j .
Hence the integrated part of the last equation becomes simply j[(0)
-()].
N o w consider the remaining integral. If we use (100) for negative
values of u, the result will be
446
[Ch. 6
(101)
(102)
This function is shown plotted in Fig. 23. It rises sharply in the vicinity of
u = 0(y = ), falling off to very small values on both sides of this point.
This means that most of the contribution to the angle at a frequency
comes from the slope of the gain in the immediate vicinity of .
Fig. 23.
(103)
Sec. 6.7]
INTEGRAL RELATIONSHIPS
447
FUNCTIONS
448
Fig. 24.
[Ch. 6
Sec. 6.7]
INTEGRAL RELATIONSHIPS
449
So far in this section we have found that for a suitably restricted net
work function, when the real part is specified over all frequencies, the
imaginary part is completely determined. Similarly, when the imaginary
part is specified over all frequencies, the real part is completely determined
(to within a constant). The question may be asked: Suppose the real partis
specified over some frequency intervals and the imaginary part over the
remainder of the entire frequency spectrum; is the function completely
determined?
Instead of considering this problem in a completely general form, let
us suppose that the real part is known for all frequencies less than and
the imaginary part for all frequencies greater than . We wish to find an
expression that will give the unknown parts of the two components. The
discussion concerning the choice of weighting functions suggests that if
we can choose a weighting function that changes character at s o that
below the term involving the real part is even while above the term
involving the imaginary part is evenour problem will be solved. What
we need is a multivalued weighting function.
Suppose we choose the following weighting function:
0
<y
0
< ,
0
y > ,
y < .
450
[Ch. 6
(104)
We have now answered the question posed at the start of this discussion,
insofar as the present problem is concerned. If we are given the real
part of a function over part of the imaginary axis and the imaginary part
over the rest of the axis, then the function is completely defined. Our
method of obtaining the result in the last equation can be extended if
there are more than two intervals over which one or the other of the two
components are known. Additional irrational factors are introduced
giving additional branch points at appropriate points on the axis. The
resulting expressions, however, become rather complicated and hence
limited in usefulness.
Let us now summarize the results of this section. Our objective is to
obtain relationships between the real and imaginary parts of a network
function F(s) (or between the gain and the phase), so that when one of these
is prescribed the other can be calculated. The point of departure is
Cauchy's integral theorem, the contour of integration consisting of the
imaginary axis with an infinite semicircular arc joining the ends. An
integrand is chosen involving F(s) or ln F(s), multiplied by a weighting
function. The contour is indented to bypass poles of the integrand
introduced by this function.
Sec. 6.8]
451
6.8
(105a)
(105b)
We shall restrict ourselves to network functions
STEP R E S P O N S E
452
[Ch. 6
Thus, requiring that w (t) -> 0 as t -> means that F(s) must have a zero
at s = 0. With this stipulation we can now write (106) as
u
(107b)
Thus the real and imaginary parts of a network function can be obtained
directly from the step response.
Converse relationships giving the step response in terms of the real or
imaginary part also exist. These can be obtained starting with the
inversion integral for the step response. Since w (t) = {F(s)/s},
we
get
_1
(108)
We are still assuming that F(s) has no poles on the j-axis, but let us not
restrict it to have a zero at the origin for this development. Then the
integrand in the last expression might have a pole at the origin. If it
were not for this pole, the Bromwich path could be taken as the j-axis.
Instead, let us take the path shown in Fig. 25, which consists of the jaxis except for a semicircular arc that bypasses the origin. As the radius
of the semicircle approaches zero, the path approaches the entire
j-axis. The three parts of the path have been labeled C1, C , and C .
Equation 108 can now be written as follows:
2
(109)
Sec. 6.8]
(a)
453
(b)
Fig. 25.
Contours of integration.
(110)
The last integral on the right involves the radius R in a complicated way.
However, we intend to let R approach zero, in which case this term
reduces to F(0)/2. Y o u should verify this. Note that if we place the addi
tional restriction that F(s) has a zero at s = 0, then this term will disap
pear. When we let R -> 0, the remaining two integrals in (110) combine to
give the principal value of the integral running from to + . Hence,
finally,
0
(111)
454
[Ch. 6
or
When we substitute the last equation into (112), we obtain the final result:
(113a)
(113b)
Up till now we have performed various mathematical manipulations to
put the relationships between F(j) and w (t) in various equivalent
forms. But now we have something new. The last equation shows that the
step response of the network can be computed when only the real part of
the network function along the j-axis is known. Note that this relation
ship does not require that R(0) = F(0) be zero. With the step response
determined, (107b) can be used to compute the imaginary part of F(j).
However, from the derivation of (107b) we know that the asymptotic
value of the step response that is to be used in (107b) must be zero.
Hence, before using w (t) as computed from (113b), we first subtract its
asymptotic value, R(0), in case it is not equal to zero. In this way F(j)
is completely determined from a knowledge of its real part alone.
Similarly, starting with the imaginary part X(), we can compute the
step response from the integral in (113a). The portion of the step response
u
Sec. 6.8]
455
IMPULSE
RESPONSE
Let us now turn to the impulse response. Everything that was done
starting from (106) can be duplicated (with appropriate changes) in terms
of the impulse response. We shall list the results and leave the details of
the development to you. It will still be required that F(s) be regular on the
j-axis, but now it need not have a zero at s = 0. Instead, application of
the inversion integral to F(s) will require that F(s) have a zero at infinity.
If we retrace the steps starting at (106), we shall get the following
equations:
(114a)
(114b)
(114c)
(114d)
The first two of these are the counterparts of (107), whereas the last two
are to be compared with (113). As a matter of fact, the last two equations
can be obtained from (113), in view of the fact that the impulse response is
the derivative of the step response. (No impulses will be involved, since
we assumed F() = 0.)
456
[Ch. 6
Equation (114d) shows that the impulse response of the network can be
computed even if only the imaginary part X() is known. Note that X()
will approach zero as approaches infinity, even though F() may not be
zero. With the impulse response computed, the real part of R()or
R() R ( ) if R() = F()0can
now be found from (114a).
Thus F(j) is determined to within the additive constant F() = R()
b y its imaginary part alone.
Similarly, starting from a knowledge of only the real part of R()or
R() R() if R() = F() 0the impulse response can be com
puted from (114c). Having found the impulse response, the imaginary part
X() is now calculated from (114b). Thus we find that a transfer function
is completely determined from a knowledge of its real part alone.
In each of the above cases, once the step response or impulse response
has been calculated from a given R() or X(), it is then necessary to find
only the Laplace transform, since {w (t)}=
F(s)/s, and {w (t)}=
F(s).
In this way one of the integrations of (107) and (114) can be avoided.
u
(5
Examples
Suppose the following is specified to be the real part of a network
function on the j-axis:
(115)
We see that this has a nonzero value at infinity, and so (114b) cannot be
used directly. If we subtract its infinite-frequency value, we get
(116)
The second line follows from the use of the exponential form of cos t.
If in the second integral in this line we replace by and appropriately
change the limits, the last line will follow.
Sec. 6.8]
457
s-plane:
where the contour consists of the entire j-axis and an infinite semi
circle to the left. The integrand satisfies the conditions of Jordan's lemma,
since the rational function in the integrand vanishes at infinity as 1/s .
Hence the contribution of the infinite arc will be zero, and the complete
integral reduces to its value along the j-axis. B y the residue theorem,
the value of the integral equals 2j times the sum of the residues at the
left half-plane poles. In the present case there are only two simple poles,
at s = 1 and s = 2 , in the left half-plane and their residues are easily
computed. Hence we get
2
The transfer function can now be found b y taking the Laplace transform.
The result will be
This function has a zero at infinity. To this we should add the infinitefrequency value of R(), which is F() and which we subtracted from the
original function at the start. Thus
458
Fig. 26.
[Ch. 6
is computed to be
* The last two lines may be obtained from integral 412, in R. S. Burington, Handbook
of Mathematical Tables and Formulas, 2nd ed., Handbook Publishers, 1940.
PROBLEMS
459
PROBLEMS
1.
For the networks shown in Fig. P1, verify that the nonzero eigenvalues of
the A matrix in the state equation are the same as the nonzero zeros of
the loop impedance matrix and of the node admittance matrix.
(b)
(a)
Fig. P I
2.
Find the even part, Ev F(s), and the odd part, Od F(s), of the following
functions from the even and odd parts of the numerator and
denominator:
(a)
3.
(b)
2
4.
460
[Ch. 6
(a) In F(s) let a = 0 for all i and insert into the expression for the
delay function. Find the value of the b coefficients if the delay is to
be a maximally flat function for the cases m = 3, m = 4, and m = 5.
(b) Repeat if a 0 and n= m 1.
i
5.
6.
</n.
21
Fig. P6
7.
Fig. P7
PROBLEMS
9.
461
Verify that the bridged-tee in Fig. 12b and the ell-networks in Figs.
12c and 12d are each constant-resistance networks when terminated in a
resistance R if Za Z = R . Verify also that under this condition the
voltage gain function is
2
10.
(b) Figures P10b and 10c show two bridged-tee networks. Show that the
first has the same y parameters as the lattice under condition (1) above,
and hence is equivalent to the lattice. Show also that the second one
has the same y parameters as the lattice under condition (2) above,
(c) If the y parameters of the lattice are expanded in partial fractions,
the result will have the form:
11.
On the right side a fraction of the finite pole has been combined with the
pole at infinity and the rest of it is combined with the pole at the
origin. Show that each of the tee networks in the twin-tee shown in
Fig. P10d has one of the sets of y parameters within the above
parentheses. Determine the range of values of and show that this
range of values exists if condition (3) above is satisfied. Thus under this
condition the twin-tee is equivalent to the lattice,
( d ) Determine the angle of the transmission zeros of the bridged-tees
and twin-tee determined by the three conditions in part (a).
Find a two-port network terminated in a 100-ohm resistor whose
voltage gain function is given by each of the following all-pass functions.
462
[Ch. 6
(b)
(c)
(a)
(b)
(c)
(d)
Fig. P10
12.
PROBLEMS
463
(b)
(a)
(c)
13.
(a)
(c)
(e)
14.
15.
(b)
(d)
(f)
(b)
(c)
(d)
16.
17.
18.
464
[Ch. 6
19.
20.
21.
22.
23.
+ )
Fig. P23
PROBLEMS
465
24.
25.
Let the real part of a function on the j axis be given by the following
functions. Find the corresponding step response using (113b) and the
impulse response using (114c).
26.
(a)
(b)
(c)
(d)
Fig. P26
27.
466
[Ch. 6
Fig. P27
2
.7.
FUNDAMENTALS OF
NETWORK SYNTHESIS
468
[Ch. 7
residues and real parts, and the relative size of coefficients. This is the
task on which we shall largely concentrate in this chapter.
In order to establish the analytical properties of network functions, it
will be necessary to introduce some additional mathematical topics. The
first two sections will be devoted to this effort. We shall not strive for
completeness of exposition but shall often be content simply with the
statement of a result with some discussion of plausibility.
7.1
TRANSFORMATION OF MATRICES
TRANSFORMATIONS
Sec. 7.1]
TRANSFORMATION OF MATRICES
469
unit matrix leads to the elementary matrix on the left below; similarly,
adding the third column to the second column of a third-order unit
matrix leads to the elementary matrix on the right:
470
[Ch. 7
(2) to interchange the third column and the second column after the second
has been multiplied by 3. The two elementary matrices that will accom
plish this are the following:
MATRICES
B is
(1)
where P and Q are
nonsingular.
- 1
- 1
Sec. 7.1]
TRANSFORMATION OF MATRICES
471
- 1
(4)
472
SIMILARITY
[Ch. 7
TRANSFORMATION
(5a)
or
(5b)
This transformation is a similarity transformation; A and B are called
similar matrices. This transformation has already been discussed in
Chapter 1, where we saw that two similar matrices have the same eigen
values. It is included here for completeness.
CONGRUENT
TRANSFORMATION
(7)
Sec. 7.1]
TRANSFORMATION OF MATRICES
473
(9)
r p
(10)
where
(11)
-1/2
(12)
-1/2
-1/2
474
[Ch. 7
7.2
(14)
Sec. 7.2]
475
when x is a real vector (i.e., the elements of x are real), and the expression
(15)
when x is a complex vector, are called quadratic forms. The reason for the
name becomes clear when we perform the indicated matrix multiplica
tions and get
(16)
when the x's are real, and
(17)
when the x's are complex. We see that these are homogeneous expressions
of degree 2 in the variables x1, x , ..., x .
The matrix A in (14) through (17) is called the matrix of the quadratic
form. We consider the x's to be variables, so that the matrix essentially
defines the quadratic form. We shall concern ourselves with quadratic
forms in which the matrix A is real and symmetric. Actually, any real
quadratic form with a real matrix can be converted into a quadratic
form with a symmetric matrix, because, if the x's and the a 's are real,
we can write
2
ij
(18)
We see that the contribution to the quadratic form of the two terms on
the left of this equation will remain unchanged if we replace both a
and a in the matrix by half their sum. Thus, if A is not symmetric, we
define the symmetric matrix B as
ij
ji
(19)
The matrix B is called the symmetric part of A. This operation leaves the
diagonal elements of A unchanged, whereas the off-diagonal elements are
476
[Ch. 7
(21)
TRANSFORMATION
OF A QUADRATIC
FORM
Let us now observe what happens to a quadratic form when the vector
x is subjected to a real, nonsingular linear transformation. Let x = Qy,
where Q is nonsingular and y is a column vector. The quadratic form
becomes
(22)
where we used the fact that Q is real to write Q* = Q'. Within the paren
theses we find a congruent transformation of A. It was observed earlier
that a real, symmetric matrix A can always be reduced to the canonical
form of (12) by means of a nonsingular, congruent transformation. Hence
the quadratic form can be reduced to
(23)
We can state this result as the following theorem:
Sec. 7.2]
477
Theorem
2. Every real quadratic form x*Ax in which A is real and
symmetric can be reduced by means of a real, nonsingular, linear trans
formation x = Qy to the canonical form given in (23) in which is the rank
of A and p is the index.
This theorem is, of course, an existence theorem, it does not give any
guidance as to how one goes about finding the appropriate linear trans
formation. One procedure for doing this is called the Lagrange reduction,
which consists of repeatedly carrying out a process similar to completing
the square. Let us illustrate this with a number of examples.
Examples
1. For simplicity, suppose x is a real vector. Let
In this process 4 x
Now set
2
2
Then
In this case the rank of A equals its order (2) and the index is 1.
2. This time let x be a complex vector and
478
[Ch. 7
In the last step, 9x x was added and subtracted in order to " complete
the square " in the preceding step. Now let
3
FORMS
It can be observed from (23) that the value of the quadratic form will
normally depend on the values of the y-variables. However, it may happen
that the value of the quadratic form will remain of one sign independent
of the values of the variables. Such forms are called definite. In particular,
a real, quadratic form x*Ax is called positive definite if for any set of
complex or real numbers x1, x , ..., x , not all zero, the value of the
quadratic form is strictly positive. Similarly, we say the quadratic form
is positive semidefinite if
2
(25)
for all x 0, provided there is at least one set of values of the variables
for which the equality holds. Since the positive property of such a quad
ratic form is not dependent on the values of the variables, it must be
associated with the matrix A of the quadratic form. The following
terminology, then, appears quite natural. A real, symmetric matrix A
is said to be positive definite or semidefinite according as the quadratic form
x*Ax is positive definite or semidefinite.
We need to find means for determining whether or not a quadratic
form is positive definite. An approach to this is obtained b y considering
the canonic form in (23). The matrix A of the form is characterized by
Sec. 7.2]
479
three integers: the order n, the rank r, and the index p. If the index is less
then the rank (but greater than zero), the matrix can be neither positive
definite nor positive semidefinite.
Suppose that the index equals the rank: p = r. Then all the signs in
(23) will be positive. There are two possibilities: (1) the rank is equal to
the order, r = n, so that A is nonsingular; or (2) r < n, so that A is singular.
Suppose r < n. Then choose y1 up to y = 0 and y + 1 to y 0. This will
cause the quadratic form to vanish but, with x = Qy, not all the x's will
be zero. For any other choice of y-variables, the quadratic form will be
positive. Hence the quadratic form satisfies (25) and is positive semidefinite. The converse is, clearly, also true.
On the other hand, if r = n (with p still equal to r), so that A is non
singular, then every nonzero choice of the y's (and hence of the x's) will
lead to a positive value of the quadratic form. The conclusion is the follow
ing theorem:
r
Theorem
3. A
rank r and index
the index equals
singular and p =
ni
in
480
[Ch. 7
easily seen from (16) with x = 0. Hence we might as well remove the
nth row and column of A and consider it to be of the (n l ) t h order.
For this new matrix (28) still applies. But the determinant of the new
matrix is the principal cofactor of the old matrix obtained by removing
the last row and column. Since permuting the variables has no effect on
the quadratic form, it is immaterial which one of the variables we call
x . It follows that all the first principal cofactors of a positive definite
matrix will be positive.
This argument can now be repeated by setting two of the variables
equal to 0, then 3, and so on, up to all but one. We shall find that all the
principal cofactors of A will be positive. In the last case, with all but one
of the variables equal to zero, we find that all the elements of A on the
principal diagonal must be positive. (These elements are the (n l ) t h
principal cofactors of A).
What we have succeeded in proving is that, if a matrix is known to be
positive definite, then its determinant and all its principal cofactors will
be positive. Actually, what we need for testing a given matrix is the
converse of this result. It happens that this is also true. The proof, how
ever, is quite lengthy and will not be given. For future reference we shall
list this result as a theorem.
n
Theorem
4. A real symmetric matrix A is positive definite if and only
if its determinant and principal cofactors are all positive. It is positive
semidefinite if and only if its determinant is zero and all its principal
cofactors are non-negative.
As an example, consider the real matrix previously used to illustrate
the Lagrange reduction. It is required to form the determinant and
principal cofactors.
We observe that the diagonal elements are all positive. Since this is a
third-order matrix, the diagonal elements are the second cofactors. The
first principal cofactors are easily formed:
They are all positive. This leaves the determinant, which is found to be
4 . This is less than zero, and hence A is not positive definite, or semidefinite.
Sec. 7.3]
HERMITIAN
ENERGY FUNCTIONS
481
FORMS
Up to this point we have been dealing with quadratic forms having real
matrices that are symmetric. If the matrix of a quadratic form is complex,
it is possible to replace the matrix by its Hermitian part without changing
the value of the form, just as a real matrix_was replaced by its symmetric
part. Let H be a Hermitian matrix (h = h ). The expression
ji
ij
(29)
is called a Hermitian form. When H is real, the Hermitian form reduces to
a quadratic form. It should be expected, then, that the properties of
Hermitian forms are analogous to those of quadratic forms. We shall
merely list a few, without extensive comment.
B y carrying out an expansion as we did in (21) for a quadratic form, it
can be shown that the value of a Hermitian form is real.
A Hermitian form of rank r can be reduced to the canonical form given
on the right side of (23) by a nonsingular linear transformation x = Qy,
where Q is generally complex.
The terms " positive definite " and " semidefinite " apply to Hermitian
forms and are defined in the same way as for quadratic forms. The theorem
relating to the determinant and principal cofactors of positive definite and
semidefinite matrices applies to a Hermitian matrix also.
7.3
ENERGY FUNCTIONS
Fig. 1.
Excited two-port.
482
[Ch. 7
(30b)
m p
(31)
Let us now compute the power supplied to the network. The complex
power supplied at port k is I pk V ]c. The total complex power delivered
at all the ports is therefore I ^ E ^ . The real part of this is the real,
average power. The imaginary part is proportional to the net average
m
Sec. 7.3]
ENERGY FUNCTIONS
483
energy stored in the network, the difference between the average energy
stored in the inductors and in the capacitors. Thus,
(32a)
(32b)
The complex power input to the network can be obtained by premulti
plying both sides of (31) by I * . The result becomes
(33)
(On the right side appear only those loop currents that are port currents,
so that the m subscript can be omitted.) We find that the complex power
input on the right equals the sum of three terms on the left. We recognize
each of these terms as a quadratic form.
For a nonreciprocal network the loop-parameter matrices are not
symmetric. However, as discussed in the last section, the value of the
quadratic form is unchanged if the matrix of the form is replaced by its
symmetric part. We shall assume that this has been done. Each of the
quadratic forms on the left side of (33) is real. Hence comparing (32) and
(33) leads to the conclusion that
(34a)
(34b)
(34c)
(35b)
(35c)
484
[Ch. 7
(36)
Recall from (56) in Chapter 2 that B T
= I is the loop transformation
that specifies the branch currents I in terms of loop currents. Thus
m p
(37)
where b is the number of branches in the network. For a general non
passive, nonreciprocal network, nothing specific can be stated about this
quadratic form.
P A S S I V E , RECIPROCAL
NETWORKS
(39a)
(39b)
Note the differences on the right-hand sides of these two expressions. The
branch-inverse-capacitance matrix is diagonal, whereas the branchinductance matrix is not necessarily diagonal. When there is no mutual
Sec. 7.3]
ENERGY FUNCTIONS
485
Fig. 2.
486
[Ch. 7
Since I + I
equals branch current 3, the term R |I
+ I |
is the
power dissipated in R and |I
+ I|
I) /2
is the energy stored in
I) . The positive semidefinite nature of T(j) is not evident from the
right-hand side. Observe, however, that the matrix L is singular only for
Li L M = 0, which is the condition of unity coupling.
To summarize the result so far obtained, the loop resistance, inductance,
and reciprocal-capacitance
matrices R , L , and D
of a passive,
reciprocal network are positive semidefinite. This result was established by
giving physical interpretations to certain quadratic forms based on a
sinusoidal steady-state analysis.
Let us now return to the original loop equations in (30) in which the
variables are Laplace transforms. Without any concern for physical
interpretation, let us premultiply both sides by I*(s). The result will be
pl
p2
pl
pl
p2
(41)
Again we find the same quadratic forms we had before, only now the
variables are loop-current transforms rather than phasors. The quadratic
forms in this equation do not have an energy interpretation like those of
(33). However, the matrices of these quadratic forms are identical with the
former ones. Hence these quadratic forms are positive semidefinite. We
therefore give them symbols similar to those of (40) and continue to call
Sec. 7.3]
ENERGY FUNCTIONS
487
(43)
(The notation on the right side has been modified in two ways. The m
subscripts have been dropped because the only loop currents that remain
in the product I*E are those that are the same as port currents. Also, the
only nonzero components in E are the port voltages. Hence I * E can be
replaced by I*V, where I and V are the port vectors.)
Let us digress here for a moment. This entire development started from
the loop equations. Alternatively, a completely dual development can
proceed on the basis of the node equations. Instead of the loop-parameter
matrices R , L , and D , the conductance, inverse-inductance, and
capacitance node-parameter matrices G , , and C , respectively, will
appear. Energy functions can now be defined in terms of these parameter
matrices and the node-voltage vector V . These will have the same form
as (42) with V in place of I and with the node-parameter matrices in
place of the loop-parameter ones. From these it is concluded that the
node conductance, capacitance, and inverse-inductance
matrices G , C ,
and of a passive, reciprocal network are positive semidefinite. An equation
similar to (43) can now be written with these new energy functions, with
V and I interchanged. This alternative development is not needed to carry
on the subsequent discussion, just as the node system of equations itself
is really superfluous. However, just as node equations provide helpful
viewpoints and often simplify computation, so also this alternative
approach may sometimes be useful. You should work out the details of
the procedure just outlined if you are interested.
m
Look again at (43). The quantities appearing on the left side are defined
in terms of loop-current variables (or branch-current variables through
the loop transformation). But on the right we find port variables. Of
course, the port voltage and current variables are related to each other.
488
[Ch. 7
(45)
or
This follows because the quadratic forms are real scalars. The result of
inserting (44a) into (43) and (44b) into (45) is
(46a)
(46b)
FUNCTION
(47)
Sec. 7.3]
ENERGY FUNCTIONS
489
Note that the quadratic forms are functions of s only through the fact that
the loop currents are functions of s. The real, positive semidefinite nature
of the quadratic forms does not depend on the current variables, but only
on the loop-parameter matrices, which are constant matrices.
The preceding expression can be separated into real and imaginary
parts after replacing s by + j. Thus
(48a)
(486)
Notice that these equations apply no matter what the value of s may be,
except at zeros of I(s). These two are extremely important equations, from
which we can draw some interesting conclusions. For later reference let
us state these results as a theorem.
Theorem
invariant,
are true.
5. Let Z(s) be the driving-point impedance of a linear timepassive, reciprocal network N. Then the following
statements
490
[Ch. 7
real functions, which we shall take up next. Part (b) leads to the historic
ally important reactance theorem of Foster. Parts (c) and (d) lead to
Cauer's results on RL and RC networks.
CONDITION ON ANGLE
(49)
where each of the coefficients is positive. Let s = + j be a point in
the right half-plane; that is, > 0, as shown in Fig. 3. Each of the
0
(a)
Fig. 3.
(b)
Sec. 7.3]
ENERGY FUNCTIONS
491
(51)
0t
(53)
(54)
When this is inserted into the preceding equation, the result will be
(55)
492
[Ch. 7
The worst case occurs when the cosine equals 1 . For this case the con
dition reduces to
or
(56)
Each side of this expresssion is the real part divided by the magnitude of
a complex quantity, which equals the cosine of the corresponding angle.
Hence
(57)
from which it follows that
(58)
Since Re s = > 0, this is identical with (50).
This completes the development of the general necessary properties of
impedance functions of passive networks. Completely similar properties
could have been developed for the admittance function by starting from
(46b) instead of (46a). Thus Theorem 5 and Eq. (50) are true when Z(s) is
replaced by Y(s). We shall now define a class of mathematical functions
having these same properties and shall investigate the detailed behavior
of this class of functions.
0
7.4
Sec. 7.4]
493
Fig. 4.
Pole of order 3.
of the pole of order n, the function has a Laurent expansion of the form
(59)
(60a)
(606)
then
(61)
494
[Ch. 7
s-plane
Fig. 5.
W-plane
(62)
is also pr; because the right haJf s-plane goes into the right half F -plane
since F (s) is positive real. Also, the right half F -plane goes into the right
2
Sec. 7.4]
495
(63)
is a pr function. Now we use 1/s and F(s) as Fi(s) and F (s) in (62), in
both possible ways, and the result follows immediately.
From the fact that the reciprocal of a pr function is itself pr, it follows
that a positive real function can have no zeros in the right half-plane;
because if it did, then its reciprocal would have poles in the right halfplane, which is impossible. Since the impedance of a passive reciprocal
network is a pr function, its reciprocalthe admittanceis also a pr
function.
From a conformal-mapping point of view, the points F(s) = 0 and
(these are the zeros and poles of the function), which are on the boundary
of the right half F-plane, cannot be images of any interior points of the
right half s-plane. Let us now inquire into the properties resulting when
other boundary points of the right half F-plane are images of boundary
points of the right half s-plane; that is, let a point on the j-axis be mapped
by a pr function F into a point on the imaginary axis of the F-plane.
If j is the point in question, then
2
(64)
where X is real (positive, negative, or zero).
Consider a neighborhood of j in the s-plane and the corresponding
neighborhood of j X in the F-plane, as shown in Fig. 6. Let si denote a
point in the right half-plane, in this neighborhood of j . Let us now
0
Fig. 6.
496
[Ch. 7
(65)
(n)
where F (j )
is the first nonvanishing derivative of F(s) at j .
As si approaches j , the dominant term on the right will be the first
term. Let us define.
(66a)
0
(66b)
(66c)
Then, in the limit, we shall find from (65) that
(67)
But the positive real condition requires that || /2 as long as || /2.
Therefore we conclude from (67) that
(68a)
(68b)
Thus the first nonvanishing derivative is the first one, and its angle is
zero at s = j . This is a very important result. For future reference we
shall state it as a theorem:
0
Theorem
6. If any point on the j-axis is mapped by a positive real function
F into a point on the imaginary axis in the F-plane, then at this point the
derivative dF/ds is real and positive.
A number of other results follow from this important theorem. Note that
if F(s) has a zero or a pole on the j-axis, the conditions of the theorem
are satisfied. In the case of a zero ( X = 0), a point on the j-axis is
mapped into the origin of the F-plane, which is on the imaginary axis.
Hence the derivative dF/ds is real and positive. This also implies that the
zero is a simple one, since at a higher order zero the first derivative will be
zero. If F(s) has a pole on the j-axis, its reciprocal will have a zero there
and the theorem will apply to the reciprocal. However, d(l/F)/ds evaluated
at a pole of F(s) is the reciprocal of the residue of F(s) at the pole. (See
0
Sec. 7.4]
497
CONDITIONS
positive
498
[Ch. 7
Any other poles on the j-axis must occur in conjugate pairs and with
conjugate residues, since F(s) is a real function. Since the residues are
real by hypothesis, the two residues are equal. Taking the principal parts
at the conjugate poles j and j together, we get
i
where k is real and positive. This function is also positive real, and, in
addition, has the property
i
(We may note that F (s) is the impedance of a capacitance, F (s) that of
an inductance, and F (s) that of a parallel tuned circuit.)
Thus we can subtract from the given function F(s) the principal parts
at all of its poles on the j-axis. The remainder function F (s) still has
property (c) of the theorem; that is,
0
(69)
The remainder function F (s) is a function that is regular in the right
half-plane and its entire boundary, the j-axis, including the point at
infinity. For such a function the minimum value of the real part through
out its region of regularity lies on the boundary. This can be proved by
using the maximum-modulus theorem (see Appendix 2) in the following
way. Let G(s) = e
This function will have the same region of regular
ity as F (s). Hence, according to the maximum-modulus theorem, the
maximum magnitude of G(s) for all 0 lies on the j-axis. Since
r
- i r ( s )
(70)
the maximum magnitude of G(s) will correspond to the smallest value of
Re[F (s)]. This proves the desired result that the minimum value of
Re [Fr(s)] for all 0 occurs on the j-axis. Since according to (69) this
value is nonnegative, the real part of F (s) must be non-negative every
where in the right half-plane; that is,
r
(71)
Sec. 7.4]
499
W e have shown that each term on the right is pr. Y o u can easily show
that the sum of two (or more) pr functions is itself pr. Hence, F(s) is
positive real. This completes the proof of the sufficiency of the stated
conditions.
Since the reciprocal of a pr function is also pr, we can restate these
necessary and sufficient conditions in terms of the zeros of F(s).
Theorem
(a)
(b)
(c)
F(s) is positive
positive
This theorem follows directly from the preceding one if we remember that
the residue of a function at a simple pole is the reciprocal of the derivative
of the reciprocal of the function.
In testing a given function to determine positive realness, it may not
always be necessary to use the necessary and sufficient conditions listed
in the preceding two theorems. It may be possible to eliminate some
functions from consideration by inspection because they violate certain
simple necessary conditions. Let us now discuss some of these conditions.
We have seen that a rational positive real function has neither zeros
nor poles in the right half s-plane. We previously defined a Hurwitz poly
nomial as one that has no zeros in the right half-s-plane. This definition
permits zeros on the j-axis. With this terminology, we see that a positive
real function is the ratio of two Hurwitz polynomials.
The factors that constitute a Hurwitz polynomial must have one of the
following two forms: (s + a) for real zeros or (s + as + b) for a pair of
complex zeros, with a being non-negative and b being positive. If any
number of such factors are multiplied, the result must be a polynomial
all of whose coefficients are non-negative. Furthermore, unless all the
factors correspond to zeros on the j-axis, all the coefficients of the poly
nomial will be strictly positive. If we introduce the added condition that
zeros on the j-axis be simple, then it is found that, when all the zeros are
on the j-axis, every other coefficient will be zero and the remaining
coefficients will be strictly postive. Even though this is a necessary
condition for a Hurwitz polynomial, it is not sufficient, as the following
counter-example readily demonstrates:
2
(72)
The polynomial on the right has no missing powers of s and ail coefficients
are positive, yet it has a pair of zeros in the right half-plane.
500
[Ch. 7
Remarks
No more than simple pole at infinity; pos
itive coefficients. Might be positive real.
Coefficient of cubic term missing. Not
positive real.
Double zero at infinity. Not positive real.
Coefficients missing in numerator looks
bad, but all even powers missing. Might
still be positive real. (In fact, it is.)
No negative or missing coefficients, but
triple pole at s = 2 might seem peculiar.
N o t ruled out. (In fact, it is pr.)
FUNCTIONS
Sec. 7.4]
501
the truth of (50) for a real function F(s), it follows that F(s) is positive
real.
Thus this angle property is not only necessary but sufficient as well. We
shall therefore state it here as a theorem.
Theorem
F(s) is positive
BOUNDED
REAL
FUNCTIONS
W-plane
F-plane
Fig. 7.
502
[Ch. 7
s=j,
F(j) takes on values in the right-half-plane or on the jX-axis,
and these fall inside or on the W-plane unit circle. Hence, if F(s) is pr
and W(s) is related to F through (74), then
(75)
N o w consider the poles of W(s). From (74), these occur where
F(s) = 1. Values of s for which this is true cannot lie in the closed
right half-plane if F(s) is pr, since that would require Re F < 0 for
Re s 0. Hence W(s) must be regular both in the right half-plane and
on the j-axis.
A function W(s) having these propertiesnamely, W(s) regular in the
closed right half-plane and | W(j)| 1is called bounded real. Thus a
bilinear transformation of a positive real function is bounded real. The
converse of this is also true; that is a bilinear transformation of a bounded
real function is positive real. Sketching the few steps in the proof is left to
you.
This relationship of bounded real and positive real functions leads to an
interesting conclusion. Suppose a pr function F(s) is written as
(76)
where ii and m are even polynomials, and n and n are odd. Then the
bilinear transformation (74) gives
2
(77)
becomes
(78)
(79)
Sec. 7.4]
503
(80)
This is simply the original F(s) with the even powers of the numerator and
denominator interchanged. Since F(s) is a bilinear transformation of a
bounded real function, it is pr. The reciprocal of F is also pr. But the
reciprocal of F is the same as the original F(s), with ni(s) and n (s) inter
changed. The conclusion is given as the following theorem:
2
Theorem
I I . If in a positive real function the even powers of the numerator
and denominator, or the odd powers, are interchanged, the result is a positive
real function.
FUNCTION
Since the real part of a pr function plays such a central role in its
properties, we should examine the behavior of the real part of such a
function on the j-axis. Remember that the j-axis real part of F is equal
to the even part evaluated at s = j ; that is,
(81)
so that statements made about the even part can easily be interpreted in
terms of the real part on the j-axis.
We already know that R() is necessarily an even function of and
non-negative for all . It is also easy to establish that the even part of
F(s) can have no poles on the j-axis. Any poles of the even part would
also have to be poles of F(s); but on the j-axis, these are simple. If we
consider F(s) expanded in partial fractions as in (71), the function F(s)
will contain the same terms, but all those involving the poles on the jaxis will have a negative sign. Hence, in forming the even part,
F(s)+ F(s), these will all cancel, leaving the function with no poles
on the j-axis. Interpreted in terms of the real part, this means that R()
must be bounded for all .
N o w let U S consider a possible zero of R(). Figure 8 shows a sketch
of R() versus in the vicinity of a zero. Because of the positive real
requirement, R() must remain positive on both sides of the zero. It
follows that a zero of R() on the -axis cannot be of odd multiplicity;
it must be of even multiplicity.
504
Fig. 8.
[Ch. 7
real
7.5
REACTANCE FUNCTIONS
Let us now turn our attention to some special types of positive real
functions. These arise from a consideration of networks containing only
two types of elements (LC, RC, RL). Historically, such networks were
studied before the more general ones, starting with the work done by
Foster in 1924.
We shall initially consider networks that have no resistance. Such net
works are referred to as lossless, or reactance, networks. In Theorem 5 we
noted that the driving-point impedance of a lossless network is purely
Sec. 7.5]
REACTANCE FUNCTIONS
505
(82b)
(a)
Fig. 9.
s-axis
(b)
becomes the imaginary z-axis, and vice versa. A similar case obtains for
the other transformation. When z is real, the argument of F(jz) is
imaginary, so by hypothesis F(jz) will also be imaginary. Hence (z) will
be real when z is real. It follows from the reflection property given in (6)
of Chapter 6 that
(83)
506
[Ch. 7
(85)
Sec. 7.5]
REACTANCE FUNCTIONS
Fig. 10.
507
Fig. 11.
zeros must hold on the entire imaginary axis (positive and negative values
of ), we conclude that the point s = 0 is either a zero or a pole of a react
ance function.
Note that, if F(s) is a pr function mapping the imaginary axis into the
imaginary axis, so is the function F(l/s). With the transformation s -> 1/s,
the point in the s-plane goes into the origin in the 1/s-plane. Hence, b y
using the immediately preceding result, we find that the point s = is
either a zero or a pole of a reactance function.
We have now discussed several properties of reactance functions. We
should also note that certain properties of general pr functions apply in
particular to reactance functions. Thus, since we have shown that poles
and zeros of a pr function that lie on the j-axis are simple and that
residues at such poles are real and positive, we conclude that all poles
and zeros of reactance functions are simple and that residues at all poles
are real and positive.
We are now in a position to consolidate our results about reactance
functions and to state necessary and sufficient conditions for a rational
function of s to be a reactance function.
508
Theorem
only if
1.
2.
3.
4.
[Ch. 7
if and
Notice that this statement involves only the poles and the residues,
not the zeros. We have already shown these conditions to be necessary; it
remains to prove that they are sufficient; that is, assuming a rational
function to satisfy the stated conditions, we must show that the function
is a reactance function. This is most easily done b y considering the partialfraction expansion of such a function. If we combine the two terms due to
conjugate poles, the most general form of the partial-fraction expansion
will be*
(86)
where the summation runs over all the poles, and all the k's are positive.
Of course, the pole at the origin or at infinity, or both, may be absent.
This expression is consistent with (71) with F (s) = 0, since in the present
case there are no other poles except these on the j-axis. The desired
result follows immediately. Each term in this expansion is imaginary for
imaginary values of s, so that (s) maps the imaginary axis into the imagin
ary axis, which makes (s) a reactance function by definition.
The alternation property of the poles and zeros forms the basis of an
alternate set of necessary and sufficient conditions, stated as follows:
r
Theorem
14. A real rational function of sis a reactance function if and only
if all of its poles and zeros are simple, lie on the j-axis, and alternate with
each other.
Again, we have already proved that a reactance function necessarily
satisfies these conditions. It remains to show that the conditions are
sufficient. A rational function that satisfies the given conditions must have
the following form:
(87)
* In the absence of condition 4 of Theorem (13) a constant term would be permitted
in the expansion of (86). Condition 4 is required to eliminate this constant, which cannot
be part of a reactance function.
Sec. 7.5]
REACTANCE FUNCTIONS
509
where
(88)
In (87) K is a positive constant, and k = 2n 2 or 2n according as (s)
has a zero or a pole at infinity. If (s) has a pole at s = 0, we take to
be zero. A factor s will then cancel. The desired result now follows
immediately. Each of the quadratic pole and zero factors in (87) is real
when s is imaginary. This means that, due to the factor s, (s) is imaginary
when s is imaginary. Hence, (s) is a reactance function, by definition.*
0
REALIZATION OF REACTANCE
FUNCTIONS
510
[Ch. 6
Network representation
Function
Fig. 12.
Impedance
Admittance
(b)
(a)
Fig. 13.
(90a)
Sec. 7.5]
REACTANCE FUNCTIONS
511
or
(90b)
(a)
(b)
Fig. 14.
Hence the values of L and C for each of the two branches can be found by
comparison of this expansion with each of the numerical terms in the
given function. The result is shown in Fig. 14b.
The two networks obtained are entirely equivalent at their terminals.
No measurements made there could distinguish one from the other.
512
[Ch. 7
The Foster forms are not the only possible networks that realize a
given function. (There are, in fact, an infinite number of alternative
structures.) Let us illustrate one possible alternative with the example
already treated, before generalizing. The impedance in (90) has a pole at
infinity. If this entire pole is subtracted from Z(s), the remaining
function will no longer have a pole at infinity and so it must have a zero
there. Thus
(b)
(a)
(c)
(d)
Fig. 15.
network whose impedance is Z1. Now the reciprocal of Z1 will have a pole
at infinity. This pole can be totally removed by subtracting s/24, leaving
Sec. 7.5]
REACTANCE FUNCTIONS
513
Fig. 16.
form:
(91)
In the example just treated the process carried out step by step is
actually a continued-fraction expansion, where each of the Z and Y
functions in (91) is of the form ks, in which k is inductance or capacitance.
The expansion deals exclusively with the pole at infinity, removing this
i
514
[Ch. 7
pole alternately from the impedance and then the remaining admittance.
The result of this process for an arbitrary reactance function will have the
network structure shown in Fig. 17a.
(b)
(a;
Fig. 17.
The result is like Fig. 17b with the first four elements being C1 =
L - , C = ^ , a n d L = WTo summarize, we have shown that the impedance and admittance of a
lossless network are reactance functions; and conversely, given any
reactance function, a number of networks can be found whose impedance
or admittance is equal to the given reactance function.
2
FUNCTIONS
Sec. 7.5]
REACTANCE FUNCTIONS
515
(87). If we denote even and odd polynomials by m(s) and n(s), respectively,
then a reactance function (s) can be written as
(92)
(93)
where (92) is used for (s). The impedance of this RLC network will
be pr and regular on the j-axis; hence its poles cannot lie in the closed
right half-plane. The polynomial m + n in (93) is therefore a strictly
Hurwitz polynomial. This is a very useful result. We shall state this result
and its converse as a theorem.
Theorem
16. IfP(s) = m(s) + n(s) is a Hurwitz polynomial, then the ratio
m/n is a reactance function. Conversely, if the ratio of the even and odd parts
of a polynomial P(s) is found to be a reactance function, then P(s) will differ
from a strictly Hurwitz polynomial by at most a multiplicative even poly
nomial.*
This theorem provides us with a means for easily determining whether
a given rational function is regular in the right half-plane, as positive
realness requires. We take the ratio of the even and odd parts (or its
reciprocal) of the denominator polynomial, then expand in a continued
fraction or a partial fraction. To illustrate, let
* A proof is outlined in problem 43 for you to work out. An alternate proof is given
in Norman Balabanian, Network Synthesis, Prentice-Hall, Englewood, Cliffs, N.J.,
1958, pp. 77-81.
516
[Ch. 7
N o w form the ratio m/n and expand in a continued fraction. The result
will be
Then
is a factor of both even and odd parts. This is an even polynomial that
has two zeros in the right half-plane and two in the left. The original
polynomial is
Sec. 7.6]
517
7.6
(95)
Let us replace each resistance in N by an inductance of equal value
(R ohms becomes R henrys). Then the loop impedance matrix of the new
network N' becomes
(96)
518
[Ch. 7
The network N' contains only capacitance and inductance, so that (s)
in the last equation is a reactance function. Thus we have found that the
impedance of an RC network can be transformed to a reactance function
b y replacing s by s and then multiplying by s.
It would be of interest to see if the converse is also true; that is, given
a reactance function (s), can we convert to the impedance of an RC
network with the opposite transformation? To do this, consider the reac
tance function to be expanded in partial fractions, as shown in (86).
N o w divide the entire result by s and replace s b y s. (This is the opposite
of the transformation just used.) The result will be
2
Theorem
then
(97)
is a reactance function.
then
(98)
is the driving-point
impedance of an RC network.
Sec. 7.6]
519
of an RC network, then
(99)
is a reactance function.
Conversely,
then
(100)
is the driving-point
admittance
of an RC network.
(101)
(102)
where the k's and 's are all real and positive. Note that we have used the
same symbols for the residues and poles in both cases, but these are general
expressions for classes of functions and the two are not supposed to be
related.
Equation (102) is not a partial-fraction expansion of Y (s).
It is, rather
an expansion of Y (s)/s,
after which the result is multiplied through by s.
If we divide (102) b y s, we find that the form is identical with (101).
This shows that an RC admittance function divided by s is an RC imRC
RC
520
[Ch. 7
pedance function. We see that the poles of both these functions are
negative real, and the residues of Z
and Y /s are all positive.
B y differentiating the last two equations along the real axis (s = ),
we obtain a result that is the counterpart of the positive-slope property
of a reactance function; that is,
R C
RC
(103a)
(103b)
RC
Theorem
19. A rational function F(s) is the driving-point impedance of an
RC network if and only if all of its poles are simple and restricted to the
finite negative real axis (including s = 0), with real positive residues at all
poles and with F(oo) real and non-negative. (This is the counterpart of
Theorem 13 for reactance functions.)
Theorem
20. A rational function F(s) is the driving-point impedance of an
RC network if and only if all the poles and zeros are simple, lie on the negative
real axis, and alternate with each other, the first critical point (pole or zero),
starting at the origin and moving down the negative real axis, being a pole.
(This is the counterpart of Theorem 14 for reactance functions.)
Theorem
21. A rational function F(s) is the driving-point admittance of an
RC network if and only if all of its poles are simple and restricted to the
negative real axis (excluding the point s = 0, but including infinity), with
Sec. 7.6]
521
Fig. 18.
Typical
Z ().
Fig. 19.
Typical
Y ().
RC
RC
Theorem
22. A rational function F(s) is the driving-point admittance of an
RC network if and only if all the poles and zeros are simple, lie on the negative
real axis, and alternate with each other, the first critical point (pole or zero),
starting at the origin and moving down the negative real axis, being a zero.
(The only difference between this theorem for admittances and Theorem
20 for impedances is the last word.)
We have already sketched the proofs of all these theorems in the pre
ceding discussion. You m a y organize the proofs as an exercise.
522
[Ch. 7
Fig. 20.
Sec. 7.6]
523
The poles are real, negative, and simple; and the residues are positive, as
the partial-fraction expansion shows. The constant term can be recognized
as a two-unit resistor. The term 3/(s + 1) represents a parallel RC branch.
B y reference to Fig. 20, the values of C and R are found to be C = \,
R = 3. The last term has the same form as the second one and can be
realized b y the same kind of branch. The complete realization is given in
Fig. 21a.
(a)
Fig. 21.
(b)
BC networks realizing a given function.
REALIZATION
524
[Ch. 7
R C
RC
RC
R C
(a)
Fig. 22.
(b)
Ladder-network realizations.
Sec. 7.7]
TWO-PORT PARAMETERS
525
What has been done for RC networks can be duplicated for RL net
works. The starting point is again a transformation that will take a reac
tance function into an RL impedance or admittance. It is immediately
found that the class of RL impedance functions is identical with the
class of RC admittance functions, and vice versa. Hence there is no need
to duplicate the detailed development. In any theorem involving RC
networks it is only necessary to replace the word " impedance " with the
word " admittance " (or " admittance " with " impedance ") to arrive
at a valid theorem for RL networks. We shall not pursue this subject
here but will suggest some of the results as problems.
7.7
TWO-PORT PARAMETERS
(104b)
526
[Ch. 7
Fig. 23.
(105a)
(1056)
(106)
Sec. 7.7]
TWO-PORT PARAMETERS
527
Since the impedance is positive real, this proves that the quadratic form
on the right is also positive real. To prove the condition for the y-matrix,
the two pairs of terminals can be connected in parallel through ideal
transformers and the overall input admittance calculated. This is left
for you as an exercise.
Let us now restrict ourselves to two-ports; the extension of the sub
sequent results to higher order multiports is simple and will become
evident. The fact that Z and Y are positive real matrices has some
interesting consequences. Let x and x be two arbitrary real numbers.
Since the quadratic forms
o c
s c
(107a)
( 1 0 7 6 )
are positive real functions, it follows that any pole of these functions on
the j-axis must be simple, and the residue at such a pole must be real and
positive, Suppose, for instance, that the z-parameters have a pole at
s = jt. Since this pole is a simple pole of the quadratic form, the residue
of
Qiis
( 1 0 8 )
(i)
(i)
12
22
(i)
22
( 1 0 9 )
Thus the residue itself is a quadratic form whose matrix is the matrix of
528
[Ch. 7
(110a)
(110b)
The first line in these expressions is already known, since zn and z
are driving-point functions and therefore positive real. The second line,
however, is a new and important result. It is known as the residue
condition.
What was done for quadratic form Qi is also valid for Q /s. Thus the
same conclusions follow for residues of Y c / . We shall state this result as
a theorem.
22
Theorem
24. At any pole on the j-axis of Z or Y$ /s of a linear, timeinvariant, passive, reciprocal two-port network, the residues of the parameters
satisfy the condition
0 C
(Ill)
where kfj is the residue of z or y / s at the j-axis pole.
(The superscript has been omitted for simplicity.)
i j
ij
In particular, if the network is lossless, all the poles z and y are on the
j-axis, and hence the residue condition (111) applies at all the poles. One
of the implications of this fact for a lossless network is that it is impossible
for z to have a pole that is not also a pole of zn and z , nor for y
to have a pole that is not also a pole of yn and y .
For if either k or
k is zero when k is not, the residue condition will be violated. On the
other hand, it is possible for zn or z (or both) to have a pole not shared
b y the other parameters. We refer to such poles as private poles of zn or
z . A similar statement applies to the y-parameters.
Let us now turn to another consequence of the positive real nature of
Z and Y . B y definition, positive realness is linked with the real part
of a function. Hence we should expect to obtain some relationship among
the real parts of the z- and y-parameters.
Let us denote these real parts
ij
21
22
22
22
21
22
22
o c
ij
s c
21
Sec. 7.7]
TWO-PORT PARAMETERS
529
12
22
21
12
22
(112a)
(112b)
(113)
and
(114)
22
o c
RESISTANCE-CAPACITANCE
s c
TWO-PORTS
As a final note, observe that what was said about lossless two-ports is
also true for RC two-ports, by virtue of the transformations previously
discussed and with appropriate and obvious modifications. Thus for RC
two-ports, the z- and y-parameters will have all their poles on the negative
real axis, and the residue condition will apply at these poles. No pole of
z can fail to be a pole of both zn and z , but zn and z can have
private poles; and similarly for the y-parameters.
As an illustration, consider the RC two-port shown in Fig. 24. The short21
22
22
530
[Ch. 7
Note, first, that all three functions have the same poles. The zeros of yn
are at (approximately) f and T h u s the zeros and poies of both
yn and y
alternate on the negative real axis, and they have the appro
priate behavior at infinity (which is?), so that both are RC admittance
functions. On the other hand, y
does not have all the properties of an
RC admittance; among other things, the residues of y /s
are not all
positive.
A test for the residue condition shows that it is satisfied at all the poles;
in fact, the residue condition is satisfied with the equals sign. To distinguish
when the residue condition is satisfied with the equals sign and when it is
not, we say the pole is compact if the residue condition at the pole is
satisfied with the equals sign. Thus, for the illustration, all the poles
(including the constant term) are compact. Verification that the real-part
condition is satisfied will be left to you.
For driving-point functions of networks containing two kinds of ele
ments only we found sets of necessary conditions that were also proved
to be sufficient. In fact, actual procedures for realizing one or more net
works from a given function were obtained. The case for a two-port is not
as simple, since a set of three parameters is involved. Although necessary
conditions on these parameters have been obtained, these conditions turn
22
21
21
Sec. 7.8]
531
7.8
Fig. 25.
or
(115b)
532
[Ch. 7
(See Chapter 3.) In the final form all impedances have been normalized to
R, which is equivalent to taking the value of R to equal 1.
Now suppose a rational positive real function Z(s) is given. The even and
odd parts of its numerator and denominator can be separated, and the
function can be written in the usual form. Then this expression can be
placed in the same form as (115b) in two possible ways, as follows:
(116)
(case A)
(117)
(case B)
(118)
For each of these two cases, formal identifications can be made b y compar
ing these expressions with (115b). Thus
Case A
Case B
(119)
21
(case A),
(120)
(case B).
Sec. 7.8]
533
Since
we get for z ,
21
(case A),
(121)
(case B).
Of course, once the z-parameters are known, the y-parameters can also be
found. (See Table 1 in Chapter 3.) The complete results are tabulated in
Table 1.
The question is, do these open-circuit or short-circuit parameters satisfy
realizability conditions for a passive, reciprocal, and lossless two-port?
The first difficulty appears to be that z is not a rational function because
of the indicated square root. However, if m\ m n\ n is a perfect square,
the apparent difficulty will disappear. Observe that mi m ni n is the
numerator of the even part of Z(s). Because Z(s) is a positive real function,
zeros of its even part on the j-axis must necessarily have even multi
plicity. There is no such requirement, however, on any other zeros. Hence,
unless some remedy can be found, it appears that z will generally be
irrational.
A remedy has been found in the following way. Suppose the given Z(s)
is augmented b y multiplying its numerator and denominator b y a strictly
Hurwitz polynomial m + n , which certainly does not change the func
tion. Thus
21
21
(122)
The new z
21
(124)
Case B:
Pole or zero of Z(s)
at s= 0
Case A:
No pole or zero of
Z(s) at s = 0
Condition
Table 1.
z-Parameters
y-Parameters
534
FUNDAMENTALS OF NETWORK SYNTHESIS
[Ch. 7
Sec. 7.8]
535
21
21
21
21
21
21
Also,
536
[Ch. 7
Case A
Case B
22
21
Sec. 7.8]
537
may be necessary to use ideal transformers. The series structure of twoports is not a very desirable one; one objection is that all but one of the
two-ports will be floating above ground.
A more desirable structure is a cascade structure. Darlington's contri
bution was to show that such a realization is possible. We shall do no more
here than outline his procedure. Observe, first, that if the given impedance
function has poles and zeros on the j-axis, these can be removed as the
branches of a ladder network. Such branches are shown in Fig. 26 as type
(a)
(b)
(c)
(d)
Fig. 26. Canonic sections for cascade realization: (a) Type A; (b) type B; (c) type C;
(d) type D.
(125)
All zeros are shown as double zeros. For a positive real function, those on
the j-axis must necessarily be double. If the other zeros are not initially
double, the function is augmented to make them so. The transmission
zeros of the lossless two-port are exactly the zeros of the even part of Z(s).
It turns out that a type C section, as shown in Fig. 26, has a pair of
transmission zeros that are either real or imaginary (depending on the dot
538
[Ch. 7
Observe that z11 has a private pole at infinity; this can be removed as a
Sec. 7.8]
539
(a)
(b)
(c)
Fig. 27.
540
[Ch. 7
where
22
7.9
21
TWO-PORTS
Let us now look back over the procedure followed in the last section
to observe if there are features of a general nature that can serve as
guides to other kinds of networks. The first step was to consider a network
structure representative of a particular class. In Fig. 25 this was a lossless
two-port terminated in a resistance. An expression was then written for
the function of interest in terms of the component parts of the structure,
as in (115). Then a rational function of the class under consideration was
manipulated to make it take on this same form, as in (117) and (118).
Finally, the pertinent functions of the component parts of the network
were identified. Of course, it must be verified that these functions satisfy
realizability conditions for the class of networks under consideration.
Furthermore, procedures must be established for realizing these functions.
In this section we shall carry out similar procedures for networks con
sisting of resistance, capacitance, and possibly active devices. A number of
considerations make such networks of practical importance. These include
Sec. 7.9J
541
the relatively large size and weight of inductors, the low cost of transistors
and other active devices, and the fact that realization of resistance,
capacitance, and active devices is obtainable in integrated circuits.
CASCADE
CONNECTION
Consider first the network shown in Fig. 28, in which two RC two-ports
Fig. 28.
Cascaded BC two-ports.
(126)
(127a)
(127b)
The first has a single finite zero, which is real. The second has two pairs
of complex zeros. Each of these is a realizable RC transfer impedance for
any value of the constant K.
* See Chapter 3.
542
[Ch. 7
Observe from (126) that the denominator is the sum of two RC impe
danceswhich is, again, an RC impedance; but the denominator of any
given rational function will be a polynomial. The situation is remedied by
dividing numerator and denominator of the given function b y an auxil
iary polynomial D(s). The degree and the zeros of this polynomial must be
chosen to make QjD an RC impedance. This is easy to do: the degree of
D(s) must be equal to, or be one greater than, the degree of Q(s) ; and its
zeros must alternate with those of Q(s). Thus, if the first function of (127)
is used as an example,
(128)
The choices
with
are both acceptable. Let us choose D(s) = (s + l)(s + 3)(s + 5). Then, by
comparison with (126), we can write
(129)
21b
Each of these two sets is now easily recognized. Two port Na is easily seen
Sec. 7.9]
543
to be a degenerate tee network with a shunt branch only (like the type B
section in Fig. 26b), as shown in Fig. 29a. As for N , in addition to a
b
(b)
(a)
(c)
Fig. 29. Realization of numerical example showing: (a) N ; (6) IV& ; (c) the overall net
work with a cascade connection.
a
l l b
l l b
22b
l l b
21a
CASCADING A N E G A T I V E
21b
CONVERTER
544
[Ch. 7
Negative
converter
l l b
(130)
(131)
where the subscripts p and n stand for " p o s i t i v e " and "negative."
The identification of z
2 2
and z
l l b
l l b
21a
22
21b
21
l l b
Sec. 7.9]
545
does not have. (The term " p o l e " includes the behavior at infinity. Thus,
if z
is nonzero at infinity, z
must also be nonzero.)
As an illustration take a fourth-order Chebyshev function with two
pairs of j-axis transmission zeros thrown in:
21b
l l b
l l b
546
[Ch. 7
21a
l l b
21b
PARALLEL CONNECTION
Negative
converter
Negative
converter
(b)
(a)
Fig. 31.
21
21b
22
22b
(132)
Sec. 7.9]
547
with a specific configuration, as shown in Fig. 31b. This does not limit the
class of transfer functions that can be handled.
When the y-parameters
of these simple networks are calculated and
substituted into (132), the result becomes
(133)
(134)
(135a)
(135b)
lb
2b
548
[Ch. 7
The value of K can be chosen with a view toward simplifying the network.
Since KP/D is to be subtracted from Q/D, K can be chosen to cancel one of
the terms. Thus, with K = 3.75/52 = 0.072, we get
Finally,
Sec. 7.9]
549
Negative
converter
Fig. 32.
~C-AMPLIFIER
CONFIGURATION
Fig. 33.
Feedback-amplifier configuration.
550
[Ch. 7
(137a)
(137b)
From the first of these we get
(138a)
(1386)
and Y
(139)
Sec. 7.9]
551
i p
ip
Fig. 34.
i n
552
[Ch. 7
In the present case, the degree of D(s) need be no more than 1. Let
D(s) = s + l . T h e n
from which
then becomes
PROBLEMS
Amplifier 2
Amplifier 1
Fig. 35.
553
Realization of BC amplifier.
PROBLEMS
Elementary Matrices. Elementary matrices are of three types, corres
ponding to the three types of elementary transformation. Use the following
notation:
r
Type 1: E
E
Type 2: E
E
Type 3: F
E
i j
i j
i + j
i + j
1. Construct the single matrix that will perform the following operations
on a matrix A:
(a) Interchange the third and second rows after the third row is multi
plied by 2. Order of A: (3,5).
(b) Add the first column to the third after multiplying the first by 3 and
the third by 4. Order of A = (2,3).
(c) Add the second row to the first after multiplying the first by 1.
Order of A: (3,3).
554
[Ch. 7
2. Find the single matrix which will perform the following operations on a
matrix A:
(a) Multiply the first row by k and add it to the third row, then
multiply the second row by 3 and interchange this row with the first.
Order of A: (4,4).
(b) Add the second column to the first, then multiply the second column
by 2 and add it to the third, then interchange the second and third
columns. Order of A = (4,3).
3. Write the single matrix which performs each of the following operations
on the rows of a matrix A:
() Multiply row 3 by 2, then add to row 1, then interchange row 1
with row 3. Order of A: (3,3).
(b) Interchange row 4 with row 2; then multiply row 1 by 5 and add to
row 3; then subtract row 3 from row 2. Order of A: (4,4).
4. Repeat Problem 3 if the operations are performed on the columns.
5. Let a matrix A of order (m,n) with m < n have rank r. Show that the
product of any elementary matrix with A will have the same rank, r.
Do this by multiplying A by each type of elementary matrix and
determining the effect of this on the determinant of appropriate sub
matrices of A.
6. Show that the inverse of an elementary matrix is an elementary matrix
of the same type.
7. Prove that a nonsingular matrix A can always be written as the product
of a finite number of elementary matrices.
8 . Prove that two matrices A and B of order (m,n) are equivalent if and
only if they have the same rank.
9. For each matrix A given below find matrices P and Q such that PAQ
is the normal form of A.
(a)
(c)
(b)
PROBLEMS
10. Reduce the following matrices to normal form:
(a)
(b)
(d)
(c)
11. Reduce the following matrices to canonic form using the Lagrange
reduction procedure:
(a)
(b)
(c)
555
556
[Ch. 7
- 1
ss
Show that
(a)
(b)
17. In the network of Fig. P17 all three inductances are mutually coupled.
Suppose it is possible to have the given inductance matrix in which all
mutual inductances are less than unity. Verify that this inductance
matrix is not positive definite or semidefinite. Setting RL = R = R = 1
for convenience, compute the natural frequencies.
2
Fig. P17
PROBLEMS
557
(b) Verify that the quadratic forms X'AX and X'BX are equal.
2
Observe how the right half F and s-planes are mapped into the W and p
planes. Show that W(p) satisfies Schwarz's lemma (see Appendix 2), then
substitute the above transformations in the lemma.
22. It is claimed that the functions below are not positive real for n >
Verify this claim and determine the value of N for each function.
0
(a)
(b)
N.
0
558
[Ch. 7
Prove that
The even part of this positive real function has a zero (double) on the
j axis. Hence, this function maps this point on the j axis to a point
on the jX axis in the Z-plane. Verify that the derivative dZ/ds at this
point is real and positive.
26. Let Z(s) = (mi + ni)/(m
+ n ) be a positive real function. Prove that
2
28. Let Z(s) = P(s)/Q(s) be a positive real function. Prove that the following
function is also positive real:
29. Let
PROBLEMS
559
30. Using the partial fraction expansions in (99) and (100), show that
(a)
(b)
(c)
(d)
lm [Z (jc)]
Im[Y (j)]
Re [Zjic(j)]
Re [Y (j)]
RC
RC
RC
$ 0
for 0, respectively.
0
for ^ 0, respectively.
is a monotonically decreasing function of for 0.
is a monotonically increasing function of for 0.
(b)
32. The results of Problem 30 can be described as a mapping of the upper
half s-plane into the lower half Z-plane and a mapping of the lower half
s-plane to the upper half Z-plane. Use this to obtain an alternative proof
that dZ (s)/ds is negative on the real axis.
RC
33. Using the approach of Problem 32 show that, for the impedance Z ( s )
of a passive RL network, dZ (s)jds
is real and positive on the real axis.
RL
RL
34. Let Z1(s) and Z (s) be RC impedance functions. Prove that Z1/Z is a
positive real function.
2
(b)
(a)
(b)
is a reactance function.
38. The admittance of a passive RC network has zeros at s = 0, 1 and 3.
At each of these points the slope of Y() equals 2. For large values of
560
[Ch. 7
RC
40. Let P(s) be a polynomial whose zeros are all real and negative. Prove
that each of the following two functions is the admittance of a passive
RC network:
41. Let P(s) be a polynomial whose zeros are all real and negative. If K is
a real, positive constant, prove that all the zeros of F(s) are real and
negative, where
42. (a) Let F(s) = P(s)/Q(s) be an RC impedance function. Prove that F(s)
is also an RC impedance function, where
PROBLEMS
561
In the second place, it must be proved that, if P(s) = m(s) + n(s) and
m/n is a reactance function then P(s) is a Hurwitz polynomial (except
for a possible even factor). The zeros of P(s) occur when
46. For each of the one-ports shown in Fig. P46 find the inverse network.
(a;
(b)
Fig. P46
Verify that the driving-point admittance of the inverse is the same as the
driving-point impedance of the given network.
47. Show that the residue condition is satisfied with the " equals " sign
(compact poles) at all the finite, nonzero poles of the y parameters of a
lossless two-port terminated in a unit resistance, as given in Table 1.
48. Show that the symmetric matrix of rational functions
562
[Ch. 7
21
12
Fig. P49
21
If in (b) it is only known that the real part condition is satisfied on the
j-axis, what additional conditions must be placed on the given functions
yn and y before the theorem will again be true?
50. Show that at a zero of z on the j-axis, z is imaginary. Hence show
that any j-axis poles of the open-circuit voltage gain
12
lx
21
are simple and with imaginary residues. Repeat for the short-circuit
current gain h (s).
21
PROBLEMS
563
(a) Find expressions for Z (s) appropriate to the Darlington case A and
case B.
(b) Observe the relationship between Z1 and Z when (1) m1 = m , and
(2) n1 = n .
2
(b)
(a)
Fig. P51
(a)
(b)
(c)
(d)
(e)
(f)
564
[Ch. 7
does not is given in the following. Let a positive real function be written
as:
1,
and
PROBLEMS
565
N
Reactive
three-port
Fig. P54
is the short-circuit admittance matrix of the above reactive three-port,
(c) Realize the following impedances in the above form. (The Darlington
realization requires transformers.)
(i)
(2)
55. Prove that the inverse of a positive real matrix is positive real.
56. Let Q(s) be a Hurwitz polynomial whose zeros are all complex. Let P(p)
be a Hurwitz polynomial associated with Q(s) in the following way, after
the transformation s =p :
2
That is, P(p) is the polynomial containing all the left-half-plane zeros
2
ofQ(P ).
Write
where A(s) and B(s) have real negative zeros only and where A/sB and
BjA are both RC impedance functions. This form of Q(s) is called the
Horowitz decomposition of Q (not to be confused with Hurwitz).
566
[Ch. 7
where
PROBLEMS
2
567
where
where
(i)
(2)
(3)
568
[Ch. 7
where
(a)
(b)
(c)
(d)
63. Let the functions in Problem 62 be transfer voltage ratios which are
to be realized in the parallel RC-negative converter network shown in
Fig. 31. Use an appropriate auxiliary polynomial and obtain a network.
64. The functions in Problem 62 are to be realized in the amplifier network
in Fig. 33. Use an appropriate auxiliary polynomial and obtain a
realization.
65. Suppose a given rational function is augmented by an auxiliary
polynomial with real negative zeros, and a partial fraction expansion is
carried out as in (131) in the text. The function can be expressed as in
(130). The result is to be realized by the cascade connection shown in
Fig. P65a, were N is RC. Two-port N is shown in dashed lines. Its
input impedance z
is z . This two-port, in turn, is made up of the
components shown in Fig. P65b. Show that the parameters of N are
a
l l b
PROBLEMS
569
(a)
(b)
Fig. P65
66. (a) Develop a procedure for the realization of a voltage transfer ratio
in the form of Fig. P22 in Chapter 3, where the two-port and the
admittance Y are RC. (b) Realize each of the functions in Problem 62
in this configuration.
67. A transfer impedance function is to be realized by a cascade of the units
shown in Fig. P26 (Chapter 3) in which each of the two-ports N and
N are to be RC. (a) Develop a procedure which will permit the
identification of RC-realizable functions z
and y ,
and similar
functions for the other units in the cascade, (b) Illustrate the procedure
with the functions a and d in Problem 62.
b
21b
21a
570
[Ch. 7
(b) Show that the numerator and denominator have a common factor
(s + k) if and only if Z ( - k ) =
Z(k).
(1)
(2)
69. Two positive real functions Fi(s) and F (s) are said to be
complementary
if their sum is equal to a positive constant K. Suppose that F(s) and
K are given. Determine the restrictions on Fi(s) and K such that F(s)
will have a complementary function. If Fi(s) and F (s) are
complementary and represent driving-point impedance functions, this
means that the series connection of the two corresponding networks has
a constant input impedance. In case Fi(s) and F (s) are admittance
functions, then the parallel connection of the corresponding networks
will have a constant input admittance. We refer to such pairs of
networks as being complementary.
2
70. Refer to Problem 69. Let Zi(s) be the driving-point impedance function
of an RC network and assume that it is regular at the origin. Show that
its complementary function Z (s) will be an RL impedance function
regular at infinity.
2
(a)
(b)
(c)
Fig. P71
71. Find complementary networks for each of the networks shown in
Fig. P71.
.8.
THE SCATTERING
PARAMETERS
572
[Ch. 8
8.1
(a)
(b)
(c)
Sec. 8.1]
573
where the subscripts i and r refer to " i n c i d e n t " and "reflected," respec
tively. The negative sign in the second equation is a result of the reference
chosen for the reflected current. Suppose we think of a real quantity r as
the "characteristic impedance" of the transmission system to the left of
the one-port terminals. Then the incident and reflected quantities are
related by
(2)
(3b)
where g = 1/r. It is now possible to define a voltage reflection coefficient p
as the ratio between reflected and incident voltage transforms, and a
current reflection coefficient as the ratio between reflected and incident
current transforms. Thus, using (3) for the incident and reflected variables,
we get
(4)
Some of the steps in this sequence used V= ZI. Just as the impedance Z
can characterize the behavior of the one-port network, so also the reflec
tion coefficient can characterize it completely. There is a one-to-one cor
respondence between Z and p given by the bilinear transformation
p = ( Z r)(Z + r) . We observe that the current and voltage reflection
coefficients are the same. It must be emphasized, however, that this is
true only for the case under consideration; namely, a real source im
pedance. When we consider the general case later, we will find that the
two reflection coefficients are different.
The wave-propagation concepts that were used in the preceding dis
cussion are artificial in the case of lumped networks. Nevertheless, it is
possible to regard (3) as formal definitions of the variables V , V and
I , I without attaching any interpretive significance to these quan
tities that reflect their intuitive origin. In the development we used r as
the characteristic impedance. However, this idea is not necessary in the
1
574
[Ch. 8
defines two new variables V and V in terms of the old ones, V and I.
The coefficients of the transformation should be chosen in such a way that
the new variables become convenient to use. The choice a11 = a = \ and
a = a = r/2 will make (5) reduce to (3). Other choices could lead to
additional formulations, which may or m a y not be useful for different
applications.
It is possible to interpret the incident and reflected variables b y refer
ence to the situation shown in Fig. lc, in which the one-port is matched
to the real source impedance. In this case V= rI. Hence, from (3a) we
find that
i
21
12
22
(6)
when matched. This tells us that when the one-port is matched to its
terminations, the voltage at the port is V and the current is I . Further
more, under matched conditions, (3b) tells us that V = 0 and I = 0;
and from Fig. 1c we observe that
i
(7)
From (4) we see that, under matched conditions, the reflection coefficient
is zero.
When the one-port is not matched, V and p are not zero. In fact, (1)
can be rewritten as
r
(8a)
(86)
that is, the reflected voltage V is a measure of the deviation of the oneport voltage, when under actual operation, from its value when matched.
r
Sec. 8.1]
575
NORMALIZED V A R I A B L E S R E A L
NORMALIZATION
The preceding discussion has been carried out b y using two pairs of
variables: the incident and reflected voltages, and the incident and re
flected currents. Since these quantities are proportional in pairs, from (2),
it should be sufficient to talk about one incident variable and one reflected
variable. However, rather than select either the voltage or current, we
use normalized variables related to both.
The normalized incident and reflected variables are defined as follows:
(9a)
(9b)
(10)
(11b)
Then the scattering variables become
(12a)
(12b)
576
[Ch. 8
(13)
(14a)
(14b)
(14c)
AUGMENTED
NETWORK
Fig. 2.
Sec. 8.1]
577
1/2
gn
(15)
We can go one more step and include the series unity resistance in with
the normalized network, as illustrated in Fig. 2b. The result is then called
the (series) augmented normalized network. Of course, it is possible to think
of the original network as being augmented by the series resistance r,
without reference to normalization. The reflection coefficient of the
original network can be expressed in terms of the input admittance
Y of the augmented network or Y
of the augmented normalized
network. It is clear from Fig. 2 that we can write
a
Then
or
(16)
(In the derivation, r was added and subtracted in the numerator.) This is
a useful expression. It can often be the simplest form for computing the
reflection coefficient.
As an example, consider the situation shown in Fig. 3. Two one-ports
having impedances given by Z =f (s)
and Z =f (s)
have reflection co
efficients p i and p 2 , respectively. It is desired to find the reflection co
efficient of the one-port shown in Fig. 3b in terms of p i and p 2 . Since p is
invariant to normalization, let the normalizing number be unity and cona
578
[Ch. 8
(a)
(b)
Fig. 3.
Illustrative example.
For this example it is clear that the reflection coefficient of the overall
network is more simply expressed in terms of the reflection coefficients of
the components than is the overall admittance in terms of the component
admittances.
REFLECTION
RECIPROCAL
COEFFICIENT
NETWORK
FOR
TIME-INVARIANT,
PASSIVE,
Sec. 8.1]
579
(17)
As for the poles of (s), they are given by the zeros of Z (s) + 1. They
cannot lie in the closed right half-plane since this would require Re Z = 1
for a point in the closed right half-plane, which is impossible for a positive
real function. Hence (s) is regular in the closed right half-plane. We see
that the positive real condition on the impedance of a one-port can be
translated into equivalent conditions on the reflection coefficient.
A bounded real function (s) was defined in Chapter 7 as a function that
is (1) real when s is real, (2) regular in the closed right half-plane, and
(3) |( j)| 1 for all .
The above discussion has shown that for a time-invariant, passive, re
ciprocal network the reflection coefficient is a bounded real function.
n
P O W E R RELATIONS
(19)
580
[Ch. 8
1
of the scattering variables are the square root of power, (voltage current) / .
We can think of the net power delivered to the one-port as being made up
of the power in the incident wave, P , less the power returned to the
source b y the reflected wave, P . Of course, under matched conditions
there is no reflection. The power delivered under these conditions is the
maximum available power, say P , from the source in series with resis
tance r. This power is easily found from Fig. lc to be
i
(20)
The last step follows from (15). With these ideas, (19) can be rewritten as
and
(21)
8.2
(22)
Sec. 8.2]
Fig. 4.
581
Fig. 5.
(23)
582
[Ch. 8
(24)
The matrix r is nonsingular and positive definite, since all r 's are assumed
positive. From Fig. 5 we can write
j
(25)
where Z is the open-circuit impedance matrix of multiport N. Suppose
now that at each of the ports the multiport is matched to the source re
sistance. This means the ratio V /I is to equal the resistance r at the j t h
port. In matrix form this becomes V = rl, or Z = r when the multiport
is matched. B y analogy with the one-port case, we introduce the incidentvoltage vector V and the incident-current vector I as equal to the
port-voltage vector and the port-current vector, respectively, when the
ports are all matched; that is,
o c
o c
(26a)
(26b)
and
(26c)
when matched.
Similarly, we introduce the reflected-voltage vector V and the reflectedcurrent vector I , as the deviation of the port-voltage vector and the portcurrent vector, respectively, from their matched values. In analogy with
(8) for the one-port case, they are written as
r
(27a)
(27b)
When the last two pairs of equations are used with (25), the incident
and reflected variables can be written as
(28)
and
(29a)
Sec. 8.2]
583
(29b)
(29c)
These expressions should be compared with (7) and (3) for the one-port.
T H E SCATTERING
MATRIX
0 C
(31a)
(31b)
Study these expressions carefully, note how S is relatively simple when
( Z + r ) premultiplies
( Z r), and S is relatively simple when it
postmultiplies ( Z r).
What is ( Z + r)? It is, in fact, the open-circuit impedance matrix of
the augmented network; that is, the multiport in Fig. 5, which includes
the series resistance at each port as part of the multiport. The inverse,
( Z + r ) , is the short-circuit admittance matrix of the augmented
multiport, which we shall label Y . In terms of Y , the two reflection
coefficients are found from (31), after some manipulation, to be
I
o c
o c
0 c
o c
o c
(32a)
(32b)
584
[Ch. 8
(33a)
(33b)
- 1
-1/2
(34)
is a real diagonal matrix, each diagonal entry of which is the square root
of the corresponding entry of matrix r. To see what will happen, let us
multiply both sides of (30a) by r and both sides of (30b) by r . Then
1/2
-1/2
(35a)
(35b)
But observe from (33)through premultiplying and postmultiplying both
sides by r that
1/2
(36)
-1/2
1/2
Sec. 8.2]
585
MATRICES
1/2
1/2
(39a)
(39b)
(39c)
where Y , the short-circuited admittance matrix of the multiport, is
Z
, then
s c
- 1
o c
(40)
(41)
We leave the details for you to work out. Compare (40) with (16), which is
the corresponding scalar result for a one-port.
The relationship between S and Y a points up an important property
that is not evident from the manner in which it was obtained. Because of
the series resistances in the augmented network, this network may have
an admittance matrix even though the original multiport has neither an
impedance nor an admittance matrix. This will be true for any passive
network. Thus an advantage of scattering parameters (as we call the ele
ments of the scattering matrix) is that t h e y exist for all passive networks,
n
586
[Ch. 8
Fig. 6.
Ideal transformer.
22
11
11
22
11
NORMALIZATION A N D THE A U G M E N T E D
MULTIPORT
1/2
(42)
Comparing these with (12) and (14) shows that the expressions for the
multiport scattering variables are identical with those for the one-port,
except that they are matrix expressions in the present case.
Sec. 8.2]
587
Finally, note that the two expressions in (28) relating incident voltage
and current to V , reduce to a single equation under normalization:
g
(43)
The normalization can again be interpreted b y appending ideal trans
formers to the ports. This is illustrated for the two-port in Fig. 7. The
(a)
(b)
Fig. 7.
588
[Ch. 8
Fig. 8.
Gyrator.
resistance r. If the port normalizing numbers r1 and r for both ports are
chosen equal to r, the matrix Z can be written very simply. Equation 40
then leads to the scattering matrix. Thus
2
8.3
Sec. 8.3]
589
Fig. 9.
I N T E R P R E T A T I O N OF SCATTERING PARAMETERS
(45)
n2
n2
r2
r2
11
1/
(46)
590
[Ch. 8
11
22
21
n2
r2
gnl
(47)
Thus S is seen to be proportional to a voltage gain, a forward voltage
gain. A further clarification is given in terms of Fig. 10. Here the two-port
21
has been replaced b y an ideal transformer of such turns ratio that the
resistance r becomes matched to r\. The output voltage under this con
dition is called V' . N o w if we take the ratio of the actual output voltage
V t o V' , the result is called the transducer voltage ratio. The calculation
given in the figure shows this quantity to be the same as the right-hand
side of (47). Hence S is the forward transducer voltage ratio. When the
source is matched as in Fig. 10, it will deliver to the network at its ter
minals (namely, the primary side of the transformer) the maximum
available power. This is seen to be P = | V | /4r .
N o w return to the original setup. The power transferred to the load
with network N in place, when it is matched and when the port 2 voltage
is V , will be P = | V | /r .
Hence the magnitude square of S in (47)
will be
2
21
m1
g1
21
(48)
Sec. 8.3]
591
12
21
n2
gn1
(49)
is the reflection coefficient at port j when all other ports are matched; that
is, terminated in their reference resistances.
The parameters off the main diagonal are called transmission coefficients,
in contrast with the reflection coefficients. They are given b y
(50)
(51)
592
[Ch. 8
(a)
(c)
(6)
But I = I + I , I = I I ,
Fig. 11. With these, we get
a
and V = V
V,
3
as observed from
The coefficient matrix here is Z . The scattering matrix can now be com
puted from (41). The result is
n
(52)
This is a very interesting result. Note first that the diagonal elements
are all zero. Hence the reflection coefficients at all ports are zero, which
means all the ports are matched. As for the transmission coefficients, con
sider Fig. 11b, in which only port 1 is excited and all 3 ports are matched.
Thus, V =I
and V = I . Hence, from (28) and (37), we find
a = V = Vg /2 and a = a = 0; from (52), b = b = 0 and b =
V /2.
But b = V = V ; hence the power in the matched load at port 3 is
2
il
r3
g1
Sec. 8.3]
593
The conclusion is that when a signal is incident at port 1, with all ports
match-terminated, none is reflected (b = 0), none is transmitted to port 2
(b = 0), but all of it is transmitted, without loss, to port 3, the power
there being the maximum available from the source at port 1.
Similar conclusions follow from (52) for the other transmissions;
namely, that a signal incident at port 2 is all transmitted to port 1, and a
signal incident at port 3 is all transmitted to port 2. In this latter case,
because of the minus sign in b = a , there is a reversal in voltage phase,
but the power transmitted is not affected.
The three-port in Fig. 11 seems to have a cyclic power-transmission
property. Power entering one port is transmitted to an adjacent port in a
cyclic order, as shown by the circular arrow. A multiport device having
this property is called a circulator, the symbol for which is shown in Fig.
11c for a three-port. For our circulator in Fig. 11, the cyclic order is 132.
Clearly, a circulator of the opposite cyclic order (123) is also possible.
Its scattering matrix must clearly be of the form
1
21
32
594
8.4
[Ch. 8
(53)
Here (42) was used to replace the normalized voltage and current vectors
b y the scattering variables. The last term in parentheses on the far right
is the difference of two conjugate quantities, since
The last step here follows because these matrix products are scalars and
the transpose of a scalar is itself. But the difference of two conjugates is
imaginary. Hence the real power will be
(54)
Sec. 8.4]
595
Without placing any restrictions on the type of network, let us take the
conjugate transpose of Q:
(57)
As discussed in Chapter 1, a matrix that is equal to its own conjugate
transpose is said to be Hermitian. Thus Q is a Hermitian matrix. Its
elements are related by q = q which requires the diagonal elements to
be real.
We are interested in particular classes of networks: active and passive,
reciprocal and nonreciprocal, lossless and lossy. There is not much of a
specific nature that can be said about active networks. Hence we shall
concentrate mainly on passive networks, which may be reciprocal or
nonreciprocal. We shall also focus on the lossless subclass of passive
networks; these also may be reciprocal or nonreciprocal.
First, for passive networks in general, the real power delivered to the
multiport from sinusoidal sources at the ports must never be negative.
Hence
ij
ji
(58)
596
[Ch. 8
In this case no power is dissipated within the multiport. Hence the realpower input shown in (56) must be identically zero for any possible vec
tor a. This is possible only if the matrix of the quadratic form vanishes;
that is,
(61)
or
- 1
from which
(62)
or
from which
(63)
where
j k
TWO-PORT N E T W O R K
PROPERTIES
Sec. 8.4]
597
In terms of poles and zeros, we conclude the following. The poles and
zeros of (s) (s)
and (s) (s) are identical, and they occur in
quadrantal symmetry. In making the assignment of poles and zeros of
(s) from those of p (s) (s), the only consideration is stability. No poles
of p (s) can lie in the right half-plane. Hence the left-half-plane poles of
(s) (s) must be poles of (s). As for the zeros, no limitation is
imposed by stability; zeros of (s) (s) can be assigned to (s) from
either the left or the right half-plane, subject to the limitation that these
zeros, plus their images in the j-axis, must account for all the zeros of
(s) ( s ) . Similar statements can be made about the poles and zeros
of (s).
Let us return again to (62) and this time set j = k = 2. The result will be
1
(67)
598
[Ch. 8
21
(69a)
or
(69b)
21
21
12
A N APPLICATION
FILTERING OR EQUALIZING
(b)
(a)
(c)
Fig. 12.
Sec. 8.4]
599
insertion between a source with a real internal impedance and a real load.
The network may be required to perform the function of filtering or
equalizing; that is, shaping the frequency response in a prescribed way.
Alternatively, the load may be an impedance Z as shown in Fig. 12b,
and the coupling network is to be designed to provide a match to the
resistive source over a band of frequencies. This case can be related to the
former one b y using Darlington's theorem. That is to say, Z can be
realized as a lossless two-port N terminated in a resistance r , as shown
in Fig. 12c. The cascade combination of N and N plays the role of the
lossless network N in Fig. 12a. A limited discussion of this matching prob
lem will be given in the next section.
Let us concentrate on the filter or equalizer problem. What is specified is
the transducer power gain as a function of frequency. We shall label this
function G( ). According to (48), G( ) is simply the magnitude square
of S . But |S |
is related to the magnitude of the input reflection
coefficient by (64a). If this expression is analytically continued, it can
be written as
L
21
21
(70)
2
where we have again replaced S by . If G( ) is specified, the righthand side is a known even function. It is now only necessary to assign the
poles and zeros of the right side appropriately to p (s).
Furthermore, the reflection coefficient and the impedance Z looking
into the input terminals of N with the output terminated in r are related
b y (46), which can be solved for Z as follows:
11
(71)
Hence, once (s) has been determined from (70), the input impedance Z
becomes known as a function of s. The task is then reduced to an appli
cation of Darlington's theorem; namely, realizing Z (s) as a lossless twoport terminated in a resistance.
To illustrate this discussion let the transducer power gain be given as
1
600
[Ch. 8
2
In the last step the denominator has been factored, putting into evidence
the left- and right-half-plane poles. The left-half-plane poles must
belong to (s), as opposed to ( s ) . In the example the zeros also are
uniquely assignable: three zeros at the origin to (s) and three to ( s ) .
The only ambiguity arises in the appropriate sign. There is no a priori
reason why (s) must have a positive sign. The conclusion is that (s)
must be the following:
1
depending on the sign chosen for (s). But these are inverse impedances,
and their realizations will be duals. In the present case the realization is
rather simply obtained by expanding in a continued fraction. Thus, b y
using the second function, we obtain
1
The network realizing this function, and its dual, are shown in Fig. 13.
These are the normalized realizations. Recall that the normalized opencircuit impedance matrix is obtained b y dividing all elements of Z b y
r r .
To undo this normalization, all branch impedances must, there
fore, be multiplied by r r . One denormalized network, with the actual
source and load resistances, is shown in Fig. 13c.
o c
Sec. 8.4]
(a)
601
(b)
c)
Fig. 13.
Illustrative example.
LIMITATIONS I N T R O D U C E D B Y PARASITIC
CAPACITANCE
(a)
(b)
(c)
Fig. 14.
602
[Ch. 8
Passive
lossless
Fig. 15.
first case, Z is the impedance Z\ looking into the left-hand port of the
dashed two-port in Fig. 14c, with R in place; and is the corresponding
reflection coefficient . For R = R , Z is the impedance Z looking into
the right-hand port of the dashed two-port in Fig. 14b, with the other
port terminated in R ; and p is the corresponding reflection coefficient .
In either case,
2
(72)
Sec. 8.4]
603
Fig. 16
(73)
(74)
(75)
As a consequence
(76)
604
[Ch. 8
The negative sign before the right-hand side suggests that we take the
logarithm of A ( s ) / instead of +A(s)/.
Thus, since ln(1 +x)->x
for
(77)
d s / s
= -j
Along the j - a x i s ,
(78)
The first step follows from the fact that the magnitude of an all-pass
function is unity on the j - a x i s , so l n | A ( j ) / ( j ) | = ln|1/|. The last step
follows because ln |1/| is an even function of , whereas the angle is an
odd function.
From Cauchy's theorem, the sum of the integrals on the left sides of the
last two equations should equal zero. Hence
(79)
Recall that s is a pole of 1/ in the right half-plane, so that its real part is
positive. The sum of all such poles will, therefore, be real and positive.
If 1 / has no poles in the right half-plane, this sum will vanish. The final
result will therefore be
k
(80)
Sec. 8.5]
COMPLEX
NORMALIZATION
605
(81)
Outside the passband, there should be a total mismatch and |p| should
ideally equal 1. More practically, although its value will be less than 1, it
should be close to 1. Hence ln l/|| will be a small positive number, ideally
zero. Therefore the integral from to will be positive, and (81)
will yield
c
(82)
(83)
This puts an upper limit on the achievable transducer power gain, | S ( j)|,
over a wide frequency band, even if we assume a constant value is possible
over the passband. The wider the frequency band of interest, the more
stringent will this limitation become for a fixed shunt capacitan.ce. Note
that the immediately preceding result is valid for Fig. 14c; it also applies
to Fig. 14b through (65) and (68).
21
8.5
COMPLEX NORMALIZATION
606
[Ch. 8
now turn our attention to this general problem. For simplicity, we shall
illustrate with a two-port, but the results will apply in matrix form to the
general multiport.
The situation to be treated is illustrated with a two-port in Fig. 17.
(b)
(a)
Fig. 17.
Two-port with general terminations: (a) general case; (6) matched case.
The terminating impedances are strictly passive with positive real parts:
(84)
These real parts play the same normalizing role as the resistive termina
tions before, and they are represented by the positive definite diagonal
matrix in (24). From Fig. 17a, we can write
when matched.
(86)
As before, the incident voltages and currents are defined as the port
voltages and currents under matched conditions; that is, V = V and I = I,
i
Sec. 8.5]
COMPLEX NORMALIZATION
607
when matched. Hence, generalizing the last equation and inserting into (85),
we obtain
(87a)
(87b)
since z + z = 2r. These should be compared with (28) for real terminations.
Note that the expression for V in terms of V is not as simple as it is in the
case of real terminations.
The condition of simultaneous match at all ports, requiring as it does
that Z be the conjugate of the terminating impedance z, is not possible
of attainment at all frequencies, but only at a single frequency. Hence the
procedure being described here is strictly applicable only at a single
frequency, which may be at any point on the j-axis. It may also be used
in narrow-band applications without excessive error.
Again we define the reflected variables as deviations from their matched
values, according to (27). When (85) and (87) are inserted there, and
after some manipulation whose details you should supply, the incident
and reflected variables become
i
o c
(88a)
(88b)
(88c)
(88d)
Again we note that the expressions for the voltages are somewhat com
plicated compared with those for the currents.
Let us now introduce voltage, current, and impedance normalizations.
To normalize currents we multiply by r , and to normalize voltages we
multiply by r . Impedances are normalized according to (39b). The
normalized incident and reflected variables become
1/2
1/2
(89a)
(89b)
and
(90a)
(90b)
608
[Ch. 8
-1/2
where z = r z r
= [z /r ]. Examine these expressions carefully. For
the case of real terminations, both quantities in (89) are the same; they
were together labeled a in (37). Likewise, both quantities in (90) are the
same for real terminations; they were collectively labeled b before. Clearly,
this is no longer true for impedance terminations. Observe that if z is
made real, the two expressions in (89) reduce to a and the two in (90)
reduce to b in (42).
Two different scattering matrices can be defined, one for the currents
and one for the voltages, even for the normalized variables. We shall
arbitrarily define a and b as the normalized incident and reflected currents.
Thus from (89) and (90) we get
n
(91a)
(91b)
We now define the scattering matrix S, as before, by the relationship
b = Sa. When (91) are inserted for a and b, we can solve for the scattering
matrix:
(92)
This should be compared with (41) for the case of real terminations. Note
that ( Z + z ) is the normalized admittance matrix of the augmented
network, Y . Hence, b y adding and subtracting z within the first
parentheses in (92), this expression can be rewritten as
1
a n
(93)
since z + z = 2U. This is the same expression as (40) for real termina
tions.
n
-1/2
(94)
Comparing this with (92) gives the relationship between the two matrices
as
(95)
Sec. 8.5]
COMPLEX NORMALIZATION
609
(96a)
(96b)
since z + z = 2U. These are not quite the same as (42) for the case of
real terminations. However, if the expression for V*nIn is formed similar
to (53), the expression for the real part will turn out to be exactly the same
in terms of the scattering matrix defined here as it was before in (54).
Hence the properties of the scattering matrix as discussed in Section
84 apply also to the scattering matrix S given in (92). This is, then,
the appropriate extension of the scattering variables and the scattering
matrix to complex normalization.
n
FREQUENCY-INDEPENDENT
NORMALIZATION
Fig. 18.
610
[Ch. 8
that is, r (s) and r (s) are the even parts of z (s) and z (s). Since r(s) is even,
r(s) = r(s); so we shall not write the argument of r except for emphasis.
Without making any claims about real power matching, let us define a
condition of the two-port in which
1
(99)
o c
(100a)
(100b)
(101a)
(101b)
Compare these with (88); note that z has been replaced by z(s). When
s = j , z(-s) = z(-j) = z[(j)] = z(j).
Another useful form is obtained by inserting V = Z
o c
I. Then
(102a)
(102b)
Sec. 8.5]
COMPLEX NORMALIZATION
611
(103)
Then
(104)
Both the numerator and denominator of the right side are even poly
nomials of s. Hence their zeros occur in quadrantal symmetry. We now
define a rational function f(s) whose zeros and poles are all the zeros and
poles of r (s) that lie in the left half-plane. The function containing all the
right-half-plane poles and zeros of r (s) will then be f(s).
Thus
j
(105)
Each of the function f(s) and f(s) is something like a "square r o o t " of
r (s), since their product gives r (s).
Before we proceed, here is a simple example. Suppose Zj is the positive
real function
j
Then
Clearly,
612
[Ch. 8
Observe that the real parts of these two functions for s = j are equal;
thus for complex normalization at a single frequency s = j , either one
will do.
We dealt here with a single element of the r(s) matrix. In matrix form
what was just described becomes
(106)
(107)
N e x t we form V'(s) I(s) and take the even part. Calling this P , we get
P = even part of
= even part of
= even part of
where
Sec. 8.5]
COMPLEX NORMALIZATION
613
since X and Y are diagonal matrices. Thus X(s) is even, and Y(s) is odd.
Therefore
(108)
Now let us define the current scattering matrix as
(109)
and insert it into (108), with the result
Recalling that r = f(s) f(s), this expression can be put in the form of
(54) uniquely as follows:
(110)
Now it is clear! We must define the scattering variables a and b as the
normalized incident and reflected current variables; and we must define
the scattering matrix S as the normalized current scattering matrix in the
following w a y :
(111a)
(111b)
(112)
[Equation 111b is justified if (109) is multiplied by f(s) and the expressions
for a(s) and S are used.] Then (110) becomes
(113)
This expression reduces, for s = j , to the power input as given in (54) for
the case of real normalization. Hence the scattering matrix defined b y
(112) and (109) has the same properties as discussed in Section 8.4 for
real normalization.
The only thing left to do is to find expressions for S in terms of Z and
o c
614
[Ch. 8
the terminating impedances. For this purpose return to (102) and use (109)
to find the current scattering matrix
(114)
(115)
When (114) is inserted into (112), the scattering matrix becomes
(116)
Compare this with (92) and (93) for the single-frequency complex nor
malization.
We see that in the case of frequency-independent normalization the
expression for S is somewhat more complicated than the expression for the
single-frequency scattering matrix.
For a one-port having an impedance Z(s), the current reflection coeffi
cient i is obtained from (114) by noting that the matrices in that expres
sion are scalars in t h e one-port case. Thus
(117)
The reflection coefficient itself is seen from (116) to differ from this expres
sion by the function f(s)/f(s), which is an all-pass function. Thus
(118)
is an all-pass function.
Examples
Consider, for example, the one-port in Fig. 19 terminated in a parallel
Sec. 8.5]
COMPLEX NORMALIZATION
Fig. 19.
615
R and C. Then
Fig. 20.
Illustrative example.
616
[Ch. 8
and
The terminating impedance z is the same as the one treated earlier just
below (105). Thus the functions f(s) and f (s)
are
2
-1
When all the above is inserted into (116), the scattering matrix becomes
Sec. 8.5]
COMPLEX NORMALIZATION
617
Observe that S
equals S
multiplied by an all-pass function; their
magnitudes on the j-axis are, thus, the same. Equation 65 and, more
generally, (66), are satisfied. You can verify that (64), (67), (68), and (69)
are also satisfied.
22
11
NEGATIVE-RESISTANCE
AMPLIFIER
(a)
(b)
Fig. 21.
This is clearly not a positive real function. Let us now form the function
Z ( s ) , as follows:
d
618
[Ch. 8
(b)
(a)
Fig. 22.
(119a)
(119b)
(119c)
Alternatively, if the termination at port 3 is included in the network,
the structure can be regarded as a two-port N', as shown in Fig. 22b,
with the real terminations R and R . Let the scattering parameters of
this two-port be labeled with a prime. B y definition,
1
(120)
The first step follows from (119a) with a = 0. Since port 3 in Fig. 22a is
2
Sec. 8.5]
COMPLEX NORMALIZATION
619
(121)
(122)
21
(123)
since |S (j)|
l . Then
12
= |S (j)|.
12
|S (j)|
12
(124)
33
620
[Ch. 8
12
33
12
(125)
where R= 1/G in Fig. 22. This is a basic " g a i n - b a n d w i d t h " limitation
dependent only on the parameters of the tunnel diode.
We turn now to a consideration of the design of the three-port net
work. Suppose the three-port in Fig. 22a is to be a reciprocal network. If
|S (j)|
= 1 over a band of frequencies, all the other scattering para
meters of the three-port (S , S , etc.) will vanish over that frequency
band. (See Problem 36.) Thus a nontrivial reciprocal three-port with the
above optimum design is not possible.
Consider the three-port circulator shown in Fig. 23a. Its scattering
12
11
23
(a)
(b)
Fig. 23.
(126)
12
Sec. 8.5]
COMPLEX NORMALIZATION
621
(127)
normalized with respect to r and z = 1/(G + sC), at its input and output,
respectively; r is the same as the port 3 normalizing resistance of the
circulator. It remains to express the scattering matrix of the overall threeport in terms of the parameters of S and S in (126) and (127). This can be
done by using the results of Problem 24. The details are left for you; the
result is
3
(128)
(129)
The reflection coefficients are both zero, indicating that the amplifier is
matched at both input and output. If (65) is used, it is observed that the
reverse transmission S' has a unit magnitude. The forward transmission
of the amplifier is related only to the output reflection coefficient of the
two-port N. Thus
12
(130)
22
22
622
Fig. 24.
[Ch. 8
in the tunnel diode at the other. The problem is not unlike that of the
filter problem illustrated in Section 8.4. From |S (j)|
it is necessary to
determine S (s).
This is related to the impedance Z looking into port 2 of
by (118), in which p plays the role of S . The two-port can then be
designed from Z.
This quick description has ignored a number of problems having to do
with the proper selection of S (s) from its magnitude squared, since this
is not a unique process. The details of this process would take us far afield
and will not be pursued any further here.
2
22
22
22
22
PROBLEMS
1. Two one-ports having impedances Z =f (s)
and Z = f ( s ) have
reflection coefficients and . Find the reflection coefficient of the
one-port shown in Fig. PI in terms of and .
a
Fig. PI
2. The sense of incident and reflected waves is related to what is taken to be
the direction of power flow. Figure P2 shows a one-port in part (a) with
PROBLEMS
623
the usual references of voltage and current. The reflection coefficient for
this one-port is 1. In Fig. P2b the current in the one-port is reversed.
(a)
(b)
Fig. P2
- 1
a n
- U.
Fig. P7
624
[Ch. 8
(a)
(b)
Fig. P8
12
S,
21
and S
22
for a
2 2 .
11
22
22
22
21
22
21
11
PROBLEMS
(b)
(a)
625
(c)
Fig. P14
port and transmitted to each of the other ports. Is this what you
would have expected without finding S?
15. The structure in Fig. P15 is a hybrid coil. It consists of a three-winding
ideal transformer from which a four-port is formed. The two transformer
Fig. P15
(a) When port 1 is excited (by a voltage source in series with its
terminating resistance), there is no transmission to port 2, and vice
versa;
(b) When port 3 is excited, there is to be no transmission to port 4, and
vice versa;
(c) All ports are matched (no reflections).
Find the scattering matrix of this four-port in terms of n and n only.
2
626
[Ch. 8
13
Fig. P17
When one of the ports is excited (by a voltage source in series with its
termination), it is assumed that equal power is delivered to the other
two-ports. Find the maximum fraction of available power that is
delivered to each port under these conditions and find the fraction of
power that is reflected.
18. The network in Fig. P18 is a lattice operating between two resistive
terminations. Compute the transducer power gain. Determine conditions
Fig. P18
on the lattice elements that will make the gain identically unity,
independent of frequency. Under these conditions find the reflection and
transmission coefficients.
19. Figure P19 shows a two-port terminated at the output by an impedance
that is not the normalizing impedance. Let be the reflection coefficient
of Z normalized to r , which is the output normalizing resistance of the
two-port. The input is match terminated; that is, r is the normalizing
2
PROBLEMS
627
Fig. P19
resistance. Find the input reflection coefficient and the voltage gain
V / Vg in terms of and the scattering parameters of the two-port.
2n
Fig. P20
628
[Ch. 8
It is desired to find the scattering matrix S' for the m-port within the
dashed lines in the figure, given by the relation b1 = S ' a 1 . Using the
relation between a , b , and , write an expression relating a , b ,
and . Insert this into the partitioned form of the scattering relations
and show that
k
21. One of the ports of the circulator in Fig. P21 is not match terminated.
Use the result of the previous problem to find the scattering matrix S'
of the two port within the dashed line.
Fig. P21
PROBLEMS
629
22. The four-port network in Fig. P22 is a hybrid that is match terminated at
two of the ports but not at the other two. Find the scattering matrix
of the two-port within the dashed line in terms of and . Under
what condition will the two-port have no reflections at either port?
3
Fig. P22
23. In order to find the relationships among the port voltages and currents
of a two-port that consists of the cascade connection of two sub-twoports, it is convenient to express the port relationships of the sub-twoports in terms of the chain matrix. It is desired to find a matrix T that
can play a similar role for two cascaded two-ports; but this time the
variables are scattering variables, rather than actual currents and
voltages. For the network shown in Fig. P23, let the desired overall
relationship be written x = Ty.
Fig. P23
(a) Determine the elements of the vectors x and y (from among the
scattering variables) such that the overall T matrix equals T i T , where
xi = Tiyi and x = T y . Specify any condition on the normalizing
impedances that may be required for complex normalization and also for
real normalization. (This can be referred to as a compatibility condition.)
2
630
[Ch. 8
(b) Express the elements of the matrix T for a two-port in terms of the
scattering parameters of that two-port.
(c) Determine a condition that the elements of T satisfy if the two-port
is reciprocal.
n ports
k ports
k ports
m ports
Fig. P24
n + k ports. A total of k ports of one are connected to k ports of the
other, leaving an (m + n) port network. Let S and S be the scattering
matrices of the two multiports. The scattering equations can be
partitioned as follows:
PROBLEMS
631
Let S' be the scattering matrix of the overall (m + n) port and write
the scattering equations as
(a) Find the compatibility condition that will permit the result b = a ,
= b . (See Problem 23.)
(b) Show that the overall scattering parameters are given by
1
(c) Compare with the more special result of Problem 20 and verify that
this result reduces to that one.
25. The four-port network within the dashed lines in Fig. P25 represents a
telephone repeater. The two-ports labeled L are low-pass filters, and
those labeled H are high-pass filters. They are reciprocal and symmetric
two-ports and hence are characterized by only two parameters each.
Fig. P25
632
[Ch. 8
and
27. Let z (s) be the complex normalizing impedance of the kth port of a
multiport.
k
28. Find the scattering matrix of the ideal transformer shown in Fig. P28
normalized to the terminating impedances. (Frequency-independent
normalization.)
PROBLEMS
Fig. P28
29. Find the scattering matrix of the gyrator in Fig. P29 normalized
(frequency independent) to is terminating impedances. Verify the
properties of reflection and transmission coefficients of a two-port.
Fig. P29
(a)
(b)
Fig. P30
633
634
[Ch. 8
(a)
(c)
(b)
Fig. P31
12
PROBLEMS
635
Fig. P37
.9.
SIGNAL-FLOW GRAPHS
AND FEEDBACK
The linear network model that has been developed up to this point
includes a number of components: resistors, capacitors, gyrators, trans
formers, etc. Each component of the network model has been characterized
b y a parameter, such as R, and by a graphical symbol, such as this -/V
for the resistor. Networks are made up of interconnections of these
components. In each voltage-current relationship, our interest often has
been directed at the scalar parameter; for example, in writing v = L di/dt,
attention has been focused on the L.
But each voltage-current relationship defines a mathematical
operation.
Instead of focussing on the parameter, one could emphasize the mathe
matical operations and the signals on which the operations are performed.
A n operational symbol could be used to represent the mathematical
operation, and these operational symbols could be interconnected into
an operational diagram. The analysis of this operational diagram would
provide an alternative means for determining transfer functions of net
works. The signal-flow graph is just such an operational diagram.
Most methods of analysis discussed in preceding chapters apply to
linear networks in general, whether passive or active, reciprocal or non
reciprocal. No special attention was directed at active, nonreciprocal
networks. Signal-flow graph analysis is particularly appropriate for such
networks. Furthermore, for active nonreciprocal networks, a number of
636
Sec. 9.1]
AN OPERATIONAL DIAGRAM
637
9.1
AN OPERATIONAL DIAGRAM
Excitation
Operator
(b)
(a)
Fig. 1.
(c)
Operational symbols.
Fig. 2.
Branch of diagram.
638
[Ch. 9
Fig. 3.
Diagram of equation.
Fig. 4.
Sec. 9.1]
AN OPERATIONAL DIAGRAM
639
(1)
The first and last equations can be represented by the operational branches
shown in Fig. 5a. The second equation expresses an equilibrium of voltages
(a)
Fig. 5.
(b)
(c)
and can be represented by the branches in Fig. 5b. All three of these can
now be combined to give the final result shown in Fig. 5c. This operational
diagram represents the network in Fig. 4 to the same extent that the
equations (1) do.
As a second example, consider the network in Fig. 6. The total output
resistance is R . The following equations can be written:
3
(2)
640
Fig. 6.
[Ch. 9
Feedback amplifier.
(e)
(a)
(d)
(b)
(e)
Fig. 7.
Sec. 9.1]
AN OPERATIONAL DIAGRAM
641
The resulting operational diagram will take the form shown in Fig. 8,
Fig. 8.
as you should verify. A comparison with the previous diagram shows the
two to be quite different. This example illustrates the interesting fact that
there is not a unique operational diagram representing a given network.
The preceding discussion has skirted a number of fundamental ques
tions. Drawing an operational diagram for a network requires, first, the
writing of equations describing the behavior of the network. This poses
the questions of what variables to select for describing the network
performance, how many of them are required, and how many equations are
required. For a systematic development, such questions must be answered.
The answers will be postponed, however, for a later section, in order to
turn to a more detailed discussion of operational diagrams.
642
9.2
[Ch. 9
SIGNAL-FLOW GRAPHS
(5)
where K is a column vector. With this expression (4) can be rewritten as
(6)
If there are additional driving functions, the scalar y will become the
vector Y and K will have several columns instead of only one.
If X is a vector of order n, then the coefficient matrix on the right side
of (6) is of order (n, n + 1). For future convenience, let us augment this
0
Sec. 9.2]
SIGNAL-FLOW GRAPHS
643
ij
ij
Then
and
644
[Ch. 9
The signal-flow graph of this set of equations will have four nodes,
labeled y , x , x , and x . These are first placed in a convenient arrange
ment, as in Fig. 9a. The branches are then inserted in accordance with the
connection matrix. Thus in the third row (corresponding to x , since the
first row containing zeros was added to create the square matrix C) and
in the fourth column (corresponding to x ) there is a nonzero entry of
value 1. Hence there will be a branch from x to x with a weight of 1.
All other branches are similarly inserted; the final result is shown in
Fig. 9b.
0
(b)
(a)
Fig. 9.
GRAPH
PROPERTIES
SIGNAL-FLOW GRAPHS
Sec. 9.2]
645
(a)
(b)
Fig. 10.
branches. In Fig. 10b node x has been split into two nodes, labeled x and
x' . One of these is a source, from which all outgoing branches leave, the
other is a sink to which all incoming branches come. Splitting a node
clearly interrupts all feedback loops that pass through that node. B y
splitting a sufficient number of nodes all feedback loops of a graph can be
interrupted. The index of a signal-flow graph is the smallest number of
nodes that must be split to interrupt all feedback loops in the graph.
In the original graph of Fig. 10a there are three feedback loops; in the
modified graph, after node x is split, there are no feedback loops. Hence
the index of the graph of Fig. 10a is one. A set of nodes, equal in number to
the index of a graph, that must be split in order to interrupt all feedback
loops is called a set of essential nodes. In Fig. 10 this set (only one node in
1
646
[Ch. 9
this case) is unique: only x is an essential node. In other cases there may
be more than one set of essential nodes.
1
INVERTING A
GRAPH
(a;
Fig. 11.
(b)
Inverting a branch: (a) original; (b) inverted.
Sec. 9.2]
SIGNAL-FLOW GRAPHS
647
the result that the node is converted to a sink node. The same process can
be carried out for a path consisting of any number of branches from a
source node to another node. The inversion is carried out one branch at a
time starting at a source node. This is illustrated in Fig. 12. (The graph
(a)
(b)
(c)
(d)
Fig. 12. Inverting a path: (a) original; (6) after inversion of branch a; (c) after inversion
of branch b (d) final.
is the same as that in Fig. 7 with general symbols for the transmittances.)
It is desired to invert the path from V to I. Branch a is inverted first,
leading to Fig. 12b. N o w node Va is a source node; so branch b can be
inverted, leading to Fig. 12c. Similarly, branches c and e can be inverted,
in order. The final graph with the inverted path is shown in Fig. 12d.
Note that the original graph in Fig. 1 2 a has three feedback loops and is
of index 1; that is, one node (V ) must be split before all loops are in
terrupted. However, the graph with the path inverted is a cascade graph,
having no feedback loops. This is inherently a simpler graph.
Besides inverting an open path, it is also possible to invert a loop. In
this case, the process is started by splitting a node on the loop, thus creating
a source and a sink node. The open path between these two is then in
verted, following which the split node is recombined. Details will be left
for you to work out.
1
REDUCTION OF A GRAPH
648
[Ch. 9
(a)
(e)
(d)
Fig. 13.
Sec. 9.2]
SIGNAL-FLOW GRAPHS
649
(8)
To eliminate x this expression is substituted into all other equations;
then (8) is disregarded; for instance, if the original equation for x is
n
(9)
the modified equation, after substituting (8) for x ,
n
becomes
(10)
n0
p0
pn
nk
nn
(11)
The last term can be transposed to the left side and the result solved
for x .
n
(12)
650
[Ch.9
This is the equation for a node without a self-loop. In terms of the graph
the equation shows that if the transmittance of each incoming branch to
x is divided by 1 c , the self-loop can be eliminated. Following this,
node x can be eliminated as before.
Division b y 1 c is possible only if c
1. We are clearly in trouble
of C = 1. A glance back at Fig. 9 shows that there is in that graph a
self-loop with a transmittance of 1. Nevertheless, the set of equations
represented by that graph are linearly independent and so can be solved.
In such cases, it is always possible to rearrange the equations in such a
way that a unity entry on the main diagonal of the connection matrix
can be avoided. Thus an unavoidable such entry can occur only if the
equations are not independentin which case we should not expect a
solution anyway.
The preceding operations can also be interpreted in terms of the con
nection matrix. This matrix has the following form, assuming node x
has no self-loop.
n
nn
nn
nn
nn
(13)
The partition indicates that the last row and column are to be eliminated.
Note that c = 0 since x has no self-loop.
N o w consider the typical equation (10) of the set, after the modification
introduced by eliminating node x . The corresponding connection matrix
will have the form
nn
(14)
Sec. 9.2]
SIGNAL-FLOW GRAPHS
651
where the external row and column have been added for comparison with
the preceding equations.
Now observe the manner in which the modified connection matrix is
obtained from the original one in (13). An entry in the last column of C,
say c , is multiplied by each entry in the last row. These products are
then added to the corresponding entry in the connection matrix. Thus
c multiplied by c is added to c , which is in the (2, 4) position of the
connection matrix. (Note that subscripts of rows and columns start
from zero.) This is repeated for each entry in the last column. This
process of successive reduction of a connection matrix is called the
node-pulling algorithm.
If node x has a self-loop, the entry c in the connection matrix is not
zero. The process of dividing all incoming transmittances to node x in the
graph by 1 c corresponds to dividing all entries of the last row in the
connection matrix b y 1 c
and then replacing the diagonal entry by
zero. The node-pulling algorithm can then be applied.
B y repeated applications of these operations, a signal-flow graph can be
reduced so that only source and sink nodes remain. If there is only one
source and one sink node, the graph reduces to a single branch from the
source to the sink. The transmittance of this branch is called the graph
gain.
As an illustration of the process, consider the set of equations
1n
1n
n3
13
nn
nn
nn
(15)
652
[Ch. 9
The corresponding reductions of the flow graph are illustrated in Fig. 15,
the corresponding steps being labeled with the same letter.
Fig. 14.
44
Sec. 9.2]
SIGNAL-FLOW GRAPHS
653
(b)
(a)
(c)
(e)
Fig. 15.
(d)
Reduction of signal-flow graph.
3. Remove self-loop at x
1 3 = 2 . Then
654
[Ch. 9
Thus we find that x = y0. It is clear that the solution can be obtained by
working on the matrix only, or the flow graph only.
The process of path inversion can also be useful in the reduction of a
signal-flow graph to determine the graph gain. This is illustrated in the
graph shown in Fig. 16a. If the path from y to x is inverted, the result
1
(b)
(a)
(c)
Fig. 16.
is as shown in Fig. 16b. The graph can now he rapidly reduced. First,
branches 1/a and 1/b are in cascade. The combination is in parallel with
branch c/a. Finally, the combination of these three is in cascade with
branch 1/d. The reduced graph is shown in Fig. 16c. The graph gain of the
original graph is thus found to be
R E D U C T I O N TO A N E S S E N T I A L
GRAPH
Sec. 9.2]
SIGNAL-FLOW GRAPHS
655
Source
node
Fig. 17.
essential nodes to all other essential nodes. There will also be self-loops at
each essential node. It is only necessary to determine what the values of
the transmittances will be. In specific cases, of course, one or more of the
branches in the essential graph might be missing.
To illustrate, consider the graph of Fig. 20 in the next subsection. This is
of index 2, although it has six feedback loops; V and I constitute a set
of essential nodes. In Fig. 17, then, the two essential nodes will be V
and I , and the source node will be V . It remains to compute the trans
mittances; for example, in the original graph there are three paths from
V to I , with the following path transmittances: G , Y R G ,
and Y Z G = G . Hence in the essential graph the branch from Vi
to I will have a transmittance C (1 Y R ) . The remaining
transmittances are evaluated similarly, with the final result shown in
Fig. 18.
1
Fig. 18.
GRAPH-GAIN
FORMULA
656
[Ch. 9
all
direct
paths
(16)
j 3
Sec. 9.2]
SIGNAL-FLOW GRAPHS
657
There is only one direct path from V\ to I, and the path gain (or trans
mittance) is G = abce. All three loops touch this path; hence = l .
The graph gain therefore equals
1
Now consider the inverted graph in Fig. 12d. This is a strictly cascade
graph, without feedback loops. Hence = 1, and all ' s = 1. The graph
gain from I to V is therefore simply the sum of direct path transmittances.
There are a total of four paths from I to V . Hence
k
Loop
Loop gains
Pj1
Then
Pj2
658
[Ch. 9
j1
j 2
Direct paths
Path gains Gk
Then
It is observed that the formula for determining the graph gain is quite
simple to apply. A source of error is the possible overlooking of one or more
feedback loops and one or more direct paths from input to output. A
systematic search procedure, however, should eliminate this possible
source of error.
The graph gain expression in (16) is a network function expressing the
ratio of a response transform to an excitation transform. Clearly, the same
network function should be obtained whether it is calculated with the use
of a signal-flow graph or from a solution of the network equations expressed
in the form of AX = Y, as in (3). However, in the latter case the net
work function will be obtained in terms of det A. Thus the graph deter
minant must be the same as det A.
Sec. 9.2]
SIGNAL-FLOW GRAPHS
659
NETWORK
where the subscript a stands for "all." The links will include all indepen
dent current sources, and the twigs will include all independent voltage
660
[Ch. 9
sources. Let us partition the current and voltage vectors to put the sources
in evidence. When Q and B are conformally partitioned, the above
expressions become
l
(19)
(20)
l g
(21a)
(21b)
Note that voltages and currents pertaining to independent sources are
not included in these equations.
The first row of (19) and of (20) are now inserted into (21) to yield
(22a)
(22b)
These equations are in the form of (4), from which a signal-flow graph can
be drawn.
The variables for which equations are written are the currents of all
links except independent current sources, and the voltages of all twigs
except independent voltage sources. Note that these variables do not
depend on the currents in voltage sources ( I ) and voltages across current
sources ( V ) . However, it may be that these latter variables are response
variables for which solutions are required. In the signal flow graph each
such variable will constitute a sink node; the corresponding equations,
are the second rows of (19) and (20), which are repeated here.
tg
lg
(23a)
(23b)
Sec. 9.2]
SIGNAL-FLOW GRAPHS
661
From (22) and (23) a signal-flow graph can be drawn. We are sure that
the equations constitute an independent set and therefore a solution can
be obtained. We also know that the equations constitute an adequate set
since any other variable can be determined once these variables are known.
However, if we were really required first to write the Q and B matrices;
then to partition them; then to write the V-I relationships and determine
all the matrices in (21); and then finally to set up (22) and (23), the value
of using a signal-flow graph would be obscured. The preceding develop
ment, in fact, constitutes an "existence proof." For a given network
we would not go through the same steps in drawing the signal-flow graph
as are needed to establish the existence. A much simpler process would be
used, illustrated by the examples to follow.
One further point must be clarified. When discussing the mixed-variable
equations in Section 2.7, the V-I relationship used was the inverse of that
given in (21), as shown in (110) in Chapter 2. For that choice the selection
of the branches of a controlled source as twig or link was given in Table 2.
With the present V-I relations these selections must be reversed. Thus,
for example, the controlled branch of a controlled current source must
now be a link, and the controlled branch of a controlled voltage source a
twig.
l
Example 1.
Consider the network in Fig. 19a. There are two controlled sources.
The graph of the network is shown in Fig. 19b, with the tree in heavy lines.
(b)
(a)
Fig. 19.
Example.
Equations for the link currents and twig voltages must now be written.
The current of any link which is not a controlled source can be expressed
in terms of its own voltage which, in turn, can be written in terms of
662
[Ch. 9
As for the voltage of any twig branch that is not a controlled source, it
can be expressed in terms of its own current, which, in turn, can be
expressed in terms of link currents. Thus for twigs 1 and 4 the equations
are
The signal-flow graph can now be drawn and is shown in Fig. 20. It has a
Fig. 20.
large number (six) of feedback loops but its index is only 2, since opening
nodes V and I will interrupt all loops. The graph-reduction procedure
can now be used to solve for any of the variables.
If it is desired to find the input impedance of the network from the
terminals of the independent voltage source, it will be necessary to solve
1
Sec. 9.2]
SIGNAL-FLOW GRAPHS
663
for the current in the voltage source. This variable does not now appear
in the graph, but a sink node for I can be easily added since I = I
I I . The graph gain relating I to V is the negative of the input
impedance.
g
Example 2 .
As a second example, consider the network in Fig. 2 1 . For the link
(b)
(a)
Fig. 21.
Network example.
664
Fig. 22.
9.3
[Ch. 9
FEEDBACK
R E T U R N RATIO A N D R E T U R N
DIFFERENCE
Consider first the signal-flow graph shown in Fig. 23a. Focus attention
on the branch with a transmittance k. This quantity is assumed to be a
specific parameter in a network, such as the or of a controlled source.*
This assumption implies that the signal-flow graph of a network can be
drawn in such a way that the desired parameter appears by itself in only
* The assumption that k is a specific network parameter is not essential to the defini
tion of return ratio. However, the assumption is made in order to provide a simple
relationship between return ratio and sensitivity of the gain to variation of a network
parameter, to be defined later.
Sec. 9.3]
FEEDBACK
665
(a)
(b)
Fig. 23.
(24)
666
[Ch. 9
Let us illustrate the calculation of the return ratio and return difference
for a more substantial graph. Look back at the signal-flow graph in Fig. 7
representing the amplifier in Fig. 6. Let the transconductance g be the
parameter of interest. This parameter appears only once in the graph, but
it is not alone. The graph can be modified as shown in Fig. 24a to put it in
(b)
(a)
Fig. 24.
(25)
(26)
B y comparing with the previous expression for F , we see that the numer
ator of F and the numerator of are the same. This result is not
accidental, but quite general, as we shall now discuss.
Consider the diagram in Fig. 25, which shows a portion of a signalflow graph containing the branch k between nodes x and x . A node has
been inserted and then split. Before the operation is performed, the equag
Sec. 9.3]
FEEDBACK
Fig. 25.
667
tions for the entire graph are as given previously in (6) and repeated here
for convenience.
(27)
For node x the equation is
b
(28)
where it is assumed that the parameter k does not appear in any of the
other a coefficients and that there is no other direct path between x
and x .
Now suppose that the equation for x and for all the other nodes is
rewritten in the form
bj
(29)
Equation 28 for x
becomes
(30)
(31)
where a is a cofactor of A.
N o w we turn to a determination
remove all source nodes. This means
(29). Then the insertion and splitting
25. This introduces a new source node
b
668
[Ch. 9
of node x only. This amounts to adding kx to the right side of (29) in the
bth row. Furthermore, as Fig. 25 shows, the term kx that appeared in
the equation for x no longer appears. Hence kx is removed from the bth
row on the left side of (29). All feedback loops in the original graph except
the ones containing branch k are included in the modified graph. Hence the
graph determinant is obtained from the old graph determinant simply
by setting k = 0. But this was called above. Hence the graph gain of
the new graph, which is just the negative return ratio, is given by
b
(32)
Sec. 9.4]
STABILITY
669
(35)
(36)
where G = R / is the graph gain when k is set equal to zero; that is,
this is the graph gain due to " l e a k a g e " paths not traversing branch k.
In the event that there are no leakage paths, meaning that all direct
paths from source to sink pass through branch k, the sensitivity becomes
simply the reciprocal of the return difference. This is the case, for example,
in the signal-flow graph of Fig. 24. Hence in this example the sensitivity
of the gain to the transconductance g is the reciprocal of the return
difference already determined in (25).
9.4
STABILITY
670
[Ch. 9
The right side follows from the change of variable (t T ) - > T . N O W , if the
upper limit is increased to infinity, the integral will not be reduced; so
the inequality will be further strengthened. Hence
Sec. 9.4]
STABILITY
671
Because of the condition of the theorem given in (38), W < and |w(t)| is
bounded for 0 t < .
For the only if portion of the proof, we start with this observation. If
(39)
then, given any 0 H < , there exists a 0 t' < such that
(40)
This portion of the proof will be by contradiction; that is, we shall assume
(38) to be invalid and show, given 0 E < and any 0 W < , that
there exists a 0 t' < such that |w(t')| > W for some |e(t)| E. Now,
choose an excitation
where sgn[x] is simply the sign of its argument. Thus e(t) is a function that
alternates between +E and E as the sign of h(t' t) changes. With this
excitation, the convolution integral in (37) yields
The final result is a consequence of the fact that x sgn x = |x|. Now let
H = W/F in (40) and insert the result into the last equation. It follows
that w(t') > W and, hence, |w(t')| > W. This completes the proof.
This theorem specifies a condition for B I B O stability in the time
domain. When H(s) = {h(t)} is a proper rational fraction, it is possible
to give an equivalent frequency domain condition. Thus
Theorem
2. If H(s) is a proper rational fraction in s, then the network will
be BIBO stable, which means that
(41)
672
[Ch. 9
if and only if all the poles of H(s) have negative real part.
The proof of the if statement is initiated with the partial-fraction
expansion
Since
|e
- s i t
| =
If all the poles of H(s) have negative real part, that is, if Re s > 0 for
i = 1, 2, ..., l, then each of the terms t e
is integrable from 0
to . A finite linear combination of integrable terms, as in (42), is also
integrable. Hence (41) is satisfied.
To prove the only if statement we turn to the Laplace-transform
integral for H(s); that is,
i
j - 1
- ( R e
For Re s 0,
|e-
s t
1; hence
s i ) t
Sec. 9.4]
STABILITY
673
Thus, if (41) is valid, then |II(s)| is bounded for all s such that Re s 0.
This means that H(s) can have no poles with non-negative real part, and
the proof is complete.
If a proper rational fraction
(43)
ROUTH
CRITERION
If the poles of H(s) are all to lie in the left half-plane, the polynomial
D(s) must be strictly Hurwitz, as defined in Section 6.2. It was observed
there that a necessary condition for a polynomial to be strictly Hurwitz
is that all the coefficients must be present and must have the same sign.
This is a useful bit of knowledge in that it provides a basis for easily
eliminating polynomials that cannot possibly be strictly Hurwitz.
However, we still need a basisa sufficiency conditionfor selecting a
strictly Hurwitz polynomial from amongst the remaining candidate
polynomials. The next theorem, which will present necessary and sufficient
conditions for a polynomial to be strictly Hurwitz, is an extension of
Theorem 16 in Chapter 7 on Hurwitz polynomials.
Suppose D(s) is a polynomial of degree n. For convenience, and certainly
without loss of generality, assume that the leading coefficient is positive.
Thus
(44)
N o w let (s) and (s) be polynomials derived by taking alternate terms
from D(s), starting with a s
and a s ,
respectively. Thus
n
n-1
(45a)
(45b)
674
[Ch. 9
(46)
Note that all the numbers must be positive; none can be zero, down to
the nth one. If all after the kth one are zero, this is an indication that
an even polynomial is a factor of both (s) and (s), and thus of D(s).
This even polynomial can have pairs of j-axis zeros or quadruplets of
complex zeros, two of which are in the right half-plane. In either case D(s)
cannot be strictly Hurwitz.
i
HURWITZ
CRITERION
2i
2i
Linear
Sec. 9.4]
STABILITY
675
(47)
Let
denote the determinant derived from by deleting the last
row and column of . Continuing in this way, let
denote the deter
minant obtained by deleting the last row and column of
. B y this
procedure we construct a total of n determinants , known as the
Hurwitz determinants, which are the basis for the next theorem.
n - 1
n k
n k + 1
Theorem
4. The polynomial
fori=l,
2,
...,n.
CRITERION
a,
a,
a,
a,
n
a
, a
, ...
a
,a
, ...
a ,
a
, ...
a ., a ,
...
n - 2
n - 4
n2
n - 4
n1
n - 3
n1
n3
and
and
and
and
,
,
, ...
-i,
,
, ...
,
,
, ...
i,
,
, ...
n
n 2
n 3
n 2
n 3
n 4
n 5
n 4
n 5
* A proof of this theorem and the preceding theorem may be found in F. R. Gantmacher,
The Theory of Matrices, Vol. 2, Chelsea Publishing Co., New York, 1959.
676
[Ch. 9
and the problem of determining those values of a and k for which D(s) is
strictly Hurwitz. We shall approach this task by using condition 2 of
the Linard-Chip art criterion. Note immediately that
(48a)
(49a)
(49b)
If conditions 2 of the theorem are to be satisfied, then from (48) and (49)
we get the following two inequality relations in a and k:
(50a)
(50b)
2
Sec. 9.5]
(a)
677
(b)
(c)
Fig. 26.
9.5
Example.
678
[Ch. 9
Fig. 27.
Nyquist contour.
For the validity of the principle of the argument, H(s) can have no
zeros or poles on the contour. Therefore we first assume that H(s) has no
zeros or poles on the imaginary axis. Similarly, since we wish to let R go
to infinity, we must make the additional assumption that H(s) is regular
and nonzero at infinity. We shall return to this assumption later and show
that the assumption relative to the zeros can be relaxed.
Let us consider the mapping of the contour by the function H; that is,
the locus of the point H(s) as s traverses the contour of Fig. 27. This may
be a curve such as the one shown in Fig. 28a. Since H is a network func
tion, it is real on the real axis, and so the map is symmetric with respect
to the real axis. Let N and N be the number of zeros and number of
poles of H(s), respectively, that lie inside the oriented Nyquist contour C.
N o w the argument principle states that
0
(51)
that is, the change in the argument of H(s) as s traverses the contour C,
which is oriented in the negative direction, is 2 times the number of poles,
minus the number of zeros of H(s) within C (taking into account the multi
plicity of each).
Let US see what the nature of the locus of H(s) must be if this change
of angle is to be nonzero. It is quite evident that the locus must go around
the origin in the H-plane if there is to be any change in the argument. This
is the case in Fig. 28b. But in Fig. 28a, there is no change in the argument
as we traverse the locus once. In other words, the locus must enclose the
Sec. 9.5]
(b)
(a)
Fig. 28.
origin in
Nyquist
enclosed
clockwise
679
the H-plane if there is a net change in the angle of II(s) over the
contour C. Clearly, if the origin is enclosed by the locus, it is
an integral number of times. Let N
denote the number of
encirclements of the origin by the locus. Then
cw
(52)
Substituting this expression into (51) yields
(53)
Thus, if the II(s) locus does not enclose the origin, we can conclude
that H(s) has as many poles as zeros in the right half-plane. But we really
want to know whether it has any poles in the right half-plane. Therefore
for this test to be useful we must know, by some other means, how many
zeros H(s) has in the right half-plane; for example, that H(s) has no zeros
in the right half-plane, which means II(s) is a minimum-phase function.
This is by no means an easy task. However, there is no need to abandon
the procedure because difficulty is encountered. What we can do is to
find another function involving D(s), the denominator polynomial of
H(s), and some other factor, this other factor being such that its zero
locations are known.
We shall suppose that a flow-graph representation of the network has
been reduced to an equivalent graph of the form depicted in Fig. 29.
Then by the graph-gain formula in (16), it follows that the transfer
function associated with this flow graph is
(54)
680
Fig. 29.
[Ch. 9
It must now be established where the poles of H(s) come from. Observe
first that the return ratio T of parameter k, and the return difference F,
are given by
(55a)
(55b)
Both of these functions have the same poles. Now the poles of the transfer
function H(s) are either (1) zeros of the return difference F(s) that are not
common with zeros of A(s) or (2) poles of A(s) that are common with zeros
of R(s). In the latter case T(s) and F(s) will not have a pole at such a pole
of A(s). Hence H(s) = kA(s)/F(s) will have this pole.
Suppose that, even though R(s) may have zeros in common with poles of
A(s), none of these lie in the right half-plane or j-axis. Hence any right
half-plane or j-axis poles of the transfer function H(s) must be zeros of
the return difference F(s). Stated differently: If all the zeros of F(s) = 1
+ k A ( s ) B ( s ) are in the left half-plane, then all the poles of the transfer
function H(s) are in the left half-plane, provided that B(s) has no right
half-plane and j-axis zeros in common with poles of A(s). (In this state
ment we cannot say "If and only if all the zeros of F(s) = 1 + kA(s)R(s)
are in the left half-plane . . . " Why?)
Under the stated condition, investigating the location of the poles of
the transfer function H(s) can be changed to investigating the location of
the zeros of the return difference F(s) = 1 +
kA(s)T(s).
In order to apply the argument principle to F(s), we must assume that
F(s) has no poles or zeros on the imaginary axis and is regular and non
zero at infinity. Now consider the locus of points F(s) on the Nyquist
contour C; a typical plot is shown in Fig. 30a, where the dashed curve
corresponds to < 0. Let N
denote the number of clockwise encircle
ments of the origin by the locus F(s). Then, as before,
cw
(56)
where No and N are the numbers of zeros and poles, respectively, of
F(s) in the right half-plane.
p
Sec. 9.5]
(a)
681
(b)
Fig. 30.
Nyquist plots.
DISCUSSION OF ASSUMPTIONS
682
[Ch. 9
j-axis. But this fact gives the information we were seeking; it tells us that
not all the zeros of F(s) are in the left half-plane, at least one being on
the j-axis.
Another earlier assumption on F(s) was that F(s) has no poles on the
imaginary axis or at infinity. This same assumption applies to T(s),
since T(s) and F(s) have the same poles. Again the validity of the assump
tion is easily observed from the behavior of the locus. Thus if the locus
T(s) becomes unbounded, then F(s) must have a pole at the corresponding
value of s.
The locations of such poles are, therefore, known. (Would this be true
if T(j) were known from experimental data?) We shall further assume
that the multiplicity of such poles is known. The principle of the argument
requires the function to which it is applied to have no poles or zeros on
the contour. But what should be done if it is discovered that F(s) has
such poles or zeros? For finite j-axis zeros, the question has been answered.
Let us turn to the case of finite j-axis poles. If the Nyquist contour is to
avoid such poles, the contour can be modified by indentations into the
right half-plane with vanishingly small semicircular arcs centered at the
pole, as shown in Fig. 31a. The corresponding change in the locus of
T(s) is shown in Fig. 31b. The solid lines show the locus for values of
s = j , with less than and greater than , where s = j is the location
of the pole. As approaches from below, the locus moves out to infin
ity at some angle. In Fig. 31b, infinity is approached in the third quadrant.
As s takes on values along the vanishingly small semicircular arc, the
locus T(s) approaches an infinite-radius circular arc of m radians, where
m is the multiplicity of the pole. The orientation on this circular part of the
locus is clockwise, as shown by the dashed curve in Fig. 31b. (You should
approximate T(s) in the vicinity of the pole by the dominant term in its
0
(a)
Fig. 31.
(b)
Sec. 9.5]
683
(a)
Fig. 32.
(b)
Loci corresponding to a pole or zero of F(s) at infinity.
In the case of a zero of F(s) at infinity, T(s) equals 1 there, so that the
locus T(s) intersects the ( 1, 0) point. In this case consider the limit of the
large circular arc of the Nyquist contour as the radius approaches
infinity. The locus T(s) will approach a vanishingly small counter
clockwise oriented arc of n radians centered at the point (1, 0), where
n is the multiplicity of the zero at infinity. This is illustrated in Fig. 32b.
NYQUIST
THEOREM
684
[Ch. 9
T(s), or F(s), on the finite j-axis and poles or zeros of F(s) = 1 + T(s)
at infinity, is called a Nyquist diagram. In (56) the number of clockwise
encirclements of the point ( 1, 0) refers to the number of encirclements by
the Nyquist diagram.
The preceding discussion will now be summarized in the form of a
theorem, called the Nyquist
criterion.
Theorem
6.
A network,
having
cw
Sec. 9.5]
685
Example
Let us now illustrate the Nyquist stability criterion b y means of an
example. We shall go through this example in some detail and show how
approximations can be incorporated in the procedure.
Consider the three-stage RC-coupled amplifier with frequency-sensitive
feedback shown in Fig. 33. In this example we shall try to show the ad-
Fig. 33
vantage of the Nyquist criterion by not computing the return ratio T(s).
Instead we shall estimate T(s) by making use of a number of approxi
mations. This network is a simplified model of a vacuum-tube network
in which many of the interelectrode capacitances have been neglected to
simplify the example.
This network can easily be modeled by a single loop flow graph of the
type we considered. The value of k will be and A(s)R(s) can be
written as a product of passive network functions, though we shall not do
so here. Thus we may properly set N = 0 and proceed in the knowledge
that No = N .
Furthermore, B(s) = 1 ; hence poles of A(s) cannot be
common to zeros of B(s). You should convince yourself of these facts.
Interest lies entirely on the j-axis; hence we shall deal mostly with
steady-state phasors instead of Laplace transforms. Remembering the
flow-graph definition of the return ratio, we must open the loop and apply
a unit signal at the right-hand node of the pair of nodes thus formed. The
signal returned to the left-hand node of that pair of nodes will be
T(j).
We shall now interpret this step in terms of the actual network rather than
the signal-flow graph. Imagine that the loop is opened at the input to the
first amplifier stage, and the voltage V is set equal to unity. Observe that
this is equivalent to replacing the first controlled voltage source by an
independent voltage source of value . Thus the condition shown in Fig.
34 is the appropriate one for computing the return ratio. Here the external
source is removed and the first controlled source is assumed to be an
independent source having a phasor voltage .
1
cw
g1
686
Fig. 34.
[Ch. 9
Notice that the reference for T is chosen to conform with the definition
that it is the negative of the returned signal to the back half of the split
node. It is easy to see how this interpretation can be used for experi
mental measurement of T(j) on the network of which this example is a
model.
In order to construct the Nyquist diagram, let us split the frequency
range 0 < into a number of bands and use suitable approximations
in each band. At very low frequencies the returned signal will be very
small due to the coupling capacitances C , C , and C . The influence of
C can be neglected in this range. There are three RC coupling networks
in the loop. Let us use the notation
1
gk
(57)
with suitable subscripts for each of the coupling networks. Then the
voltage ratio of each stage will be
(58)
with appropriate subscripts. Hence in this range the return ratio will be
given by
(59)
Sec. 9.5]
687
(The negative signs disappear because of the reference for T.) The
asymptotic phase of each of the factors in (59) as -> 0 will be /2 radians.
Thus the total asymptotic phase of T(j) will be 3/2 radians, the magni
tude approaching zero. Hence the low-frequency portion of the locus of
T(j) looks like the curve in Fig. 35.
Fig. 35.
Fig. 36.
g k
Midband approximation.
688
[Ch. 9
At high frequencies Fig. 36 can still be used, except that the effect of
C must now be included. Since C is in parallel with R s, the third factor
in (60) should be modified and replaced by the following:
gk
gk
(61)
Hence the angle of T will asymptotically approach /2. The high end
of the T(j) locus therefore takes the form shown in Fig. 37.
Fig. 37.
We can now estimate the T(j) locus for 0 < to have roughly
the shape shown in Fig. 38. To improve the approximation, we should
Fig. 38.
estimate a few points on the curve. Suppose, for simplicity, that the three
significant half-power frequencies of the interstage circuits are either
identical or widely separated. In the first case we know that at the common
half-power frequency each circuit contributes an angle of 45 and a 3-dB
attenuation. This point, marked in Fig. 38, must be 9 dB less than 20
3
Sec. 9.5]
689
log T . Similarly, we can find the frequency at which each circuit contri
butes a 60 angle. This is the frequency at which each of the denominator
factors in (59) contributes 30, which is easily found to be approximately
= 0 . 5 8 . At this frequency each factor will be down about 4 dB.
Therefore T(j ) will be down 12 dB from 20 log T . The frequency ,
marked in Fig. 38, is the point where the locus crosses the negative real
axis. The other points , , , are similarly computed. The widely
separated case is left as a problem. (See Problem 25.)
Once the locus for the positive range of is known, the diagram can
be completed by symmetry about the real axis. The complete locus for the
example is shown in Fig. 39.
m
Fig. 39.
Complete T-locus.
690
[Ch.9
PROBLEMS
1. Determine the index of each of the signal-flow graphs in Fig. PI. The
numbers on the branches are numerical values of branch transmittances.
2. By reducing each of the signal-flow graphs of Fig. PI, determine the
graph gain.
3. Determine the graph gain of each of the signal-flow graphs of Fig. PI
by applying the graph gain formula.
(a)
(b)
(c)
(d)
(e)
(f)
PROBLEMS
691
(h)
(g)
(j)
(i)
Fig. P I
4. Each of the signal-flow graphs in Fig. PI has one source node and one
sink node, (a) Invert the path from source to sink in each graph.
(b) Determine the graph gain of the inverted graph and from this, that of
the original graph.
5. Reduce each of the signal-flow graphs in Fig. PI to an essential signalflow graph. Then evaluate the gain using the graph gain formula.
Compare with the values from Problem 2 or 3.
6. In the graph of Fig. 12a, invert the loop V V V V .
Find the graph
gain I/V and verify that it is the same as found in the text.
a
7. Draw a signal-flow graph for the networks of Fig. P7. Reduce the
signal-flow graph to find the transfer function V /V .
2
(a)
692
[Ch. 9
(b)
(c)
(a)
(c)
(b)
(d)
Fig. P7
10. Find the transfer impedance V /I for the network of Fig. 21 with
signal-flow graph in Fig. 22 by (a) reducing the signal-flow graph alone,
3
(b)
(a)
(c)
(d)
(e)
Fig. P11
693
694
[Ch. 9
(b) operating on the connection matrix alone, (c) applying the graph
gain formula alone, and (d) using node equations.
11. Determine a signal-flow graph for each of the networks of Fig. P 1 1 a to
d. Use the triode model shown in Fig. P11e. Then evaluate the graph
gain(s).
12. Set up a signal-flow graph for the network of Fig. P12 to find the
transfer function Y (s) = I /V .
Find this function by reducing the
signal-flow graph. Also, apply the graph gain formula to the original
graph, and compare the answers obtained.
21
Fig. P12
13. Find the voltage ratio V2/V1 for the general ladder network of Fig. P13
by first setting up the signal-flow graph. From the signal-flow graph
show that the transmission zeros occur at series impedance poles or shunt
impedance zeros.
Fig. P13
14. Determine a signal-flow graph for each of the networks of Fig. P14a to c.
Use the transistor model shown in Fig. P14d. Then evaluate the
graph gain.
15. Find the gain V / V\ for the "pseudo-tuned" amplifier of Fig. P15
using signal-flow graphs.
2
PROBLEMS
(a)
(b)
(c)
(d)
Fig. P14
695
696
[Ch. 9
Fig. P15
16. For each of the networks of Fig. P16a to d, use (40) to determine the
sensitivity of the transfer function V /V to (a) c, (b) r, and (c) g.
Use the transistor model of Fig. P16. The nominal values are c =
100 F , r = 103, and g = 0.5.
0
17. The return ratio and return difference of a branch are defined in the
text. The same quantities can be defined for a node. The return ratio
of node j , represented by T , is the negative graph gain of the graph
obtained by splitting the node and removing all other source and sink
nodes. The return difference of node j is defined as F = 1 + T . The
partial return ratio of node j , Tj, is defined as the return ratio of node j
when all higher numbered nodes are removed from the graph. This is
obviously dependent on the ordering of the nodes. The partial return
difference of node j is also defined; it is Fj = 1 + Tj.
(a) Suppose the nodes of a graph are numbered in a particular order.
Now remove all nodes above the one numbered k. Next, draw a reduced
graph in general form, retaining only nodes k and k 1. Determine
the partial return differences and find the product F' Fjk_ . Finally,
interchange the numbers of nodes k and k 1 and again find the product
F F _ 1 . Show that this product is independent of the node
numbering.
(b) Show that the product of the partial return differences of all nodes
of a graph is a unique property of the graph, independent of the node
numbering.
j
18. Which of the following impulse response functions are associated with
BIBO stable networks:
(a)
(b)
(c)
(d)
(e)
(f)
PROBLEMS
697
(b)
(a)
(c)
(d)
(e)
Fig. P16
19. Use (i) the Routh criterion and (ii) the Linard-Chipart criterion to deter
mine which of the following polynomials in s have zeros only in the
left half-plane:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
698
[Ch. 9
(l)
(m)
20. Suppose H(s) is a rational fraction and is regular at infinity; thus, H(s)
equals a constant (which may be zero) plus a proper rational fraction.
Then prove that a network having H(s) as its transfer function is BIBO
stable if and only if all the poles of H(s) have negative real part.
21. Use the Linard-Chipart criterion to determine the value of for
which the networks in Fig. P21 are BIBO stable.
(a)
(b)
Fig. P21
22. Use the Linard-Chipart criterion to determine the values of p and for
which the networks in Fig. P22 are BIBO stable. In the py parameter
plane, shade the stability regions.
23. Use the Linard-Chipart criterion to determine the values of r and g
for which the oscillator networks shown in Fig. P23 to c are not BIBO
stable. Assume a ficticious voltage source input to be in series with the
transistor base lead. Use the transistor model shown in Fig. P23d. In
the r-g parameter plane, shade the instability regions.
24. Draw the Nyquist diagram for each of the following return ratio
functions:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
PROBLEMS
(a)
(b)
Fig. P22
(a)
(b)
(d)
(c)
Fig. P23
699
700
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
[Ch. 9
= 1000,
= 100,000
Sketch the Nyquist diagram carefully for this case. Find the maximum
value that T can be, if the network is to be stable.
m
26. Sketch the Nyquist diagram for the network of Fig. P26. Find values
of RfLf for which the network is stable. What is the maximum value
of under this condition, if the network is to remain stable for small
variations of parameter values (R , G , R , L in particular) from the
design value?
e
Fig. P26
PROBLEMS
27. Sketch the Nyquist diagram for the network of Fig. P27 and give the
condition for stability.
Fig. P27
28. Draw a signal-flow graph for the network given in Fig. P28. By
reducing the graph, calculate the transfer voltage ratio.
Fig. P28
criterion:
701
702
[Ch. 9
30. With reference to Problem 29, draw the inverse Nyquist diagram for
each of the return ratio functions listed in Problem 24. Then, indicate
the values of k for which the associated feedback network is known to
be stable by the inverse Nyquist criterion. Assume A(s) has only left
half-plane zeros.
31. Consider all possible combinations of A(s) and B(s) derivable from the
following list. Apply (i) the Nyquist criterion and (ii) the inverse
Nyquist criterion (of Problem 29) to determine the values of k for which
each network is known to be stable.
(a)
(b)
In the first case the determinant is that of the loop impedance matrix
and is formed by short-circuiting the input terminals. In the second
case the determinant is that of the node admittance matrix and is
formed with the input terminals open-circuited. The zeros of these
determinantswhich are, the poles of the respective transfer functions
will be different. Hence, the stability properties under the two kinds
of excitation need not be the same.
If a network is stable when its terminals are short-circuited, it is
said to be short-circuit stable. Similarly, if a network is stable when its
terminals are open-circuited, it is said to be open-circuit stable. It is
possible for a network to be both open-circuit and short-circuit stable.
PROBLEMS
703
(b)
(a)
Passive
reciprocal
Passive
reciprocal
(d)
(c)
Negative
converter
(e)
(f)
Fig. P32
704
[Ch. 9
34. In Fig. 20, nodes V\ and V constitute a set of essential nodes. Reduce
this graph to an essential graph with V\ and V as essential nodes and
compare with Fig. 18.
4
35. Signal-flow graphs for two feedback amplifiers are shown in Fig. P35;
(b)
(a)
Fig. P35
is to be adjusted so as to make the graph gains of the two graphs
equal. Determine the sensitivity S of the gain of each of the two graphs
to the transmittance . Compare and determine which type of amplifier
is less sensitive to changes in .
2
.10.
LINEAR TIME-VARYING
AND NONLINEAR NETWORKS
706
[Ch. 10
(la)
(lb)
(2)
R E D U C T I O N TO NORMAL FORM
(3)
Sec. 10.1]
707
for t t0.
N o w we can proceed in two ways to put (2) into normal form. For the
first way, set
(5)
then
(6)
Upon substituting this expression into (2) and rearranging terms, the
desired form in (3) is obtained with
(7)
(8)
and
(9)
(10)
708
[Ch. 10
In this case
(11)
After this expression is substituted into (2) and the terms in the resulting
equation are regrouped, we obtain the normal-form equation (3), with
(12)
and
(13)
The vector e is again given b y (9).
The second way of choosing the state vector leads to an A matrix, set
forth in (12), which requires the derivative of the parameter matrix of
(4). Since our interest is in a solution of the state equation for
tt0,
we must assume that this parameter matrix is differentiable for t t .
This is an assumption that does not have to be made when the state vector
is chosen by the first method.
In addition, the second way of choosing the state vector yields A and
B matrices that, in general, are functionally more complicated than those
obtained when the state vector is chosen by the first method. This is most
evident when
0
Sec. 10.1]
709
and
(14)
(15)
The last step follows from (113e) in Chapter 4, which is repeated here:
C t
C l
Ct
710
[Ch. 10
Thus
(16)
The last step follows from (113d), in Chapter 4, which is repeated here:
L l
t l
L t
l t
L t
t t
t l
L t
L l
l t
L t
Ll
L l
(17)
Fig. 1.
Sec. 10.1]
711
(You should do this.) This is, of course, the result that would be obtained
by using the methods of Section 4.5 through (129a) and (129b).
As indicated in Fig. 1, || is less than 1. Therefore
Observe that the elements of x are the capacitor charge and inductor
flux linkage. Now, b y (7) and (8), we find
712
[Ch. 10
Observe that the A and B matrices in this case are functionally more
complicated than before. Thus by example, as well as b y the previous
general discussion, we see that the better choice of x is as given in (5)
linear combinations of charges and flux linkages as elements of the state
vector. This will again be found to be true when we turn our attention to
nonlinear networks.
(18a)
(18b)
Sec. 10.2]
713
exists,
(22)
A solution for x is obtained by integrating. The result is
(23)
B y (19) and the assumption that Y(t ) is nonsingular, the initial vector
x(to) is given by the equation
0
(24)
714
[Ch. 10
To obtain x(t) we now premultiply (23) by Y(t). If, at the same time, we
introduce x(t ) as given by (24), we get
0
(25)
(26)
In Chapter 4 it was seen that the state-transition matrix is a function of
t T for time-invariant networks; this is not true in the more general case
of time-varying networks. The solution for x can be expressed in terms of the
state-transition matrix by inserting (26) into (25) to get
(27)
EQUATION
SOLUTION
Before the state vector x(t) can be determined from (27), it is necessary
to find the state-transition matrix (t, T), or, equivalently, Y(t). This is the
task to which we now turn our attention. We shall consider the solution
of the homogeneous matrix equation 21 when, in particular, we set
Y(t ) = U. This is no real restriction since (24) requires only that Y(to)
have an inverse.
First, consider the corresponding scalar equation
0
Sec. 10.2]
715
d is commutative.
to
(31)
This follows as a generalization of
716
[Ch. 10
that we get
(32)
The last step follows from the fact that the summands do not depend on
the summation index i. B y combining this result with (30), we find, in
this special case, that
(33)
Hence (28) is the solution of the homogeneous equation.
When the solution of the homogeneous equation is given b y (28), we
find that the state-transition matrix (t, T) has a particularly simple form.
We know that (t, T) = ( t , t )|
and that (t, t ) = Y(t) Y(to)~ =
Y(t), since Y(to) = U. Thus, by replacing t with in the integral
appearing in (28), the relation for (t, t ) = Y(t), we get
1
t0
(34)
Example
Before turning our attention to the solution of (21) under less restric
tive conditions, let us consider as an example a network for which the
preceding commutativity condition is satisfied. The network illustrated
in Fig. 2 with two time-varying resistors is easily shown to satisfy the
following differential equation for the state vector [q
]':
5
Sec. 10.2]
Fig. 2.
717
Time-varying network.
with
718
[Ch. 10
Furthermore, it may be
equation for almost all
integrate both sides of
equation
(35)
Oscillations,
Sec. 10.2]
719
Theorem
2. Given any Y ( t ) , the homogeneous differential equation 21
has a unique continuous solution, in the extended sense, equal to Y(t ) at
time t , if A is a locally integrable function for t t .
0
Fig. 3.
720
[Ch. 10
j[2u()
u(1)]dr
exists for all finite t. In addition, the theorem tells us the solution we have
found is the unique continuous solutionwe need search for no other.
We must still consider one assumption relative to the solution of the
homogeneous state equation. This is the assumption that Y ( t ) exists
for all finite t t . B y reference to Chapter 1, you may verify that
-1
(36)
where y
denotes the (l, k)th element of Y and
cofactor of Y. From (21) we know that
lk
l k
(37)
where a
lm
n
Since
y
mk
lk
= | Y | , we get
ml
k=l
(39)
ll
tr A. Thus
(40)
Sec. 10.2]
721
(41)
-1
are satisfied,
hence
[tr A()]
d =
AND
UNIQUENESS
Theorem
3. Given any x ( t ) , the state equation 18a has a unique solution,
in the ordinary sense, equal to x(t ) at time t , if A and Be are continuous
functions of t for t t < .
0
(42)
* See the footnote on page 718.
722
[Ch. 10
Theorem
4. Given any x ( t ) , the state equation 18a has a unique continuous
solution, in the extended sense, equal to x(t ) at time t , if A and Be are
locally integrable functions for t t .
0
Negative
converter
k=1
Fig. 4.
Sec. 10.2]
723
(43)
is a solution for every finite value of . Several different solutions are
shown in Fig. 5. The primary observation to be made is that there is no
unique solution.
Fig. 5.
j A()
o
dT =j'
PERIODIC N E T W O R K S
724
[Ch. 10
Assume that A is periodic with period T; that is, A(t + T) = A(t) for all t.
First of all, we find that, if Y(t) is a solution of (21), then so is Y(t + T);
that is, if there is a shift of the time variable by an amount equal to the
period of A, a solution of (21) will remain a solution. To see this, note that
Y() satisfies
(44)
-1
(46)
(47)
Sec. 10.2]
725
(49)
i j
(50)
j1
j1
j - 2
(51)
where P is given by (49) and Q(t) is a matrix to be determined. In fact,
this expression can be inverted to solve for Q(t). Thus we have
(52)
Now we can use the relationship in (47) to find out something about
Q(t). Substitution of (51) into (47) yields
726
[Ch. 10
P ( t _ t 0 )
P ( t - t 0 )
P ( t - t 0 )
(56)
with X(t ) = U. As for Q, substituting for Y(t) from (51) into (21) leads to
0
(57)
with Q(t ) = Y(t0).
0
Sec. 10.3]
10.3
727
We are now ready for a study of the properties of the solution of the
state equation for a time-varying network. This will be done by compari
son with some "reference" network about whose solution we have some
knowledge.
T H E GRONWALL
LEMMA*
In this study we shall make strong use of the following result from
mathematical analysis and will therefore discuss it first:
Lemma.
If
(58)
of t for t t
and
(59)
This result is called Gronwall's Lemma; its proof follows. From (58) we see
that
(60)
After multiplying both sides by the non-negative function (t) and adding
(t)/{y +
Note that
0. Thus
() () + () 0
(t)/{ +jf
728
[Ch. 10
in (61), we get
which is equivalent to
(62)
B y combining (58) and (62), we get the indicated inequality (59). This
completes the proof.
N o w we turn to a consideration of the state equation, which is repeated
here:
Suppose that the right-hand side is close, in some sense not yet defined,
t o the right-hand side of the homogeneous
reference
equation
Suppose also that all solutions of the reference equation either approach
zero as t tends to infinity or are bounded. It would be useful to be able to
infer as a consequence that all solutions of the original state equation
either approach zero or are bounded. The next several theorems state precise
conditions under which such inferences are valid. The conditions imposed
b y the theorems on the difference between A(t) and A(t) and on B(t) e(t)
establish the sense in which we view the state equation as being close to
the reference homogeneous equation.
Sec. 10.3]
ASYMPTOTIC P R O P E R T I E S RELATIVE TO A
REFERENCE
729
TIME-INVARIANT
5.
If all solutions of
(63)
(65b)
The double bars denote the norm of the matrix or vector enclosed. Refer
to Chapter 1 to refresh your memory on the properties of norms, since
they will be used extensively in the rest of the chapter.
This theorem is quite valuable because it is fairly easy to establish
whether or not all solutions of the reference equation 63 are bounded.
Recall that e
) y ( t ) for any y(t ) is a solution of (63). It is then an
immediate consequence of the constituent-matrix expansion of e
)
that all solutions will be bounded if none of the eigenvalues of A has
positive real part.
The proof of Theorem 5 is as follows. Rewrite the state equation as
A(t-t0
A ( t - t 0
730
[Ch. 10
A ( t -
A ( t -
The second inequality is obtained b y letting t tend to infinity. The righthand side of this relation is finite because of the conditions (65) of the
theorem. Thus ||x(t)|| is bounded, and, by implication, x(t) is bounded as
t->. The theorem is thus proved.
Observe that this theorem, like the others to follow, does not provide
a constructive method b y which the stability properties of the solution
can be discovered. Instead, given a time-varying state equation as in (64),
the theorem tells us that we must first verify that the norm of Be is
integrable over infinite time. Then we must look for a constant matrix A
* The term "triangle inequality," heretofore applied to the relation
shown this to be a valid inequality; you should do so. It is a natural extension of the
first inequality, if you think of the integral in terms of its defining sum under a limiting
process.
Sec. 10.3]
731
6.
If all solutions of
(63)
(64)
provided
(66)
That is, if A has no eigenvalues in the right-hand plane and on the j-axis;
then the norm of Be need not be integrable over infinite time. It is only
necessary for Be to be bounded.
The proof is as follows. We start with the equation
732
[Ch. 10
Taking the norm of both sides of this equation and using the usual norm
inequalities, we obtain
( t _ )
This inequality shows that ||z(t)|| and hence z(t) are bounded for
tt .
N o w set w = x z. Differentiation yields dw/dt = dx/dt dz/dt. Since
the differential equations for x and z determine dx/dt and dz/dt, we easily
establish that w satisfies the differential equation
0
Sec. 10.3]
733
7 . If all solutions of
The constants and have the same meanings as those given to them in
the proof of the last theorem. The proof of this theorem is left to you.
As an illustration of the second of the previous three theorems, consider
the filter network shown in Fig. 6. The state equation is easily shown to be
Fig. 6.
Time-varying network.
734
[Ch. 10
then
is bounded.
Thus all of the conditions of Theorem 6 are satisfied. Hence, by the
theorem, x(t) = [q (t) q (t) q4(t)]' is bounded as t -> . Observe that we have
established this fact without explicitly computing the solution to the
state equation or (63).
2
We now turn our attention to networks that are close to being described
by a homogeneous equation with periodic coefficients. We shall call this
Sec. 10.3]
8.
735
If all solutions of
(67)
where A(t)
is periodic,
provided
(68a)
(68b)
The proof of this theorem rests on the fact that any solution of (67)
may be expressed as Q ( t ) e
y ( t ) for some y(to), where Q(t) is non
singular and periodic, and P is a constant matrix. To start, the state
equation is first rewritten as
P(t_to)
Let y(t) denote the solution of the reference equation 67, with y(t ) =
x(to); then, using the transition matrix Q ( t ) e
Q ( ) - associated with
(67), the state equation may be put in the equivalent integral equation
form
0
P(t)
-1
Taking the norm of both sides of this equation and applying the usual
inequalities associated with norms, we get
(69)
736
[Ch. 10
-1
P ( t - )
Then, b y the conditions (68) of the theorem, ||x(t)|| is bounded and hence
x(t) is bounded as t-> . Thus the theorem is proved.
Again, the conditions of the theorem can be relaxed if A(t) is further
restricted. Instead of asking only that the solutions of the periodic
homogeneous reference equation 67 be asymptotically bounded, we ask
that they approach zero as t-> . Then conditions (68) can be relaxed.
This is stated more precisely in the next theorem.
Theorem
where A(t)
9.
If all solutions of
is periodic,
provided
(70)
A S a key item, the proof, the details of which are left for you, requires
that all of the characteristic exponents have negative real part. This
Sec. 10.3]
737
Theorem
10.
where A(t)
is periodic,
If all solutions of
( t T )
||A(t) A(t)||
The proof, which is similar to that for Theorem 7, is left for you.
To illustrate the first of the last sequence of theorems, consider the
network in Fig. 7. The state equation is
Fig. 7.
reveals that
738
[Ch. 10
and
and
Note that the max-magnitude vector norm is a more convenient norm for
this problem than the sum-magnitude norm, since the sum-magnitude
norm of B(t) e(t) is a more cumbersome expression. In particular,
and
Sec. 10.3]
739
Thus the conditions of Theorem 8 are satisfied and the state vector
x = [q
] ' is bounded as t->.
Unlike in the previous example,
it was necessary to solve the homogeneous equation dy/dt = A(t)y. Thus
the fact that A(t) is not constant requires more work in the application of
the theorem than the corresponding case when A(t) is constant.
1
ASYMPTOTIC P R O P E R T I E S RELATIVE TO A G E N E R A L T I M E - V A R Y I N G
REFERENCE
(71)
Two cases have been considered: (1) A(t), a constant matrix and (2)
A(t), a periodic matrix. When A(t) is not limited to these cases, but
assuming a solution of (71) to have the appropriate asymptotic property,
it may be tempting to suppose that the general solution of the state
equation will have similar properties if similar conditions are satisfied
b y A(t) and B(t) e(t). Such a supposition would be incorrect. Counter
examples can be found to demonstrate this; one such counter example
follows.
Suppose
and
Then set
740
[Ch. 10
It is clear that y =
[y
||A(t) A(t)||
-vt
dt =j\e
dt = 1/v <
and
||B(t) e(t)|| dt = 0 < . We arbitrarily chose to use the maxmagnitude vector norm and the associated matrix norm in this example.
The solution of the state equation is
You should verify this statement. Let us turn our attention to the inte
gral in this last expression. Set t = 2n + / 2 ; then, since e
is
positive for 0 t .
-
s i n
Sec. 10.3]
741
-1
Let y(t) denote the solution of (71), with y(t ) = x(t ), and let Y(t)Y()
be the transition matrix associated with (71). Then the state equation has
the equivalent integral equation form
0
Taking the norm of both sides and applying the norm inequalities as
previously, we get
-1
-1
-1
742
[Ch. 10
-1
then det[Y(t)] is bounded away from zero, and ||Y(t) || is bounded for
t t0. Equivalently, ||Y() || with to t is bounded for t t0.
With this added condition, the proof proceeds as with previous proofs,
making use of Gronwall's lemma at the end. Having sketched the proof
the details are for you to fill inwe state the last theorem in this section.
1
Theorem
provided
(72a)
(72b)
(72c)
The last two conditions are the same as before. The modification lies
in condition (72a). This is the price we pay for not limiting A(t) to a
constant or periodic matrix. Condition (72a) is quite severe; so much so,
in fact, that a network which is known to have an asymptotically bounded
state vector by one of the previous theorems may fail to satisfy (72a).
For example, suppose A(t) is a constant matrix and the conditions of
Theorem 5 are satisfied. Then, in addition, suppose all solutions of the
reference homogeneous equation 63 approach zero as t -> certainly a
more demanding condition than that of boundedness required b y Theorem
Sec. 10.3]
743
5 and the above theorem. In this case condition (72a) cannot be fulfilled.
You should verify this.
Now let us apply this theorem to an example. Consider the network
in Fig. 8. All elements are time-varying. The state equation is found to
be
Fig. 8.
Time-varying network.
as you may verify. Note that the solution, for all y(t0), is bounded as
t -> . Since tr A(t) = 2 e
3t
, we find tr A(t)
dt = | ( e
3 t
3t0
) and
we find
744
[Ch. 10
and as a consequence
B y this theorem, we may now make the following statement: For all
excitation voltages e(t) for which condition (72c) is satisfied, the state
[q (t) (t)]' of the network of Fig. 8 is bounded as t-> .
2
10.4
Sec. 10.4]
745
(73b)
(73c)
(73d)
(73e)
(73f)
(73g)
(73h)
The details of the formulation will be different depending on the specific
variables which are selected as components of the state vector. It is possible
for these to be voltage and current, or charge and flux linkage. We shall
formulate the branch relationships so as to make linear combinations of
capacitor charges and linear combinations of inductor flux linkages the
elements of the state vector. We know that
(74a)
(746)
where the elements of qct and qcl are the charges on the twig capacitors
and link capacitors, respectively. We shall suppose that the charges are
nonlinear functions of the capacitor voltages. Thus
(75a)
(75b)
where f and f are vector-valued functions of v and v . In addition,
they may be functions of time. After substituting (74a) and (74b) in (73b)
and rearranging terms, we obtain
Ct
Cl
C t
C l
(76)
746
C t
[Ch. 10
+ Q
C C
C l
in
(77)
Substituting v
C l
C t
, which
(79)
where g
is a vector-valued function of its arguments q + Q q
and v .
Before considering (76) and (79) further, we turn our attention to the
branch relationships for the inductors. We know that
C t
C t
C C
C l
(80a)
(80b)
where the elements of and are the flux linkages through the twig
inductors and link inductors, respectively. We shall suppose that the
flux linkages are nonlinear functions of the inductor currents. Thus
L t
L l
(81a)
(81b)
where f and f are vector valued functions of i and i . Also they may
be functions of time. N e x t we substitute (80a) and (80b) into the topolog
ical expression (73g) and rearrange terms to establish the following:
Lt
Ll
L t
L l
(82)
The linear combination of flux linkages
terms of the relations in (81) as follows
L l
L L
L t
may be expressed in
(83)
Sec. 10.4]
Substituting i
L t
747
(84)
L L
L t
L l
in terms
(85)
(86a)
(86b)
Were it not for the presence of the resistor branch variables i and v ,
we would have the desired pair of first-order vector differential equations
for q + Q q and \ Q _ Lt. Therefore we turn our attention next
to the branch relations for the resistors.
Suppose the relationship between the resistor voltages and resistor
currents may be expressed by the implicit vector equation
R l
Ct
C C
C l
L l
R t
(87)
(88)
(89)
748
[Ch. 10
Rl
ct
C c
C l
x l
LL
L t
(90a)
(90b)
Substitution of these two expressions for v
desired differential equations
R t
and i
R l
(91a)
(91b)
We take the elements of the state vector to be linear combinations of
capacitor charges and inductor flux linkages. Thus,
(92)
(93)
(94)
where h is a vector-valued function of x and e, which is determined by
the right-hand sides of the differential equations (91) and by the defining
Sec. 10.4]
749
relations (92) and (93) for x and e. You should determine the actual
expression for h. Furthermore, h may also be an explicit function of time;
when it is necessary to make this fact evident, we shall write (94) as
(95)
Since x is a state vector for the network, either of these last two differential
equations for x will be called the state equation.
We have shown a method for developing the state equation for a non
linear network in which the branch-variable relationships can be expressed
in the form specified in (74), (80), and (87) and for which the algebraic
equations (78), (84), and (89) have solutions as assumed. For any specific
problem, you should not take the final expression and substitute the
appropriate numerical values. Rather you should repeat the steps in the
procedure for the particular problem at hand. The results given are
general; in a specific case linearity of some components may permit
simplifications that come to light only when the detailed steps of the
formulation procedure are followed.
Let us now apply this formulation procedure to the network shown in
Fig. 9a. With the usual low-frequency model for the triode, the network
may be redrawn as in Fig 9b. There is a nonlinear inductor and a non
linear controlled source. A network graph and normal tree are shown in
Fig. 9c. There are no degeneracies, so the fundamental cut-set matrix is
easily determined and used to express the Kirchhoff equations as in (73).
Thus
750
[Ch. 10
from which
Sec. 10.4]
(a)
(b)
Twigs
Links
(c)
Fig. 9.
751
752
ig
v ]'
i ]'
10
in terms of the
10
[Ch. 10
is obviously
Sec. 10.4]
753
The second equality results from using the Kirchhoff equation giving
[i
i3 I4] in terms of the link currents. Next, we substitute the previ
ously established expressions for [i
i
i
i9 i ]' and [i ] to obtain
2
10
11
754
[Ch. 10
With x = [q
q
q
] ' and e = [v ], the right-hand side is the vectorvalued function h(x, e, t) of the state equation 95. It may not be evident
that this right-hand side is indeed a function of x and e, since only
dependence on the elements of x and e is evident. Therefore we shall show
that the second row on the right-hand side, as a typical row, is a function
of x and e. It is obvious that
2
1 1
OUTPUT
EQUATION
L t
Sec. 10.4]
elements of i
and i . B y (85), i
(part of x) and ij (part of e). Thus
Q'
i ,
Ll
L t
is given in terms of
755
Ll
Ll
establishes
x l
and
as functions of
Lt
z z
L l
Q'
LL
and i
Lt
LL
Lt
L l
L t
and iJ.
L t
L t
L t
L t
L t
L l
L l
L l
L l
L l
L L
L t
L t
C t
C t
C l
C l
C t
C t
C l
C l
C t
C C
C t
C l
C l
756
[Ch. 10
C t
C t
C l
C l
C C
C l
C l
R t
R l
R t
R l
R t
R l
C t
10.5
Once a state equation has been written for a nonlinear network, the
next task is to solve this equation. For convenience, in the equation
dx/dt = h(x, e, t), we shall set h(x, e, t) = f ( x , t). This is proper, since,
when e(t) is given, h[x, e(t), t] is an explicit function of x and t only. Thus
the vector differential equation that will be the center of attention in this
section is
(98)
Sec. 10.5]
757
UNIQUENESS
The search for conditions under which (98) is known to have a solution,
possibly a unique solution, is a very important task; because, amongst
other reasons, only if a solution, usually a unique solution, is known to
exist does it become meaningful to seek an approximate solution. Further
more, the existence or uniqueness, if one exists, is b y no means certain.
We discovered this for time-varying networks through an example in
Section 10.3.
As an additional example, in this case only of nonuniqueness, consider
the network in Fig. 10. You should find it easy to verify that the state
equation is
Negative
converter
Fig. 10.
is a solution for all t0. Therefore this network does not have a unique
solution when q (to) = 0.
1
* Almost any text on nonlinear analysis will provide considerable information on this
subject; for example, Nicholas Minorsky, Nonlinear Oscillations, D. Van Nostrand Co.,
Princeton, N.J., 1962.
758
[Ch. 10
The conditions for a solution to exist for all t t0, the time interval of
interest to us, are usually established in two parts. First, given t t0,
conditions are established whereby a solution exists in an interval
t t t , where t is determined by properties of the function f(x, t) in
the interval t t t . Then conditions are established by which a
solution can be extended from one interval of time to the next until the
solution for all t to is insured.
A solution of (98) in the ordinary sense requires that x satisfy the differ
ential equation for all finite t t0. It is often possible to find a function
that satisfies (98) for almost all t, although dx/dt fails to exist for some
discrete values of t. To admit this type of function as a solution, we con
sider the integral equation
1
(99)
associated with (98). We shall call any solution of (99) a solution of (98)
in the extended sense. You should convince yourself that (99) can possess
a solution for t t and yet (98) will not have a solution in the ordinary
sense for t t .
Apropos of existence, we state the following theorem without proof:*
0
Theorem
12. Given any t t and x(ti), the differential equation 98
possesses at least one continuous solution in the extended sense for t > t t ,
equal to x(ti) at time t , if f(x, t) is continuous in x for fixed t t and
locally integrable in t for t t and fixed x, and if ||f(x, t) || is bounded in
any bounded neighborhood of the origin 0 by a locally integrable function of
t for t t .
1
The conditions of this theorem are more extensive than needed to estab
lish existence of a solution for t > t t ; however, they are appro
priate to the statement of the next theorem on the extension of the solution
to the entire half-line t t .
2
Theorem
13. Suppose the conditions of Theorem 12 on existence apply,
then any solution of (98) in the extended sense, equal to x(t ) at time t ,
may be extended to yield a defined solution for all t t , if ||f(x, t)||i
(t)(||x|| ), for t t , where (t) is non-negative and locally integrable
and where (v) is positive and continuous for v > 0 and satisfies
0
(100)
for some u > 0.
0
Sec. 10.5]
759
Other extension theorems exist* and may be more useful in a given situa
tion. One of these other theorems, because it is extremely simple, is given
in the problem set as a theorem to be proved.
Before continuing, let us indicate under what additional restrictions
the solution in the extended sense is also a solution in the ordinary sense.
Applying the definition of a derivative to the integral equation (99)
yields
Differential
760
[Ch. 10
2/3
2/3
2 / 3
2/3
2/s
2/3
2/3
2/3
Fig. 11.
Sec. 10.5]
761
Obviously, the derivatives exist and are continuous in x \ and x for almost
all t; d f / d x
does not exist for t = 0. Further, | d f / d x \ 3|t|
,
\df /dx2\
1, |df /dx | 1, and |df /dx | 5; therefore, for all x and hence
in some neighborhood of every x, the derivatives are bounded by nonnegative locally integrable functions of t for t 1 . Thus the Lipschitz
condition is satisfied, and, by the uniqueness theorem, the solution will be
unique.
It can in fact be shown, as you should verify, that, when the summagnitude vector norm is used;
2
1/3
762
[Ch. 10
Then we might anticipate that, if all solutions of the latter equation are
bounded or approach zero as t-> , the solutions of the former equation
will do likewise. It turns out that this inference is true if A(t) is a constant,
or periodic, matrix. It will be left as a task for you to show that a theorem
similar to those to be given does not exist when A(t) is an arbitrary timevarying matrix.
For networks that are close to being described by a time-invariant,
homogeneous differential equation (namely, A(t) = A) we shall present a
sequence of useful theorems. The first theorem describes conditions for a
bounded response.
Theorem
equation
(101)
Further,
with an initial vector x(t ) such that ||x(t )|| , where is a constant
depends on f(x, t), are bounded as t tends to infinity if
0
that
(102a)
(102b)
(102c)
where is a suitably
chosen positive
constant.
Sec. 10.5]
763
A ( t - )
where e
is the transition matrix associated with (101). Upon taking
the norm of both sides of this equation and applying the usual norm
inequalities, it is found that
Since all solutions of (101) are bounded, a positive constant exists such
that | | e
| | for to T t and for all t t . Using this bound, that
of (102a), and ||x(t )|| , we get
A(t)
t t
0
764
[Ch. 10
indicated in (102c). (Note: the conclusion of the theorem becomes valid for
all initial states only if =
+.
As the next theorem shows, the conclusions of the previous theorem can
be strengthened, if all solutions of (101) approach zero as t tends to infinity.
Theorem
16. Suppose all solutions of the reference equation 101 in which A
is a constant matrix, approach zero as t tends to infinity.
Furthermore,
suppose f(x, t) = Ax + B (t)x + f(x, t) + g(t). Then all solutions of the
nonlinear equation 98 with initial vector x(t ) such that ||x(t )|| , where
is a constant that depends on f(x, t), approach zero as t tends to infinity if
0
(103a)
(103b)
(103c)
constants.
Taking the norm of both sides and applying the usual norm inequalities
yields
Sec. 10.5]
765
or, equivalently,
provided
vt
V/2
_
vt
The second inequality follows from the first upon letting t approach
infinity in the first integral and using (103b) with = v in the second
integral. Multiplication by e~ yields
vt
Now set Y < v/2; then ||x(t) || is uniformly bounded, by virtue of (103a), and
less than
for all t t0. Hence, to satisfy the condition that ||x()|| < for t0 T t
and all t t , it is merely necessary to select such that
0
766
[Ch. 10
Further, with reference to the bound on ||x(t)||, we see that x(t) must
approach zero as t tends to infinity since ||x(t)|| is bounded by a decaying
exponential.
Other theorems are possible. The two we have given establish the type
of conditions that are usually needed to get a proof. You will find it to be
a relatively easy task to vary these conditions and still get a valid proof.
For example, in the last theorem, condition (103c) can be replaced by
||f(x, t)|| ||x|| for ||x|| and sufficiently small. This variation is
significant because it permits a small linear term to be part of f(x, t).
In the two theorems given, if we replaced A by A(t), where A(t) is
periodic, we would still have true theorems. The proofs vary only slightly
from those given. You should determine what changes are needed.
As an illustration of the second theorem, consider the network in
Fig. 12 for t 0. With q = x1 and = x , the state equation is
2
Fig. 12.
First of all, we shall express this equation in the form to which the theorem
applies. Thus
Sec. 10.5]
767
higher degree, then the Routh criterion or one of the other similar criteria
could have been used to show whether or not all the roots had negative
real part.) Thus all the solutions of (101) approach zero as t>. B y
using the sum-magnitude vector norm, we get
and
Now
768
[Ch. 10
10.6
NUMERICAL SOLUTION
(104)
BACKWARD-DIFFERENCE
(k)
FORMULA
where y (t ) = d y(t)/dtk| .
Often y(t) can be approximated by the
polynomial obtained when r(t) is neglected. Of course, if y(t) is a polyi
t=ti
Sec. 10.6]
NUMERICAL SOLUTION
769
i2
(k)
ik
(k)
ik
(106)
where r(t) is a remainder that is, of course, different from r(t); the a (t)
are polynomials in t of degree no greater than j . B y neglecting r(t), a
polynomial that approximates y(t) is obtained.
The coefficients a (t) are not as easy to evaluate as might be desired.
However, if the terms in (106) are rearranged to express y(t) in terms of
sums of and differences between, the various y ( t ) , the new coefficients are
easy to evaluate. This will become evident as the representation evolves.
Let us first define a set of functions, which are sums and differences of
the values of y(t) at different points in time, as follows:
k
ik
(107a)
(107b)
(107c)
Each function, after the first one, is the difference between the preceding
function at two successive instants in time, divided b y a time difference.
Thus these functions are known as divided differences. You should note
that, since the first divided difference is just y(t ) b y definition, all succes
sive divided differences are sums and differences of y(t) at various points
in time, divided b y time intervals.
Our next task is to express y(t) in terms of sums and differences of its
values at t , t , ...; that is, in terms of the divided differences. We shall
i
i1
770
[Ch. 10
i1
i1
(108c)
where
(109)
is viewed as the error in approximating y(t) b y the polynomial
(110)
Sec. 10.6]
NUMERICAL SOLUTION
771
i1
ij
(111a)
(1116)
(111c)
These are defined as the backward differences. Let us assume in all that
follows that differences between adjacent values of time are equal; that is,
t t = h for k = i j + 1, ..., i. Then, starting with (111a), we obtain
k
k-1
The second line follows from the first by using the previously derived
expression for Vy(t ), and the third line follows from the second by using
(107c). Continuing in this manner, it is found that
i
(112a)
(112b)
(112c)
772
[Ch. 10
or, equivalently,
(113)
(114)
(115)
FORMULAS
i + 1
(116)
i + 1
i+1
(117)
Sec. 10,6]
NUMERICAL SOLUTION
773
(118)
where b = 1, and, for k > 0,
0
(119)
The second integral in (119) is obtained from the first b y the change of
variable T = (t t )/h and the fact that t
= t lh. Upon evaluating this
integral for several values of k and substituting the result into (118),
we find that
i
il
(120)
We have, in a sense, achieved the goal. If we know x(t ), x(t1), ...,
and x(tj), then we can use the state equation t o determine x(to), x(ti),
..., and x(tj). The backward differences V x(tj), V x(tj), ..., and V x(t ) are
computed next, and (118) is used t o evaluate x ( t ) . The steps are then
repeated t o evaluate x(t + ) starting with x(t ), x(t ), ..., and x(tj+i).
Continuing in this way, the value of x(t) at t = t , t , ... is established.
We shall give this subject greater attention later.
Equation 118 states the dependency of ( t + ) on the immediately
preceding value of x (namely, x(t )) and on x(t ), Vx(t ), ..., and V x(t ). We
can just as easily establish the dependency of x ( t + ) on x(t ) for l O
and on x(t ), Vx(t ), ..., and V x(t ). This is accomplished b y integrating
x(t) from t
= t lh to t
= t + h. The integral relation obtained
thereby is
0
j+1
j + 1
j + 2
i l
i - l
i + 1
(121)
Upon substituting (116), we obtain
(122)
774
[Ch. 10
(123)
For subsequent use, values of the coefficients b (l) that appear in the very
general relation (122), are given in Table 1.
k
Table 1.
Values of b ()
k
Nothing has been said about the error arising from using the truncated
version of Newton's backward-difference formula. A good treatment of
truncation errors is to be found in most texts on numerical analysis.*
However, it should be noted that the error in (122) is proportional to
h . Thus, in an intuitive sense, if h is sufficiently small and j is sufficiently
large, the error should be quite small.
j+2
CLOSED
FORMULAS
The relations obtained from (122) for the various values of l are known
as open formulas because x(t + ) depends not on x at t
but only on x at
previous points in time. On the other hand, the closed formulas to be
derived next exhibit a dependency relationship between x(t + ) and x at
t
as well as preceding points in time.
Set y(t) = x(t) and then replace t by t
in Newton's backwardi
i+1
i + 1
i + 1
* See, for example, R. W. Hamming, Numerical Methods for Scientists and Engineers,
McGraw-HiU Book Co., New York, 1962.
Sec. 10.6]
NUMERICAL SOLUTION
775
(124)
i + 1
= t + h leads to
i
(125)
(126)
(127)
Values of c (l), for the completely general closed formula (126), have been
computed and are given in Table 2.
k
Table 2.
Values of c (l)
k
776
[Ch. 10
As in the case of the open formulas, the error in (126) due to truncating
Newton's backward difference formula for x(t) is proportional to h .
At first glance it might appear as though closed formulas are of little
value, since the state equation cannot be used to compute x(t + ) until
-(ti+ ) is known, and (126) cannot be used to determine x(t + ) until
(ti+ ) is known. However, the closed formulas are useful in numerical
solution of the state equation by what are known as predictor-corrector
methods, which we shall examine shortly.
j+2
EULER'S
METHOD
(128a)
(128b)
The value of x at t = t inserted into the right-hand side of the second
equation gives the value of the derivative x(t ). When this is inserted into
the right-hand side of the first equation, the result is the value of x at
t . Alternate use of these two expressions leads to the values of x at
t + kh for all values of k. This numerical procedure is called Fuler's
method.
We shall not devote much more attention to this elementary method
because the error is significantly larger than that associated with other
methods. There is the further undesirable feature that the error may grow
significantly as time progresses. This is best illustrated by example.
Consider the scalar equation
i
i + 1
Sec. 10.6]
NUMERICAL SOLUTION
777
Exact solution
Numerical solution
Fig. 13.
THE M O D I F I E D
EULER
METHOD
i + 1
(129)
which, if V x(t +i) = x(t + ) x(t ) is substituted, becomes
i
(130)
i + 1
i+1
i+1
i+1
i+1
i+1
i+1
778
[Ch. 10
This is the modified Euler method; as the words used were meant to
imply, it is a predictor-corrector method. It is just one method, perhaps the
most elementary one, from a large class of such predictor-corrector
methods, some others of which will be considered.
To illustrate the calculations required in the modified Euler method,
suppose that
as the predicted value of x(h). The state equation then gives the corres
ponding value of x(h) as x(h) = [0.729 2.600]'. When (130) of the modi
fied Euler method is used, the first corrected value of x(h) becomes
[0.914 1.740]'. The state equation now gives a new value of x(h),
and (130) yields the second corrected value of x(h). These are
Repetition of this step yields a new x(h) and a third corrected value of
x(h); they are
METHOD
The open formula (118) for some j , such as the particular case of
j = 3 given here,
(131)
Sec. 10.6]
NUMERICAL SOLUTION
779
forms the basis for the Adams method. The state equation is used, of course,
to evaluate x(t ) from x(t ).
Like other methods that use the open and closed formulas with j > 1,
the Adam's method is not self-starting; that is, it is not sufficient to know
just the state equation and x(t ); the values of x ( t ) , ..., and x(t ) must also
be known, Only with this added information can the first complete set of
backward differences be calculated, at time t , and (118) used to evaluate
i
x(t + ).
j
(132)
This series method clearly requires that f ( x , t) be sufficiently differentiable
( j times in x and t) at t ; only then can x ( t o ) be evaluated; for example,
(k)
t0] + f [ x ( t 0 ) , t0].
t
To illustrate the Adams method, let us consider the very simple state
equation
with x(0) = 0. Set h = 0.1 and take j = 3. Thus (131) will be the particular
case of (118) used here.
The starting values will be obtained using a truncated Taylor series.
First,
780
[Ch. 10
B y using this series, x(h), x(2h), and x(3h) are evaluated; thus x(h) = 0.0905,
x(2h) = 0.1637, x(3h) = 0.2222. With these values (131) can be used to
compute subsequent values of x(ih). These values along with those of the
several backward differences are shown in Table 3 for values of i up to 10.
For comparison, the exact values, (ih)e ,
of x at t = ih are also tabulated.
-(ih)
Table 3
Exact
x(ih)
0.0000
0.0000
1.0000
0.0905
0.0905
0.8143
-0.1857
0.1637
0.1637
0.6550
-0.1593
0.0264
0.2222
0.2222
0.5186
-0.1364
0.0229
-0.0035
0.2681
0.2681
0.4022
-0.1164
0.0200
-0.0029
0.3032
0.3032
0.3033
-0.0989
0.0175
-0.0025
0.3293
0.3292
0.2196
-0.0837
0.0152
-0.0023
0.3476
0.3475
0.1491
-0.0705
0.0132
-0.0020
0.3595
0.3594
0.0899
-0.0592
0.0113
-0.0019
0.3659
0.3658
0.0408
-0.0491
0.0101
-0.0012
10
0.3679
0.3678
M O D I F I E D ADAMS
Numerical
x(ih)
x(ih)
V x(ih)
V x(ih)
x(ih)
METHOD
i+1
i+1
(133)
which is (126) with j = 3 and I = 0, is used to compute corrected values of
x ( t ) . This method is called the modified Adams method; like the modified
Euler method, it is a predictor-corrector method.
i+1
Sec. 1 0 . 6 ]
MILNE
NUMERICAL SOLUTION
781
METHOD
(134)
Note that only two backward differences are computed, as when truncat
ing the open formula (122) at j = 2. However, the accuracy is the same as
that achieved when truncating the open formula at j = 3 . The closed
formula (126) with j = 3 and l = 1 is used to compute the corrected values
of x ( t ) . Since c (l) = 0, only two, rather than three, backward differen
ces must be computed. Thus the equation may be written without the
V x(t + ) term as
i+1
(135)
A second Milne method uses the open formula (122) with j = 5 and
l = 5 to predict x ( t ) , and the closed formula (126) with j = 5 and l = 3
to correct x ( t ) . The fact that b (5) and c (3) are zero reduces the comput
ing effort to that expended when (122) and (126) are truncated at j = 4.
i+1
i+1
PREDICTOR-CORRECTOR
METHODS
i+1
782
[Ch. 10
with the same accuracy as the corrector; and the amount of computing
needed to predict x(t + ) is the same as with the original predictor.
You should make particular note of the fact that most of the predictorcorrector methods based on (122) and (126) are not self-starting. The
exceptions are (122) with j = 0 and l=0 and (126) with j = 0 or 1 and
l=0.
i
RUNGE-KUTTA
METHOD
Sec. 10.7]
LIAPUNOV STABILITY
783
We have said very little about errors in the numerical solution of the
state equation except to point out the dependence on h of the error due to
truncation of Newton's backward-difference formula. There are other
errors, and you should be aware of how they occur.
There are errors that occur because arithmetic is done with numbers
having a limited number of significant digits. This type of error is known
as roundoff error, since the word "roundoff" denotes eliminating insigni
ficant digits and retaining significant digits.
The truncation and roundoff errors occurring at each step in the calcula
tion affect not only the error of the numerical solution at that step but also
the error at subsequent steps; that is, the error propagates.
Another source of error is properly viewed as a dynamic error and occurs
in the following manner. The equations used to obtain a numerical
solution of the state equation may exhibit more dynamically independent
modes than the state equation. If any of the additional modes are unstable
then the numerical solution may depart radically from the actual solution.
We shall not consider errors any further but refer you to books on
numerical analysis. (See Bibliography.)
10.7
LIAPUNOV STABILITY
In the case of linear networks, for which general analytic solutions of the
state equation exist and can be determined, it is possible to examine the
solution and study its properties. In particular, it is possible to make
784
[Ch. 10
DEFINITIONS
It is observed that at those points at which f(x, t) = 0 for all t t0, the
time rate of change of x is identically zero. This implies of course that if
the state starts at or reaches one of these points, it will remain there.
Clearly these points are distinctive; they are therefore, given the special
name of singular
points.
The subject of stability is concerned with the behavior of the state
equation solution relative to a singular point. It is a matter of convenience
in definitions and theorems to make the singular point in question the
origin. To see that this is always possible, let x be a singular point. Next,
set y(t) = x(t) x . Then x(t) = x corresponds to y(t) = 0. The equivalent
identification x(t) = y(t) + x substituted in the state equation 137 gives
5
(138)
Since y and x differ only by the constant vector x , either determines the
state of the network. If we view y as the state vector, then (138) is the
state equation and the origin is a singular point. Thus, without loss of
generality, we shall assume that (137) establishes the origin as a singular
point.
In what follows we shall be using norms without reference to a particu
lar norm; however, the illustrations will pertain to the Euclidean norm.
We shall let Sp denote the spherical region ||x|| < in vector space and
B denote the boundary of the spherical region S . Thus B is the sphere
s
Sec. 10.7]
LIAPUNOV STABILITY
785
(b)
(a)
(c)
Fig. 14. Illustrations of stability definitions: (a) Stable origin: (6) asymptotically
stable origin; (c) unstable origin.
Definition.
The origin is stable, if for each R < E there is an R such
that any trajectory x originating in S remains in S R . [The point at which
the trajectory originates is x(t ).]
+
786
[Ch. 10
More precisely,
Definition.
The origin is asymptotically stable if it is stable and if for any
e > 0 there exists a t such that the trajectory remains in S for t > t .
e
This definition states in a precise way that x(t) approaches zero as t tends
to infinity, by requiring the existence of a value of time after which the
norm of the solution remains less than any abritrarily small number.
Observe that both stability and asymptotic stability are local, or in-thesmall, properties, in that their definitions permit r > 0 to be as small as
necessary to satisfy the defintion. On the other hand, if, when R =
+,
the origin is asymptotically stable for r = + , then the origin is said to
be asymptotically
stable in-the-large, or globally asymptotically
stable. In
other words, x(t) approaches zero as t tends to infinity for all x(t ).
Since not all networks are stable, the concept of instability must be
made precise; this is done by the next definition.
0
Definition.
The origin is unstable, if for some R < E and any r R there
exists at least one trajectory originating in S that crosses B R .
r
(139)
* A region is said to be open if it contains none of its boundary points.
Sec. 10.7]
LIAPUNOV STABILITY
787
The last form follows since dx/dt = f(x, t) along a trajectory of the network.
Thus it is meaningful to talk about the time rate of change of V along a
trajectory of the network. A very important class of functions is defined
on the basis of the sign of this rate of change.
Definition.
A positive definite function V(x, t) is called a Liapunov
function ifdV/dt 0 along trajectories in D,
On the basis of these definitions we can now discuss the question of
stability and the conditions under which a singular point is stable.
STABILITY
THEOREMS
Fig. 15.
Liapunov function.
curves are projected vertically into the x x plane, they will form a set of
concentric constant-V contours with the value of V decreasing toward
zero as the contours are crossed in approaching the origin. This is illustra
ted in Fig. 15. Since V is a Liapunov function, V must be nonincreasing
along a trajectory. Hence a trajectory originating inside the contour
V(x) = C can never cross that contour. The trajectory is thus constrained
to a neighborhood of the origin. This is very nearly the condition laid
1
788
[Ch. 10
(140)
Because x was a vector x of least norm for which U(x) = C and because
V(x, t) U(x) for all t t0, V(x, t ) < C for all x such that ||x|| < ||x || = r.
Thus for all x(t ) in S ,
r
(141)
Next, because dV/dt 0,
(142)
Clearly, (140) is in contradiction with (141) and (142). Thus t does not
exist and x is contained in S .
This proof can be given the following geometrical interpretation when
x is a 2-vector, as illustrated in Fig. 16. The relation W(x) = C defines a
closed contour K contained in S plus its boundary B . The relation
U(x) = C defines a closed contour K contained in the closed region
1
* Recall from the definition for a time-varying positive definite function V(x, t)
that W(x) is a time-invariant positive definite function such that V(x, t) > W(x) for all
t
Sec. 10.7]
LIAPUNOV STABILITY
Fig. 16.
789
Theorem
18. The origin is asymptotically stable if in an open region D
containing the origin there exists a Liapunov function such that V(x, t) U(x)
for all t t , where U(x) is a positive definite function, and such that
dV/dt is positive definite.
0
790
[Ch. 10
t t0. Let w be the least value of W(x) for x such that /2 ||x||
Then, under the restriction /2 ||x|| R,
R.
Theorem
19. The origin is asymptotically stable in-the-large if there exists
a Liapunov function defined everywhere (the state equation has a solution)
such that V(, t) U(x) for all t t , where U(x) is a positive definite
function, such that W(x) is radially unbounded* and such that dV/dt is
positive definite.
0
The proof does not depart very much from those given for the previous
two theorems. Therefore it is left for you as an exercise.
Examples
Let us now illustrate some of the preceding theory with examples.
Consider the network shown in Fig. 17. It is described by the state
equation
Sec. 10.7]
LIAPUNOV STABILITY
Fig. 17.
791
Nonlinear network.
Though we shall not do so, it can be shown that the solution of this
state equation exists and is unique for all initial-state vectors x(t );
that is, the positive scalar E that appears in the stability definitions
is infinite.
A decision as to stability requires the discovery of a Liapunov function.
There is no algorithm that can be followed to arrive at one. Experience is
a guide; so, to gain some experience, let us see if the positive-definite
function
0
The second line is obtained by substituting dxi/dt and dx /dt from the
state equation. Clearly dV/dt is positive definite. Thus V is a Liapunov
function that, furthermore, satisfies the additional restrictions of the
theorem on asymptotic stability in-the-large. You may feel we forgot to
consider the problem of selecting U(x); however, a moment's reflection
will convince you that, when V is time invariant, the supplemental
condition calling for the existence of a U(x) with the properties indicated
is trivially satisfied by setting U(x) = V(x). We now know that the origin
is asymptotically stable in-the-large.
It is not always possible to establish asymptotic stability in-the-large.
Often only local stability properties can be certified. As an example,
2
792
Fig. 18.
[Ch. 10
consider the network illustrated in Fig. 18. It has the state equation
for which a unique solution may be shown to exist for all initial-state
vectors. We shall use the same positive-definite function as before to be a
trial Liapunov function; that is,
The second line follows by substitution of dx /dt and dx /dt from the
state equation. Observe that dV/dt is positive definite if 6x tanh x
x
is positive for x 0. This is approximately equivalent to |x |
< 5.99999. Now, in the open region |x | < 5.99999 (all values of x are
allowed) dV/dt is bounded from below by the positive-definite function
1
which we may take to be W(x). This is just one W(x) function; other
positive-definite functions m a y also be found and used for W(x).
All the conditions of the theorem on asymptotic stability have been
satisfied; therefore the origin is locally asymptotically stable. In fact, it
may be shown, though we shall not do so, that all trajectories originating
inside the disk of radius 5.99999 in the x x plane approach the origin
asymptotically.
1
Sec. 10.7]
LIAPUNOV STABILITY
INSTABILITY
793
THEOREM
(a)
(b)
(c)
Dr
Points common
to Dr and SR
Fig. 19.
helpful in the
For any arbi
in both S and
exists a region
Unstable origin.
794
[Dh. 10
contained entirely within D such that U(x) is greater than V[x(t ), t ] > 0;
denote it by D . Let w > 0 denote the greatest lower bound assumed by
W(x) at the points common to both D and S . Now
0
Negative
converter
Fig. 20.
Sec. 10.7]
LIAPUNOV STABILITY
and dx /dt,
2
795
we get
The second line was obtained from the first b y adding and subtracting
x . Clearly,
2
LIAPUNOV FUNCTION
CONSTRUCTION
Observe the nature of the stability theorems. We are not given a definite
prescription to follow; that is, there is no defined set of steps at the end of
which we can reach an unambiguous conclusion as to whether or not the
network is stable. Rather, the theorems provide a "hunting license."
They ask that we seek a Liapunov function whose value remains bounded
by a time-invariant positive-definite function. Finding an appropriate
Liapunov function is a creative, inductive act, not a deductive one.
The functional form of a Liapunov function is not rigidly fixed. On the
one hand, this is an advantage, because it affords a greater opportunity to
establish stability by trying numerous potential Liapunov functions. On
the other hand, it is a disadvantage, because there are no guidelines in
picking a potential Liapunov function from the countless positive-definite
functions one can think of. We shall now discuss a particular Liapunov
function for time-invariant networks and establish, as a consequence, an
alternative set of conditions for stability. Following this, we shall discuss
a method of generating Liapunov functions.
Suppose that the network under consideration is described by the timeinvariant state equation
(144)
Let us make explicit note of the fact that f(x) is a real-valued vector
function of x. In searching for a possible Liapunov function, consider
(145)
796
[Ch. 10
(146)
(147)
(148)
Sec. 10.7]
LIAPUNOV STABILITY
797
Negative
converter
Fig. 21.
Upon adding F'(x) to F(x) and taking the negative of the resulting matrix,
we find
* See Section 7.2 for a discussion of the conditions for positive definiteness of a matrix.
Then see Problem 18 in Chapter 7 for the particular criterion for positive definiteness
used here.
798
[Ch. 10
(149)
where
is the gradient of V. From (149) it is clear that the sign of dV/dt is deter
mined by the sign of the gradient of V, since f(x, t) is known. Hence,
instead of looking for a Liapunov function V the sign of whose derivative
will be suitable, we can look for a gradient function V V that, in (149),
makes dV/dt have the appropriate sign. Then the Liapunov function V
itself can be determined by the line integral of the gradient from 0 to x:
(150)
Sec. 10.7]
LIAPUNOV STABILITY
799
(151)
(152)
As you see, the problem of finding a Liapunov function b y picking an
arbitrary positive-definite function V(x), and then ascertaining whether or
not dV/dt is non-negative or positive definite, has been replaced b y pick
ing a function V V(x) such that dV/dt is non-negative or positive definite,
as determined from (149), and D(x) is symmetric, and then ascertaining
whether or not V(x) is positive definite. Usually there is less guesswork
involved in using this latter method, which is known as the variablegradient method. However, this method of finding a Liapunov function is
not really decisive; if the V(x) found from the selected gradient function
does not turn out to be positive definite, it cannot be concluded that the
* The requirement that D(x) be symmetric is equivalent to requiring that the curl
of VF(x) be zero. It is a known theorem of vector analysis that, when the curl of a vector
is zero, the vector is the gradient of a scalar function. For further discussion see H. Lass,
Vector and Tensor Analysis, McGraw-Hill Book Co., New York, 1950, p. 297.
800
[Ch. 10
origin is not stable in some sense. It only means that a suitable Liapunov
function has not yet been found.
It is usual practice to start by selecting the gradient of V to be of the
form
(153)
ij
Negative
converter
k= 1
Fig. 22.
Nonlinear network.
Sec. 10.7]
Let us pick
LIAPUNOV STABILITY
1 1
801
Let
2 1
= b and
2 2
and
1 2
1 2
802
[Ch. 10
Before taking the line integral of VV(x), we must verify that dV/dt is
positive definite for the chosen and , and the resultant derived
by the symmetry condition of D(x). To verify that dV/dt is positive
definite, we must show that the two previously stated inequalities hold.
First = b, so is positive. Second,
2 1
2 1
2 2
1 2
2 1
1 2
2 2
is
which you may easily verify is positive definite provided a > 0, a condi
tion already imposed, and a + ab > 2 b . This latter inequality is satisfied
when the previous requirement a > 2b is satisfied. Thus, for a > 2b, a
Liapunov function has been constructed such that dV/dt is positive
definite in a suitably small neighborhood of the origin. This implies that
the origin is asymptotically stable.
In this section we have introduced the basic concepts of Liapunov
stability relative to a singular point, have proven some basic theorems on
stability, and have given two methods by which to guide a search for a
Liapunov function. Much more is known. The concept of stability can be
extended in a useful way to stability relative to a set of points. More
sophisticated theorems on stability and instability exist. Also, other
guidelines for selecting a Liapunov function for specific classes of problems
are known. These advanced topics are treated in the books on stability
listed in the bibliography.
2
PROBLEMS
803
PROBLEMS
1. Derive the state equations for the time-varying networks shown in Fig.
P1 by using the state vector of (5). Repeat with the state vector of (10)
Assume the element parameters are in Farads, Ohms, or Henrys.
(a)
(b)
(c)
Fig. P1
2 . Suppose
804
[Ch. 10
matrices A(t) commutes with its integral A() d for all t t0:
J to
(a)
(b)
(d)
(c)
4. For those matrices A(t) of Problem 3 that commute with their integral
J^A()
as a matrix.
5. Which of the following state equations, with t0 = 0, have a solution
in the extended sense? Of those that do, which have a solution in the
ordinary sense? Of those that have a solution only in the extended sense,
indicate whether or not the associated homogeneous equation has a
solution in the ordinary sense.
(a)
(b)
PROBLEMS
(c)
(d)
(e)
6.
Q(t)e
p t
when
(a)
(b)
(c)
J ^0
805
806
[Ch. 10
(a)
(b)
(c)
(d)
A t
(a)
PROBLEMS
807
(b)
(c)
(d)
(a)
(b)
(c)
(a)
808
[Ch. 10
(b)
(c)
(d)
(e
(f)
15. Verify that the network illustrated in Fig. 8 has the state equation
given on p. 743.
16. Consider the following state equations:
(a)
(b)
PROBLEMS
809
17. Let Y(t) Y() be the state-transition matrix associated with (71).
Prove the theorem obtained after replacing condition (72a) in Theorem
11 by
Does this new theorem apply to the example following Theorem 11?
If so, find a value for . This new theorem is less restrictive than
Theorem 11, but it may be more difficult to apply. Explain why.
18. Show that the state equations in the examples of Section 10.7 are those
for the networks illustrated in Figs. 17, 18, 20, 21, and 22.
19. Derive the state equations for each of the networks in Fig. P19.
(a)
(b)
(c)
Fig. P19
810
[Ch. 10
2 0 . Derive the state equations for the amplifier depicted in Fig. P20a. Use
the transistor model shown in Fig. P20b.
(a)
(b)
Fig. P20
21. Derive the state equations for the amplifier depicted in Fig. P21. Use
the transistor model shown in Fig. P20b.
22. Derive the state equations for the amplifier shown in Fig. P22. Use
the transistor model depicted in Fig. P20b.
23. Derive the state equations for the amplifier depicted in Fig. P23. Use
the transistor model shown in Fig. P20b.
24. (a) Show that f in (81a) must be identically zero or that f and g
in (85) must be differentiable functions in order to express v and
v as functions of x, e, and de/dt.
(b) Show that f , in (75b), must be identically zero or that f and
g in (79) must be differentiable functions in order to express i and
i as functions of x, e, and de/dt.
Lt
Lt
L l
L l
L t
Cl
C t
C l
Cl
C t
PROBLEMS
811
Fig. P21
Fig. P22
Fig. P23
25. Consider the networks that can be represented as the interconnection
of a capacitor subnetwork, an inductor subnetwork, and a resistor and
independent source subnetwork as shown in Fig. P25a. Formulate the
state equation in terms of the port parameters of each subnetwork
812
[Ch. 10
Capacitor
subnetwork
Resistor and
independent
source
subnetwork
Inductor
subnetwork
Capacitor
subnetwork
Resistor and
independent
source
subnetwork
Inductor
subnetwork
(a)
(b)
Fig. P25
26. Using Theorem 13, determine which of the following state equations
has a solution for all t 2.
(a)
PROBLEMS
813
(b)
(c)
(d)
(e)
28. Using the theorem proved in Problem 27, determine which of the follow
ing state equations has a solution defined for all t 1:
(a)
(c)
(b)
814
[Ch. 10
(d)
(e)
30. Prove that f (x, t) satisfies a Lipschitz condition if the partial derivatives
dfi (x, t)/dxj exist, are continuous in x for almost all t t0, and are
bounded in magnitude in some neighborhood of every point by nonnegative locally integrable functions of t t . Show by counterexample
that these conditions are not necessary and therefore are only sufficient.
0
31. Verify that each of the functions f (x, t) in Problems 28 and 29 satisfies
a Lipschitz condition in the neighborhood ||x x|| < 1 of the arbitrary
point x by finding a (t) that satisfies the conditions set forth in
Theorem 14.
32. For each of the following state equations, verify whether the conditions
of Theorems 15 and/or 16 are satisfied. Set t = 1.
0
(a)
PROBLEMS
815
(b)
(c)
(d)
(e)
with an initial vector x(t ) such that ||x(t )|| < , which is a constant
depending on f (x, t), are bounded as t tends to infinity, if
0
816
[Ch. 10
-1
(a) When A has distinct real eigenvalues and, in some cases, equal real
eigenvalues, there exists a P such that
PROBLEMS
817
(a)
(b)
(d)
(c)
818
[Ch. 10
Solve for the elements of P in terms of the elements of A and the real
and imaginary parts, and , of the eigenvalues,
(b) Polar coordinates in the z z plane are r and , where r = z
+ z and tan = z /z . Verify that
2
(c) Solve for r and and sketch a typical set of trajectories in the
z z plane, when (i) is positive and (ii) is negative.
(d) For
1
40. With reference to Problem 36: The singular point x is called (i) a node
if A has distinct, real eigenvalues of the same sign, (ii) a log-node if
A has equal, real eigenvalues, (iii) a saddle if A has real eigenvalues of
opposite sign, and (iv) a focus if A has conjugate complex eigenvalues
with nonzero real part. The singular point is said to be stable, if both
eigenvalues are negative real or have negative real part; otherwise, the
singular point is said to be unstable. Let = det A and = tr A.
Divide the plane into regions corresponding to the different
classifications for a singular point such as stable node.
s
41. With reference to Problem 36: The x x plane is known as the phase
plane, and a typical set of trajectories in the phase plane is called a
phase portrait. For each of the following state equations find the singular
points and, using the results of Problems 36 through 40, sketch a phase
1
PROBLEMS
819
(a)
(b)
(d)
(c)
(e)
Sketch a phase portrait valid throughout the phase plane using the
local phase portraits just completed. You will find it useful to keep in
mind that x , for i = 1 and 2, is increasing along a trajectory in that
region of the phase plane where f (x) > 0 and decreasing along a tra
jectory in that region of the phase plane where f (x) < 0. Furthermore,
the trajectory is vertical [horizontal] where it crosses the curve
f (x) = 0[f (x) = 0].
i
(b)
(c)
(d)
(e)
(f)
820
[Ch. 10
Next, obtain the solutions by using the modified Euler method. Assume
three-digit accuracy is adequate. For the state equations in (c) and (d)
compare the two numerical solutions and the exact solutions
evaluated at ih.
43. Repeat Problem 42 by using the Adams method with j = 3 and the
modified Adams method with j = 3 in place of Euler's method and the
modified Euler method, respectively. Obtain starting values by using
a truncated Taylor series.
44. Obtain a numerical solution of the following state equation by using
the Adams method with j = 1, 2, 3, 4, 5. Set h = 0.1 and compute
x(ih) for i = 1, 2, ..., 15. Obtain the starting values by using a truncated
Taylor series. Compare the numerical solutions and the exact solution
evaluated at ih.
Next, obtain the solution by using the corrector in the modified Euler
method and (122), with j = 1 and l = 1 as the predictor. Obtain the
starting values by using the preceding numerical solution. Compare the
number of times the corrector had to be applied at each time step in
each of the two methods.
47. Create three predictor-corrector pairs from (122) and (126) and use them
to obtain a numerical solution of the following state equations. Set h = 0.1
and compute x(ih) for i = 1, 2, ..., 10. Assume three-digit accuracy is
adequate. Obtain starting values by using a truncated Taylor series.
(a)
(b)
PROBLEMS
821
then
ij
x(h-h)
i+1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
4 9 . By using each of the predictors in Problem 48, obtain a numerical solution
of the following state equations. Set h = 0.2 and compute x(ih) for
i = 1, 2, ..., 10. Assume three-digit accuracy is adequate. Obtain
starting values using a truncated Taylor series.
(a)
(b)
then
822
[Ch. 10
ij
ik
i+1
Let t
i+1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(n)
(0)
where x ( t )
denotes the nth iterate; the zero-th iterate x ( t )
is
the solution of a predictor. An alternate method for solving the corrector
uses the well known Newton-Raphson iteration; in this case the
iterates are given by
i+1
i+1
(n-1)
where g ( x ( t )
) is the Jacobian matrix of g(x) evaluated at the
(n l)-st iterate for x(t ).* This latter method for solving the
x
i+1
i+1
PROBLEMS
823
(ii)
(iii)
(iv)
(0)
(1)
(1)
(1)
(1)
(i)
i+1
(1)
(0)
( 0 )
.]
(0)
(c) Repeat (a) using the solution of the Euler predictor for x ( t ) ;
that is
i+1
(0)
(i)
(ii)
(iii)
(iv)
824
[Ch. 10
+1
i+1
i+1
Set h + = 2h + if
i
Set h + = h + if
i
i+1
PROBLEMS
i+1
825
54. Which of the following scalar functions are positive definite for all x?
Let t = 1.
0
(a)
(b)
(c)
(d)
(e)
(f)
55. Consider a linear, time-varying RLC network with no external excitation.
The state equation (not in normal form) is
(a)
(b)
(c)
826
[Ch. 10
(a)
(b)
(c)
(d)
59. Describe how the stability theorems 17, 18, and 19 establish that a
solution of the state equation 98, equal to x(t ) at time t , may be
extended to yield a defined solution for all t t . Indicate any
restrictions on x(t ). Find a state equation, of order greater than 1, to
which these results apply and to which Theorem 13 does not apply.
0
PROBLEMS
827
(a)
(b)
(c)
61. Apply Theorem 21 to each of the state equations in Problem 58. How
do the results, here obtained with Theorem 21, compare with those
obtained in Problem 58?
62. Use Theorem 21 to show that the origin is asymptotically stable in-thelarge when
(a)
(b)
828
[Ch. 10
(c)
64.* Prepare a program flow chart and a set of program instructions, in some
user language such as FORTRAN IV, for a digital computer to obtain a
numerical solution of the state equation 104 by the
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Euler method
Modified Euler method
Adams method (j = 4)
Modified Adams method (j = 4)
Milne method
Predictor-corrector pairs of Problem 47
Predictors of Problem 48
Predictor-corrector pairs of Problem 50
Appendix
GENERALIZED FUNCTIONS
830
GENERALIZED FUNCTIONS
[App. 1
Sec. A.1.1]
831
A1.1
where a {0} and b are known functions. Since the convolution of functions
has the same algebraic properties as scalar multiplicationassociativity,
commutativity, and so onit is tempting to suppose that there exists a
uniquely defined inverse operation to convolution, which permits us to
express the solution as z = b//a; that is, the convolution quotient of b by a.
(Observe the double slant line used as the symbol.) We shall show that the
notion of a convolution quotient is meaningful.
To set the background for defining convolution quotients, let us turn
again to the algebraic equation (1), when and are known integers, and
consider its solution. It is the uniqueness of that solution that gives
meaning to the quotient, /, by which the solution is expressed. Suppose
(1) has more than one solution; then let i and denote two distinct
solutions. Since oci and both equal , then a( ) = 0. But is
assumed to be a nonzero integer; hence i = 0 and, b y contradiction,
the solution of (1) must be unique. If the solution is an integer, then that
is the value assigned to /. Of course, (1) may not have an integer solu
tion. However, because the solution is unique, we could say that / is
the quantity, called a quotient, that denotes the unique solution; in this
way / is made meaningful. In fact, history has given the name rational
number to the quotient of two integers. Now, / is a rational-number
solution of = that is unique to within the equivalence / = /,
where is any nonzero integer.
2
832
GENERALIZED FUNCTIONS
[App. 1
If we can show that it has a unique solution, then that solution ascribes
meaning to the convolution quotient, 6 / , that expresses the solution.
Suppose (2) has more than one solution; then let z and z be two distinct
solutions. Since a * z and a * z both equal b, then a * (z z ) = {0}.
There is a theorem due to Titchmarsh* that states the following:
1
Theorem
I . If f and g are continuous and f * g = {0}, then either f = {0}
or g = {0} or both.
Since the function a {0} in the previous result a * (z z ) = {0}, then
z z must be the zero function {0}. But this is a contradiction of the
assumption that z and z are distinct. Hence the solution of (2) must be
unique.
If the solution of (2) is a continuous function, then b//a is identified
with that function. Of course, there may be no continuous function that
solves (2), just as = may not have an integer solution; for example,
if b(0) 0, then the solution cannot be a continuous function. (Why?)
However, because the solution is unique, we could extend the meaning of
the term " f u n c t i o n " by saying b//a is the quantity, called a convolution
quotient, that denotes the unique solution. In this way b//a becomes
meaningful.
N o w suppose the function c {0} and consider the convolution equation
1
(3)
or, equivalently,
Sec. A1.2]
833
A1.2
(5b)
(5 c)
You should verify that the generalized functions on the right are indepen
dent of the specific convolution quotients that characterize each of the
generalized functions on the left; for example, suppose b//a = b''//a' and
d/c =
d'/c';then
and
834
GENERALIZED FUNCTIONS
[App. 1
(6a)
(6b)
(6c)
(6d)
(6e)
(6f)
(6g)
(6h)
(6i)
Sec. A1.2]
835
The cancellation law is also valid; that is, if e {0} and f {0}, then
(7)
(8a)
(8b)
(8c)
(9)
836
GENERALIZED FUNCTIONS
[App. 1
with (9) yields (b//a) * (x//y x'//y') = {0}/a or x//y = x'//y', since b//a
{0}//a. Thus a * d//b * c is the unique, generalized-function solution of (9).
The solution of (9) indicates that the convolution quotient of the general
ized function d//c by the generalized function b//a, that is, (d//c)//(b//a),
has the following meaning:
(10)
(ii)
n
(12a)
(12b)
(12c)
A1.3
Sec. A1.3]
837
(13)
The verification of condition 2 is left to you; however, to illustrate how this
is done, we shall verify that addition of scalars has its counterpart. If
<-> a//a and <-> a / a , then
(14)
We leave the task of verifying condition 2 to you; however, we shall
illustrate how this is done. If b<->b * a//a and c<->c * a/a, then
838
GENERALIZED FUNCTIONS
[App. 1
CERTAIN CONTINUOUS
FUNCTIONS
A function that will appear often is the constant function {1}, which we
shall denote henceforth by u; that is,
(15)
This is the well-known unit step function with its value specified as 1 at
t = 0 as well as at t > 0.
Repeated convolution of u with itself yields an often-encountered set of
functions. Thus
(16)
(17)
Sec. A1.3]
839
(18a)
(18b)
(18c)
or, equivalently,
(19)
Y o u should verify this integral relation, thus verifying (18b).
The one-to-one relationships between continuous functions and general
ized functions, such as given in (18), are useful in solving some convolution
equations. As an example, consider
Now, by (18c) we know that this is the same as the continuous function
{cos t}.
840
GENERALIZED FUNCTIONS
LOCALLY I N T E G R A B L E
[App. 1
FUNCTIONS
Sec. A1.3]
841
be denoted b y u , is defined as
(21)
where
(22)
Although when first introduced in (17), n was a positive integer, we can
now set n = 0 and find
(23)
(24a)
842
GENERALIZED FUNCTIONS
[App. 1
(24b)
(24c)
We shall discuss one of these and leave to you the examination of the
others. Focus attention on the right side of (24c). It is the generalizedfunction solution of the convolution equation
(25)
There is a term in this equation unlike any we have previously encoun
tered, u * z. From (23) we see that u * z, in more precise terms, is u * z//u.
Now, if z is a continuous or locally integrable function, u * z//u is, by (14)
or (20), just z.* With this in mind, the convolution equation, defined on
the set of continuous or locally integrable functions, corresponding to
(25) is
0
(26)
B y (24c) we know z(t) = cos t is the solution of (26). The other relations
in (24) may be examined in a similar manner.
A1.4
* Note that this is the same result as that obtained by convolving an impulse function
with z. We shall discuss this identification of u with the impulse function in the next
section.
0
Sec. A1.4]
843
(28)
Note that (28) has the same solution, z(t) = cos t, as the convolution
equation (26) that corresponds to (25).
Before going on, let us use this idea of u being an n-fold integration
to help assign a value to a generalized function at time t. Let a denote an
arbitrary generalized function. Suppose u * a stands in a one-to-one
relation with the ordinary function b, which possesses an nth derivative
for <t < . Then, for <t < , we shall assign the value b (t) to
a at time t. The value assigned to a in this manner is unique.*
As an example, take a = u and determine a value to be assigned to it.
Now u * u =u //u
is by (14) the same as the continuous function u,
which is differentiable for 0 <t. Since du(t)/dt = 0 for t > 0, we assign
the value zero to u for t > 0.
N o w we turn to an interpretation of u for negative n. Alternatively,
we shall examine u- for positive n. For convenience, let p = u- . N e x t
suppose that a is a function that is n 1 times continuously differentiable
and that possesses a locally integrable nth derivative. Then
n
(n)
(29)
(k)
where a
denotes the kth derivative of a with respect to t. This relation
is easily established by induction. We shall start the proof and leave its
completion to you. Let n = 1; then (29) becomes
844
GENERALIZED FUNCTIONS
[App. 1
or, equivalently,
or
(1)
which is true if a
is locally integrable. Thus (29) is valid for n = 1;
the proof is completed by induction.
If a is sufficiently differentiable and (0) = a (0) = ... = a (0)
= 0,
(29) shows that the generalized function p * a stands in a one-to-one
relation with the function a . Hence the generalized function p should
be viewed as a differential operator. If a is not sufficiently differentiable
or if one or more of the a (0), k = 1, ..., n1, are not zero, then
p * a does not stand in a one-to-one relation with an ordinary function.
In this case p * a exists only as a generalized function; we shall refer to
p * a as the generalized nth derivative of a.
Let us now apply some of these results to an example. The ordinary
differential equation
(1)
(n-1)
(n)
(:)
(30)
Sec. A1.4]
845
which, by (24a) is the same as 2 { e ) . This agrees with our knowledge that
the solution of the differential equation (30) is 2 { e } for z(0) = 2.
In finding the ordinary function standing in a one-to-one relation with
the generalized function in (31), it was useful to have had the relation
in (24a). In the solution of other differential equations, which will be
considered in the next section, it would be helpful to have relationships
between generalized functions expressed in terms of p and the corre
sponding ordinary functions. Such relationships are shown in Table 1.
- t
Table 1.
Ordinary
Function
Generalized
Function
Ordinary
Function
Generalized
Function
846
GENERALIZED FUNCTIONS
THE IMPULSE
[App. 1
FUNCTION
Most of the preceding effort has been devoted to considering the general
ized functions that have a one-to-one relationship with ordinary functions
or the properties of operators on ordinary functions. We shall now turn to
a consideration of the relationship of generalized functions to the impulse
function and its derivatives. We shall find that p = u is properly inter
preted as the impulse function. To verify this, let a be a continuous func
tion; then
0
(32)
This is what is called the sifting property associated with the impulse
function .
The generalized function p is properly interpreted as the first derivative
of the impulse function. To show this, let the function a possess a continu
ous first derivative. Then by using (29) we have
(34)
If u(0) 0, p * a is a generalized function that does not stand in a one-toone relation with an ordinary function. However, as previously shown,
p can be assigned the value 0 for all t > 0. This is a useful fact. Since p is
the generalized first derivative of p , which is interpreted as the impulse
function, let denote a symbolic function corresponding to p; then for
t > 0 (34) is equivalent in a formal sense to
0
( 1 )
(35)
n
Sec. A1.5]
INTEGRODIFFERENTIAL EQUATIONS
847
A1.5
INTEGRODIFFERENTIAL EQUATIONS
848
GENERALIZED FUNCTIONS
[App. 1
or, equivalently,
(38)
(39)
In the event that l n, it is easy to verify that scalars 0, ..., l-n and
v , ..., v
exist such that
0
n 1
(40)
n
is factorable as
(41)
Sec. A1.5]
INTEGRODIFFERENTIAL EQUATIONS
n
n - 1
n 1
2
849
(42)
Assume that the 1 are distinct; the right side may then be expressed as a
sum of convolution quotients. Thus
(43)
where
(44)
The right side of (43) is the partial-fraction expansion of the left side,
when the are distinct. You should verify (44). If the are not distinct,
the partial-fraction expansion is more complicated and cumbersome in
notation. However it is quite similar to the ordinary variable case. We
shall not consider this case in detail. To complete the case of distinct ,
substitution of (43) into (42) and of that result into (40) establishes the
following:
i
(45)
1 - n
_ i t
850
GENERALIZED FUNCTIONS
[App. 1
The last line is obtained from Table 1. B y substituting this result in (38),
we get
A1.6
Sec. A1.6]
851
some right half-plane to the ratio of two functions that are bounded and
regular in that right half-plane; for example, l/(s + 2) and (s + l)/(s + 3)
are bounded, regular functions of s for Re (s) > 2 ; their ratio [l/(s + 2)]/
[(s + l)/(s + 3)] = (s + 3)/(s + l)(s + 2) is regular, except at the pole
s = 1, for Re (s) > 2 . Thus (s + 3)/(s + l)(s + 2) is a function in M.
Next, let C denote the set of all continuous functions c such that
(46)
- 1
2.
functions
G.
The relation
-1
generalized function
in G. Furthermore if a//b is a
in G, then
(48)
NOW let I denote the set of all locally integrable functions b such that
852
GENERALIZED FUNCTIONS
[App. 1
the right half-plane Re (s) > e for any > 0. Thus u * b//u is in G. There
fore
(49)
n+1
Since u and u
n+1
u//u .
=
l/s
n + 1
( n )
(n)
Appendix
THEORY OF FUNCTIONS
OF A COMPLEX VARIABLE
A2.1
ANALYTIC FUNCTIONS
854
[App. 2
(1)
We may interpret this statement in the complex plane as follows. Let
> 0 be a given number. We consider a circular neighborhood of F(s )
as in Fig. 1, where all the points within the circle of radius around
0
Fig. 1.
(2)
exists and is finite.
Implicit in this definition is the assumption that s may approach so in
any direction, or may spiral into it, or follow any other path. The limit
in (2) must exist (and be unique) independently of how s approaches s .
It is this fact that makes differentiability in the complex plane a very
strong requirement. In consequence, differentiable functions of a complex
variable are extremely "well-behaved," as contrasted with real functions,
which can be "pathological."
It can be shown (this is only one of many " i t can be s h o w n ' s " that we
shall meet in this appendix) that the usual rules for derivatives of sums,
products, quotients, etc., carry over from the real case, with no changes.
0
Sec. A2.1]
ANALYTIC FUNCTIONS
855
So does the chain rule for the function of a function; and all the familiar
functions have the same derivatives as on the real line, except that the
variable is now complex. We summarize these results below.
Let F1(s) and F (s) be two differentiable functions. Then
2
(3)
(4)
(5)
(6)
(7)
If a function F of a complex variable is differentiable at the point s
and at all points in a neighborhood of s , we say that F(s) is regular at s .
Notice that the statement "F(s) is regular at s " is a very much
stronger statement than "F(s) is differentiable at s ." A function F(s)
that has at least one regular point (i.e., a point at which the function is
regular) in the complex plane is called an analytic function. A point s
at which the analytic function F(s) is not regular is a singular point of the
function. F(s) is said to have a singularity at s . In particular, a point
at which the derivative does not exist is a singular point.
Although the requirement of regularity is a very strong condition and
therefore the class of analytic functions is a " v e r y s m a l l " subset of the
set of all functions, almost all functions that we meet in physical applica
tions are analytic functions. An example of a nonanalytic function is
|s| . This function has a derivative at s = 0 and nowhere else. Hence it has
no regular points. The function s(= j) is another simple example of
a nonanalytic function. The function F(s) = l/(s 1) is a simple example
of an analytic function. Its region of regularity consists of the whole plane
exclusive of the point s = 1. The point s = 1 is a singular point of this
function
The singularities of an analytic function are extremely important, as
we shall see. For the present we can only distinguish between two kinds
of singularities. The point so is an isolated singularity of F(s), if so is a
0
856
[App. 2
(8)
The denominator becomes zero whenever
(9)
and so these points are singular points of F(s). The origin is a limit point of
these singularities.
The famous French mathematician Augustin Cauchy (who originated
about half of complex function theory) gave the following necessary and
sufficient condition for the differentiability of a function of a complex
variable: The function
(10a)
(10b)
at this point.
The necessity is proved by letting s approach s in (2) by first letting
approach and then letting approach for one computation, and
reversing the order for another computation. Equating the two derivatives
0
Sec. A2.2]
MAPPING
857
(11a)
(11b)
Thus the real and imaginary parts of an analytic function are harmonic
functions. The converse of this statement is also true. Every harmonic
function (in two dimensions) is the real part of an analytic function, and
the imaginary part of another analytic function. This fact makes analytic
functions of considerable interest in two-dimensional potential theory.
A2.2
MAPPING
858
[App. 2
s-plane
2
s -plane
Fig. 2.
MAPPING
Sec. A2.2]
859
(12)
(13)
Since this derivative exists, we may take the limit along C1, and since the
derivative is nonzero, we may write
(14)
Then from (13),
(15a)
and
(15b)
The point F(s) is on the curve C'1, which is the image of C\ under the
mapping F(s). Thus the left side of (16) is the angle of the tangent to
C'1 at F(so). Thus from (16), the curve C'1 has a definite tangent at F(so),
making an angle + 1 with the positive real axis. An identical argument
gives the angle of the tangent to C at F(so) to be + . Thus the angle
between the two tangents, taken from C[ to C , is ( 1), which is the
2
860
[App. 2
same (in magnitude and sign) as the angle between the curves C1 and C
at so measured from C1 to C .
Incidentally, we see from (15a) that the local magnification (i.e., the
increase in linear distance near s ) is independent of direction and is given
b y the magnitude of the derivative. Thus, locally, the mapping by an
analytic function [when F'(s )0]
produces a linear magnification
|F'(s )| and a rotation arg F'(s ), thus preserving shapes of small figures.
An auxiliary consequence is that the images of smooth curves are also
smooth curves; that is, they cannot have "corners."
We have not yet defined some point-set-topological concepts about
regions and curves that are really needed to clarify the earlier discussions.
Let us proceed to rectify this omission, although we cannot be completely
precise without introducing very complex ideas, which we do not propose
to do. Therefore we shall take a few concepts such as path, continuous
curve, etc., to be intuitively obvious.
A simple arc is a continuous path in the complex plane that has no
crossover or multiple points. A simple closed curve is a path in the complex
plane that, if cut at any one point, becomes a simple arc. If the end points
of a simple arc are joined, we form a simple closed curve.
An open region is a set of points in the complex plane each of which
has a neighborhood all of whose points belong to the set. The region
" i n s i d e " a simple closed curve, not counting the curve itself, is an
example. If we add the points on the boundary of an open set to the open
set itself, the combined region is called a closed region. An open or closed
region is said to be connected if any two points in the region can be con
nected by a line all points on which are in the region.
In the preceding paragraph the word " i n s i d e " was put in quotation
marks. Although we have a strong intuitive feeling that the inside of a
closed curve is well defined, nevertheless this requires a proof. The
Jordan curve theorem gives the desired result. It states that every simple
closed curve divides the complex plane into two regions, an "inside" and an
"outside" the curve itself being the boundary of these two regions. If we start
at some point on the curve and traverse it in a counterclockwise sense,
the region to the left of the curve will be called the inside; that to the right,
the outside.
If we do not permit a closed curve to pass through infinity, then the
" i n s i d e " region, as just defined, will be bounded; that is, all points in the
region will satisfy the condition |s| M, where M is a fixed positive
number. On the other hand, if the closed curve goes through infinity,
then neither the inside nor the outside is bounded.
The question arises as to what is meant by a closed curve passing
through infinity. The path consisting of the imaginary axis, for example,
2
Sec. A2.2]
MAPPING
861
is such a curve. But this may appear to be a simple arc rather than a
closed curve. The Reimann sphere will serve to clarify this point.
Consider a sphere placed on the complex plane with its " s o u t h p o l e " at
the origin, as illustrated in Fig. 3. Now consider joining by a straight line
North pole
South pole
Fig. 3.
each point in the plane to the "north p o l e " of the sphere. These lines will
all intersect the sphere, thus setting up a one-to-one correspondence
between the points in the plane and those on the sphere. Each point in the
finite plane will have its counterpart on the sphere. As we go further and
further away from the origin of the plane in any direction, the point of
intersection of the lines with the sphere will approach closer and closer to
the north pole. Thus the north pole corresponds to infinity. On the sphere
infinity appears to be a unique point. Both the real and the imaginary
axes become great circles on the sphere, and a great circle appears like a
simple closed curve.
The concept of the Reimann sphere serves another purpose; it permits
us to look upon "infinity" as a single point, whenever this is convenient.
We refer to infinity as the point at infinity.
Very often we wish to talk about the behavior of a function at the point
infinity. A convention in mathematics is that no statement containing the
word " i n f i n i t y " is to be considered meaningful unless the whole statement
can be defined without using this word. This convention is introduced
to avoid many inconsistencies that would otherwise arise. The behavior of
a function at the point infinity is defined as follows.
That behavior is assigned to the function F(s) at s = , as is exhibited
by the function
862
[App. 2
A2.3
INTEGRATION
Fig. 4.
two points P1 and P are connected by a simple arc C. The path is divided
into intervals by the points s ; the chords* joining these points are labeled
s . Suppose we multiply each of the chords by the value of a function
F(s) evaluated at some point s * of the interval and then add all these
products. Now we let the number of intervals increase with a simultaneous
decrease in the lengths of the chords. We define the definite integral of F(s)
as the limit of this sum as the number of intervals goes to infinity while
the length of each chord goes to zero. More precisely,
2
(17)
provided the limit on the right exists.
* Here the chords are taken to be expressed as complex numbers. Thus ks =
sk-1.
Sec. A2.3]
INTEGRATION
863
Note that in addition to the lower and upper limits P1 and P , we have
indicated that in going from P1 to P we shall follow the path C. It is
conceivable that a different answer will be obtained if a different path is
followed. It would not be necessary to write the limits on the integration
symbol if we were to always show the path of integration on a suitable
diagram together with the direction along the path. Because the path, or
contour, is inseparable from the definition of an integral, we refer to it as a
contour integral.
To determine the conditions under which the definite integral in (17)
exists, we must first express this integral as a combination of real
integrals. With F(s) = U +jX, and after some manipulation, (17) becomes
2
(18)
Each of the integrals on the right is a real line integral; if these integrals
exist, then the contour integral will exist. From our knoweldge of real
integrals we know that continuity of the integrand is a sufficient condition
for the existence of a real line integral. It follows that the contour integral
of a function F(s) along a curve C exists if F(s) is continuous on the curve.
CAUCHY'S INTEGRAL THEOREM
The question still remains as to the conditions under which the integral
between two points is independent of the path joining those points.
Consider Fig. 5, which shows two points P1 and P joined by two simple
2
paths C1 and C . Note that the directions of these paths are both from
P1 to P . The combined path formed by C1 and the negative of C
2
864
[App. 2
(19)
This is a very powerful and important theorem, but we shall omit its
proof.
A word is in order about the connectivity of a region in the complex
plane. Suppose we connect any two arbitrary points P1 and P that lie
in a region b y two arbitrary simple arcs C1 and C also lying in the region.
The region is said to be simply connected if it is possible to slide one of
these arcs along (distortion of the arc is permitted in this process) until
it coincides with the other, without ever passing out of the region.
Cauchy's theorem is proved ab initio for just such a region. The hatched
region between the two closed curves in Fig. 6 is called doubly connected.
2
(a)
Fig. 6.
(b)
A doubly-connected region.
Sec. A2.3]
INTEGRATION
865
to the composite curve consisting of the inner and outer curves and the
"canal." The canal is traversed twice, but in opposite directions, so that
its contribution to the complete contour integral is zero. If we denote the
outside and inside curves by C1 and C , respectively, both in the counter
clockwise direction, then Cauchy's theorem will lead to the result that
2
(20)
As a matter of fact, if we choose any other closed path between the inner
and outer ones in Fig. 6, the same reasoning will tell us that the integral
around this path in the counterclockwise direction will be equal to each
of the integrals in (20).
This reasoning leads us to conclude that the value of a contour integral
around a simple closed curve will not change if the contour is distorted,
so long as it always stays inside a region of regularity.
Turn again to Fig. 5. The points P1 and P are in a simply connected
region R throughout which a function F(s) is single valued and regular.
Let P1 be a fixed point that we shall label s0, and P a variable point
that we shall label s. We have stated that the integral from s to s is
independent of the path of integration so long as the paths remain in the
region of regularity. Hence we can define the function G(s) as
2
(21)
where z is a dummy variable of integration. This function is a singlevalued function of the upper limit s for all paths in the region of regularity.
It is easy to show that G(s) is regular in R and that its derivative is F(s).
We call it the antiderivative of F(s). (For each s we get a different antiderivative.)
Actually it is not necessary to assume that F(s) is regular in the region.
Instead it is sufficient to assume that F(s) is continuous in R and that its
closed-contour integral for all possible simple closed curves in R is zero.
However, Morera's theorem, which we shall discuss later, states that a
function satisfying these conditions is regular.
In evaluating a definite integral in real variables we often look for an
antiderivative of the integrand. The same procedure is valid for complex
variables; that is, if an antiderivative of F(s) is C(s), then
0
(22)
866
[App. 2
Let us now consider a simple closed curve C within and on the boundary
of which a single-valued function F(s) is regular. It is possible to express
the value of the function at any point so inside the curve in terms of its
values along the contour C. This expression is
(23)
(24)
(25)
(26)
Sec. A2.3]
INTEGRATION
867
(27a)
(27b)
(28a)
868
[App. 2
C in the s-plane; let this region, including the curve C, be R. The map of the
the curve C may take one of the forms shown in Fig. 7. Note that the maps
(a)
Fig. 7.
(c)
(b)
Sec. A2.4]
INFINITE SERIES
869
(286)
A2.4
INFINITE SERIES
Let f2(s),f (s), ... be an infinite sequence of functions and consider the
sum of the first n of these:
2
(29)
(30)
for all values of n greater than N . The value of the integer N will depend
on the number and on the point s .
We say that the sequence is uniformly convergent in a closed region
if the same integer N can be used in the role of N for all points in the
j
870
[App. 2
Theorem
2 . If a sequence of continuous functions S (s) converges uni
formly to a limit function F(s) in a region R, then the integral of F(s) along
any simple arc C in the region R can be obtained by first finding the integral
along C of a member S (s) of the sequence and then taking the limit as
n -> ; that is,
n
(31)
(k)
* All of these theorems have to do with conditions under which two limit operations
can be interchanged. They are of the general character
This interchange is permissible if both limits (separately) exist and one of them (say
x -> a) exists uniformly with respect to the other variable.
Sec. A2.4]
INFINITE SERIES
871
TAYLOR S E R I E S
(32)
The partial sums of a power series are polynomials in (s s ); hence
they are regular in the entire finite complex plane (this implies that they
are continuous as well). If we can now determine the region of uniform
convergence, we can use Theorems 1 through 4 to deduce properties of
the limit function.
Suppose that a power series converges for some point s = s1. It is easy
to show that the series will converge absolutely (and hence it will also
converge) at any point inside the circle with center at s and radius
|s1 s0|- The largest circle with center at s within which the series con
verges is called the circle of convergence, the radius of the circle being the
radius of convergence. It follows that a power series diverges (does not
converge) at any point outside its circle of convergence, because if it
does converge at such a point s , it must converge everywhere inside the
circle of radius |s s |, which means, that the original circle was not
its circle of convergence.
Let R be the radius of convergence of a power series and suppose that
R1 is strictly less than R . Then it can be shown that the given series is
uniformly convergent in the closed region bounded b y the circle of radius
R1< Ro with center at s .
Suppose now that a power series converges to a function F(s) in a
circle of radius R . This means that the sequence of partial sums S (s)
will have F(s) as a limit function. Since S (s) is a continuous function,
it follows from Theorem 1 that F(s) is also continuous everywhere inside
the circle. Furthermore, since the partial sums are regular in the region
of uniform convergence, it follows from Theorem 3 that F(s) is regular
in the region. Thus a power series represents an analytic function that is
regular inside its circle of convergence.
Two other important conclusions about power series follow from
Theorems 2 and 4. According to Theorem 2, since the partial sums of a
power series satisfy the conditions of the theorem, a power series that
converges to F(s) can be integrated term by term and the resulting series will
0
872
[App. 2
converge to the integral of F(s) for every path inside the circle of convergence.
Similarly, according to Theorem 4, a power series may be differentiated
term by term, and the resulting series will converge to the derivative of F(s)
everywhere inside the circle of convergence. The circles of convergence of
both the integrated series and the differentiated series are the same as that
of the original series.
We saw that a power series converges to an analytic function that is
regular within the circle of convergence. The converse of this statement,
which is more interesting, is also true. Every analytic function can be
represented as a power series about any regular point s0. The desired
result is Taylor's theorem, which states: Let F(s) be regular everywhere in a
circle of radius R about a regular point s . Then F(s) can be represented as
0
(33)
(34)
The circle of convergence of the power series is the largest circle about s
in which F(s) is defined or is definable as a regular function.
This series is referred to as a Taylor series. The theorem is proved b y
starting with Cauchy's integral formula given in (23) and expanding
(z s)
as a finite number of terms in inverse powers of (z s0) (after
adding and subtracting s to the denominator of the integrand), together
with a remainder term. Use of the integral formulas for the derivatives
of an analytic function given in (28) leads to a polynomial in (s s )
plus a remainder term. The proof is completed by noting that the remaind
er term vanishes as the order of the polynomial in (s s ) approaches
infinity.
An important consequence of Taylor's theorem is that the circle of
convergence of any power series passes through a singular point of the
analytic function represented by it, because by Taylor's theorem, the
radius of convergence is the distance from the point so to the nearest
singular point.
To find the power-series representation of a function, it is not necessary
to use the formulas given in Taylor's theorem. But independent of the
method used to find the power series representation, we shall end up with
Taylor's series, with the coefficients satisfying Taylor's formula. This fact
0
Sec. A2.4]
INFINITE SERIES
873
If
have positive radii of convergence and if their sums coincide for an infinite
number of distinct points having the limit point s , then a = b for all n;
that is, they are identical.
In particular, the conditions of the theorem are satisfied if the two
series agree in a neighborhood of so or along a line segment (no matter how
small) that contains s . This result is proved by induction on n. Thus
the representation of an analytic function by a power series about a given
regular point s is unique.
0
LAURENT SERIES
Fig. 8.
the region between them. The point so may be a regular point or a singular
point of F(s). Also there may be other singular points of F(s) inside the
inner circle. The annular region can be made simply connected b y the
device of "digging a canal" discussed in a preceding section. If we now
apply Cauchy's integral formula, we get
(35)
874
[App. 2
(36)
(37)
These can be checked by noting that the expression
(38)
is an identity for all values of w except w = 1. Equation 36 is obtained by
adding and subtracting s in the denominator on the left and then writing
it in the form of (38) with
0
(39)
A similar case obtains for (37), except that w is now
(40)
Now let us use (36) in the first integral in (35) and (37) in the second
integral. Each integral will give a finite number of terms plus a remainder
term. It can be shown, as in the proof of Taylor's theorem, that the
remainder terms vanish as n-> . The final result is
(41)
or
(42)
where a in the last expression is given by
k
(43)
INFINITE SERIES
Sec. A2.4]
875
FUNCTIONS D E F I N E D B Y SERIES
One of the results that we noted previously is that a power series defines
an analytic function that is regular inside its circle of convergence.
We shall now use this fact to define some specific functions. Up until now
* This property of Laurent series can be interpreted as saying that the series of posit
ive powers in (s s0) converges everywhere inside C of Fig. 8 and the series of negative
powers converges everywhere outside of C1, the two converging simultaneously in the
annular region between C1 and C .
2
876
[App. 2
(44)
The last form is obtained b y inserting s = + j in the series; expanding
the powers of s; collecting terms; and finally identifying the real power
series representing e , cos , and sin . We are not completely free in
choosing a defining series for e , because it must reduce to the correct
series when s is real.
To determine the radius of convergence of the defining series we can
resort to various tests for the convergence of series (which we have not
discussed). Alternatively, since the series represents an analytic function,
we can use the Cauchy-Riemann equations. In the latter case we find that
there are no singular points in the entire finite plane, since the CauchyRiemann equations are satisfied everywhere. Hence the series converges
everywhere. (The same result is, of course, obtained by testing the series
for convergence.)
We can now follow the same procedure and define other transcendental
functions in terms of series. However, it is simpler to define the trigono
metric and hyperbolic functions in terms of the exponential. B y definition,
then,
(45)
(46)
From the behavior of the exponential we see that the sines and cosines,
both trigonometric and hyperbolic, are regular for all finite values of s.
The singular points of tan s occur when cos s = 0; namely, for an infinite
Sec. A2.5]
MULTIVALUED FUNCTIONS
877
number of real values of s at the points s = (2k l)/2 for all integral
values of k. Similarly, the singular points of tanh s occur when cosh s = 0;
namely, at an infinite number of imaginary values of s at the points
s = j(2k l)/2 for all integral values of k.
The trigonometric and hyperbolic functions of a complex variable
satisfy practically all of the identities satisfied by the corresponding real
functions.
A2.5
MULTIVALUED FUNCTIONS
878
[App. 2
(51)
Therefore from the definition of the logarithm one of the values of this
function is
(52)
This particular value, which is unique by virtue of (50) is known as the
principal value of the logarithm function. We signify this conventionally
b y writing a capital " L " in log F(s); similarly, arg F(s) always means the
principal value given in (50). Thus we can write, for ail values of the
log function,
(53)
where k is an integerpositive, negative, or zero.
Thus there are an infinite number of values for the logarithm function,
one for each value of k. Because of this difficulty, we might try to simplify
by using only the principal value, Log F(s). Before considering Log F(s),
let US first consider the behavior of the function Log s in the complex plane.
Log s is given b y
(54)
where
(55)
We notice that the angle is undefined at s = 0. Therefore this equation
does not define Log s at s = 0. But no matter how we define Log 0, Log s
will not be continuous at s = 0, since the imaginary part of Log s takes
on all values from to in any neighborhood of s = 0. Therefore s = 0
is a singular point of Log s. Even though we restrict ourselves to the
Sec. A2.5]
MULTIVALUED FUNCTIONS
879
principal value, Log s is discontinuous at any point on the negative real axis;
for, the imaginary part of Log s here is , but there are points arbitrarily
close to it at which the imaginary part is very nearly . Thus Log s is
not regular at any point on the negative real axis, including s = 0, .
(The behavior at is identical to the behavior at 0, since Log 1/s =
Log s, as you can verify.)
B R A N C H P O I N T S , CUTS, A N D R I E M A N N SURFACES
Fig. 9.
one side of it to the other, Log s is regular in the rest of the complex plane.
In fact, we have
(56)
at all other points of this " c u t " plane. Thus Log s is an antiderivative
of 1/s. We can show that
(57)
880
[App. 2
provided the path of integration does not go through the " c u t " negative
real axis.
Similar remarks apply to the other values of the logarithm function.
The restriction to principal values is unnecessary. The only thing we
need to do is to restrict the imaginary part of log s to some 2 range.
For (57) to apply, we have to add a suitable multiple of j 2 to the right
side. It is not even necessary that the cut be along the negative real axis.
We may cut the plane along any radius vector by defining
(58)
E v e n this is unnecessary. Any simple path from s = 0 to s = will do.
Thus b y suitable restrictions we can make the function log s singlevalued and regular in any neighborhood. The only exceptional points
are s = 0, . No matter what artifice we employ, we cannot make log s
regular and single valued in a deleted neighborhood of s = 0, . (Since
these points are singular points, we have to delete them from the neighbor
hood if we hope to make the function regular.) Thus these two singular
points are different in character from the ones we have met so far.
Therefore we give them a different name. They are called branch points.
Precisely, a branch point is defined as follows.
The point s is a branch point of the function F(s) if s is an isolated
singular point and there is no deleted neighborhood of s in which F(s)
is defined or is definable as a single-valued regular function.
We now see that the plane has to be cut along a simple path from one
branch point of log s to the other branch point. Each value of log s so
obtained is called a branch of the function. Thus Log s is a branch of log s.
Riemann introduced an artifice that allows us to consider the complete
log function and treat it as a single-valued function. This important
concept is known as the Riemann surface. It is quite difficult to define
this term precisely, and we shall not attempt it. Instead let us describe
a few Riemann surfaces. For the function log s the Riemann surface has
the following structure. We consider the s-plane to consist of an infinite
number of identical planes. One of these is the plane in which arg s is
restricted to its principal value. There are an infinite number of sheets
above this and another infinity below. All of these planes are cut along the
negative real axis. All of these have the same origin and , so that the
sheets are all joined together at these points. Each sheet is also joined to
the ones immediately above and below, along the negative real axis. The
upper edge of the negative real axis of each sheet is joined to the lower
edge of the negative real axis of the sheet immediately above it. The whole
Riemann surface looks somewhat like an endless spiral ramp.
0
Sec. A2.5]
MULTIVALUED FUNCTIONS
881
(63)
We may make this function single-valued by restricting the angle of s as
before; that is,
(64)
and defining the "positive square r o o t " as
(65)
where |s| is a real positive number.
882
[App. 2
Again we find that G1(s) is not continuous on the negative real axis,
including s = 0, . The points s = 0, are seen to be branch points of
this function G(s). The Riemann-surface concept may be introduced as
follows. We need two sheets of the Riemann surface, both cut along the
negative real axis. To make G(s) continuous and regular on this surface,
we "cross-connect" the two sheets along the negative real axis. The
upper edge of the negative real axis of each sheet is connected to the lower
edge of the negative real axis of the other sheet. (Obviously, it is useless
to attempt to draw a picture of this in three dimensions.) On this Riemann
surface, G(s) is regular and single valued except at s = 0, .
We see that the branch points of the function log s are somewhat
different from the branch points of s . In one case we have an infinite
number of branches, and in the other case we have only a finite number.
Therefore we sometimes distinguish between these by calling the former
a logarithmic singularity (or a logarithmic branch point) and the other an
algebraic singularity (or an algebraic branch point).
For example, we can extend this discussion to other algebraic irrational
functions,
1/2
(66)
in an obvious way.
CLASSIFICATION OF M U L T I V A L U E D F U N C T I O N S
Sec. A2.6]
883
A2.6
884
[App. 2
integrated around a closed contour inside which the function has one
singular point, the value of the integral will be 2j times the coefficient
of the first negative power term in the Laurent series. None of the other
terms in the series contributes anything; they all " w a s h out." We call
this coefficient the residue. Note that the function is regular on the
contour.
If the contour in question encloses more than one singular point (but a
finite number), we can enclose each singular point in a smaller contour
of its own within the boundaries of the main contour. B y "digging canals"
in the usual way, we find the value of the integral around the original
contour to be equal to the sum of the integrals around the smaller con
tours, all taken counterclockwise. Now we consider a Laurent series
about each of the singular points such that no other singular points are
enclosed. According to the preceding paragraph, the value of the integral
about each small contour is equal to 2j times the corresponding residue.
Hence the integral around the original contour is equal to 2j times the
sum of the residues at all of the singular points inside the contour ; that is,
(68)
(69)
The function on the left is regular at s , and the series on the right is the
Taylor series representing it in the neighborhood of s . Hence, b y using
the formula for the Taylor coefficients, we get
0
(70)
Sec. A2.6]
885
where so is a simple pole of F(s), in the nontrivial case H(s) has a simple
zero at so and G(s) is regular and nonzero at s0. In this case we may write
(73)
(74)
Thus the residue at a simple pole is the reciprocal of the derivative of the
reciprocal function.
One of the important applications of the residue theorem is the following
identity theorem for Laurent series:
If the two Laurent
series
886
[App. 2
(77)
Since the positive and negative series are power series, they converge
uniformly for |s s | R , ( > 0) and |s s | R1 + , respectively.
Therefore in the annular region R1 + | s s | R the Laurent
series are uniformly convergent. We now multiply both sides of (77) b y
(s s0)* , where k is an integerpositive, negative, or zeroand inte
grate along a circular path C lying in the region of uniform convergence
and enclosing s0. B y the residue theorem we get
0
-1
Sec. A2.6]
887
Path of
integration
Simple
pole
(a)
(b)
Consider the semicircular path shown in Fig. 10b around a simple pole
at s0. The Laurent expansion of F(s) about s has the form
0
(78)
Note that the direction of the path is counterclockwise around the pole
when we are to indent the contour in such a way that the pole is inside.
We can also indent the contour to exclude the pole. Then the value
obtained will be the negative of that obtained here. The series in this
equation can be integrated term b y term; let (s s ) = re and let C
represent the semicircle. On the semicircle varies from 0 to . The integral
of F(s) on the semicircle becomes
j
(79)
(80)
that is, the integral around half a circle about a simple pole will have one
half the value of an integral around a complete circle. In fact, b y the same
reasoning, if the contour is a fraction k of a circular arc, the contribution
will be k(2ja ).
1
888
[App. 2
J O R D A N ' S LEMMA
Fig. 11.
* That is, the limit is approached at the same rate for all angles of s within this range.
The range is |arg s\ < / 2 for a semicircle in the right half-plane and |arg s\ > /2 for a
semicircle in the left half-plane. In the e- language, the magnitude of the difference
between sF(s) and the limit (in this case 0), can be made less than e, so long as |s| > N(e),
where N(e) is independent of arg s in the appropriate range.
Sec. A2.6]
889
will be no contribution from the infinite arc; for example, if F(s) is a ratio
of two polynomials, the degree of the denominator must exceed that of the
numerator b y 2 or more.
Let t be a real variable and suppose the integrand has the form
(83)
Then it can be shown that for t > 0 the infinite arc in the left half-plane
will not contribute to the integral, nor will the arc to the right for t < 0,
provided that G(s) vanishes uniformly as the radius of the semicircle
approaches infinity. This result is called Jordan's lemma. The presence of
the exponential loosens the restriction on the remaining part of the
integrand. Thus, if G(s) is a ratio of two polynomials, it is enough that the
degree of the denominator exceed that of the numerator b y 1 (or more).
As an example of the evaluation of integrals consider
(84)
(85)
(86)
(87)
where the contour C is the closed contour shown in Fig. 12. The inte
grand has a simple pole on the original contour so that we indent the
contour around the pole as shown. The complete contour consists of two
portions of the j-axis and two semicircles, the radius of one of which will
890
Fig. 12.
[App. 2
approach zero while the other will approach infinity. Since the integrand
is regular everywhere inside the contour, the closed-contour integral will
vanish. We can write
(88)
The integrand satisfies Jordan's lemma, so that the last integral in this
equation will vanish. The value of the integral on C is j times the
residue of the integrand at s = 0 , according to (80). To calculate the residue
we use (71) and find it to be unity. Hence
0
(89)
(90)
But by the improper integral in (86) we mean precisely the left side of the
last equation. Hence, finally,
(91)
Sec. A2.6]
891
P R I N C I P L E OF THE A R G U M E N T
(92)
(93)
We see that this function has a simple pole at the zero of F(s) with a
residue n. The function Fi(s) can now be treated in the same way and
the process repeated until all the zeros of the original function F(s) have
been put into evidence. Each zero will lead to a term like the first one on
the right side of (93).
Now suppose that F(s) has a pole of order m at a point s in R. Then
we can write
2
(94)
The desired function is seen to have a simple pole at the pole of F(s),
892
[App. 2
with a residue that is the negative of its order. Again the same process
can be repeated and each pole of F(s) will lead to a term like the first one
on the right side of the last equation. The only singularities of F'(s)/F(s)
in the region R will lie at the zeros and the poles of F(s). Hence, by the
residue theorem, the value of the desired contour integral will be
(95)
where the n are the orders of the zeros of F(s) in R and the m are the
orders of the poles.
Note, however, that
j
(96)
Sec. A2.7]
PARTIAL-FRACTION EXPANSIONS
893
shown in Fig. 10b. Let F(s) have a zero (or a pole) of order k at s .
we may write
Then
(97a)
(97b)
If so is a pole, we let k be a negative integer in these expressions, thus
including a zero of order k and a pole of order k simultaneously in the
discussion. As we let the radius of the circle approach zero, log Fi(s) will
not contribute anything to the integral, since it is regular at s . Thus it
is sufficient to consider
0
A2.7
PARTIAL-FRACTION EXPANSIONS
894
[App. 2
(102)
The function F (s) is regular at the singularity s ; hence it will not
contribute anything to the principal part F (s).
This means that the
principal part F (s) will be the same whether we expand F (s) or the
original function F(s).
We now repeat this process with F (s) and keep repeating with each
singularity. At each step we subtract the principal part of the Laurent
expansion until ail the singular points are exhausted. The regular part of
the last Laurent expansion will have no other singularities in the finite
plane. Hence it must be an entire function. In this fashion we have
succeeded in obtaining a representation of F(s) that has the form
p1
p2
p2
r1
r2
(103)
Each of the terms in the summation is the principal part of the Laurent
series of F(s) expanded about one of its singularities. The last term is an
entire function. If F(s) is regular at infinity, this term will be a constant.
If F(s) has a pole of order n at infinity, this term will be a polynomial of
degree n. Finally, if F(s) has an essential singularity at infinity, this term
will be an infinite power series. The representation of an analytic function
given in (103) is called a partial-fraction
expansion.
Suppose a function has an infinite number of poles and no essential
singularities in the finite plane (this makes it a meromorphic function).
ANALYTIC CONTINUATION
Sec. A2.8]
895
A2.8
ANALYTIC CONTINUATION
N o w let US consider two functions F1(s) and F (s), which are respec
tively regular in overlapping regions R1 and R , the common region being
R0, as shown in Fig. 13. (The regions need not be circular as shown here.)
The two functions Fi(s) and F (s) determine each other uniquely. This
follows from the identity theorem since only one function can be regular
in R1 (or R ) and have the same values in R .
Suppose we were starting with the function Fi(s) in R1 and could find
a function F (s) in R with the property just described. We would say
that Fi(s) has been analytically coninued beyond its original region into
region R . But we might just as well consider F (s) to be the original one
and Fi(s) its analytic continuation into region R1. For this reason we say
that each of them is but a partial representation, or an element, of a single
function F(s) that is regular in both R1 and R .
Consider now the problem of starting with one element Fi(s) of a
2
896
Fig. 13.
[App. 2
Sec. A2.8]
ANALYTIC CONTINUATION
897
Appendix
THEORY OF LAPLACE
TRANSFORMS
(i)
The function K(t, s), which is a function of two variables, is called the
kernal of the transformation. Note that the integral transform no longer
depends on t; it is a function of the variable s on which the kernel depends.
898
Sec. A3.1]
LAPLACE TRANSFORMS
899
and is
(2)
(3)
Since the defining equation contains an integral with infinite limits,
one of the first questions to be answered concerns the existence of Laplace
transforms. A simple example of a function that does not have a Laplace
transform is e . Let us therefore state a few theorems (a few of which we
shall also prove) concerning the convergence of the Laplace integral.
Since s appears as a significant parameter in (2) we may expect the con
vergence to depend on the particular value of s. In general, the integral
converges for some values of s and diverges for others.
In all of the theorems to follow we shall consider only integrable
functions f(t) without specifically saying so each time. As a first theorem,
consider the following:
If the function f(t) is bounded for all t 0, then the Laplace integral
converges absolutely for Re(s) > 0.
To prove the theorem, note that the condition on f(t) means |f(t)| < M
for all t 0, where M is a positive number. Then for > 0 we shall get
et
(4)
In the limit, as T approaches infinity, the right-hand side approaches
M/. Hence
(5)
The familiar sine and cosine functions, and other periodic functions
such as the square wave, satisfy the conditions of the theorem. Before
commenting on this theorem, let us consider one more theorem.
900
[App. 3
(6)
(7)
(8)
or
(9)
- ( s - s 0 ) T
(10)
Sec. A3.1]
LAPLACE TRANSFORMS
901
since b y this theorem, whenever the integral converges for some point in
the s-plane, it converges at all points to the right. Thus we can define
an abscissa of convergence such that the Laplace integral converges for
all s with > and diverges for all s with < . The stronger result,
which we have not proved, is that the region of convergence is also the
region of absolute convergence. The behavior of the Laplace integral is thus
somewhat analogous to the behavior of power series. The function f(t)
plays the role of the coefficients of the power series, and the function e
plays the part of (s s0) . Just as a power series may have any behavior
on the circle of convergence, the Laplace integral may also have any
behavior on the abscissa of convergence. The only difference concerns the
existence of a singular point on the circle of convergence, which we shall
examine a little later.
With infinite series, we have many tests for convergence. All of these
have analogues in Laplace transforms. We shall be content to state just
two of these. The analogue of the ratio test is the following:
If |f(t)| M e for some constant M and some number c, for all t (or
only for t greater than some T ) , then the Laplace integral converges absolutely
for > c.
We see this result immediately since
c
- s t
c t
(11)
(13)
and b y no c = . In this case we have established that the Laplace
integral converges absolutely for > and diverges for < .
Many functions that are not of exponential order have Laplace trans
forms. However, we can state the following necessary and sufficient
condition, which shows that the integral of a transformable function is of
exponential order.
0
902
[App. 3
of
convergence
(14)
satisfies
(15)
for any c = + .
The proof of this result depends on the Stieltjes integral, and so we
cannot give it here. We can use this theorem to get an analogue for the
Cauchy root test for power series.
Let g(t) be the function defined in (14). If
0
(16)
(17)
for every > 0.
This region is shown in Fig. 1. We may take to be the abscissa of
convergence if the integral converges at this point. Otherwise, is a
point arbitrarily close to and to the right.
In the case of functions of exponential order, however, the region of
uniform convergence coincides with the region of convergence; that is,
we may take = 0 in the theorem above.
0
903
Axis of
convergence
Fig. 1.
For functions
the half-plane
of exponential
A3.2
converges for > , then the function F(s) defined by the integral is regular
in the half-plane > . In fact, the derivative of F(s) is given by
c
(19a)
904
[App. 3
and in general
(19b)
Given any point s with > , we can surround this point with a circle
that is entirely within the region of uniform convergence, since in (17)
is arbitrary. Now, because of the uniform convergence, the limit operations
of integration and differentiation can be interchanged. Hence
c
(20)
This leads to (19a). The convergence of (19a) is easily established for
functions of exponential order. For the general case we integrate by parts.
Thus the Laplace integral defines a regular function within the halfplane of convergence. However, although the function F(s) is defined
b y the integral only in the half-plane of convergence, we can use the
technique of analytic continuation to extend the function across the
abscissa of convergence whenever it may be continuable. [In practice this
is merely a formality, the "analytic continuation" being merely an
extension of the formula for F(s).] It is this more general analytic function
that is referred to as the Laplace transform. If F(s) is the Laplace trans
form of f(t), we refer to f(t) as the determining function and F(s) as the
generatingfunction.
In this more general concept of a Laplace transform the generating
funtion will, in general, have singularities. They will have to lie in the
half-plane or at . Here we may revert to the power-series
analogy again. The function defined by a power series always has a
singular point on the circle of convergence. We may ask whether F(s)
has a finite singular point on the abscissa of convergence . Here the
analogy breaks down. In general, there may be no singular point on
= . The following example has been given by Doetsch:
c
(21)
For this function the abscissa of convergence is zero. However, its trans
form satisfies the difference equation
(22)
so that F(s) is an entire function.
905
(23)
(24)
for all s in this sector. These two conditions fix T1 and T and therefore
the value of the integral
2
(25)
906
[App. 3
so that
(26)
Since s approaches in the sector |arg(s s )| / 2 , its real part
has to exceed 1 eventually. If we put together the three conditions
(23), (24), and (26) and restrict s by
0
we get
(27)
so that
(28)
Sec. A3.3]
OPERATIONS ON FUNCTIONS
907
where the + and indicate, as usual, the right- and left-hand limits.
Implicit here is the assumption that these limits exist, which we shall
assume.
There cannot exist two different normalized determining functions
f1(t)
and f (t) with the same Laplace transform F(s).
The proof of this result is too complicated to be given here. If we do not
normalize the functions, we can only conclude that the two functions
f1(t) and f (t) differ at most b y a null function.
2
(30)
This linearity is quite useful in both direct and inverse Laplace trans
formations.
(31a)
(31b)
908
[App. 3
(32a)
where
(32b)
If
and
(33)
have finite abscissae of convergence 1 and , then the product F1(s) F (s)
is also a Laplace transform
2
(34a)
where
(34b)
(35)
Sec. A3.3]
OPERATIONS ON FUNCTIONS
909
D I F F E R E N T I A T I O N A N D INTEGRATION
if
(37a)
then
(37b)
where the abscissae of convergence are the same and the path of integration is
restricted to the sector of uniform convergence.
This result is proved by integrating by parts in the s-domain, noting
that F(s) approaches 0 as s approaches . More important operations
than the proceding ones, as far as the application to network theory is
concerned, are differentiation and integration in the t-domain. These
operations are found to resemble duals of the ones above.
Let f (t) be differentiable (and therefore continuous) for t > 0, and let the
910
[App. 3
derivative f'(t) be transformable. Then f (t) is also transformable and with the
same abscissa of convergence. Further
(38a)
where
(38b)
and
(38c)
Let
(39)
(40)
where G(s) and F(s) are Laplace transforms of g(t) and f (t) respectively.
The first part follows as before. Equation 40 follows from (38b) on
observing that g(0+) = 0 b y (39).
These results can easily be extended to higher order derivatives and
integrals of f(t) b y repeated applications of these theorems.
I N I T I A L - V A L U E A N D F I N A L - V A L U E THEOREMS
Sec. A3.3]
OPERATIONS ON FUNCTIONS
911
if
(41)
where the limit on the right is to be taken in the sector
(43)
where the limits are to be taken with |arg s| (-/2) .
It is first of all not clear that the last limit exists. If it does, we do not
see what it might be. If we can interchange the limit and the integral,
however, we shall get
(44)
If we assume uniform convergence of the Laplace integral for f'(t) in
a region including s = 0, then this interchange can be made. In such a
case, however, the abscissae of convergence of both f'(t) and f(t) must be
negative. This is possible only if f(t) approaches 0 as t approaches ;
that is,
(45)
But in this instance the whole theorem will be devoid of content. Hence,
in order to establish the theorem, the interchange of the limit and the
912
[App. 3
SHIFTING
(47)
If we make the substitution x = t + a, and then change the dummy
variable of integration back to t, we shall get
(48)
If we assume that f (t) vanishes for t < 0, then f(t a) will vanish for
t < a and the lower limit of the integral can be replaced b y zero. To
indicate that f(t a) is zero for t < a we can write it in the form
f(ta)u(ta).
The function u(x) is the unit step, defined as zero for
negative x and unity for positive x. This leads to the following result:
If [f (t)] = F(s) and a is nonnegative real, then
(49)
with the same abscissa of convergence.
This result is called the real shifting or translation theorem, since f(t a)
is obtained b y shifting f(t) to the right b y a units.
The operation of multiplying f(t) b y e leads to a similar result. This
is called the complex shifting theorem.
at
Sec. A3.4]
913
(50)
with the abscissa of convergence + Re(a).
This theorem follows directly from the definition of the Laplace trans
form.
c
A3.4
(51)
(52)
or simply
(53)
.
c
914
[App. 3
However we do know that F(s) is regular for > . Hence, in such a case,
we take the path of integration to be a vertical line to the right of all the
singular points of F(s). Such a path is known as a Bromwich path after the
famous mathematician T. J. I'A. Bromwich, who made many significant
contributions to the theory of Laplace transformation. The abbreviation
" B r " is sometimes used on the integral sign, instead of the limits, to
signify this contour.
We saw in Section 6 of Appendix 2 that the residue theorem can often be
used to evaluate integrals of this type. In order to use the residue theorem
we have to close the contour. Let us consider the two closed paths shown
st
in Fig. 2. If the integrand F(s)e satisfies Jordan's Lemma on either of
the semicircular arcs, we can evaluate the integral by the residue theorem.
If Jordan's lemma is satisfied on the arc to the right; that is, if
c
(which will be true, for instance, if t < 0), the integral on C1 of Fig. 2 is
Fig. 2.
f(t)
= residues
of
st
F(s)e
Sec. A3.4]
915
F(s) =
L{f(t)}
916
F(s) =
L{f(t)}
[App. 3
BIBLIOGRAPHY
1. MATHEMATICAL B A C K G R O U N D
Complex Variable Theory
Churchill, R. V., Introduction to Complex Variables and Applications, McGraw-Hill Book Co.,
New York, 1948.
Hille, E., Analytic Function Theory, Guin and Co., New York, vol. I, 1959.
Knopp, K., Theory of Functions, Dover Publications, New York, vol. I, 1945, vol. II, 1947.
LePage, W. R., Complex Variables and the Laplace Transform for Engineers, McGraw-Hl Book
Co., New York, 1961.
Spiegel, M. R., Theory and Problems of Complex Variables, McGraw-Hill Book Co., New York,
1964.
Computer Programming
Healy, J. J., and Debruzzi, D. J., Basic FORTRAN IV Programming, Addison-Wesley Publish
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McCalla, T. R., Introduction to Numerical Methods and FORTRAN Programming, John Wiley
& Sons, New York, 1967.
McCracken, D. D., A Guide to ALGOL Programming, John Wiley & Sons, New York, 1962.
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1965.
McCracken, D. D., FORTRAN with Engineering Applications, John Wiley & Sons, New York,
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917
918
BIBLIOGRAPHY
McCracken, D. D., and Dom, W. S., Numerical Methods and FORTRAN Programming, John
Wiley & Sons, New York, 1964.
Organick, E. I., A FORTRAN IV Primer, Addison-Wesley Publishing Co., Reading, Mass.,
1966.
Differential Equations
Ayres, F., Jr., Theory and Problems of Differential Equations, McGraw-Hill Book Co., New York,
1952.
Bellman, R., Stability Theory of Differential Equations. McGraw-Hl Book Co., New York,
1953.
Coddington, E. A., and Levinson, N., Theory of Ordinary Differential Equations, McGraw-Hl
Book Co., New York, 1955.
Coppel, W. A., Stability and Asymptotic Behavior of Differential Equations, D. C. Heath and Co.,
Boston, 1965.
Hartman, P., Ordinary Differential Equations, John Wiley & Sons, New York, 1964.
Lefschetz, S., Differential Equations: Geometrical Theory, 2nd edition, John Wiley & Sons,
New York.
Sansone, G., and Conti, R., Nonlinear Differential Equations, revised edition, The Mac_1lan Co.,
New York, 1964.
Stmble, R. A., Nonlinear Differential Equations, McGraw-Hill Book Co., New York, 1962.
Matrix Algebra
Ayres, R., Jr., Theory and Problems of Matrices, McGraw-Hl Book Co., New York, 1962.
Bellman, R., Introduction to Matrix Analysis, McGraw-Hl Book Co., New York, 1960.
Gantmacher, F. R., The Theory of Matrices, Chelsea Publishing Co., New York, vol. I, 1959, vol.
II, 1959.
Hohn, F. E., Elementary Matrix Algebra, The Macmlan Co., New York, 1958.
Pease, M. C, III, Methods of Matrix Algebra, Academic Press, New York, 1965.
Perlis, S., Theory of Matrices, Addison-Wesley Publishing Co., Cambridge, Mass., 1952.
Numerical Analysis
Beckett, R., and Hurt, J., Numerical Calculations and Algorithms, McGraw-Hl Book Co., New
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Hanmiing, R. W., Numerical Methods for Scientists and Engineers, McGraw-Hl Book Co., New
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Henrici, P., Discrete Variable Methods in Ordinary Differential Equations, John Wey & Sons,
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Henrici, P., Elements of Numerical Analysis, John Wey & Sons, New York, 1.964.
BIBLIOGRAPHY
919
Householder, A. S., Principles of Numerical Analysis, McGraw-Hill Book Co., New York, 1953.
Kelly, L. G., Handbook of Numerical Methods and Applications, Addison-Wesley Publishing Co.,
Reading, Mass., 1967.
Macon, N., Numerical Analysis, John Wiley & Sons, New York, 1963.
Scheid, F., Theory and Problems of Numerical Analysis, McGraw-Hill Book Co., New York, 1968.
Wilkinson, J. H., Rounding Errors in Algebraic Processes, Prentice-Hall, Englewood Cliffs, N.J.,
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Kim, W. H., and Chien, R. T-W., Topological Analysis and Synthesis of Communication Networks,
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Seshu, S., and Reed, M. B., Linear Graphs amd Electrical Networks, Addison-Wesley Publishing
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3.
LOOP, N O D E - P A I R , M I X E D - V A R I A B L E EQUATIONS
Desoer, C. A., and Kuh, E. S., Basic Circuit Theory, McGraw-Hl Book Co., New York, 1966.
Guillemin, E. A., Theory of Linear Physical Systems, John Wiley & Sons, New York, 1963.
Seshu, S., and Balabanian, N., Linear Network Analysis, John Wiley & Sons, New York, 1959.
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STATE E Q U A T I O N S
Calahan, D. A., Computer-Aided Network Design, McGraw-Hill Book Co., New York, 1968.
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920
BIBLIOGRAPHY
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N E T W O R K R E S P O N S E A N D T I M E - F R E Q U E N C Y RELATIONSHIPS
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Guillemin, E. A., The Mathematics of Circuit Analysis, John Wiley & Sons, New York, 1949.
Kuo, F. F., Network Analysis and Synthesis, 2nd edition, John Wiley & Sons, New York, 1966.
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NETWORK SYNTHESIS
8.
SCATTERING PARAMETERS
Carlin, H. J., and Giordano, A. B., Network Theory, Prentice-Hall, Englewood Cliffs, N.J., 1964.
Kuh, E. S. and Rohrer, R. A., Theory of Linear Active Networks, Holden-Day, San Francisco, 1967.
9.
SIGNAL-FLOW GRAPHS
Horowitz, I. M., Synthesis of Feedback Systems, Academic Press, New York, 1963.
Huggins, W. H., and Entwisle, D. R., Introductory Systems and Design, Blaisdell Publishing Co.,
Waltham, Mass., 1968.
Lorens, C. S., Flowgraphs for the Modeling and Analysis of Linear Systems, McGraw-Hill Book
Co., New York, 1964.
Mason, S. J., and Zimmerman, H. J., Electronic Circuits, Signals, and Systems, John Wiley &
Sons, New York, 1960.
Robichaud, L. P. A., Boisvert, M., and Robert, J., Signal Flow Graphs and Applications,
Prentice-Hall, Englewood Cliffs, N.J., 1962.
Truxal, J. G., Control System Synthesis, McGraw-Hill Book Co., New York, 1955.
Ward, J. R. and Strum, R. D., The Signal Flow Graph in Linear Systems Analysis, Prentice-Hall,
Englewood Cliffs, N.J., 1968.
BIBLIOGRAPHY
10.
921
SENSITIVITY
Bode, H. W., Network Analysis and Feedback Amplifier Design, D. Van Nostrand Co., New York,
1945.
Horowitz, I. M.,Synthesis of Feedback Systems, Academic Press, New York, 1963.
Kuh, E. S., and Rohrer, R. A., Theory of Linear Active Networks, Holden-Day, San Francisco,
1967.
Truxal, J. G., Automatic Feedback Control System Synthesis, McGraw-Hl Book, Co., New York,
1955.
11.
STABILITY
Liapunov's Method
Hahn, W., Theory and Application of Liapunov's Direct Method, Prentice-Hali, Englewood
Cliffs, N.J., 1963.
LaSalle, J., and Lefschetz, S., Stability by Liapunov's Direct Method, Academic Press, New York,
1961.
Liapunov, A. M., Stability of Motion, Academic Press, New York, 1966.
Ogata, K., State Space Analysis of Control Systems, Prentice-Hali, Englewood Cliffs, N.J., 1967.
Zubov, V. I., Methods of A. M. Liapunov and Their Application, P. Noordhoff, Groningen, The
Netherlands, 1964.
12.
T I M E -V A R Y I N G A N D
NONLINEAR
NETWORK
ANALYSIS
Blaquiere, A., Nonlinear System Analysis, Academic Press, New York, 1966.
Butenin, N. V., Elements of the Theory of Nonlinear Oscillations, Blaisdeli Publishing Co., New
York, 1965.
Cunningham, W. J., Introduction to Nonlinear Analysis, McGraw-Hl Book Co., New York, 1958.
d'Angelo, H., Time-varying Networks, Aliyn and Bacon, Boston, 1969.
Desoer, C. A. and Kuh, E. S., Basic Circuit Theory, McGraw-Hl Book Co., New York, 1966.
Minorsky, N., Nonlinear Oscillations, D. Van Nostrand Co., Princeton, N.J., 1962.
Stern, T. E., Theory of Nonlinear Networks and Systems, Addison-Wesley Publishing Co., Reading,
Mass., 1965.
INDEX
924
INDEX
Corrector, 781
Cotree, 71
impedance products, 200
Current, basis set, 92
gain, 157
link, 91
loop, 93
mesh, 95
reference, 35
shift, 100
standard reference, 36
twig, 91
Cut-set, 83
fundamental, 86
orthogonality relations, 87
rank, 87
relation to incidence matrix, 87, 88
relation to loop matrix, 87, 88
Darlington theorem, 538, 599
Datum node, 63
Definite integral evaluation, 886
Delay function, 399, 459
Determinant, 11
Binet-Cauchy theorem, 14, 192
cofactor expansion, 12
derivative of, 14
major, 14, 191
of matrix product, 14
pivotal condensation, 17
principal cofactor, 13
ascending, 566
principal minor, 13
of signal flow graph, 656
Determining function, 904
Differential equations, solution by varia
tion of parameter, 246
Divided differences, 769
Driving-point admittance, 154, 162
Driving-point functions, topological formu
las, 196, 202
Driving-point impedance, 154, 162
angle condition, 490
properties, 489
Dual networks, 119
Duhamel integrals, 355
Dynamicaliy independent variables, 230
Dynamic equations, 230
Eigenvalue, distinct, 262
of a matrix, 26
multiple, 265
Eigenvector, 26
Electric network, 58, 90
topology of, 69
Elementary transformations, 468, 553
Ell networks, constant-resistance, 410
Embedding, 837
Energy functions, 485, 487
Equal ripple approximation, 415, 422
Equicofactor matrix, 184
Equivalent matrices, 470
Essential graph, 654
Essential node, 645
Euler's method, 776
Even part, of rational function, 396
Exponential order, 901
Feedback, 664
Fialkow condition, 225
Filter, Butterworth, 415
Final-value theorem, 910
Focus, singular point, 818
Forest, 72
Foster forms, 510, 522
Foster reactance theorem, 509
Free response, 348
numerical evaluation of, 367
Frequency response, 398
Function, algebraic irrational, 883
analytic, 855
branch of, 880
complex valued, 853
continuous, 853
defined by a series, 875
exponential, 876
hyperbolic, 876
trigonometric, 876
differentiable, 854
entire, 882
of exponential order, 901
generalized, see Generalized functions
integral, 882
logarithm, 877
principal value of, 878
of a matrix, 258
meromorphic, 883
multivalued, 853, 877
polynomial, 882
rational, 392, 883
INDEX
regular at a point, 855
single-valued, 853
singular point of, 855
Fundamental cut-set, 85
loop, 79
Gain-bandwidth product, 444
Generalized functions, 833
algebra of, 833
Laplace transform of, 850
as operators, 842
table of, 845
Generating function, 904
Gewertz method, 429
Graph, see Linear graph
Gronwall lemma, 727
Grounded multiport network, 159
Gyrator, 45, 211
scattering parameters of, 588, 632
925
Kirchhoffs laws, 58
current, 58, 90
voltage, 59, 95
Kronecker delta, 13
Ladder network, 406
Lagrange interpolation formula, 264
Lagrange reduction, 477
Laplace transform, 898
abscissa of convergence, 901
analytic properties of, 903
convergence properties, 899
determining function, 904
of generalized function, 850
generating function, 904
inversion integral, 913
region of convergence, 900
table of transform pairs, 915
uniqueness of determining function, 906
Lattice, 408
constant resistance, 410
Laurent series, 873
principal part of, 875
regular part of, 875
Liapunov function, 787
Liapunov stability, 783
Linard-Chipart criterion, 675
Linear equations, 20
consistency, 20, 22
926
INDEX
Linear graph, 70
branch, 70
closed path, 71
coforest, 72
connected, 71
cotree, 71
co-two-tree, 202
cut-set, 83
dual graphs, 119
forest, 72
fundamental cut-set, 85
fundamental loops, 79
hinged, 84
hinged node, 85
internal nodes, 71
isomorphic, 77
link, 71
loop, 71
mesh, 88
node, 70
oriented, 70
path, 71
planar, 88
three-tree, 205
tree, 71
twig, 71
two-tree, 194
Link, 71
Lipschitz condition, 750
Local integrability, 719
Locally integrable function, 840
Log-node, singular point, 818
Loop equations, 61, 105
Loop-impedance matrix, 105
cofactors, 20, 202
determinant, 200
Loop matrix, complete, 77
fundamental, 79
orthogonality relations, 80, 87
rank of, 81
relation to cut-set matrix, 87, 88
relation to incidence matrix, 81
Loop parameter matrices, 106
of passive, reciprocal network, 486
Loop transformation, 92
Lossless network, 504
in pr function synthesis, 531
transmission zeros, 537
two-port cascade structure, 537
Magnitude function, 398
Mapping, 858
Mason's formula, 655
Matrix, adjoint, 16
ascending principal cofactors, 556
bounded, 32, 726
canonical, 473
characteristic equation, 27
characteristic polynomial, 27
characteristic value, 26
characteristic vector, 26
congruent, 472
conjugate of, 9
conjugate transpose of, 9
constituent matrices of, 268
definite, 478, 557
derivative of, 8
differential equation, 247
eigenvalue of, 26
eigenvector of, 26
elementary, 553
equivalent, 470
index of, 474
integral of, 8
inverse, 15
nonsingular, 16
norm of, 32
nullity of, 24
order of, 3
partitioning, 6
semidefinite, 278
similar, 28
singular, 16
trace of, 9
transformation of, 468
transpose of, 9
types of, column, 4
diagonal, 10
Hermitian, 11
identity, 10
nuli, 9
orthogonal, 472
row, 4
skew symmetric, 11
symmetric, 11
Matrix exponential, 248
inverse of, 249
Laplace transform evaluation of, 258
numerical evaluation of, 373
Maximally flat approximation, 415
Maximum modulus theorem, 867
Mesh transformation, 95
INDEX
Milne method, 781
Minimal polynomial, 262
Minimum modulus theorem, 868
Minimum-phase functions, 399
Minimum-phase networks, 406
Minimum-reactance function, 428
Minimum-susceptance function, 428
Mittag-Leffler expansion, 895
Mixed-variable equations, 131
Miyata method, 431
Modified Adams method, 780
Modified Euler method, 777
Morera's theorem, 867
Multiport networks, 159, 176
Multiport parameters, positive real property,
526
Multiterminal networks, 158
Natural frequencies, 68, 394
number of, 231
number of nonzero, 233
Negative converter, 50, 216
active RC synthesis with, 543, 568
compensating a, 228
Negative-resistance amplifier, 617
Network classification, 36
active, 38, 122
linear, 36
nonreciprocal, 39, 122
passive, 37, 484
reciprocal, 38, 484
time invariant, 37
Network components, 39
Network function, 153, 392
topological formulas for, 191
Network response, complementary func
tion, 68
complete, 69
forced, 68
free, 68
natural, 68
particular integral, 68
steady state, 68
transient, 68
Newton's backward difference formula, 768
Node, singular point, 818
Node-admittance matrix, 111
cofactors of, 192, 197
determinant of, 192
Node equations, 62, 110
927
928
INDEX
INDEX
Residue condition, 528
Residue theorem, 883
Resistance-integral theorem, 439
Resistor, 39
Resolvent matrix, 269
algorithm, 271
Resolving equation, 277
Resolving polynomials, 276
Return difference, 665
of a node, 696
Return loss, 602
Return ratio, 665
of a node, 696
Riemann sphere, 861
Riemann surface, 880
RL network function, 525
Routh criterion, 673
Runge-Kutta method, 782
Scattering matrix, 585, 608, 613
bound on elements, 595
Hermitian, 595
positive semidefinite, 595
properties, 594
unitary, 596
Scattering parameters, 571, 585
interpretation, 589
match-terminated, 589
reflection coefficients, 591
transducer power gain, 590
transducer voltage ratio, 590
transmission coefficients, 591
Scattering relations, of augmented multi
port, 583, 586
of augmented one-port, 576
for equalizing network, 598
incident variables, 572
of matched terminated one-port, 574
for matching network, 598, 600
of multiport, 580
of one-port, 572
reflected variables, 572
of symmetric two-port, 586, 624
Schwartz's lemma, 869
Sensitivity, 668
Separatrices, 817
Sequence, 869
Series, infinite, absolute convergence of,
870
convergence of, 870
929
Laurent, 873
partial sum of, 869
power, 871
Taylor, 871
uniform convergence of, 870
Short-circuit admittance parameters, 162
Fialkow condition, 225
topological formula for, 206
Short-circuit current gain, 164
Short-circuit stable, 702
Sifting property, of impulse function, 846
Signal-flow graph, 636, 642
branch transmittance, 643
cascade branch, 645
determinant, 656
equivalent graphs, 648
essential graph, 654
essential nodes, 645
feedback branch, 644
feedback loop, 644
feedback node, 645
gain, 651
gain formula, 655
index of, 645
inversion, of a branch, 646
of a loop, 647
of a path, 646
Mason's formula, 655
node-pulling algorithm, 651
node splitting, 645
nontouching loops, 565
reduction, 647
to essential graph, 654
self-loop, 645
sink node, 644
source node, 644
Similarity transformation, 28, 472
Similar matrices, 472
Simple closed curve, 860
Singular point (of analytic function), 855
algebraic, 882
essential (nonisolated), 856
compact, 530
isolated, 855, 875
logorithmic, 882
pole, 875
saddle point, 818
Singular point (of state equation), 784
Source, controed, 49
dependent, 49
930
INDEX
independent, 47
accompanied, 99
Stability, 668
asymptotic, 786
BIBO, 670
bounded-input, bounded-output, 670
Hurwitz criterion, 674
in-the-large, 790
inverse Nyquist criterion, 701
Krasovsk's theorem, 796
Liapunov theory, 783
Linard-Chipart criterion, 675
Nyquist criterion, 677
open-circuit, 702
Routh criterion, 673
short-circuit, 702
variable-gradient method, 799
State equations, 63, 229, 240, 245, 281
of linear, time-invariant networks, 291
with controlied sources, 302
multiport formulation, 306
normal form, 241
output equation, 245, 281, 314
of RLC networks, 292
topological formulation, linear networks,
280
State-transition matrix, 247, 249
of time-varying network, 714
State variables, 240
Step response, 351, 451
Stop band, 415
Superposition integrals, 355
numerical evaluation of, 362
Superposition principle, 341, 353, 357, 360
Surplus factor, 412
Sylvester's inequality (law of nullity), 30,
81
Synthesis, active RC structures, 543
cascaded RC two-ports, 541
Cauer forms, 523
Darlington form, 531, 538
Foster forms, 510, 522
ladder forms, 512, 523
lossless three-port, 563
lossless two-port, 538
reactance function, 509
Tangent function, 424
Taylor series, 871
Taylor's theorem, 872
INDEX
linear, 72
normal, 235
star, 72
two-, 194
three-, 205
Twig, 71
Twin-tee, 408
Two-port networks, 160
compact pole, 530
parallel-connected, 171
parallel-series connected, 219
series-connected, 171
series-parallel connected, 218
Two-port parameters, 525
positive real property, 526
real-part condition, 529
residue condition, 528
Uniform convergence, 869, 870
Variable-gradient method, 799
931
04576
BALABANIAN
BICKART
Seshu