Robust Control Notes

Robust Control: EE 575
Murti V. Salapaka
Electrical Engineering Department
Iowa State Univeristy
murti@iastate.edu
November 9, 2006
PRELIMINARIES 1
Preliminaries
PRELIMINARIES 2
Why Feedback?
0.5
10
d
y
u +
+
Why use feedback?
Consider the problem described by the block diagram. It depicts a
simplied version of cruise control problem in automobiles. The car transfer
function between the fuel ow u and the speed y is given by a constant 10.
This is the model when the road is at.
PRELIMINARIES 3
When the road has a gradient the discrepancy is modelled by adding 0.5d
to the fuel ow (downhill is positive d.)
Typically not much is known about the gradient of the road; thus the fuel
ow has to be designed assuming a at road. Thus if we want y to track a
given reference r we determine the ow as 0.1r which results in y = r when
no gradient is present.
The speed in the presence of the gradient y is given by
y = 10(0.5d + 0.1r) = r + 5d.
Under no disturbance d the open loop controller K = 0.1 yield ideal
tracking.
However, under the disturbance (road gradient) the performance can be
unsatisfactory.
PRELIMINARIES 4
Why Feedback?
0.5
10
d
y
u +
+
K
r
+
-
Consider the closed loop conguration shown in the gure.
In the above setup
y =
5
1 + 10K
d +
10K
1 + 10K
r.
PRELIMINARIES 5
It is clear that if K is chosen large then
5
1+10K
0 and
10K
1+10K
= 1 and thus
y r
making it insensitive to d. For example choose K = 100 then the
contribution of d to y is
5
1001
d which is much smaller than 5d that exists for
the open loop case.
Note that now even when d = 0, y is not equal to r as was the case with the
open loop design.
Evaluate the performance of the closed and the open loop when the plant
model by the constant 10 is uncertain.
Note that to implement the closed loop design one needs to sense the
speed of the car so that it can be fed back. This involves sensors.
PRELIMINARIES 6
Furthermore the sensors are typically noisy and they do not yield the exact
measurement of the car speed. Note that such effects of sensor noise are
absent from the open loop design.
Feedback controllers can stabilize unstable plants. However, bad controller
design can lead to unstable closed loop systems even when the plant is
stable.
SISO STABILITY 7
Primary Reasons For Feedback Control
The primary reasons for feedback are
Model uncertainty
Signal Uncertainty
Stabilization
SISO STABILITY 8
Single Input Single Output Interconnections:
Stability
SISO STABILITY 9
SISO Feedback Interconnection
Consider the unity negative feedback interconnection shown in Figure (a).
Denition 1. The interconnection in Figure is said to be well posed if for any
signals r and d there exist unique signals e
1
and e
2
that satisfy the
loop-conditions implied by the interconnections.
Note that
e
1
= d +Ke
2
e
2
= r Ge
1
SISO STABILITY 10
That is
_
I K
G I
__
e
1
e
2
_
=
_
d
r
_
The following Theorem follows immediately:
Theorem 1. The interconnection is well posed if and only if there exists
some s
0
such that G(s
0
)K(s
0
) + 1 ,= 0.
Let G(s) =
n
g
d
g
and K(s) =
n
K
d
K
where n
g
, d
g
and n
K
, d
K
are coprime
polynomial pairs (no common factors).
It is evident that if the interconnection is well posed ( we will assume this
throughout unless mentioned otherwise) then
_
e
1
e
2
_
=
_
I K
G I
_
1
_
d
r
_
SISO STABILITY 11
and thus
_
e
1
e
2
_
=
1
I +GK
_
I K
G I
__
d
r
_
Denition 2. The interconnection is stable if the map
_
d
r
_

_
e
1
e
2
_
is bounded input bounded output.
The following theorem follows immediately
Theorem 2. The interconnection is stable if and only if
1
1+GK
,
G
1+GK
and
K
1+GK
have no poles in the right half complex plane.
Theorem 3. The interconnection is stable if and only if the polynomial
d
G
d
K
+n
G
n
K
has no zeros in the right half complex plane.
SISO STABILITY 12
Proof:() Suppose d
G
d
K
+n
G
n
K
has no zeros in the right half complex
plane. Note that
_
e
1
e
2
_
=
1
d
G
d
K
+n
G
n
K
_
d
G
d
K
d
G
n
K
d
K
n
G
d
G
d
K
__
d
r
_
As the poles of all transfer functions are included in the zeros of the
polynomial d
G
d
K
+n
G
n
K
we have that all transfer functions are stable.
() Suppose there is a s
0
with Re(s
0
) 0 and (d
G
d
K
+n
G
n
K
)(s
0
) = 0. If the
interconnection is stable then
1
1+GK
,
G
1+GK
and
K
1+GK
have all the poles in the
strict left half plane. This implies that
d
G
d
K
d
G
d
K
+n
G
n
K
,
n
G
d
K
d
G
d
K
+n
G
n
K
and
d
G
n
K
d
G
d
K
+n
G
n
K
have no poles in the right half plane.
This implies that d
G
(s
0
)d
K
(s
0
) = n
G
(s
0
)d
K
(s
0
) = d
G
(s
0
)n
K
(s
0
) = 0 as the
unstable pole at s
0
has to be cancelled by the respective numerator
polynomials.
SISO STABILITY 13
Note that as d
G
(s
0
)d
K
(s
0
) = 0 at least one of the terms d
G
(s
0
) or d
K
(s
0
) has
to be zero. Lets assume that d
G
(s
0
) ,= 0. In this case d
K
(s
0
) = 0. We also
have that d
G
(s
0
)n
K
(s
0
) = 0. As we have assumed that d
G
(s
0
) ,= 0 we have
n
K
(s
0
) = 0. Thus we have that d
K
(s
0
) = n
K
(s
0
) = 0 which is a contradiction
as we assumed that n
K
and d
K
are coprime polynomials (no common
factors).
Similar conclusion can be reached if one assumes that d
K
(s
0
) ,= 0 in which
case n
G
(s
0
) = d
G
(s
0
) = 0.
In case both d
G
(s
0
) = d
K
(s
0
) = 0 then as n
k
(s
0
)n
G
(s
0
) +d
G
(s
0
)d
K
(s
0
) = 0 it
follows that n
K
(s
0
)n
G
(s
0
) = 0. This will again lead to the conclusion that either
the plant or the controller representation is not coprime leading to a
contradiction.
This proves the theorem.
SISO STABILITY 14
1. I +L with L = GK has all zeros in the strict left half plane
2. There are no unstable pole-zero cancellations while forming the product
GK =
n
G
n
K
d
G
d
K
. That is there no s
0
in the right half plane with
n
G
(s
0
)n
K
(s
0
) = d
G
(s
0
)d
K
(s
0
) = 0.
Proof:() Let the interconnection be stable. This implies that n
G
n
K
+d
G
d
K
has no zeros in the right half plane. This implies that I +L =
n
G
n
K
+d
G
d
K
d
G
d
K
has
no zeros in the right half plane and thus (1) is satised. Also as
n
G
(s
0
)n
K
(s
0
) +d
G
(s
0
)d
k
(s
0
) ,= 0 for all elements s
0
in the right half plane it
follows that there can be no unstable pole zero cancellation in forming the
product GK. This establishes (2).
() Assume (1) and (2) are satised. Then it follows that I +L =
n
G
n
K
+d
G
d
K
d
G
d
K
has no zeros in the right half plane. Suppose there exists a s
0
in the right half
plane such that n
G
(s
0
)n
K
(s
0
) +d
G
(s
0
)d
k
(s
0
) = 0. Then this unstable pole has
to be cancelled by the numerator i.e. d
G
(s
0
)d
K
(s
0
) = 0. This in turn would
SISO STABILITY 15
imply n
K
(s
0
)n
G
(s
0
) = 0 and an unstable pole-zero cancellation will ensue.
This is a contradiction to (2).
This proves the theorem. .
SISO STABILITY 16
Nyquist Plots
Consider a transfer function H(s). In the Nyquist plot of H, the imaginary
part of H(j) is plotted against the real part of H(j).
Consider the transfer function G(s) = s a. We will consider two cases
What happens to the phase of G(s) when s is traversed on a circle in the
clockwise direction that does not contain a.
What happens to the phase of G(s) when s is traversed on a circle in the
clockwise direction that contain a.
SISO STABILITY 17
Argument Principle
a
s
1
s
1
a -
a
s
1
a
-
s
1
In the case when a is is outside the contour (a circle in the gure) then
H(s) =

(s a) remains less than 360 deg as s is made to traverse the
circle in the clockwise direction.
In the case when a is is inside the contour (a circle in the gure) then
H(s) =

(s a) is equal to 360 deg as s is made to traverse the circle in
SISO STABILITY 18
the clockwise direction starting from s
1
and returning to s
1
. As s is made
traverse the circle in the clockwise direction the point G(s) traverses around
the origin in the clockwise direction.
Similarly the contour of G(s) encircles the origin in the counterclockwise
direction if G(s) has a pole inside the countour that s traverses (note that
(s a) =
(
1
sa
).
0.5 1 1.5 2 2.5 3 3.5
0.5
1
1.5
2
2.5
3
3.5
1.5 1 0.5 0 0.5
1.5
1
0.5
0
0.5
real sa
im
g

s
a
0,0
SISO STABILITY 19
The Argument Principle: The contour map of a complex function G(s) will
encircle the origin Z P times in the clockwise direction when the contour
itself is traversed in the clockwise direction where Z and P are the number of
zeros and poles respectively of G(s) that are inside the contour.
SISO STABILITY 20
Nyquist Plot: Argument Principle used to Determine Stability
The closed loop poles are the zeros of 1 +KG(s). Let the number of RHP
zeros of 1 +KG be Z.
The poles of L := KG are same as the poles of 1 +KG = 1 +L which can
be determined as K and G are known quantities. Let the number of right
hand plane poles of L be P.
SISO STABILITY 21
Consider a contour that covers the entire RHP (called the Nyquist
contour;shown above).
The map of 1 +L will encircle the origin N = Z P times where P is a
known quantity.
This implies that L will encircle the origin N = Z P times.
For stability we need Z = 0.
SISO STABILITY 22
1. The Nyquist plot of L encircles the 1 point in the counter-clockwise
direction N number of times where N is equal to the poles of L = GK.
2. There are no unstable pole-zero cancellations while forming the product
GK =
n
G
n
K
d
G
d
K
. That is there no s
0
in the right half plane with
n
G
(s
0
)n
K
(s
0
) = d
G
(s
0
)d
K
(s
0
) = 0.
SISO STABILITY 23
Bode Plots
Bode plot for a given frequency response function H(j) consists of two
subplots
the gain plot where the log
10
[H(j)[ is plotted against log
10
for positive
the phase

H(j) is plotted against log
10
for positive .
SISO STABILITY 24
Bode Plots
Given a plant that is stable the bode plot can be obtained by following the
following steps
Give G an input u(t) = Asin(t) and obtaining the steady state output
y(t). If the system is linear then y(t) will be a sinusoid of the same
frequency .
Let y(t) = y
sin(t +
).
Obtain the ratio [
y
A
[. This will be the magnitude of the frequency
response G(j) at frequency .
Set

(G(j)) =
.
Repeat the steps for various frequencies to obtain G(j).
Note that Spectrum Analyzers obtains the frequency response by
SISO STABILITY 25
essentially following the above steps and often provide G(j) as a complex
number (Example: HP 3565 A).
If the plant is not stable then rst it needs to be stabilized by some
controller. The closed-loop system can now be used in the steps given
above. In steady state all the internal signals in the plant controller
interconnection will be sinusoidal with the same frequency as the frequency
of the sinusoidal input to the closed loop system. The input and the output
sinusoids of the plant G can be employed to determine G(j).
SISO STABILITY 26
Bode plot of s
Bode Diagram
Frequency (rad/sec)
P
h
a
s
e

(
d
e
g
)
M
a
g
n
it
u
d
e

(
d
B
)
20
10
0
10
20
30
40
10
1
10
0
10
1
10
2
89
89.5
90
90.5
91
y = 20 log
10
[G(j)[ = 20 log
10
[j[ = 20 log
10
[[ = 20x,

(G(j)) = 90 deg.
SISO STABILITY 27
Bode plot Contd: plot of 1/s
Bode Diagram
Frequency (rad/sec)
P
h
a
s
e

(
d
e
g
)
M
a
g
n
it
u
d
e

(
d
B
)
40
30
20
10
0
10
20
10
1
10
0
10
1
10
2
91
90.5
90
89.5
89
BODE PLOT OF 1/s
y = 20 log
10
[G(j)[ = 20 log
10
[j[ = 20 log
10
[[ = 20x

(G(j)) = 90 deg.
SISO STABILITY 28
Bode plot Contd: plot of s + 2
Bode Diagram
Frequency (rad/sec)
P
h
a
s
e

(
d
e
g
)
M
a
g
n
it
u
d
e

(
d
B
)
5
10
15
20
25
10
1
10
0
10
1
0
30
60
90
BODE PLOT OF s+2
0
1
Asymptotes
G(j) = j + 2 = 2 if [[ 2
= j if [[ > 2.
SISO STABILITY 29
Bode plot of 1/(s + 2)
Bode Diagram
Frequency (rad/sec)
P
h
a
s
e

(
d
e
g
)
M
a
g
n
it
u
d
e

(
d
B
)
25
20
15
10
5
10
1
10
0
10
1
90
60
30
0
0
1
Bode plot for 1/(s+2)
Asymptotes
G(j) =
1
j+2
=
1
2
if [[ 2
=
1
j
if [[ > 2.
SISO STABILITY 30
Bodes Criterion For Stability
Typical Case
Let K be a positive scalar constant. A typical case is that the closed loop
poles are all in the LHP for small enough K.
As K is increased at least one of the closed-loop poles migrates into the
RHP . The value of K when atleast one of the poles is on the imaginary
axis is when KG is neutrally stable.
SISO STABILITY 31
At this value of K = K
n
1 +K
n
G(j
180
) = 0 and
[K
n
G(j
180
)[ = 1 and

(K
n
G(j
180
) =

(G(j
180
) = 180.
Note that
180
is determined by G alone.
Any value of K less than the neutral value leads to a stable closed loop
system.
This leads to the following conclusions: For all values of K that lead to
stable closed loop maps K < K
n
which is true if and only if
[KG(j
180
)[ < [K
n
G(j
180
)[ = 1.
Thus the rule in this case is that K leads to a stable closed loop map if
[KG(j
180
)[ < 1 where
180
is dened by G(j
180
) = 180.
SISO STABILITY 32
Assumption is that [G(j)[ = 1 for a unique value of .
Note that G(j) is the frequency response of the system.
SISO STABILITY 33
Gain Cross Over Frequency
+
+
+
+
+
-
K G
r
d
n
y
y
m
u
v
Denition 3. Gain crossover frequency for the unity feedback conguration
shown is dened to be the frequency
c
which satises
L(j
c
) = 1
where L := GK.
SISO STABILITY 34
Phase Cross Over Frequency
+
+
+
+
+
-
K G
r
d
n
y
y
m
u
v
Denition 4. Phase crossover frequency for the unity feedback conguration
shown is dened to be the frequency
180
which satises
(L(j
180
)) = 180
where L := GK.
SISO STABILITY 35
Stability Margins
(Gain Margin) The factor by which the gain can be raised before instability
occurs. This is given by
GM := [
1
L(j
180
)
[
where
180
is the phase crossover frequency. Clearly the the closed loop
system is unstable if GM < 1. Typically a GM > 2 is desired.
(Phase Margin) The phase that can be added at the gain crossover
c
frequency before instability occurs. That is
PM :=

(L(j
c
)) + 180
where
c
is the gain crossover frequency. The closed loop system is
unstable if GM is negative.
SISO STABILITY 36
Stability Margins On the Nyquist Plot
Bode Diagram
Frequency (rad/sec)
P
h
a
s
e

(
d
e
g
)
M
a
g
n
i
t
u
d
e

(
d
B
)
40
30
20
10
0
10
Gm = 7.9637 dB (at 0.41248 rad/sec), Pm = 48.448 deg (at 0.20372 rad/sec)
10
2
10
1
10
0
10
1
270
225
180
135
90
45
0
PM
1/GM
180

c

SISO STABILITY 37
Nyquist Diagram
Real Axis
I
m
a
g
i
n
a
r
y

A
x
i
s
1 0.5 0 0.5 1 1.5 2 2.5 3
2
1.5
1
0.5
0
0.5
1
1.5
2
L(j
180
)

L(j
c
)

1(1/GM)
PM
Phase and Gain margins for L = 0.12
(s0.5)
(s+0.1)(s+0.2)
on the Nyquist plot.
Note that P = 0 and thus N has to be zero for stability.
SISO STABILITY 38
Gain-Phase Relationship For Minimum Phase Systems
Suppose G is a LTI system that is such that G(s) is analytic in the RHP (that is
it is stable) and is minimum phase (that is it has no time delays or RHP zeros).
Then the following identity holds
G(j
0
) =
_

d ln [G(j)[
d ln
. .
N()
ln
+
0

0
d.
Thus the phase for such plants is completely determined by its gain [G(j)[.
Also, any other system which has the same gain as [G(j)[ has at least as
much phase as

G(j). That is why the system G is termed minimum phase.
It is clear that ln
+
0
takes large values near =

0
and thus N() can
SISO STABILITY 39
be approximated by N(
0
). Thus
G(j
0
)
_

N(
0
) ln
+
0

0
d =

2
N(
0
).
Note that N() is the slope of magnitude in the bode plot (that is
N() =
d ln |G(j)|
d ln
and in the bode plot log
10
[G(j)[ is plotted against
log
10
).
SISO PERFORMANCE 40
Single Input Single Output Systems:
Performance Measures
SISO PERFORMANCE 41
Unity Negative Feedback Conguration
y
m
= y +n where y
m
is the measured signal which is typically corrupted by
noise n.
e = y r where e is the error signal. Note that e ,= y
m
r as is done in
most treatments. v = y
m
r is the input to the controller. The error signal is
the difference between to be what is desired (r) and what the actual output
is (that is y).
u is the controller output
SISO PERFORMANCE 42
G
d
d is a disturbance (typically has low frequency content).
G is the plant
K is the controller.
SISO PERFORMANCE 43
Important Closed Loop Transfer Functions
y = Gu +G
d
d, u = K(r y
m
), y
m
= y +n.
This implies that y = GK(r y
m
) +G
d
d = GKr GKy GKn +G
d
d
Thus (I +GK)y = GKr GKn +G
d
d.
Thus the output y is given by
y = (I +GK)
1
GK
. .
T
r (I +GK)
1
GK
. .
T
n + (I +GK)
1
. .
S
G
d
d.
We have dened two important closed loop transfer functions
Sensitivity transfer function S = (I +GK)
1
Complimentary transfer function T = (I +GK)
1
GK.
SISO PERFORMANCE 44
Note that S +T = (I +GK)
1
(I +GK) = I.
Note that the error e = y r = (T I)r Tn +SG
d
d Thus
e = Sr Tn +SG
d
d.
Note that we have shown that S +T = 1.
It is worthwhile remembering that
The sensitivity transfer function S is the map between the reference and the
error. Thus small sensitivity S would imply good tracking.
Small sensitivity S would imply good disturbance rejection.
The complimentary transfer function is the map between the noise n and
the error. Thus small complimentary sensitivity T would imply good noise
SISO PERFORMANCE 45
rejection. Note that the noise n is absent in the open loop designs and thus
closed-loop designs should be careful to minimize the effects of n typically
caused by the sensor (otherwise the closed-loop can yield worse
performance than the open-loop).
Remember: Minimize S for good tracking and good disturbance
rejection, minimize T for good noise rejection.
We have shown that S +T = 1. Thus it is clear that it is not possible to
achieve small S and small T in the same frequency region.
The reference trajectories to be tracked have low frequency content.
The noise n effects only in the high bandwidth region (in the low bandwidth
region as the noise is random there is time to average out the effect of
noise).
SISO PERFORMANCE 46
Thus S needs to be low in the low frequency region.
T needs to be low in the high frequency region.
Thus a tradeoff can be made between S and T as the objectives on S and
T are in different frequency regions.
SISO PERFORMANCE 47
Shaping Closed Loop Transfer Functions
Typical Requirements on Sensitivity Transfer Function S.
Minimum bandwidth frequency
B
dened as the frequency where S(j)
crosses 0.707 from below.
S(j) not to exceed certain prespecied values at given frequencies
=
1
, . . . ,
n
(maximum tracking error requirement at certain
frequencies).
S is to have a maximum peak magnitude M (robustness requirement as we
will see later).
Mathematically the requirements can be captured by choosing and
SISO PERFORMANCE 48
appropriate upper bound w
p
(j) such that
[S(j)[
1
[w
p
(j)[
.
The above condition holds if and only if
[S(j)w
p
(j)[ 1
which holds if and only if
sup
[S(j)w
p
(j)[ 1.
For any function f(s) analytic in the RHP the H
norm is dened as
|f|
H
= sup
[f(j)[.
SISO PERFORMANCE 49
Thus the specications on the sensitivity transfer function S takes the form
|w
p
S|
H
1.
SISO PERFORMANCE 50
Weight Selection on S
Suppose the weight needs to capture the following specications
|S|
H
M
p
.
[S(j)[ m
p
for
p
.
Let w
p
(s) =
s/M
p
+
p
s+
p
m
p
. Then
|Sw
p
|
H
1
imposes all the needed conditions.
SISO PERFORMANCE 51
Typical Requirements on Sensitivity Transfer Function S.
Minimum bandwidth frequency
B
dened as the frequency where S(j)
crosses 0.707 from below.
S(j) not to exceed certain prespecied values at given frequencies
=
1
, . . . ,
n
(maximum tracking error requirement at certain
frequencies).
S is to have a maximum peak magnitude M (robustness requirement).
SISO PERFORMANCE 52
Weight Selection For S
Bode Diagram
Frequency (rad/sec)
P
h
a
s
e
(
d
e
g
)
M
a
g
n
itu
d
e
(
d
B
)
60
50
40
30
20
10
0
10
20
10
1
10
0
10
1
10
2
10
3
10
4
10
5
0
45
90
For example with M
p
= 6, m
p
= 1e 3 and
P
= 2827 (
P
= 2f where the
bandwidth is f = 450Hz.) we have
w
p
=
s/M
p
+
p
s +
p
m
p
=
0.1667s + 2827
s + 2.827
.
SISO PERFORMANCE 53
The bode plot of
1
w
p
is shown.
SISO PERFORMANCE 54
Typical Requirements on Complimentary Sensitivity Transfer Function T. Note
that the weight on T should ensure that T is small at high frequencies.
[T(j)[ < 1/A
for all <

T

[T(j)[ < A
h
for all >
T
+
where typically 1/A
1 and thus does not conict with the sensitivity

weighting, A
h
is small forcing T to be small in the high frequency region. A
typical weighting function has the form
w
T
=
s + (1/A
)
T
A
h
s +
T
.
SISO PERFORMANCE 55
The specications on T can be achieved by imposing
[T(j)[
1
[w
T
(j)[
for all
|w
T
T|
H
1
Typical Requirements on KS. The weight on KS is to restrict the magnitude of
the control signal u = KS(r G
d
d). Thus we need
[KS(j)[
1
[w
u
(j)[
which is satised if and only if
|w
u
KS|
H
1
SISO PERFORMANCE 56
Thus the requirements on the closed loop maps translate into the following
conditions
|w
p
S|
H
1
|w
T
T|
H
1
|w
u
KS|
H
1
Note that the search of a controller that satises the above constraints is not
what the standard H
software solves. Instead the problem of nding a

controller to satisfy the stacked constraint
_
_
_
_
_
_
w
p
S
w
T
T
w
u
KS
_
_
_
_
_
_
H
1
SISO PERFORMANCE 57
is solved where the H
norm for a vector valued transfer function f : C C

n
is dened as
|f|
H
:= sup
(f(j)).
SISO PERFORMANCE 58
Generalized Plant: The LFT Framework
P
K
z
w
u
v
w : exogenous variables. This consists of all external signals including the
reference signal.
z : regulated variables. These are the signals which have to be controlled.
SISO PERFORMANCE 59
For example the error signal, the control signals.
v : measured variables. These consist of the inputs to the controller.
Usually the sensor output is fed to the controller.
u : control variable. This is the output of the controller.
SISO PERFORMANCE 60
Generalized Plant: Example
SISO PERFORMANCE 61
+
+
+
+
+
-
K G
r
d
n
y
y
m
u
v
+
+
G
+
+
-
+
d
n
r
K
v
u
z
+
-
P
SISO PERFORMANCE 62
w = [r n d]
,
z = y r = Gu +d r = [I 0 I G]
_
w
u
_
v = r y
m
= r y n = r Gu d n = [I I I G]
_
w
u
_
SISO PERFORMANCE 63
Generalized Plant: Example Contd.
+
+
G
+
+
-
+
d
n
r
K
v
u
z
+
-
r
n
d
u
v
z
-I 0 I G
I -I -I -G
P
K
SISO PERFORMANCE 64
Generalized Plant For The Stacked Problem
+
-
K G
r y
u
G
-
r
K
v
u
P
v
Wp Wu WT
z
3
z
2
Wp
z
1
z
1
WT
2
z
Wu
3
z
SISO PERFORMANCE 65
The transfer function between r and z
1
is W
P
S.
2
is W
T
T.
3
is W
u
KS.
Thus the above setup describes the performance objectives.
The regulated outputs are given by
z =
_
_
z
1
z
2
z
3
_
_
=
_
_
W
p
(r Gu)
W
T
Gu
W
u
u
_
_
,
CASE STUDY 66
and the generalized plant, P is described by
_
z
v
_
=
_
_
_
_
W
p
W
p
G
0 W
T
G
0 W
u
_
_
_
I G
_
. .
=P
_
r
u
_
.
CASE STUDY 67
Nanopositioning: A Quick Introduction to H
Control Design
CASE STUDY 68
Case Study: Nanopositioning
CASE STUDY 69
Serpentine Stage: Unassembled View
CASE STUDY 70
CASE STUDY 71
Serpentine Stage: Working Principle
CASE STUDY 72
Evaluation Stage
CASE STUDY 73
Piezo Actuators
CASE STUDY 74
Block Diagram
CASE STUDY 75
Control Implementation
CASE STUDY 76
Obtaining a Model
10
0
10
1
10
2
10
3
40
30
20
10
0
10
20
30
Comparasion of Model and Experimental Data
Frequency in Hz
M
a
g
n
i
t
u
d
e

(
d
B
)
10
0
10
1
10
2
10
3
400
300
200
100
0
Frequency in Hz
P
h
a
s
e

i
n

D
e
g
The frequency response of the plant with the input being the low voltage
signal to the amplier for the piezo actuators and the output being the LVDT
sensor voltage was obatined. HP 3563 A control system analyzer was
CASE STUDY 77
employed.
This system analyzer stores a complex number corresponding to each
frequency . Each complex number H(j) is the frequency response of the
system at frequency .
Matlab routine invfreqs can be used to t a model to the frequency data.
CASE STUDY 78
Matlab Code
freq=load(freq.txt); % denes the frequency vector
mag=load(mag1.txt); %denes the magnitude in dB (corresponding to the
freq vector)
pha=load(phase1.txt);
freqr=freq*2*pi; mag=10.
(mag/20)
;
phar=unwrap(pha*pi/180);
H=mag.*exp(i*phar);
[num,den]=invfreqs(H,freqr,2,4);
Ht=freqs(num,den,freqr);
magt=abs(Ht);
CASE STUDY 79
phat=unwrap(angle(Ht))*180/pi;
gure;
subplot(2,1,1)
hold on;
plot(freq,mag,freq,magt);
title(Comparasion of Model and Experimental Data);
xlabel(Frequency in Hz);
ylabel(Magnitude);
subplot(2,1,2);
hold on;
plot(freq,phar*180/pi,freq,phat);
CASE STUDY 80
xlabel(Frequency in Hz);
ylabel(Phase in Deg);
G=tf(num,den);
CASE STUDY 81
The transfer function is given by
G(s) =
97030.7242(s
2
1.44e004s + 1.06e008)
(s
2
+ 23.43s + 2.312e006)(s
2
+ 3729s + 2.369e007)
The poles and zeros are at
(1.0e + 003)
_
_
1.8647 + 4.4958i,
1.8647 4.4958i,
0.0117 + 1.5206i,
0.0117 1.5206i
_
_
, (1.0e + 003)
_
7.1993 + 7.3616i,
7.1993 7.3616i
_
Presence of right half plane zeros.
CASE STUDY 82
CASE STUDY 83
+
-
K G
r y
u
G
-
r
K
v
u
P
v
W
p
W
u
W
T
z
3
z
2
W
p
z
1
z
1
W
T
2
z
W
u
3
z
CASE STUDY 84
1
is W
P
S.
2
is W
T
T.
3
is W
u
KS.
The regulated outputs are given by
z =
_
_
z
1
z
2
z
3
_
_
=
_
_
W
p
(r Gu)
W
T
Gu
W
u
u
_
_
,
CASE STUDY 85
and the generalized plant, P is described by
_
z
v
_
=
_
_
_
_
W
p
W
p
G
0 W
T
G
0 W
u
_
_
_
I G
_
. .
=P
_
r
u
_
.
CASE STUDY 86
Weight Selection
CASE STUDY 87
Frequency (rad/sec)
20
0
20
40
60
10
0
10
2
10
4
10
6
100
50
0
50
100
P
h
a
s
e

M
a
g
n
i
t
u
d
e

(
d
e
g
)

(
d
B
)
weighting transfer functions
W
1

W
2

W
1

W
2

CASE STUDY 88
The transfer function, W
p
, is chosen such that it has high gains at low
frequencies and low gains at high frequencies. This scaling ensures that
the sensitivity function is small at low frequencies, thus guaranteeing good
tracking at the concerned frequencies. W
p
was chosen to be a rst order
transfer function,
W
p
= W
1
(s) =
0.1667s + 2827
s + 2.827
.
This transfer function is designed so that its inverse (an upper bound on the
sensitivity function) has a gain of 0.1% at low frequencies (< 1 Hz) and a
gain of 5% around 200 Hz.
The weighting function W
p
puts a lower bound on the bandwidth of the
closed loop system but does not allow us to specify the roll off of the open
loop system to prevent high frequency noise amplication and to limit the
bandwidth to be below Nyquist frequency.
Piezoactuators do not have any backlash or friction and therefore have very
CASE STUDY 89
ne resolution. The resolution of the device, therefore, depends on the
experimental environment and it is limited by thermal and electronic noise.
In any closed loop framework the high resolution of the piezoactuators may
be compromised due the introduction of the sensor noise (in this case the
LVDT) into the system. Clearly this effect is absent in the open loop case.
In the H
paradigm these concerns of sensor noise rejection are reected

by introducing a weighted measure of the complementary sensitivity
function, T, (which is the transfer function between the noise and the
position y).
In this case this weight was chosen to be
W
T
= W
2
=
s + 235.6
0.01s + 1414
CASE STUDY 90
which has high gains at high frequencies (note that noise is in the high
frequency region).
There is another interesting interpretation of this weighting function. It
decides the resolution of the device. Resolution is dened as the variance
of the output signal y, when the device is solely driven by the noise n; i.e.,
resolution is equal to the variance of Tn.
W
T
that guarantees lower roll off frequencies gives ner resolution. In this
way, the trade-off between conicting design requirements of high
bandwidth tracking (characterized by low S, T 1) and ne resolutions
(characterized by low T) are translated to the design of weighting transfer
functions W
p
and W
T
.
The transfer function, KS was scaled by a constant weighting W
u
= 0.1, to
restrict the magnitude of the input signals such that they are within the
saturation limits. This weighting constant gives control signals that are at
CASE STUDY 91
most six times the reference signals.
CASE STUDY 92
Matlab Code Dening the weights wbp=2*pi*450;
Mp=6;mp=1e-3;
mth=1e-2;mtl=1/6;wbt=0.5*wbp;
Mu=1e4;
muv=1/10; red=200;
nump=[1/Mp wbp];denp=[1 wbp*mp];
numt=[1 mtl*wbt];dent=[mth*1 wbt];
numu=[0 muv];denu=[0 1];
sysWp=tf(nump,denp);
sysWt=tf(numt,dent);
sysWu=tf(numu,denu);
CASE STUDY 93
P=[sysWp -sysWp*G;0 sysWt*G;0 sysWu;1 -G];
ssP=minreal(ss(P));
[aP,bP,cP,dP]=ssdata(ssP);
pckP=pck(aP,bP,cP,dP);
qt=1;gmin=0.1;gmax=15;tol=1e-3;epr=1e-12;epp=1e-8;rm=2;
nc=1;nm=1;
[K,cl,gf,ax,ay,hx,hy]=hinfsyn(pckP,nm,nc,gmin,gmax,tol,rm,epr,epp,qt);
[aK,bK,cK,dK]=unpck(K);
ssK=ss(aK,bK,cK,dK);
tfK=tf(ssK);
zpkK=zpk(tfK);
CASE STUDY 94
The optimal value returned is 2.416. Note that this implies that
_
_
_
_
_
_
W
p
S
W
T
T
W
u
KS
_
_
_
_
_
_
H
2.416.
Thus it is not guaranteed that
|W
p
S|
H
1, |W
T
T|
H
1 and |W
u
u|
H
1.
The controller transfer function is given by
277030168.45
(s + 1.414e5)(s
2
+ 23.43s + 2.31e6)(s
2
+ 3729s + 2.37e7)
(s + 1.15e7)(s + 1.414e5)(s + 5643)(s + 2.827)(s
2
+ 3135s + 3.66e7)
.
Another Matlab routine is based on the hinfopt and hinf routines. The code to
use these functions is
CASE STUDY 95
% Assuming that the weights (sysWp, sysWt, sysWu) and the plant transfer
function G are dened as tf objects
ssG=ss(G);
TSS=augtf(ssG,sysWp,sysWu,sysWt);
[gammaopt,ssf,sscl]=hinfopt(TSS);
% gammopt is the optimal gamma value, ssf is the optimal controller (an ss
object) and sscl is the optimal closed loop map (again an ss object)
CASE STUDY 96
This yields an optimal value of 1/0.4102 = 2.438. Note that the gammaopt
value returned by the hinfopt command is the reciprocal of the gamma
value returned by the hinfsyn command. Also, the optimal controllers as
provided by the hinfsyn and hinfopt commands are not the same. The H
optimal controllers are not unique.

The results provided are for the controller associated with the control tool
box (the function hinfsyn).
CASE STUDY 97
Controller and Closed loop Transfer Functions
CASE STUDY 98
100
0
100
controller transfer function
10
0
10
2
10
4
10
6
100
0
100
400
200
0
10
2
10
3
10
4
10
5
10
6
10
7
10
8
1000
500
0
P
h
a
s
e

M
a
g
n
i
t
u
d
e

(
d
e
g
)

(
d
B
)
closed loop transfer function
P
h
a
s
e

M
a
g
n
i
t
u
d
e

(
d
e
g
)

(
d
B
)
CASE STUDY 99
Hysteresis
CASE STUDY 100
4 2 0 2 4
30
20
10
0
10
20
30
input (V)
P
o
s
i
t
i
o
n

(
m
)
1.5 0 1.5
20
0
20
Reference (V)
P
o
s
i
t
i
o
n

(
m
)
(a)
(b)
open loop
max. out. hyst. max. inp. hyst.
1 0.74 m ( 7.2%) 0.14 V (5.8%)
2 2.09 m ( 9.3%) 0.36 V (7.5%)
3 3.46 m ( 9.8%) 0.56 V (7.7 %)
4 4.93 m ( 10.0 %) 0.73 V (7.6 %)
1
2
3
4
closed loop
max. out. hyst.:
62.3 nm (0.14 %)
max. inp. hyst.:
2 mV (0.07 %)

CASE STUDY 101
Creep
CASE STUDY 102
Imaging: Closed and Open Loop
CASE STUDY 103
CASE STUDY 104
Imaging: Closed and open loop
CASE STUDY 105
FUNDAMENTAL LIMITATIONS 106
Reading Assignment
Read the paper
S. Salapaka, A. Sebastian, J. P. Cleveland and M. V. Salapaka, High
Bandwidth Nano-positioner: A Robust Control Approach, Review of Scientic
Instruments, Vol. 73, no. 9, pp. 3232-3241.
Fundamental Limitations For Single-input
Single-output Systems
Denition 5. (Analytic functions, holomorphic functions) Let be a domain
in C and let f be a function dened on . Then f is said to be analytic or
holomorphic at s
0
in C if
df
ds
(s
0
) exists. It is analytic or holomorphic in if it
analytic or holomorphic at all elements of .
Denition 6. (Entire functions) A function is said to be entire if it is analytic
on C.
Example 1. Rational functions on the complex plane are analytic everywhere
on the the complex plane except at the poles.
Denition 7. (Rectiable curve, simple curve, closed curve) A set in the
complex plane C is a rectiable curve if there exists a continuously
differentiable function : [a, b] R C such that = ([a, b]).
A rectiable curve is a simple curve if it does not intersect itself. That is the
associated function is such that (x) ,= (y) if x ,= y for all x, y (a, b).
A rectiable curve is closed if (a) = (b).
Denition 8. (Contour) A contour is a collection of rectiable curves
j
such that the nal point of
j
is the initial point of
j+1
. Closed and simple
contours are analogously dened as the corresponding denitions for curves.
Denition 9. (Integral) For a function f that is continuous on the domain S
the integral along a rectiable curve S is dened as
_
f(s)ds :=
_
b
a
f((x))
d
dx
(x)dx,
where ([a, b]) = .
The integral over a contour is dened as
_
f(s)ds :=
n
j=1
_
b
j
a
j
f(
j
(x))
d
j
dx
(x)dx
where
j
=
j
([a
j
, b
j
]), j = 1, . . . n forms the contour .
Denition 10. (Positively oriented contour)
Consider a simple, closed contour formed by rectiable curves
j
=
j
([a
j
, b
j
]), j = 1, . . . n.
Let x
0
be such that x
0
[a
j
, b
j
] such that
d
j
dx
(x
0
) ,= 0
If the vector obtained by rotating the tangent vector at x
0
given by
d
j
dx
(x
0
) by
90 degrees anticlockwise points to the inside of the contour then the closed
simple contour is positively oriented.
Maximum Modulus Theorem
Theorem 6. (Maximum Modulus theorem) Suppose that is a non-empty,
open, connected set in the complex plane and F is a function that is analytic
in . Suppose that F is not equal to a constant. Then [F[ does not attain its
maximum value at an interior point of
A simple application of the above theorem is the following fact for a stable
transfer function F:
|F|
H
= sup
R
[F(j)[ = sup
Re(s)0
[F(s)[.
Cauchys Theorem
Theorem 7. (Cauchys theorem) Consider the simply connected domain S
that contains the simple, closed contour that is positively oriented. If f is
analytic in S then
_
f(s)ds = 0.
Also, for any point s
0
S
1
2j
_
f(s)
(s s
0
)
ds = f(s
0
).
Weighted Cauchys Theorem
Theorem 8. Let F be analytic and of bounded magnitude on
s C[Re(s) 0. Let s
0
= x +jy be a point such that x > 0. Then
F(s
0
) =
1
F(j)
x
x
2
+ ( y)
2
d.
Proof: Consider the Nyquist Contour D(r) of radius r that includes s
0
. From
Cauchys theorem we have that
F(s
0
) =
1
2j
_
D(r)
F(s)
(s s
0
)
ds.
Note that s
0
= x +jy is in the strict left half plane and thus is not inside the
Nyquist Contour. This implies that the function
F(s)
s+s
0
is analytic inside D(r).
Thus using Cauchys theorem it follows that
1
2j
_
D(r)
F(s)
(s +s
0
)
ds = 0.
Subtracting the two integrals
F(s
0
) =
1
2j
_
D(r)
F(s)(
1
(ss
0
)

1
(s+s
0
)
)ds
=
1
2j
_
r
r
F(j)(
1
(js
0
)

1
(j+s
0
)
)jd+
1
2j
_
/2
/2
F(re
j
)(
1
(re
j
s
0
)

1
(re
j
+s
0
)
)rje
j
d
=
1
2j
_
r
r
F(j)(
2x
(js
0
)(j+s
0
)
)jd+
1
2j
_
/2
/2
F(re
j
)(
2x
(re
j
s
0
)(re
j
+s
0
)
)rje
j
d
=
1
_
r
r
F(j)
x
x
2
+(y)
2
d+
1
_
/2
/2
F(re
j
)(
x
(re
j
s
0
)(re
j
+s
0
)
)re
j
d
=: I
1
(r) +I
2
(r)
Note that as r , I
1
(r)
1
F(j)
x
x
2
+(y)
2
d.
Note that
[I
2
(r)[
1
_
/2
/2
[F(re
j
)[(
x
|(e
j
r
1
s
0
)| |(e
j
+r
1
s
0
)|
)r
1
d
1
r
|F|
H
_
/2
/2
(
x
|(e
j
r
1
s
0
)| |(e
j
+r
1
s
0
)|
)d
Const
1
r
Thus I
2
(r) 0 as r .
Thus
F(s
0
) = lim
r
I
1
(r) =
1
F(j)
x
x
2
+ ( y)
2
d.
This proves the lemma.
All-pass and Minimum Phase Transfer Function
Denition 11. (All pass transfer function) A stable proper rational function G
is all pass if
[G(j)[ = 1, R.
It can be shown that if G is an all pass transfer function then s
0
is a pole of G if
and only if s
0
is a zero. Thus all pass functions have the form
G(s) =
n
i=1
s +s
n
s s
n
.
Denition 12. (Minimum-phase transfer functions) A proper rational function
is minimum phase if all its zeros are in the strict left half plane.
All-pass/Minimum Phase Factorization
Theorem 9. (All-pass/minimum phase factorization) Every stable proper
rational function G admits a factorization of the form
G = G
ap
G
mp
where G
ap
is all pass and G
mp
is minimum phase.
Proof: Let G(s) = K
(sz
1
)...(sz
n
)
(sp
1
)...(sp
k
)
. As G is stable it is clear that p
i
are all in the
left half plane. Without loss of generality assume that z
1
, z
2
, . . . , z
m
are the
zeros in the right half plane (we will assume that there are no zeros on the j
axis). Then it is clear that
G(s) = [
m
i=1
s z
i
s +z
i
]
. .
G
ap
[K
m
i=1
(s +z
i
)
n
i=m+1
(s z
i
)
k
i=1
(s p
i
)
]
. .
G
mp
,
where as z
i
is in the strict right half plane, z
i
is in the strict left half plane.
Clearly G
ap
is all pass and G
mp
is minimum phase.
A Lemma
Lemma 1. Let G(s) be a stable proper transfer function with the factorization
G = G
ap
G
mp
with G
ap
being all-pass and G
mp
being minimum phase. Let
s
0
= x +jy be in the strict right half plane. Then
log [G
mp
(s
0
)[ =
1
log [G(j)[
x
x
2
+ ( y)
2
d.
Proof: Let F := log(G
mp
). As G
mp
is analytic in the right half plane and has
no zeros in the right half plane it follows that F is analytic in the right half
plane. Applying Lemma 8 it follows that
F(s
0
) =
1
F(j)
x
x
2
+ ( y)
2
d.
Taking the real part on both sides we have
Re(F(s
0
)) =
1
Re(F(j))
x
x
2
+ ( y)
2
d. (1)
Note that G
mp
= e
F
= e
Re(F)
e
jImg(F)
. Thus [G
mp
[ = e
Re(F)
and
log [G
mp
[ = Re(F).
It follows from (1) that
log [G
mp
(s
0
)[ =
1
log [G
mp
(j)[
x
x
2
+ ( y)
2
d.
Noting that [G
mp
(j)[ = [G(j)[ it follows that
log [G
mp
(s
0
)[ =
1
log [G(j)[
x
x
2
+ ( y)
2
d.
Fundamental Limitations on Performance
Let L := GK.
We have seen that typical performance requirements need S =
1
1+L
to be
small for good tracking and disturbance rejection.
It is desired that T = I S be small for good noise rejection.
Given a certain set of objectives it is desirable to evaluate the feasibility of the
specications that are targeted.
Cautionary Example
The importance of fundamental limitations is highlighted by the following
example that concerns the design of X-29 aircraft. Considerable design effort
was directed towards designing a controller that provides a phase margin of at
least 45 degrees. However, a simple argument based on a result to be
developed that utilizes the presence of an unstable pole and a right half plane
zero would have indicated the infeasibility of such a requirement. Clearly
utilization of results that yield such an analysis can lead to signicant
economy in time, effort and cost.
Waterbed Effect I
Theorem 10. (Waterbed effect I) Let L have relative degree two and let L
have N
p
poles in the right half given by p
1
, . . . , p
N
p
. If the closed-loop system
is stable then S =
1
1+L
satises
_

0
ln [S(j)[d =
N
p
i=1
Re(p
i
).
Proof:Note that the poles of L are the zeros of S. Thus p
i
are the right half
plane zeros of S. Thus
S
ap
(s) =
N
p
i=1
s p
i
s +p
i
.
From Lemma 1 it follows for any x > 0 that
ln [S
mp
(x)[ =
1
ln [S(j)[
x
x
2
+
2
d =
1
_

0
ln [S(j)[
2x
x
2
+
2
d.
Thus it follows that
_

0
ln [S(j)[
x
2
x
2
+
2
d =

2
xln [S
mp
(x)[.
Therefore
lim
x
_

0
ln [S(j)[
x
2
x
2
+
2
d = lim
x
2
xln [S
mp
(x)[
which implies that
_
0
ln [S(j)[d = lim
x
2
xln [S
mp
(x)[
=

2
(lim
x
xln [S(x)[ lim
x
xln [S
ap
(x)[)
=

2
(0 +
N
p
i=1
Re(p
i
)),
where lim
x
xln [S(x)[) = 0 follows from the hypothesis that L has relative
degree atleast two.
Waterbed Effect II
Theorem 11. (Waterbed effect II; Weighted Sensitivity Integral) Let L have
N
p
poles in the right half given by p
1
, . . . , p
N
p
. Let z = x +jy be any zero of L
in the strict right half plane (that is x > 0). If the unity feedback system is
stable then S =
1
1+L
is such that
_

0
ln [S(j)[(
x
x
2
+ ( y)
2
)
+ (
x
x
2
+ ( +y)
2
)d = ln(
N
p
i=1
z +p
i
z p
i
).
Proof:
Note that the poles of L are the zeros of S. Thus p
i
are the right half plane
zeros of S. Thus
S
ap
(s) =
N
p
i=1
s p
i
s +p
i
.
Using Lemma 1 it follows that
ln [S
mp
(z)[ =
1
ln [S(j)[
x
x
2
+(y)
2
d.
=
1
0
ln [S(j)[(
x
x
2
+(y)
2
+
x
x
2
+(+y)
2
)d.
This implies that
1
0
ln [S(j)[(
x
x
2
+(y)
2
+
x
x
2
+(+y)
2
)d = ln [S
mp
(z)[ = ln [
S(z)
S
ap
(z)
[
= ln
1
|S
ap
(z)|
= ln(
N
p
i=1
z+p
i
zp
i
)
Bounds on Weighted S and T Transfer Functions
Theorem 12. Supoose L has right half plane poles and zeros at p
1
, . . . , p
N
p
and z
1
, . . . , z
N
z
respectively. If the closed-loop system is stable then
1. |w
p
S|
H
max
j
[w
p
(z
j
)[
N
p
i=1
z
j
+p
i
z
j
p
i
.
2. |w
T
T|
H
max
i
[w
T
(p
i
)[
N
z
j=1
z
j
+p
i
z
j
p
i
.
Proof:
Note that for any z
j
|w
p
S|
H
= sup
R
[w
p
(j)S(j)[ = sup
R
[w
p
(j)S
mp
(j)[
= sup
Re(s)0
[w
p
(s)S
mp
(s)[
[w
p
(z
j
)S
mp
(z
j
)[
= [w
p
(z
j
)
S(z
j
)
S
ap
(z
j
)
[ = [w
p
(z
j
)[
N
p
i=1
z
j
+p
i
z
j
p
i
.
The third equality above from from the maximum-modulus theorem
(Theorem 6) and the last equality follows by noting that as z
j
is a zero of L,
S(z
j
) =
1
1+L(z
j
)
= 1. This proves the rst part of the theorem. The proof for T
is similar.
Note that the terms
z
j
+p
i
z
j
p
i
1 for all relevant i and j
s.
Bandwidth Limitations For Typical Weights
Note that for achieving the objectives of
|S|
H
M and
[S(j)[ m
p
for all
B
an appropriate weight is
w
p
=
s/M
p
+
B
s +
B
m
p
.
The following corollary takes the limiting case of m
p
= 0 and M
p
= 2.
Corollary 1. Let z be any right half plane zero of L. Let
w
p
=
s/M
p
+
B
s +
B
m
p
.
Then for the performance objective
|w
p
S|
H
1
to be achieved the following conditions have to be satised
If z is real then
B
< z
1 1/M
p
1 m
p
.
In particular if M
p
= 2 and m
p
= 0 then
B
< z/2.
If z is purely imaginary and M
p
= 2 and m
p
= 0 then
B
< [z[
3
2
.
Proof: From Theorem 12 we have that
|w
p
S|
H
max
j
[w
p
(z
j
)[
N
p
i=1
z
j
+p
i
z
j
p
i
[w
p
(z)[
N
p
i=1
z
j
+p
i
z
j
p
i
[w
p
(z)[.
Thus if the performance specication
|w
p
S|
H
1,
has to be achieved then it is necessary that [w
p
(z)[ < 1. Thus
[z/M
p
+
B
[ < [z +
B
m
p
[ has to be satised. If z is real then this implies that
B
< z
1 1/M
p
1 m
p
,
whereas if z is purely imaginary with M = 2 and m
p
= 0 then
B
< [z[
3
2
.
This proves the corollary.
Note that the weight on T should ensure that T is small at high frequencies.
[T(j)[ < 1/M
T
for all <
T

[T(j)[ < m
T
for all >
T
+
where typically 1/M
T
1 and thus does not conict with the sensitivity
weighting, m
T
is small forcing T to be small in the high frequency region. A
typical weighting function has the form
w
T
=
s + (1/M
T
)
T
m
T
s +
T
.
The specications on T can be achieved by imposing
|T|
H

1
w
T
(j)
|w
T
T|
H
1
Corollary 2. Let p be any right half plane pole of L. Let
w
T
=
s + (1/M
T
)
T
T
where we have set m
T
= 0. Then for the performance objective
|w
T
T|
H
to be achieved the following conditions have to be satised

If p is real then
T
> p
M
T
M
T
1
.
In particular if M
T
= 2 then
T
> 2p
If p is purely imaginary then
T
> [p[
M
T
_
M
2
T
1
.
In particular if M
T
= 2 then
T
> 1.15[p[
Proof: From Theorem 12 we have that
|w
T
T|
H
max
i
[w
T
(p
i
)[
N
z
j=1
z
j
+p
i
z
j
p
i
[w
T
(p)[
N
z
j=1
z
j
+p
i
z
j
p
i
[w
T
(p)[.
Thus if the performance specication
|w
T
T|
H
1,
has to be achieved then it is necessary that [w
T
(p)[ < 1. Thus
[p + (1/M
T
)
T
[ < [
T
[ has to be satised.
The rest of the proof follows from this condition.
Bandwidth Limitation: Crossover Frequency
Let z be the zero of L(s) in the right half plane closest to the j axis. Then
from Corollary 1 it follows that the crossover frequency has to be chosen
such that
c
<
z
2
.
Let p be the pole of L(s) in the right half plane farthest from the j axis.
Then from Corollary 2 it follows that the crossover frequency has to be
chosen such that
c
> 2p.
This would imply that
c
has to satisfy
2p
c

z
2
.
This would necessarily imply that z > 4p to achieve good performance. In
case this is not satised no controller will yield satisfactory performance.
Bandwidth Limitation Imposed by Disturbance Rejection
The error due to disturbance is given by
e = SG
d
d.
If G
d
is appropriately scaled then the objective of disturbance rejection is
captured by
[e()[ 1, whenever [d()[ 1.
In other words the objective is to ensure
|SG
d
|
H
1.
A typical G
d
has low frequency content. Let
d
be the value such that
[G
d
(
d
)[ = 1 that is
d
is the frequency at which G
d
crossed the 0dB line from
above.
From Theorem 12 it follows that
|SG
d
|
H
[G
d
(z)[
where z is any right half plane zero of L(s). Thus for good disturbance
rejection it is needed that [G
d
(z)[ < 1.
|SG
d
|
H
1 implies that
[S(j)[
1
[G
d
()[
.
Note that [G
d
())[ > 1 for all <
d
. Thus it follows that [S(j)[ < 1 for all
<
d
. This would imply that
B
>
d
. Thus good disturbance rejection
requires that the controller ensure that
B
>
d
.
Bandwidth Limitation Imposed by Input Bounds
Suppose we need the following condition to be satised:
[e(j)[ 1 and [u(j)[ 1, whenever [d(j)[ 1 and r = n = 0.
Note that
e = y r = G
d
d +Gu r.
Assuming the needed condition (that is [e()[ < 1 when d() < 1 and
r = n = 0) is satised we have for any [d()[ < 1 that
[Gu[ = [e G
d
d[ [G
d
d[ [e[
[G
d
[ [d[ 1
Clearly the above condition holds for any d with [d[ = 1 which implies that
[G[ [G[ [u[ [G
d
[ 1.
Bandwidth Limitation Imposed by Input Bounds for Unstable
Plants
When the plant is unstable more restrictive conditions can be derived due to
disturbance rejection. Note that the map between the control signal u and the
disturbance d is given by
u = KSG
d
d = G
1
TG
d
d.
From Corollary 2 it follows that if p is a right half plane pole
T
> p
M
T
M
T
1
> p.
It follows that
[T[ > 1, < p.
As u = TG
1
G
d
d if the condition [u[ < 1 whenever [d[ < 1 needs to be
satised then
[G[ > [G
d
[, < p.
SISO ROBUST STABILITY 146
Limitation Imposed by Reference Tracking
Assume that the references r to be tracked are well modeled as r = R r where
[ r()[ 1. The performance objective is that
if [ r()[ < 1, for all then [e()[ < 1 and [u()[ < 1 for all <
r
.
If the above condition is satised and [ r[ < 1 then
[Gu[ = [R r +e[ [R r[ [e[ [R r[ 1 for all <
r
.
The above relationship is also satised for any r that is such that [ r()[ 1,
and [G r[ = [G[. Thus we have
[G[ [u[ [R[ 1 for all <
r
As [u[ < 1 it follows that
[G[ [R[ 1 for all <
r
.
Robust Stability for SISO systems
Introduction
A control system is said to be robust if it is insensitive to the differences
between the actual system and the model used to design the controller.
The differences between the model and the actual plant is called the model
uncertainty.
In the robust control paradigm the key concept is to design controllers that
fulll the specications even for the worst case uncertainty. The approach that
is pursued is
Characterize the uncertainty mathematically.
Analyze and synthesize controllers that achieve Robust stability (RS), that
is analyze and synthesize controllers that ensure stability of the closed loop
for all plants in the uncertainty set.
Analyze and synthesize controllers that achieve Robust performance (RP),
that is analyze and synthesize controllers that ensure stability and
performance of the closed loop for all plants in the uncertainty set.
Sources of Uncertainty
Nonlinearities: Note that a central design criteria for the robust control
paradigm is the use of linearity. However, most plants exhibit nonlinear
behavior. This leads to uncertainty.
Uncertain parameters: Some of the parameters are uncertain in the model.
Measurement equipment: Note that the measurement device has nite
resolution and the equipment used to obtain the frequency response has
limited capabilities. Thus often it is impossible to ascertain the model and
high frequency where even the model order and structure cannot be
determined.
Undermodeling: Often the detailed and precise model of the plant is of very
high order making it unsuitable for engineering purposes. Thus a lower
order model is chosen resulting in uncertainty.
Implementation: The controller implemented might not be the same as the
one obtained by the design procedure. For example, often the design is
performed in continuous time and implementation digital. The involved
delays and approximations lead to uncertainty.
Classes of Uncertainty
Parametric uncertainty: The model order and structure is assumed to be
known. However, specic parameters that are real are uncertain in the
model. Parametric uncertainty is quantied by assuming that the parameter
lies in a certain region [
min
,
max
].
Unmodeled or undermodeled dynamics uncertainty: Here the model order
and the structure is not certain. Such types of uncertainty results from
either delibrate undermodeling or from a lack of physical understanding and
unknown dynamics at higher frequecies.
Lumped uncertainty: This class of uncertainty can accommodate the above
two types of uncertainty by lumping them into a single description.
Notation

LTI
:= linear time invariant plants.

LTV
:= linear time varying plants.

NL
:= nolinear plants.
G
nom
: The nominal plant assumed to be LTI.
We will identify to be
LTI
unless otherwise stated. Also, we represent by
G
p
any element of ; G
p
is the perturbed plant in contrast to G
nom
which is
the nominal plant.
Typical Uncertainty Characterization
+
+
G
u y
w
A
nom
Additive uncertainty:
G
p
= G
nom
+w
A
w
A
is a weight usually chosen to be stable and minimum phase
is scaled to lie is a set. For example
H
[||
H
1.
Multiplicative Uncertainty
+
+ u y
w
I
G
nom
Multiplicative uncertainty:
G
p
= G
nom
(I +w
I
)
w
I
H
[||
H
1.
Inverse Multiplicative Uncertainty
+
+
u y
w
iI
G
nom
Inverse Multiplicative uncertainty:
G
p
= G
nom
(I +w
iI
)
1
w
iI
H
[||
H
1.
Parametric Uncertainty
The uncertain parameter is assumed to lie inside an interval [
min
,
max
].
Thus can be represented by
= (1 +r
)
where =

min
+
max
2
, r
=

min
max
min
+
max
and [1, 1].
Example 2. (Gain uncertainty) Let
= k
p
G
0
(s)[k
min
k
p
k
max
Dene k =
k
min
+k
max
2
, r
k
=
k
min
k
max
k
min
+k
max
and [1, 1]. Then
G
p
(s) = kG
0
(s)
. .
G
nom
(s)
(1 +r
k
).
G
p
is in the multiplicative uncertainty form.
Example 3. (Time constant uncertainty) Let
=
1
p
s + 1
G
0
(s)[
min

max
.
Dene =

min
+
max
2
, r
=

min
max
min
+
max
and [1, 1]. Then
G
p
(s) =
1
p
s + 1
G
0
(s) =
1
s +r
s + 1
G
0
(s) =
1
s + 1
G
0
(s)
. .
G
nom
(s)
(1 +w
iI
(s))
1
;
where
w
iI
=
r
s
1 +s
.
G
p
is in the inverse multiplicative form.
Example 4. Consider a plant with an uncertain zero
:=
s +a
s
2
+ 3s + 1
[a
min
a a
max
.
Multiplicative uncertainty form
Dene a =
a
min
+a
max
2
, r
a
=
a
max
a
min
2
and [1, 1]. This implies that
G
p
(s) =
s +a
s
2
+ 3s + 1
=
s +a +r
a
s
2
+ 3s + 1
= (
s +a
s
2
+ 3s + 1
)
. .
G
nom
(s)
(I +w
I
)
where w
I
=
r
a
s+a
.
Additive uncertainty form
Dene a =
a
min
+a
max
2
, r
a
=
a
max
a
min
2
and [1, 1].
This implies that
G
p
(s) =
s +a
s
2
+ 3s + 1
=
s +a +r
a
s
2
+ 3s + 1
=
s +a
s
2
+ 3s + 1
. .
G
nom
(s)
+w
A
where w
A
=
r
a
s
2
+3s+1
.
Remarks
Either multiplicative or additive uncertainty forms can be used to represent
the uncertain set .
Multiplicative uncertainty form represents relative error:
G
p
G
nom
G
nom
= w
I
.
Additive uncertainty form represents the absolute model error:
G
p
G
nom
= w
A
.
Robust Control Oriented Modeling
The modeling suited for the robust control paradigm involves the following
steps:
1. Obtain the model class . G
p
is any particular element of .
2. Obtain the nominal model G
nom
(s).
3. Obtain the bound on the deviation of the actual behavior of the plant from
the nominal behavior.
For the additive uncertainty characterization the deviation is given by the
function
A
() = max
G
p
[G
p
(j) G(j)[.
For the multiplicative uncertainty characterization the deviation is given
by the function
I
() = max
G
p
G
p
(j) G(j)
G(j)
.
4. Obtain the weight that describes the deviation, that is, choose a rational
weight (w
A
(s) and w
I
(s) for additive and multiplicative uncertain form
respectively) that has low order stable and such that [w
(j)[
(j)
(where = A or = I).
Obtaining a Model Class
The following methods can be utilized
1. A model derived based on the understanding of the plant. The plant model
for example could be derived based on physical principles and rough
estimates on the parameters of the model can be derived.
Advantages: The resulting model is typically simple and captures the
qualitative dynamics well.
Disadvantages: Not always possible or difcult to obtain. For example in
the nanopositioning example the serpentine stage is quite intricate and
obtaining a model of the system based on physical arguments is difcult.
2. Evaluate the frequency response of the system at various experimental
conditions and obtain the frequency response repeated number of times.
For example, if the frequency response is obtained about different bias
voltages to the the piezo different plots are obtained for the nanopositioning
stage. Also, the gain at DC depends on the history of the applied voltage
due to hysteresis. Also, the plant is slightly time varying due to creep and
other effects.
Advantages: Not hard to obtain as it does not involve much analysis.
Disadvantages: The resulting model can be quite involved and might not
capture the physics of the plant.
Obtaining a Nominal Model
Once the model class is obtained one has to choose a nominal model and
the associated uncertainty has to be determined. The following approaches
can be taken to identify the nominal model.
1. A simplied model obtained by ignoring delays and higher order dynamics.
For example, if the model class was determined to be
G
p
(s) =
s + 20
(s + 1)(0.1s + 1)
e
s
with [
min
,
max
] then one can choose
G
nom
(s) =
20
s + 1
The advantages are the simplicity of the nominal model that can lead to
easier controller design. The disadvantage is the large uncertainty that
might result.
2. If the model class is characterized by multiple parameters then choose the
nominal model to the one with the parameters taken to be central values of
the ranges involved. (see the examples derived earlier).
3. At every choose G(j) as the point on the Nyquist plot that leads to the
smallest uncertainty. This leads to the smallest uncertainty however, it
needs considerable effort, the resulting nominal plant can be of very high
order and the nominal model might not capture the essential features of the
system.
4. Typically a judicious combination of the above three methods provides the
best alternative.
Determining the Uncertainty Bound
Once the nominal model is xed, then the uncertainty bound (j) has be
determined. Note that for the additive uncertainty case the bound is dened as
A
(j) = sup
G
p
[G
p
(j) G
nom
(j)[
whereas for multiplicative uncertainty form we have
I
(j) = sup
G
p
G
p
(j) G
nom
(j)
G
nom
(j)
It is evident that the above formulae cannot always be utilized to generate the
bound mainly because the sup is over an innite number of plants and it has to
be evaluated over all R. Different techniques are employed depending
upon the data available.
Example 5. (Model is known with parameters uncertain) In this case, one
possible method of evaluating the bound is to rst grid the parameter region
and obtain the frequency plot for each parameter vector on the grid. Let G
k
denote the k
th
model. A grid is obtained on the frequency region . Let the
corresponding frequency vector be =
1
, . . . ,
n
. For additive uncertainty
A
(j
i
) = max
k
[G
k
(j
i
) G
nom
(j
i
)[
i

and in the case of multiplicative uncertainty we have
I
(j
i
) = max
k
G
p
(j
i
) G
nom
(j
i
)
G
nom
(j
i
)
i
.
We will obtain the multiplicative uncertainty description of the following class:
=
k
s + 1
e
s
, 2 k, , 3.
The nominal model is chosen as
G
nom
(s) =
2.5
2.5s + 1
.
The attached Matlab code does the appropriate gridding of the parameter
space and the frequencies.
Robust Stability Condition for Additive Uncertainty
+
+
G
w
A
nom
K
-
G
p
We will consider the following model class
:= G(s) +w
A
(s)(s) [||
H
1.
where w
A
(s) is assumed to be a stable, proper rational transfer function. We
will denote by L = KG and by L
p
= KG
p
where G
p
.
Assumption 1. We will assume that the nominal model G(s) is such that the
unity feedback conguration shown in the gure above (with = 0) is stable.
Theorem 13. The closed loop system shown in Figure is robustly stable (that
is the for all G
p
) if and only if
|w
A
KS|
H
< 1
where S := (I +L)
1
is the sensitivity transfer function corresponding to the
nominal plant.
Proof: By assumption we have that the with = 0 the closed loop system is
stable. Let the number of encirclements of 1 by the Nyquist plot of L be N.
Note that as and w
A
are assumed to be stable, it follows that the number of
poles in the right half plane of any G
p
K = GK +w
A
K in is not greater
than the number of rhp poles of L = GK.
() We are given that the number of encirclements by the Nyquist plot of L is
N. Suppose |w
A
KS|
H
< 1. Thus
|w
A
KS|
H
< 1
|w
A
KS|
H
|w
A
KS|
H
||
H
< 1 if ||
H
1
[(w
A
K)(j)[ < [1 +L(j)[ for all , ||
H
1
[(L
p
L)(j)[ < [1 +L(j)[ for all
Thus the distance of the Nyquist plot of L from 1 is greater than the
distance of the perturbed open loop gain L
p
from L. As L encircles 1
point N times L
p
also encircles 1, N times.
Note that if the number of rhp poles of L is P then as the nominal system is
stable the number of rhp zeros Z of 1 +L is zero we have N = P (thus
Z P = N i.e N counterclockwise encirclements).
Let Z
p
and P
p
denote the number of rhp poles zeros and poles of 1 +L
p
. As
the number of counterclockwise encirclements of 1 point of L
p
is N we
have Z
p
P
p
= N. Thus Z
p
= P
p
N. However, we have already seen
that P P
p
as the weight w
A
and are stable. Thus
Z
p
= P
p
N P N = 0.
This implies Z
p
= 0. This in turn implies P
p
= P and that there are no
unstable pole-zero cancellations in the product G
p
K.
Thus there are no unstable pole-zero cancellations in the product G
p
K and
the number of counterclockwise encirclements of the 1 point on the
complex plane is equal to the number of unstable poles of G
p
K.
From Theorem 5 the interconnection is stable.
We will not prove that |w
A
KS|
H
< 1 is a necessary condition.

Robust Stability Condition for Multiplicative Uncertainty
+
+
w
I
G
-
K
G
p
:= G(s)(I +w
I
(s)(s))[ ||
H
1.
where w
I
p
= KG
p
where G
p
.
is the for all G
p
) if and only if
|w
I
T|
H
< 1
where T := L(I +L)
1
is the complimentary sensitivity transfer function
corresponding to the nominal plant.
Proof: By assumption we have that the with = 0 the closed loop system is
stable. Let the number of encirclements of 1 by the Nyquist plot of L be N.
Note that as and w
I
are assumed to be stable, it follows that the number of
poles in the right half plane of any L
p
= G
p
K = GK(1 +w
I
) in is not
greater than the number of rhp poles of L = GK.
() We are given that the number of encirclements by the Nyquist plot of L is
N. Suppose |w
I
T|
H
< 1. Thus
|w
I
T|
H
< 1
|w
I
T|
H
|w
I
T|
H
||
H
< 1 if ||
H
1
[(w
I
GK)(j)[ < [1 +L(j)[ for all R if ||
H
1
[(L
p
L)(j)[ < [1 +L(j)[ for all R
Thus the distance of the Nyquist plot of L from 1 is greater than the
distance of the perturbed open loop gain L
p
from L. As L encircles 1
point N times L
p
also encircles 1, N times. Note that if the number of rhp
poles of L is P then as the nominal system is stable the number of rhp
zeros Z of 1 +L is zero we have N = P (thus Z P = N i.e N
counterclockwise encirclements).
Let Z
p
and P
p
denote the number of rhp poles zeros and poles of 1 +L
p
. As
the number of counterclockwise encirclements of 1 point of L
p
is N we
have Z
p
P
p
= N. Thus Z
p
= P
p
N.
However, we have already seen that P P
p
as weight w
I
and are stable.
Thus Z
p
= P
p
N P N = 0. This implies Z
p
= 0.
Thus there are no unstable pole-zero cancellations in the product G
p
K and
the number of counterclockwise encirclements of the 1 point on the
complex plane is equal to the number of unstable poles of G
p
K.
From Theorem 5 the interconnection is stable.
We will not prove that |w
I
T|
H
< 1 is a necessary condition.

The M Conguration
M
Theorem 15. Consider the interconnection depicted in the Figure above

where M and are two LTI stable causal systems such that ||
H
1.
Then the interconnection is stable if and only if |M|
H
< 1.
Proof: It follows from the Nyquist criterion that as M and are stable (no rhp
poles) the unity feedback interconnection of Figure is stable if and only if
[1 +M(j)(j)[ does not encircle or touch the point 0.
Thus robust stability is achieved if and only if
[1 +M(j)(j)[ > 0, , such that ||
H
1. (2)
If |M|
H
< 1 then 1 [M(j)[ [(j))[ > 0 for any ||

H
1 (as
[M(j)[ [(j))[ < 1). Thus [1 +M(j)(j)[ > 0 for all . Thus robust
stability is ensured if |M|
H
< 1.
Suppose |M|
H
1 then we can construct a such that is a stable

proper transfer function with ||
H
1 and there exists an where

1 +M(j)(j) = 0. This would violate the condition (2) and thus there is
no robust stability.
Robust Stability Condition for Inverse Multiplicative
Uncertainty
+
+
y
w
iI
G
K
-
G
p
:= G(s)(I +w
iI
(s)(s))
1
[ ||
H
1.
where w
iI
p
= KG
p
where G
p
.
is the for all G
p
) if and only if
|w
iI
S|
H
< 1
where S := (I +L)
1
is the sensitivity transfer function corresponding to the
nominal plant.
Proof: We will apply Theorem 15 to obtain the result. Note that the equivalent
M seen by is
M =
w
iI
1 +GK
= w
iI
S.
SISO ROBUST PERFORMANCE 184
Thus robust stability holds if and only if
|w
iI
S|
H
< 1.
Note that the other robust stability conditions could have been derived in this
manner.
Robust Performance for SISO systems
Robust Performance
+
+
w
I
G
-
K
w
p
d
+
+
y
In the robust performance problem the following are the objectives
Robust stability
Performance for all the plants in the model class .
Consider the feed-back loop shown where the plant class is described by
multiplicative uncertainty. The robust stability criteria was determined to be
RS |w
I
T|
H
< 1.
The performance desired in the above setup is that of tracking and/or
disturbance rejection (which are the same if G
d
= I. Thus the performance
requirement is
|w
p
S
p
|
H
< 1
where
S
p
= (I +L
p
)
1
with L
p
= G
p
K where
G
p
:= G(1 +w
I
)[ ||
H
1|.
The performance condition translates to the condition that for all ||
H
1
|w
p
1
1 +GK(1 +w
I
)
|
H
< 1 |
w
p
S
1 +w
I
T
|
H
< 1 (3)
We summarize the above observations as a Lemma.
Lemma 2. Necessary and sufcient conditions for robust performance are
1. |w
I
T|
H
< 1.
2. |
w
p
S
1+w
I
T
|
H
< 1, stable with ||

H
1.
Theorem 17. A necessary and sufcient condition for robust performance for
the interconnection in the Figure is
|[w
p
S[ +[w
i
T[|
H
< 1.
Proof: () Let
|[w
p
S[ +[w
i
T[|
H
< 1. (4)
Then the robust stability requirement |w
I
T|
H
< 1 is satised. Let be xed

with ||
H
1.
From (4) it follows that
for all , [w
p
S[ +[w
I
T[ < 1 which implies
[w
p
S[ < 1 [w
I
T[ < 1 [w
I
T[[[ < [1 +w
I
T[
thus
[w
p
S[
[1 +w
I
T[
< 1 .
Thus all conditions of Lemma 2 are satised and robust performance
follows.
() Suppose there exists a
0
such that
[(w
p
S)(j
0
)[ +[(w
I
T)(j
0
)[ > 1.
(Case 1) If [(w
I
T)(j
0
)[ 1 then
|w
I
T|
H
[(w
I
T)(j
0
)[ > 1
and therefore there is no robust stability and therefore no robust
performance. Thus the proof is complete.
(Case 2) Suppose [(w
I
T)(j
0
)[ < 1.
Note that as [(w
p
S)(j
0
)[ +[(w
I
T)(j
0
)[ > 1, we have
[(w
p
S)(j
0
)[ > 1 [(w
I
T)(j
0
)[
|(w
p
S)(j
0
)|
1|(w
I
T)(j
0
)|
1
Construct a transfer function that is stable with ||
H
1 such that
[1 [(w
I
T)(j
0
)[[ = [(1 +w
I
T)(j
0
)[.
This is indeed possible and is left as an exercise
It follows that
|
w
p
S
1+w
I
T
|
H

|(w
p
S)(j
0
)|
|(1+w
I
T)(j
0
)|
=
|(w
p
S)(j
0
)|
|(1(|w
I
T)(j
0
)|)|
1
.
From Lemma 2 we have that there is no robust performance.
This completes the proof.
Summary
Nominal Performance |w
p
S|
H
< 1.
Robust Stability |w
I
T|
H
< 1.
Nominal performance and robust stability | max([w
p
S[, [w
I
T[)|
H
< 1
(follows from the above two conditions).
Robust Performance |[w
p
S[ +[w
I
T[|
H
< 1.
It can be shown that
1
2
([w
p
S[ +[w
I
T[) ([w
p
S[
2
+[w
I
T[
2
)
1
2
2([w
p
S[ +[w
I
T[)
and
max([w
p
S[, [w
I
T[) ([w
p
S[
2
+[w
I
T[
2
)
1
2
2 max([w
p
S[, [w
I
T[).
Thus the following lemma holds:
Lemma 3. The following hold:
1. | max([w
p
S[, [w
I
T[)|
H

_
_
_
_
w
p
S
w
I
T
_
_
_
_
H
2| max([w
p
S[, [w
I
T[)|
H
2.
1
2
|[(w
p
S[ +[w
I
T)[|
H

_
_
_
_
w
p
S
w
I
T
_
_
_
_
H
2 |([w
p
S[ +[w
I
T[)|
H
.
Proof: Follows from the fact that
_
_
_
_
w
p
S
w
I
T
_
_
_
_
H
= sup
([w
p
S[
2
+[w
I
T[
2
)
1
2
.
SISO CONTROLLER SYNTHESIS 195
Thus the conclusion is that be solving an appropriately scaled stacked H
problem one can achieve the objectives of robust performance. Note that we
have employed the stacked framework to obtain robust stability and nominal
performance for the nanopositioning example.
Optimal Controller Synthesis For SISO
Systems
Coprime Factors: Single Input Single Output Case
+
+
+
G
u
y
K
+
v
1
v
2
22
Figure 1:
Let the plant G
22
be given by G
22
=
N(s)
M(s)
and K =
Y (s)
X(s)
where
N(s), M(s), X(s), and Y (s) are polynomials in s. We assume there are no
common factors in the ratios being formed.
We assume that G
22
and K are rational functions of s.
Notice that for the SISO positive feedback conguration shown in the gure
we have
1 G
22
K = 1
NY
MX
=
MX NY
MX
.
By the Nyquist stability criterion the feedback interconnection is stable if and
only if
MX NY
has no zeros in the right half plane.
In other words we need
R := (MX NY )
1
to be stable transfer function for the interconnection to be stable.
Note that K can be written as K =
Y
1
X
1
where
Y
1
:= Y R and X
1
= XR
and
MX
1
NY
1
= 1.
Denition 13. We say two rational stable functions X and Y are coprime
over stable systems if they do not have common unstable factors.
We will use the term coprime to mean coprime over stable systems.
We make the above arguments precise with the the following lemma.
Lemma 4. Let G
22
=
N
M
be the plant with N and M being coprime. K is a
stabilizing controller for the feedback interconnection shown in Figure 1 if and
only if there exist stable coprime factors Y and X such that K =
Y
X
satisfying
MX NY = 1.
Proof:From the Nyquist stability criterion (Theorem 5) the closed-loop system
is stable if and only if
1. There is no rhp pole zero cancellation while forming the product GK =
NY
MX
2. S = (I GK)
1
=
1
1
NY
MX
=
MX
MXNY
is stable.
We will rst show that the above two conditions are equivalent to the condition
that MX NY have no rhp zeros.
Indeed let MX NY have no rhp zeros. Clearly then MX and NY have no
common rhp zeros. Thus there can be no rhp pole zero cancellation while
forming the product GK. Also, as MX NY has no rhp zeros (MX NY )
1
is stable. This implies S =
MX
MXNY
is stable. Thus the two conditions for
stability of closed loop map are satised.
Now, suppose (1) and (2) are met. Then MX and NY have no common rhp
zeros and
MX
MXNY
is stable. Suppose that MX NY has a rhp zero at z.
Then M(z)X(z) = N(z)Y (z). Also as
MX
MXNY
is stable it has to be true that
the rhp zero z of the denominator MX NY be cancelled by the rhp zero of
the numerator MX. Thus M(z)X(z) = 0. This would imply
M(z)X(z) N(z)Y (z) = 0 and thus MX NY has a rhp zero at z which
implies that MX and NY have a common rhp zero. This would contradict (1).
Thus MX NY has no rhp zeros.
Thus, it follows that stability of the closed loop system is equivalent to
MX NY having no zeros in the rhp. Or equivalently closed loop system is
stable if and only if R = (MX NY )
1
is stable. Clearly
K = Y/X =
Y R
XR
= Y
1
/X
1
where Y
1
= Y R and X
1
= XR are coprime.
Furthermore
MX
1
NY
1
= (MX NY )R = 1.
This proves the theorem. .
Parametrization of Stabilizing Controllers for SISO Systems
In the SISO case the derivations of the class of stabilizing controllers is
straightforward.
Theorem 18. Let P =
N
M
and let K =
Y
1
X
1
be a stabilizing controller with
N and M coprime and Y
1
and X
1
coprime with
MX
1
NY
1
= 1.
Then all stabilizing controllers are given by
K =
Y
1
MQ
X
1
NQ
where Q is any stable rational function.
Proof:: () Let
K =
Y
1
MQ
X
1
NQ
.
It follows that
M(X
1
NQ) N(Y
1
MQ) = MX
1
NY
1
= 1.
From Lemma 4 it follows that the closed-loop is stable.
() Suppose K is a stabilizing controller. We infer from Lemma 4 that K =
Y
X
with Y and X such that
MX NY
is stable Dene Q to satisfy the relation
Y
X
=
Y
1
MQ
X
1
NQ
which holds if
Q =
X
1
Y Y
1
X
MX NY
which is stable.
Let K be a stabilizing controller. Then it follows that there exists a stable Q
such that K =
Y
1
MQ
X
1
NQ
.
S =
1
1GK
=
1
1
N
M
Y
1
MQ
X
1
NQ
=
M(X
1
NQ)
M(X
1
NQ)N(Y
1
MQ)
=
MX
1
MNQ
MX
1
MNQNY
1
+NMQ
= MX
1
MNQ
It follows that
T = 1 S = 1 MX
1
+MNQ = NY
1
+MNQ.
and
KS =
Y
1
MQ
X
1
NQ
M(X
1
NQ) = M(Y
1
MQ).
Note that the stacked H
problem is given by
= inf
K stabilizing
_
_
_
_
_
_
w
P
S(K)
w
T
T(K)
w
u
KS
_
_
_
_
_
_
H
= inf
Q stable
_
_
_
_
_
_
w
P
M(X
1
NQ)
w
T
N(Y
1
+MQ)
w
u
M(Y
1
MQ)
_
_
_
_
_
_
H

Generalized Plant
G
11
G
21
G
22
G
12
K
w z
y
u
v
2
v
1
+
+
+
+
Figure 2:
Many control design issues can be cast into the framework shown in Figure 2.
Lets assume that v
1
= v
2
= 0. Then
z = G
11
w +G
12
u
y = G
21
w +G
22
u
MIMO SYSTEMS 207
When we substitute u = Ky we have
y = G
21
w +G
22
Ky y = (I G
22
K)
1
G
21
w. Thus
z = [G
11
+G
12
K(I G
22
K)
1
G
21
]w
= [G
11
+G
12
M(Y
1
MQ)G
21
]w
.
The map bewteen z and w is afne linear in the parameter Q.
MIMO SYSTEMS 208
Multiple Input Multiple Output Systems
MIMO SYSTEMS 209
Linear Systems
We present notions of stability, causality and well-posedness of
interconnections of systems.
MIMO SYSTEMS 210
Notation
Signals
L
n
= x : x = (x
1
, x
2
, . . . , x
n
) with x
i
L.
Signal-norms
For any x in L
n
let
|x|
p
=
_
_

i=1
[x
i
(t)[
p
_1
p
1 p < and
|x|
= sup
t
max
i
[x
i
(t)[
Signal Space Let
L
n
p
= x[x L
n
, |x|
p
<
MIMO SYSTEMS 211
Let L
mn
p
denote the spaces of mn matrices with each element of the
matrix in L
p
.
MIMO SYSTEMS 212
Truncation Operator
Let P
denote the truncation operator on L

mn
which is dened by
P
(x(t)) = x(t) if t
= 0 if t >
.
MIMO SYSTEMS 213
Shift Operator
Let S denote the shift map from L
n
to L
n
dened by
S
(x(t)) = x(t ).
MIMO SYSTEMS 214
Causal Systems
Denition 14. [Causality] A linear map T : L
n
L
m
is said to be causal
if
P
t
T = P
t
T P
t
for all t.
MIMO SYSTEMS 215
Time Invariant Systems
Denition 15. [Time invariance]
A map T : L
n
L
m
is time invariant if S
T = T S
where S
is the shift
operator.
MIMO SYSTEMS 216
Induced Norms
Let T be a linear map from (o
n
p
, |.|
p
) to (o
m
p
, |.|
p
). The p-induced norm of T
is dened as
|T |
pind
:= sup
x
p
=0
|T x|
p
|x|
p
.
MIMO SYSTEMS 217
Input-output Stability
Denition 16. [Stability] A linear map T : (X, |.|
X
) (Y, |.|
Y
) is said to be
stable if it is bounded. That is T is stable if
|T | = sup
x=0
|T x|
Y
|x|
X
= M < .
Note that
T : (L
n
p
, |.|
p
) (L
m
p
|.|
p
) is said to be L
p
stable if it is a bounded operator.
MIMO SYSTEMS 218
Convolution Maps
Denition 17. [Convolution maps] T : L
n
L
m
is linear, time invariant,
convolution map if and only if y = T u is given by
_
_
_
_
y
1
y
2
.
.
.
y
m
_
_
_
_
=
_
_
_
_
T
11
T
12
. . . T
1n
T
21
T
22
. . . T
2n
.
.
.
.
.
.
.
.
.
.
.
.
T
m1
T
m2
. . . T
mn
_
_
_
_
_
_
_
_
u
1
u
2
.
.
.
u
n
_
_
_
_
,
where y = (y
1
y
2
, . . . , y
m
) L
m
and u = (u
1
, u
2
, . . . , u
n
) L
n
, T
ij
: L L
is described by
(T
ij
x)(t) =
_

x()T
ij
(t )d
where T
ij
(t) termed the impulse response of the system T .
Further T is also causal if T
ij
(t) = 0 for all t < 0 and thus for any x L we
MIMO SYSTEMS 219
have
(T
ij
x)(t) =
_

0
x()T
ij
(t )d
The operation given by Equation is often written as y
i
= T
ij
u
j
.
Note that any linear time invariant system can be described as a convolution
map via its impulse response.
MIMO SYSTEMS 220
Transforms
Denition 18. For a linear, time invariant, causal, convolution map
T : L
n
L
m
the s-transform of T is dened as
T(s) :=
_

0
T(t)e
st
dt,
where
T(t) :=
_
_
_
_
T
11
(t) T
12
(t) . . . T
1n
(t)
T
21
(t) T
22
(t) . . . T
2n
(t)
.
.
.
.
.
.
.
.
.
.
.
.
T
m1
(t) T
m2
(t) . . . T
mn
(t)
_
_
_
_
.

T is analytic inside the open right half plane (open unit
disc) and continuous on the boundary if the matrix T(t) L
mn
1
(
mn
1
).
MIMO SYSTEMS 221
Transfer Function
Denition 19. [Rational, proper transfer functions] Let g(s) =
n(s)
d(s)
where
n and d are polynomials in s. Then g(s) is said to be a rational transfer
function. Furthermore if deg(d(s)) deg(n(s)) then g(s) is a proper transfer
function. If deg(d(s)) = deg(n(s)) then g(s) is a bi-proper transfer function. If
deg(d(s)) > deg(n(s)) then g(s) is a strictly proper transfer function.
Example 6. e
s
is not a rational transfer function.
s
s+1
is a strictly proper
rational function.
s+2
s+1
is a bi-proper transfer function.
Note that a rational function g(s)
is proper if and only if g() = d a constant.
is strictly proper if and only if g() = 0 a constant.
is bi-proper if and only if g() = d ,= 0 where d is a constant.
MIMO SYSTEMS 222
Also note that every proper rational function can be written as
g(s) = g() +g
sp
(s)
where g
sp
(s) is strictly proper (example:
s+2
s+1
= 1 +
1
s+1
).
MIMO SYSTEMS 223
Finite Dimensional Systems: Proper Transfer Matrices
Denition 20. [Finite dimensional system]
If the s-transform of any linear, time invariant, causal, map T : L
n
L
m
is
such that

T
ij
(s) is a proper transfer function then T represents a nite
dimensional system.

T(s) is said to be a proper transfer matrix. If every
entry

T
ij
(s) is a strictly proper transfer function then

T(s) is a strictly proper
transfer matrix. If

T is square with both

T(s) and

T
1
(s) being proper then
T(s) is a bi-proper transfer matrix.

We use the term FDLTIC as an abbreviation for nite-dimensional, linear, time
invariant, causal. Note that FDLTIC maps are characterized by proper transfer
matrices.
Analogous to the scalar case, any proper transfer matrix G(s) can be written
as
G(s) = G() +G
sp
(s)
MIMO SYSTEMS 224
where G
sp
is strictly proper transfer matrix.
MIMO SYSTEMS 225
Coprime Factors of Transfer Functions
Denition 21. [Coprime Factors] Consider a transfer function
g(s) = n(s)/d(s) where n(s) and d(s) are polynomials in s. If n(s) and d(s)
have no common factors then we say that n(s) and d(s) are coprime and
g = nd
1
is a coprime factorization of g.
Example 7. Let g(s) =
1
s1
. Then n = 1 and d = (s 1) gives a coprime
factorization of g. Note that g(s) =
(s2)
s
2
3s+2
. However,
n = (s 2) and d = s
2
3s + 2 is not a coprime factorization because of the
common factor (s 2).
MIMO SYSTEMS 226
Degree of Transfer Functions
Denition 22. [Degree of a transfer function] Given any proper transfer
function g(s) =
n(s)
d(s
let r(s) be the greatest common divisor of the polynomials
n(s) and d(s). Then n(s) = n
m
(s)r(s) and d(s) = d
m
(s)r(s). Clearly n
m
(s) and
d
m
(s) will have no common divisor and therefore are coprime and
g(s) =
n
m
(s)r(s)
d
m
(s)r(s)
.
d
m
(s) is called the characteristic polynomial of g(s).
The degree of g(s) is the degree of d
m
(s).
Example 8. Let g(s) =
s
2
1
4(s
3
1)
:=
n
d
. The gcd of n(s) and d(s) is s 1 =: r(s).
Thus n
m
= s + 1 and d
m
= 4s
2
+ 4s + 1 with d
m
being the characteristic
polynomial. The degree of g(s) is thus two (not three).
MIMO SYSTEMS 227
Characterization of Coprimeness of Polynomials
Theorem 19. [Aryabhatta, Bezout, Diophantine] Polynomials n(s) and d(s)
are coprime if and only if there exist polynomials n
c
(s) and d
c
(s) such that
n(s)n
c
(s) +d(s)d
c
(s) = 1.
Proof:() Suppose there exist polynomials n
c
(s) and d
c
(s) such that
n(s)n
c
(s) +d
m
(s)d
c
(s) = 1. (5)
Also assume that n(s) = n
m
(s)r(s) and d(s) = d
m
(s)r(s) where r(s) is a
nonconstant polynomial, d
m
(s), n
m
(s) are polynomials. Then we have from
(5) that
r(s)[n
m
(s)n
c
(s) +d
m
(s)d
c
(s)] = 1.
MIMO SYSTEMS 228
This would imply that product of two non-trivial (i.e.nonconstant) polynomials
is 1 which is not possible. This proves that n(s) and d(s) are coprime.
() Assume that n(s) and d(s) are coprime. Then using the Euclidean
Algorithm one can construct polynomials n
c
(s) and d
c
(s) such that
n(s)n
c
(s) +d(s)d
c
(s) = 1.
MIMO SYSTEMS 229
Poles of a Transfer Matrix
Denition 23. [Characteristic polynomial, poles] The characteristic
polynomial of a transfer matrix G(s) is the least common denominator of all
minors of G(s) where the minors are reduced to be coprime transfer functions.
The degree of G(s) is the degree of the characteristic polynomial. The poles
are the roots of the characteristic polynomial.
Example 9. Consider the transfer matrix
G(s) =
1
1.25(s + 1)(s + 2)
_
s 1 s
6 s 2
_
.
The minors of order 1 are the four elements of the transfer matrix G(s) all of
which have the same denominator (s + 1)(s + 2). The minor of order two is the
determinant of the matrix:
det(G(s)) =
(s 1)(s 2) + 6s
1.25
2
(s + 1)
2
(s + 2)
2
=
1
1.25
2
(s + 1)(s + 2)
.
MIMO SYSTEMS 230
Note that we have reduced the second order minor to be coprime. The lcd of
all the minors is (s + 1)(s + 2) and thus the characteristic polynomial is given
by
(s) = (s + 1)(s + 2).
The poles are given by s = 1 and s = 2.
MIMO SYSTEMS 231
Normal Rank of a Transfer Matrix
Denition 24. [Normal rank] The normal rank of a transfer matrix G(s) is
the maximum rank of the transfer matrix over the variable s.
Example 10. Consider
G(s) =
1
s + 2
_
s 1 0
0 2(s 2)
_
.
The maximum rank of this matrix is two even though at s = 1 and s = 2 the
rank drops to one. Thus the normal rank of the transfer matrix is two.
MIMO SYSTEMS 232
Zeros of a Transfer Matrix
Denition 25. [Zeros] z
i
is a zero of a transfer matrix G(s) if rank of G(z
i
) is
less than the normal rank of G(s).
Example 11. Consider
G(s) =
1
s + 2
_
s 1 0
0 2(s 2)
_
.
Thus the normal rank of the transfer matrix is two. The rank of this matrix at
s = 1 and s = 2 drops to one. Thus these are the zeros of the transfer matrix.
MIMO SYSTEMS 233
Unimodular Matrices
Denition 26. [Unimodular matrices] A square polynomial matrix function
P() = P(0) +P(1) +. . . +P(k)

k
, is said to be unimodular if the
determinant of

P() is a non-zero constant independent of .
Theorem 20. Let

T() be a mn matrix of rational functions of (a function
is a rational function of if it can be written as a ratio of two polynomials of ).
Then there exist

L,

U and

M such that

T =

L

M

U where

L and

U are
unimodular with appropriate dimensions and

M has the structure
M =
_
_
_
_
_
_
_
_
_

1
1
0 . . . 0
.
.
.
.
.
.
.
.
.
.
.
.

r
r
0 . . . 0
0 . . . 0 0 . . . 0
.
.
.
.
.
.
.
.
. 0 . . . 0
0 . . . 0 0 . . . 0
_
_
_
_
_
_
_
_
_
.
MIMO SYSTEMS 234

i
,

i
are coprime (that is they do not have any common factors) monic
(leading coefcient is one) polynomials, which are not identically zero for all
i = 1, . . . , r with the following divisibility property:
i
() divides
i+1
() without
remainder and

i+1
() divides

i
() without remainder.
M is called the Smith-Mcmillan form of

T().
MIMO SYSTEMS 235
Zeros and Poles
Theorem 21. [Zeros and poles of

T] The zeros of

T() are the roots of
r
i=1

i
(). The poles of

T() are the roots of
r
i=1

i
().
Proof:Left to the reader.
MIMO SYSTEMS 236
State Space Characterization of Finite Dimensional Systems
Lemma 5.
Let T be a FDLTIC system; T : L
n
L
m
. Then there exist real matrices
A, B, C and D such that if y = Tu for some u L
n
then
x = Ax(t) +Bu(t)
y(t) = Cx(t) +Du(t)
x(0) = 0,
(6)
Proof:See C. T. Chen.
The representation of the map T as given in (6) is called a state space
representation of T. A convenient notation employed to denote the system
described by (6) is
_
A B
C D
_
.
MIMO SYSTEMS 237
Controllability
Denition 27. [Controllability] The dynamical system described by
x = Ax(t) +Bu(t)
y(t) = Cx(t) +Du(t)
x(0) = x
0
,
(7)
is said to be controllable if for any initial condition x(0) = x
0
, t
1
> 0 and nal
state x
1
there exists a piecewise continuous input u(.) such that the solution of
(7) satises x(t
1
) = x
1
. Otherwise the system or the pair (A, B) is said to be
uncontrollable.
MIMO SYSTEMS 238
Tests For Controllability
Theorem 22. The following are equivalent
(A, B) is controllable.
The matrix
W
c
(t) :=
_
t
0
e
A
BB
e
A
d
is positive denite for any t > 0.
The controllability matrix
C = [B AB A
2
B . . . A
n1
B]
has full row rank.
MIMO SYSTEMS 239
The matrix [AIB] has full row rank for all in the complex plane.
Let and x be any eigenvalue and corresponding left eigenvector of A (that
is x
A = x
) then x
B ,= 0.
For any given set of n complex numbers F can be chosen such that
A+BF has the given set as its eigenvalues.
MIMO SYSTEMS 240
Stabilizability
Denition 28. [Stabilzability]
The pair of real matrices A and B with A R
nn
and with B R
nm
is a
stabilizable pair if there exists a real matrix K such that
Real(
i
(A+BK)) < 0, where
i
denotes the i
th
eigenvalue.
MIMO SYSTEMS 241
Tests for Stabiliability
(A, B) is stabilizable.
The matrix [AIB] has full row rank for all in the complex plane with
Re() 0.
For all and x that satisfy x
A = x
and Re() 0, x
B ,= 0.
F can be chosen such that A+BF is stable.
MIMO SYSTEMS 242
Observability
Denition 29. Observability The dynamical system described by
x = Ax(t) +Bu(t)
y(t) = Cx(t) +Du(t)
x(0) = x
0
,
(8)
is said to be observable (or the pair (A, C) is observable) if any initial condition
x
0
can be determined uniquely from the output trajectory y(t), t [0, t
1
] where
t
1
> 0 is arbitrary with y(t) being generated by Equation (8) with u() = 0.
MIMO SYSTEMS 243
Test for Observability
(A, C) is observable.
The matrix
W
o
(t) :=
_
t
0
e
A
Ce
A
d
is positive denite for any t > 0.
The observability matrix
O =
_
_
C
CA
CA
2
.
.
.
CA
n1
_
_
has full column rank.
MIMO SYSTEMS 244
The matrix
_
AI
B
_
has full column rank for all in the complex plane.
Let and y be any eigenvalue and corresponding right eigenvector of A
(that is Ax = x ) then Cx ,= 0.
For any given set of n complex numbers L can be chosen such that A+LC
has the given set as its eigenvalues.
Note that (A, C) is observable if and only if (A
, C
) is controllable.
MIMO SYSTEMS 245
Detectability
Denition 30. [Detectability] The pair of real matrices A and C is
detectable if there exists a real matrix L such that Real(
i
(A+LC)) < 0
MIMO SYSTEMS 246
Tests for Detectability
(A, C) is detectable.
The matrix
_
AI
B
_
has full column rank for all in the complex plane
with Re() 0.
For all and x that satisfy Ax = x and Re() 0, Cx ,= 0.
L can be chosen such that A+LC is stable.
Note that (A, C) is detectable if and only if (A
, C
) is stabilizable.
MIMO SYSTEMS 247
Minimal Realization
Denition 31. The triplet (A, B, C) is minimal if (A, B) is controllable and
(A, C) is observable.
Lemma 6. Let

T be a proper transfer function matrix. Then it admits a
realization
_
A B
C D
_
such that (A, B, C) is minimal. Such a realization is
called a minimal realization of

T.
MIMO SYSTEMS 248
Continuous Time Stability Characterization
Theorem 26. Suppose T is a FDLTIC system. Then the following statements
are equivalent.
1. T is L
p
stable for any p, 1 p .
2. If
_
A B
C D
_
is any state-space description of T such that (A, B) is
stabilizable and (A, C) is detectable then Real(
i
(A)) < 0 for all i where
i
(A) denotes the i
th
eigenvalue of A.
3.

T(s) the s-transform of T has all its poles outside the closed right half
plane(that is if s
0
is a pole of

T then Real(s
0
) < 0.)
Proof:See C. T. Chen.
MIMO SYSTEMS 249
This theorem establishes the fact that for FDLTIC systems stability in L
p
sense
implies stability in L
q
sense for any p and q such that 1 p and
1 q . Thus for FDLTIC systems we can use the term stability to mean
stability in L
p
sense for any 1 p .
MIMO SYSTEMS 250
Properties of State Space Realizations
Suppose G
1
and G
2
have a state space realizations
_
A
1
B
1
C
1
D
1
_
and
_
A
2
B
2
C
2
D
2
_
respectively. Then
G
1
G
2
=
_
_
A
1
B
1
C
2
0 A
2
B
1
D
2
B
2
C
1
D
1
C
2
D
1
D
2
_
_
=
_
_
A
2
0
B
1
C
2
A
1
B
2
B
1
D
2
D
1
C
2
C
1
D
1
D
2
_
_
.
G
1
+G
2
=
_
_
A
1
0
0 A
2
B
1
B
2
C
1
C
2
D
1
+D
2
_
_
MIMO SYSTEMS 251
Suppose G(s) =
_
A B
C D
_
is square and D is invertible then
G
1
=
_
ABD
1
C BD
1
D
1
C D
1
_
.
MIMO INTERCONNECTIONS 252
Internal Stability
Denition 32. Consider the state space dynamics described by
x(t) = Ax(t) +Bu(t)
y(t) = Cx(t) +Du(t)
x(0) = x
0
.
The above system is said to be internally stable if for any x
0
R
n
,
x(t) 0 as t with u = 0.
Lemma 7. The system described by the state space equations above is
internally stable if and only if all the eigenvalues of the matrix A are in the
open left half plane s C[Real(s) < 0.
Stability of Interconnections of Multiple-input
Multiple-output Systems
Interconnections
G
K
u
y
22
Figure 3:
Consider the interconnection represented by the block diagram in Figure 3.
Suppose G
22
and K are described by the state space descriptions
G
22
_
_
_
x
22
(t) = Ax
22
(t) +B
2
u(t)
y(t) = C
2
x
22
(t) +D
22
u(t)
x(0) = x
0
and K
_
_
_
x
K
(t) = A
K
x
K
(t) +B
K
y(t)
u(t) = C
K
x
K
(t) +D
K
y(t)
x
K
(0) = x
0
(9)
Well-posedness: State Space Viewpoint
Before analyzing the interconnection one has to address the issue of
well-posedness. For the interconnection described by Figure 3 with the state
space descriptions of G
22
and K given by
_
A B
2
C
2
D
22
_
and
_
A
K
B
K
C
K
D
K
_
respectively we dene
Denition 33. The interconnection described by Figure 3 governed by the
(9) is well posed if there exist unique signals x
22
(t), x
K
(t), y(t), u(t) that satisfy
(9) for every initial condition x(0) := [x
22
(0) x
K
(0)]
T
. Such a condition should
hold for all corresponding matrices in a neighbourhood of the given matrices
A, A
K
, B
2
, B
K
, C
2
, C
K
, D
22
, and D
K
.
Lemma 8. The interconnection described by Figure 3 governed by the (9) is
well posed if and only if
det(I D
22
D
K
) ,= 0.
Proof: () Assume that (I D
22
D
K
) is singular. Using ( 9) note that
y = C
2
x
22
+D
22
u
u = C
K
x
K
+D
K
y
(10)
_
I D
K
D
22
I
_
. .
M
_
u
y
_
=
_
0 C
K
C
2
0
_
. .
N
_
x
22
x
K
_
(11)
The matrix M is singular as (I D
22
D
K
) is assumed singular. Thus it is not
onto (use the fact that dim(range(M)) = rank(M)). Let (u
0
y
0
) be a vector
that is not in the range space of M. It is evident that the matrix on the right
hand side of the equation can be made full rank by a small perturbation if
needed. Choose (a b) such that N(a b)
T
= (u
0
y
0
)
T
. Thus for the initial
conditions x
22
(0) = a and x
K
(0) = b there is no solution to the (11).
() Assume that det(I D
22
D
K
) ,= 0 and let x
22
(0) and x
K
(0) be given initial
conditions.
Note that ( 9) is satised if and only if
x
22
(s) = (sI A)
1
x
22
(0) + (sI A)
1
B
2
u(s) (12)
x
K
(s) = (sI A
K
)
1
x
K
(0) + (sI A
K
)
1
B
K
y(s) (13)
y(s) = C
2
x
22
(s) +D
22
u(s)
= C
2
[(sI A)
1
x
22
(0) + (sI A)
1
B
2
u(s)] +D
22
u(s)
= C
2
(sI A)
1
x
22
(0) + [C
2
(sI A)
1
B
2
+D
22
] u(s)
= e
1
(s) +G(s) u(s) (14)
u(s) = C
K
x
K
(s) +D
K
y(s)
= C
K
[(sI A
K
)
1
x
K
(0) + (sI A
K
)
1
B u(s) +D
K
y(s)]
= C
K
(sI A
K
)
1
x
K
(0) + [C
K
(sI A
K
)
1
B
K
+D
K
] y(s)
= e
2
(s) +K(s) y(s) (15)
where e
1
(s) = C
2
(sI A)
1
x
22
(0) is determined by x
22
(0),
e
2
(s) := C
K
(sI A
K
)
1
x
K
(0) is determined by x
k
(0),
G
22
= C
2
(sI A)
1
B
2
+D
22
and K = C
K
(sI A
K
)
1
B
K
+D
K
are the
transfer functions of G
22
and K. Note that by substituting into (14) the relation
(15) we have
y(s) = e
1
(s) +G
22
(s)[ e
2
(s) +K(s) y(s)]
(I G
22
(s)K(s)) y(s) = e
1
(s) +G
22
(s) e
2
(s)
As (I D
22
D
K
) is invertible it follows that the transfer function
(I G
22
(s)K(s))
1
exists (as it has full normal rank) and is proper. Thus y(t)
is uniquely determined from e
1
(t) and e
2
(t) that in turn are xed for the x
22
(0)
and x
K
(0) given.
Note that u(t) is uniquely determined causally from the equation (15) as y(s)
is uniquely determined.
Similarly x
22
(t) and x
K
(t) are uniquely determined from the equation (12) and
(13) as y(t) and u(t) are uniquely determined.
Well-posedness: Input-output Viewpoint
+
+
+
G
u
y
K
+
v
1
v
2
22
Figure 4: Parametrization of stabilizing controllers for G
22
.
Consider the interconnection shown in Figure 4 where G
22
and K are
input-output maps whose realizations are not available. In this case the
well-posedness property is dened in terms of the input signals v
1
and v
2
.
Denition 34. The interconnection shown in Figure 4 is said to be well
posed if for any signals v
1
and v
2
there exist unique signals y and u satisfying
the relations imposed by the interconnection and y and u should be
determinable from v
1
and v
2
causally.
Theorem 27. The interconnection in Figure 4 is well posed if and only if
I

G
22
()

K()
is invertible.
Proof:Left to the reader.
Note that in the above interconnection it is not needed that G
22
or K be nite
dimensional.
Also, it is needed that the two well-posedness denitions should coincide
when state space realizations of G
22
and K are available.
Note that when G
22
and K have state space realizations
_
A B
2
C
2
D
22
_
and
_
A
K
B
K
C
K
D
K
_
then
I

G
22
()

K() = I D
22
D
K
.
This also implies that when G
22
and K have state space realizations then well
posedness implies all transfer functions between any input-output pair is a
proper transfer function. This follows from the causality requirement in well
posedness.
From Theorems 27 and 8 it follows that the denitions agree.
Internal Stability of Interconnections
+
+
+
G
u
y
K
+
v
1
v
2
22
22
.
Suppose G
22
and K have minimal state space realizations
_
A B
2
C
2
D
22
_
and
_
A
K
B
K
C
K
D
K
_
. It follows that the state space equations are described by
G
_
_
_
x
22
(t) = Ax
22
(t) +B
2
u(t)
e
1
(t) = C
2
x
22
(t) +D
22
u(t)
x(0) = x
0
K
_
_
_
x
K
(t) = A
K
x
k
(t) +B
K
y(t)
e
2
(t) = C
K
x
k
(t) +D
K
y(t)
x
K
(0) = x
0
(16)
Also
u = v
1
+e
2
y = v
2
+e
1
.
Thus the interconnection is internally stable if and only if the combined state
x := (x
22
x
K
)
T
converges to zero for any initial condition x(0) with the inputs
v
1
= v
2
= 0.
Lemma 9. Let
_
A B
2
C
2
D
22
_
and
_
A
K
B
K
C
K
D
K
_
be minimal state space
representations of G
22
and K. Consider the input and the output of the well
posed interconnection as
w =
_
v
1
v
2
_
and z =
_
u
y
_
respectively. Let the map between w and z be denoted by H(G
22
, K). Then
the state space description as given by Equation 16 is a stabilizable and
detectable realization of H.
Proof:
We will rst establish Controllability of the interconnection realization (16).
Suppose x(0) := [x
22
(0) x
K
(0)] is a given initial condition and x
d
:= [x
1
x
2
]
is the desired nal condition to be reached at time t
.
From controllability of realizations of G
22
and K it follows that there exist
signals u(t) and y(t) such that
x
1
= x
22
(t
) = e
At
x
22
(0) +
_
t
0
e
A(t
)
B
2
u(tau)d
x
2
= x
K
(t
) = e
A
K
t
x
K
(0) +
_
t
0
e
A
K
(t
)
B
K
y(tau)d
Let
e
1
= C
2
x
22
(t) +D
22
u(t)
e
2
= C
K
x
K
(t) +D
K
y(t)
Dene the signals v
1
(t) := u(t) e
2
and v
2
(t) = y(t) e
1
.
Clearly, the signals v
1
, v
2
and u, y form one input output pair that satisfy the
interconnection relationship. From well posedness with v
1
and v
2
as the input
u(t) and y(t) as dened above are the only possible signals.
Thus with v
1
and v
2
as dened it is clear that the interconnection intial
condition x(0) is driven to the desired state x
d
.
The proof of observability is left to the reader.
An immediate consequence of the above lemma and Theorem 26 is the
following theorem
Theorem 28. The well posed interconnection described by Figure 5 with the
minimal state space descriptions
_
A B
2
C
2
D
22
_
and
_
A
K
B
K
C
K
D
K
_
for G
22
and K respectively is internally stable if and only if the map H(G
22
, K) is a
stable map.
Proof:Follows from Lemma 9 and Theorem 26.
Corollary 3. The well posed interconnection described by Figure 5 with the
minimal state space descriptions
_
A B
2
C
2
D
22
_
and
_
A
K
B
K
C
K
D
K
_
for G
22
and K respectively is internally stable if and only if the maps
(I G
22
K)
1
, (I G
22
K)
1
G
22
, (I KG
22
)
1
, and (I KG
22
)
1
K are
stable.
Proof:Note that
u Ky = v
1
y G
22
u = v
2
_
u
y
_
=
_
I K
G
22
I
_
1
_
v
1
v
2
_
=
_
(I KG
22
)
1
(I KG
22
)
1
K
(I G
22
K)
1
G
22
(I G
22
K)
1
_
. .
H(G
22
,K)
_
v
1
v
2
_
.
From Theorem 28 it follows that the interconnection is internally stable if all
the elements of H(G
22
, K) are stable maps.
This motivates the following denition of internal stability when no state space
representations are available.
Denition 35. Consider Figure 5 where G
22
and K are linear time invariant
systems (possibly innite dimensional). Then the interconnection is internally
stable in the L
2
sense if the following maps
(I G
22
K)
1
(I G
22
K)
1
G
22
(I KG
22
)
1
(I KG
22
)
1
K
are in H
.
In other words (5) is internally stable if
_
I K
G
22
I
_
1
H

Stability Theorem For MIMO Systems
Theorem 29. Let n
G
22
and n
K
be the number of rhp poles of G
22
and K
respectively in the interconnection shown in Figure 5. Then the
interconnection is internally stable if and only if the following conditions are
satised:
1. The number of rhp poles of L := G
22
K is equal to n
G
22
+n
K
.
2. The matrix transfer function (I G
22
K)
1
is stable.
Proof:Note that
_
I K
G
22
I
_
. .
T
_
u
y
_
=
_
v
1
v
2
_
.
Consider a stabilizable and detectable realization of G
22
and K
G
22
_
_
_
x
22
(t) = Ax
22
(t) +B
2
u(t)
e
1
(t) = C
2
x
22
(t) +D
22
u(t)
x(0) = x
0
K
_
_
_
x
K
(t) = A
K
x
K
(t) +B
K
y(t)
e
2
(t) = C
K
x
K
(t) +D
K
y(t)
x
K
(0) = x
0
(17)
We will rst obtain a stabilizable and detectable realization of T
1
. Note that
v
1
= u e
2
v
2
= y e
1
Thus a state space realization of T :=
_
I K
G
22
I
_
is described by
x =
_
A 0
0 A
K
__
x
22
x
K
_
+
_
B
2
0
0 B
K
__
u
y
_
_
v
1
v
2
_
=
_
0 C
K
C
2
0
__
x
22
x
K
_
+
_
I D
K
D
22
I
__
u
y
_
Thus T admits a state space realization
T =
_
_
A
..
A 0
0 A
K
B
..
B
2
0
0 B
K
0 C
K
C
2
0
. .
C
I D
K
D
22
I
. .
D
_
_
.
Thus
T
1
=
_
A+BD
1
C BD
1
D
1
C D
1
_
=:
_
A B
C D
_
where
D = D
1
=
_
I D
K
D
22
I
_
1
=
_
I + (I D
22
D
K
)
1
D
22
D
K
(I D
22
D
K
)
1
(I D
22
D
K
)
1
D
22
(I D
22
D
K
)
1
_
=
_
I 0
0 0
_
+
_
(I D
22
D
K
)
1
D
22
D
K
(I D
22
D
K
)
1
(I D
22
D
K
)
1
D
22
(I D
22
D
K
)
1
_
=
_
I 0
0 0
_
+
_
D
K
I
_
(I D
22
D
K
)
1
_
D
22
I
_
Thus
A = A+BD
1
C =
_
A B
2
C
K
0 A
K
_
+
_
B
2
D
K
B
K
_
(ID
22
D
K
)
1
_
C
2
D
22
C
K
_
The following are equivalent
(A, B, C, D) is stabilizable and detectable.
(A, B
2
, C
2
, D
22
) and (A
K
, B
K
, C
K
, D
K
) are stabilizable and detectable.
The above can be proven using the stabilizability and detectability
characterization provided (see Theorem 23 and Theorem 25).
A realization of L = G
22
K is given by
_
A
L
B
L
C
L
D
L
_
where
A
L
=
_
A B
2
C
K
0 A
K
_
, B
L
=
_
B
2
D
K
B
K
_
, C
L
= (C
2
D
22
C
K
), D
L
= D
22
D
K
.
Also
S = (I L)
1
=
_
A
S
B
S
C
S
D
S
_
where
A
S
= A =
_
A B
2
C
K
0 A
K
_
+
_
B
2
D
K
B
K
_
(I D
22
D
K
)
1
_
C
2
D
22
C
K
_
B
S
=
_
B
2
D
K
B
K
_
(I D
22
D
K
)
1
C
S
= (I D
22
D
K
)
1
_
C
2
D
22
C
K
_
D
S
= (I D
22
D
K
)
1
The following are equivalent:
(A
S
, B
S
, C
S
, D
S
) is stabilizable and detectable
(A
L
, B
L
, C
L
, D
L
) is stabilizable and detectable
The above can be proven using the stabilizability and detectability
characterization provided.
Now we prove the theorem
() Suppose the interconnection is internally stable.
This implies T
1
is a stable transfer function. (A, B, C, D is a stabilizable and
detectable realization of T
1
as established earlier. Thus there can be no
unstable pole zero cancellations in forming the transfer matrix T
1
. Thus
A = A
S
has all eigenvalues in the open left half plane.
This implies that the realization (A
S
, B
S
, C
S
, D
S
) of S is a stabilizable and
detectable realization and that S is stable.
This implies (A
L
, B
L
, C
L
, D
L
) is stabilizable and detectable (as A
S
is stable
there can be no unstable pole zero cancellations).
This implies (1) in the theorem statement. We have established (2) that S is
stable.
() Assume (1) and (2) to be true. Then from (1) it follows that
(A
L
, B
L
, C
L
, D
L
) is stabilizable and detectable which in turn implies
(A
S
, B
S
, C
S
, D
S
) is stabilizable and detectable. As S is stable, A
S
= A has all
eigenvalues in the open left half plane. Thus T
1
the interconnection matrix is
stable.
Nyquist Stability Criterion For MIMO Interconnections
Theorem 30. Let L = G
22
K be such that there are no unstable pole zero
cancellations while forming the product. Let the number of rhp poles of L be
denoted by P
ol
. The closed-loop interconnection of G
22
and K is internally
stable if and only if the Nyquist plot of det(I L(s))
1. makes P
ol
anticlockwise encirclements of the origin
2. does not pass through the origin.
Proof: Condition (1) of Theorem 29 is satised as there are no pole-zero
cancellations in forming the product L = G
22
K. Condition (2) of Theorem 29 is
that S = (I L(s))
1
be stable, that is, it should have no poles in the rhp.
As in the proof of Theorem 29 let G
22
and K have state space stabilizable and
detectable realizations
_
A B
2
C
2
D
22
_
and
_
A
K
B
K
C
K
D
K
_
respectively.
A realization of L = G
22
K is given by
_
A
L
B
L
C
L
D
L
_
where
A
L
=
_
A B
2
C
K
0 A
K
_
, B
L
=
_
B
2
D
K
B
K
_
, C
L
= (C
2
D
22
C
K
), D
L
= D
22
D
K
.
As there are no rhp pole zero cancellations in the product G
22
K it follows that
above is a stabilizable and detectable realization of L. It follows that a
stabilizable and detectablle realization of S is given by
S = (I L)
1
=
_
A
S
B
S
C
S
D
S
_
where
A
S
= A = A
L
+B
L
(I D
L
)
1
C
L
=
_
A B
2
C
K
0 A
K
_
+
_
B
2
D
K
B
K
_
(I D
22
D
K
)
1
_
C
2
D
22
C
K
_
B
S
=
_
B
2
D
K
B
K
_
(I D
22
D
K
)
1
C
S
= (I D
22
D
K
)
1
_
C
2
D
22
C
K
_
D
S
= (I D
22
D
K
)
1
Thus the S is a stable map if and only if all eigenvalues of the A
S
are in the
lhp. Thus the stability of S is characterized by the zeros of the polynomial
cl
(s) = det(sI A
L
B
L
(I D
L
)
1
C
L
).
We will use the following result which is called the Schurs formula:
det
_
P
11
P
12
P
21
P
22
_
= det(P
11
)det(P
22
P
21
P
1
11
P
12
)
= det(P
22
)det(P
11
P
12
P
1
22
P
21
)
.
From above it follows that
cl
(s)det(I D
L
) = det(sI A
L
. .
P
22
B
L
(I D
L
. .
P
11
)
1
C
L
)det(I D
L
)
= det(sI A
L
)det[(I D
L
) C
L
(sI A
L
)
1
B
L
]
=
ol
(s)det(I L(s))
This implies that
det(I L(s)) =

cl
(s)
ol
(s)
c
STABILIZING CONTROLLERS PARAMETRIZATION 284
where c is a constant and
cl
and
ol
(s) are polynomials in s. Let N be the
number of clockwise encirclements of the origin of det(I L(s)) as s varies
over the Nyquist contour. Let P
ol
and Z be the number of zeros in the rhp of
ol
(s) and
cl
(s). From the Argument principle we have
N = Z P
ol
.
Stability of S is guaranteed if and only if Z = 0. Thus S is stable if and only if
the Nyquist plot of det(I L(s)) should encircle the origin P
ol
times in the
counterclockwise direction without touching the origin.
.
Parametrization of Stabilizing Controllers
Coprime Factorization for MIMO Systems
Denition 36. [rcf, lcf, dcf] Stable FDLTIC systems M and N are right
coprime if there exist stable FDLTIC systems X and Y such that the
s-transforms satisfy the identity
X(s)M(s)

Y (s)N(s) = I. (18)
Stable FDLTIC systems

M and

N are left coprime if there exist stable FDLTIC
systems

X and

Y such that the s-transforms satisfy the identity
M(s)X(s)

N(s)Y (s) = I. (19)
Suppose T = NM
1
=

M
1

N where N and M are right coprime and

M and
N are left coprime. Then the pair N and M form a right coprime factorization
(rcf) of T and the pair

M and

N form a left coprime factorization (lcf) of T.
A doubly-coprime factorization (dcf) of a FDLTIC system T is a set of stable
FDLTIC maps M, N,

M and

N such that T = NM
1
=

M
1

N and
_

X

N

M
__
M Y
N X
_
= I. (20)
Note that the dcf identity is a compact way of expressing
X(s)M(s)

Y (s)N(s) = I
M(s)X(s)

N(s)Y (s) = I
NM
1
=

M
1

N
Y X
1
=

X
1
Y
Interconnection of FDLTIC Systems
+
+
+
G
u
y
K
+
v
1
v
2
22
22
.
Lemma 10. Let G
22
be a FDLTIC system which has a dcf given by
G
22
= NM
1
=

M
1

N where
_

X

N

M
__
M Y
N X
_
= I. (21)
A FDLTIC controller K stabilizes the closed loop map shown in Figure 6 if and
only if K has a rcf K = Y
1
X
1
1
such that the map
_
M Y
1
N X
1
_
has a stable inverse.
Proof:
+
+
+
u
y
+
v
1
v
2
M N
Y X
-1
-1
Figure 7:
() Suppose an rcf of K is given by Y
1
X
1
1
and suppose
_
M Y
1
N X
1
_
1
is
stable. It is clear that the Figure 6 is the equivalent to Figure 7. Note that the
map from (, ) to (v
1
, v
2
) is given by
_
M Y
1
N X
1
_
. Because the inverse
of this map is stable it follows that the map from (v
1
, v
2
) to (, ) is stable. But
|y|
p
= |N +v
2
|
p
|N|
pind
||
p
+|v
2
|
p
and
|u|
p
= |Y
1
+v
1
|
p
|Y
1
|
pind
||
p
+|v
1
|
p
. Thus the map from (v
1
, v
2
) to
(u, y) is stable and therefore the closed loop map is stable.
()Let FDLTIC controller K be such that the closed loop map in Figure 6 is
stable. Thus the map from (v
1
, v
2
) to (u, y) is stable. Every FDLTIC system
admits a dcf (see Lemma 13), and therefore it admits a rcf also. Let a rcf of K
be given by K = Y
1
X
1
1
. From the dcf of G
22
it is follows that

XM

Y N = I.
Multiplying both sides of this equation by we have =

X(u)

Y (y v
2
) and
thus ||
p
|

X|
pind
|u|
p
+|
Y |
pind
(|y|
p
+|v
2
|
p
). This implies that the
map from (v
1
, v
2
) to (, ) is stable. Thus the inverse of the map
_
M Y
1
N X
1
_
is stable.
Youla Parametrization
Theorem 31. Let FDLTIC system G
22
admit a dcf as given in Lemma 10.
Then K is a FDLTIC stabilizing controller for the closed loop system in Figure
6 if and only if
K = (Y MQ)(X NQ)
1
= (

X Q

N)
1
(
Y Q

M),
for some FDLTIC stable system Q.
Proof:() Multiplying both sides of (21) by
_
I Q
0 I
_
from the left and by
_
I Q
0 I
_
from the right we have
_

X Q

N
Y +Q

M

N

M
__
M Y MQ
N X NQ
_
= I, (22)
where Q is a stable FDLTIC map. From Lemma 10 it follows that
K = (Y MQ)(X NQ)
1
is a stabilizing controller. It also follows from (22)
that (Y MQ)(X NQ)
1
= (

X Q

N)
1
(
Y Q

M) (follows from the
observation that the (1, 2) element of the product in (22) is zero).
() Suppose K is a stabilizing controller. Then from Lemma 10 we know that
there exist stable FDLTIC systems Y
1
and X
1
such that K = Y
1
X
1
1
and
_
M Y
1
N X
1
_
1
is stable. Thus it folows that
_

X

N

M
__
M Y
1
N X
1
_
=
_
I QD
0 D
_
is stable with a stable inverse where D =

NY
1
+

MX
1
and
Q := (

XY
1

Y X
1
)D
1
. Therefore D is stable with a stable inverse. Thus
D
1
is a stable system and therefore Q is also stable. Multiplyng both sides of
the above equation by
_
M Y
N X
_
we have
_
M Y
1
N X
1
_
=
_
M (Y MQ)D
N (X NQ)D
_
.
By comparing entries in the above equality we have the result that
K = (Y MQ)(X NQ)
1
. This proves the theorem. .
Generalized Plant
G
11
G
21
G
22
G
12
K
w z
y
u
v
2
v
1
+
+
+
+
Figure 8:
Many control design issues can be cast into the framework shown in Figure 8.
G is the generalized plant.
w is the exogenous input
u is the control inputs
y is measured output
z is the regulated output.
K is the controller which maps the measured outputs y to control inputs u
when v
1
and v
2
are zero.
Both K and G are linear systems. With respect to the interconnection of
systems G and K in Figure 8, the rst issue that needs to addressed is the
existence and uiqueness of signals z, u and y for given input signals
w, v
1
and v
2
.
Well-posedness of Interconnection
Denition 37. The interconnection in Figure 8 is well posed if for arbitrary
inputs w, v
1
and v
2
, u and y can be uniquely determined from w, v
1
and v
2
in a
causal manner.
An equivalent denition in the case G =
_
A B
C D
_
and K =
_
A
K
B
K
C
K
D
K
_
are FDLTIC is
Denition 38. The interconnection in Figure 8 is well posed if for arbitrary
initial conditions x
G
(0) and x
K
(0) the dynamics
x
G
(t) = Ax
G
(t) +B
_
w
u(t)
_
y(t) = Cx
22
(t) +D
_
w
u(t)
_
and
x
K
(t) = A
K
x
K
(t) +B
K
y(t)
u(t) = C
K
x
K
(t) +D
K
y(t)
(23)
where w = v
1
= v
2
= 0 has trajectories x
G
(t) and x
K
(t) uniquely dened that
satisfy (23). The above condition must hold for arbitrarily small perturbations
of the state space matrices.
Well Posedness
Let us assume that in Figure 8 G and K are FDLTIC systems. Also assume
that a stabilizable and detectable state-space description of G is described by
G =
_
G
11
G
22
G
21
G
22
_
=
_
_
A B
1
B
2
C
1
C
2
D
11
D
22
D
21
D
22
_
_
.
This notation is a convenient way of writing
G
11
=
_
A B
1
C
1
D
11
_
, G
12
=
_
A B
2
C
1
D
12
_
, G
21
=
_
A B
1
C
2
D
21
_
and G
22
=
_
A B
2
C
2
D
22
_
.
Theorem 32. With the state space representations of G and K as given
above, the interconnection is well posed if and only if
det(I D
22
D
K
) ,= 0.
Proof:: Left to the reader (Follows similar arguments as provided in the proof
of well posedness of G
22
and K interconnection.
Input Output Map
Note that for the interconnection in Figure 8 the existence and uniqueness of
z, u and y is sufcient for the well-posedness of the interconnection. The
signals satisfy the relation
_
_
I G
12
0
0 I K
0 G
22
I
_
_
_
_
z
u
y
_
_
=
_
_
G
11
0 0
0 I K
G
21
0 0
_
_
_
_
w
v
1
v
2
_
_
. (24)
We will suppose throughout that the interconnection is well-posed. This is
guaranteed if the map G
22
is strictly causal. Let H(G, K) be such that
_
_
z
u
y
_
_
= H(G, K)
_
_
w
v
1
v
2
_
_
.
The interconnection described by H(G, K) is often referred to as the closed
loop map.
Stability of Closed Loop Maps
Denition 39. [Stability of closed loop maps] The closed loop map
described by Figure 8 is
p
stable if |H(G, K)|
pind
< . In such a case K is
said to be a stabilizing controller in the
p
sense.
Lemma 11. There exists a FDLTIC system K which stabilizes the closed
loop in Figure 8 if and only if (A, B
2
) is stabilizable and (A, C
2
) is detectable. If
F and L are such that A+B
2
F and A+LC
2
are stable matrices then a
controller with a state space realization given by
K =
_
A+B
2
F +LC
2
+LD
22
F L
F 0
_
, (25)
stabilizes the closed loop system depicted in Figure 8.
Proof:() If (A, B
2
2
) is detectable then there exist
matrices F and L are such that A+B
2
F and A+LC
2
are stable. Let K be a
controller with a state space realization given in (25). It can be shown that the
closed loop system has a state-space description given by
_

A

B
C

D
_
where
A =
_
A B
2
F
LC
2
A+B
2
F +LC
2
_
,
which has the same eigenvalues as the matrix
_
A+LC
2
0
LC
2
A+B
2
F
_
.
Thus

A is stable from which it follows from Theorem 26 that the closed loop
map is stable.
() If (A, B
2
) is not stabilizable or (A, C
2
) is not detectable then some
eigenvalues of

A will remain unstable for any FDLTIC controller K. Details are
left to the reader.
The controller K given above is called the Luenberger controller
Lemma 12. Suppose (A, B
2
2
) is detectable. Then
FDLTIC system K stabilizes the closed loop system depicted in Figure 8 if and
only if it stabilizes the closed loop system depicted in Figure 6.
Proof:() The closed loop map depicted in Figure 8 is described by the
equations
z = G
11
w +G
12
u
y = G
21
w +G
22
u
u = Ky +Kv
2
+v
1
.
(26)
The description of the closed loop map depicted in Figure 6 is given by
y = G
22
u
u = Ky +Kv
2
+v
1
.
(27)
It is thus clear (substitute w = 0 in (26)) that if the map from (w, v
1
, v
2
) to
(z, u, y) in (26) is stable then map from (v
1
, v
2
) to (u, y) in (27) is stable.
() Suppose K is a stabilizing controller for the closed loop map in Figure 6.
Let
_
A
K
B
K
C
K
D
K
_
be a stabilizable and detectable state-space description of
K. By assumption
_
A B
2
C
2
D
22
_
is a stabilizable and detectable state-space
description of G
22
. Suppose,
_
A B
C D
_
is a state-space description of the
closed loop map obtained by employing the aforementioned state-space
descriptions of G
22
and K. Then one can show that (A, B) and (A, C) are
stabilizable and detectable. Thus from Theorem 26 it follows that A is stable.
If
_

A

B
C

D
_
is a description of the closed loop map in Figure 8 obtained by
using the descriptions
_
A
K
B
K
C
K
D
K
_
for K and
_
A B
C D
_
for G
22
then by
computing

A one can verify that

A = A. Thus

A is stable and therefore from
Theorem 26 it follows that the closed loop system in Figure 8 is stable.
Youla Parametrization of Stabilizing Controllers
Theorem 33. Suppose (A, B
2
2
) is detectable. Let
FDLTIC system G
22
admit a dcf as given in Lemma 10. Then K is a FDLTIC
stabilizing controller for the closed loop system in Figure 8 if and only if
K = (Y MQ)(X NQ)
1
= (

X Q

N)
1
(
Y Q

M),
for some FDLTIC stable system Q.
Proof:Follows immediately from Theorem 31 and Lemma 12.
Youla Parametrization of Closed-loop Maps
- By using the above parametrization we can show that
K(I G
22
K)
1
= (Y MQ)

M.
The map from w to z in Figure 8 is given by
= G
11
+G
12
K(I G
22
K)
1
G
21
.
Thus we have the folowing theorem
Theorem 34. Let G be FDLTIC system and let G
22
admit a dcf as given in
Lemma 10. is a map from w to z in Figure 8 for some FDLTIC, K which
stabilizes the closed loop if and only if
= H UQV,
where
H = G
11
+G
12
Y

MG
21
U = G
12
M
V =

MG
21
and Q is some stable FDLTIC system.
We now present a result which is a generalization of Theorem 34.
Theorem 35. [Youla parametrization] Let G be a FDLTIC system and let
G
22
admit a dcf as given in Lemma 10. is a map from w to z in Figure 8 for
some linear, time invariant, causal K which stabilizes the closed loop in the
sense if and only if

= H UQV,
where
H = G
11
+G
12
Y

MG
21
U = G
12
M
V =

MG
21
and Q is some
stable system.
The parameter Q is often referred to as the Youla parameter. The difference
between Theorem 34 and Theorem 35 is that in Theorem 35 the controller K
is not restricted to be nite-dimensional. The proof of this theorem is similar to
the one presented for Theorem 34 except that an analogous result for coprime
factorization over
stable systems is utilized.

MIMO PERFORMANCE 311
Existence of Coprime Factors
Lemma 13. Let T be a FDLTIC map with a state space description
_
A B
C D
_
. Suoppse(A, B) is stabilizable and (A, C) is detectable. Then
there exists a dcf of T.
Proof:In the denition of dcf let
X =
_
A+BF L
C +DF I
_
, Y =
_
A+BF L
F 0
_
,
X =
_
A+LC B +LD
F I
_
,

Y =
_
A+LC L
F 0
_
,
M =
_
A+BF B
F I
_
, N =
_
A+BF B
C +DF D
_
,
M =
_
A+LC L
C I
_
, and

N =
_
A+LC B +LD
C D
_
.
Then it can be shown that T = NM
1
=

M
1

N and (20) is satised.
Multiple Input Multiple Output
Interconnections: Performance
Denition 40. [Group] A group is a set G with a binary operation
(.) : GG G dened which has the following properties.
1. (a.b).c = a.(b.c); associativity property.
2. There exists an element e in G such that a.e = e.a = a for all a in G. e is
called the identity.
3. For every a in G there exists an element a
1
in G such that
a.a
1
= a
1
.a = e. a
1
is called the inverse of a.
Denition 41. [Subgroup] If H is a subset of a group G the H is a subgroup
if H is a group with the binary operation inherited from G.
Lemma 14. H is a subgroup of the group G if the identity element e is in H, a
belongs to H implies a
1
is in H and a and b belong to H implies a.b belongs
to H.
Lemma 15. A group G has a unique identity element. Also, every element in
G has a unique inverse.
Denition 42. [Abelian group] A group G is an abelian group if for any two
elements in G, a.b = b.a.
Denition 43. [Homomorphism] Let G and H be two groups. : G H is
a homomorphism between the two groups if (a.b) = (a).(b), for all a, b in G.
Lemma 16. A homomorphism : G H sends identity of G to the identity
of H and sends inverses to inverses.
Denition 44. [Isomorphism] An isomorphism is a homomorphism which is
one to one and onto.
Denition 45. [Fields] A eld K is a set that has the operations of addition
(+) : K K K and multiplication (.) : K K K dened such that
1. multiplication distributes over addition
a.(b +c) = a.b +a.c,
2. K is an abelian group under addition with identity written as 0 for addition.
3. K0 is an abelian group under multiplication with identity being 1.
Lemma 17. If in a eld K elements a ,= 0 and b ,= 0 then ab ,= 0.
Vector Space
Denition 46. A set V with two operations addition (+) : V V V and
scalar multiplication (.) : V K V, where K is a eld dened is a vector
space over the eld K if
1. V is an abelian group under addition.
2. multiplication distributes over addition
.(b +c) = .a +.b, for all in K, for all a, b in V.
The elements of the eld K are often called as scalars. The vector space is
called a real vector space if the eld K = R and the vector space is called a
complex vector space if the eld K = C.
Denition 47. [Algebra] V a vector space is an algebra if it has an operation
vector multiplication () : V V V dened such that this operation
distributes over vector addition.
Denition 48. [Units] If A is an algebra then x in A is an unit if there exists
some y in A such that x y = y x = 1.
Lemma 18. If A is an algebra with an associative vector multiplication and U
is the set of units in A then U is a group under vector multiplication.
From now on we will restrict the eld to be either the set of real numbers R or
the set of complex numbers C. Thus when we say K we mean either R or C.
Normed Vector Space
Denition 49. A normed linear space is a vector space X with a function
[[ [[ : X R dened such that
1. [[x[[ 0 and [[x[[ = 0 if and only if x = 0.
2. [[x[[ = [[ [[x[[ for any scalar and vector x in X.
3. [[x +y[[ [[x[[ +[[y[[.
Denition 50. [Induced Norm] Let (X, | |
X
) and (Y, | |
Y
) be normed
vector spaces. Let A : (X, | |
X
) (Y, | |
Y
) be a map. The induced norm of
the operator A is dened by
|A|
ind
= sup
x=0
|A(x)|
Y
|x|
X
.
Example 12. Let A be a mn matrix. Thus A : R
n
R
m
. Let the norm on
R
m
and R
n
spaces be the norm (|x|
= max
i
[x
i
[ where
x = (x
1
, . . . , x
k
)
T
.) Then the innity induced norm is given by
|A|
ind
= max
1im
n
j=1
[a
ij
[.
Proof: Note that
|Ax|
= max
i
[
n
j=1
a
ij
x
j
[
max
i
n
j=1
[a
ij
[ [x
j
[
|x|
max
i
n
j=1
[a
ij
[
Thus
max
x=0
|Ax|
|x|
max
i
n
j=1
[a
ij
[.
Suppose
i
0
= argmax
i
n
j=1
[a
ij
[.
Let x
j
= sgn(a
ij
). Then it follows that
A x
= max
i
[
n
j=1
a
ij
x
j
[
[
n
j=1
a
i
0
j
x
j
[
=

n
j=1
[a
i
0
j
[
= max
i
n
j=1
[a
ij
[
Therefore
max
x=0
|Ax|
|x|
max
i
n
j=1
[a
ij
[.
Lemma 19. Let x c
n
and y c
m
.
1. Suppose n m. Then |x|
2
= |y|
2
if and only if there exists a matrix
U c
nm
such that x = Uy and U
U = I.
2. Suppose n = m. Then [x
y[ |x|
2
|y|
2
. Moreover the equality holds if and
only if x = y for some c or y = 0.
3. |x| = |y| if and only if there is a matrix c
nm
with ||
2ind
1 such
that x = y. Furthermore |x| < |y| if and only if ||
2ind
< 1.
4. |Ux|
2
= |x|
2
for any unitary matrix U.
Lemma 20. Let A and B be matrices with appropriate dimensions. Then
1. (A) |A| where | | is any induced norm.
2. |AB| |A| |B| where | | denotes any induced norm.
3. |UAV |
2ind
= |A|
2ind
where U and V are unitary matrices.
Theorem 36. Let A c
mn
. Then there exists unitary matrices U c
mm
and V c
nn
such that
A = UV
such that
=
_

1
0
0 0
_
with
1
= diag(
1
,
2
, . . . ,
p
) where p = minm, n and
1

2
. . .
p
0.
i
are called the singular values of A.
Example 13. Let A c
mn
. Thus A : c
n
c
m
. The two induced norm of A
is its maximum singular value.
Proof: Note that
max
x=0
|Ax|
2
|x|
2
= max
x=0
|UV
x|
2
|x|
2
= max
x=0
_
(UV
x)
(UV
x)
|x|
2
= max
x=0
UV
x
|x|
2
= max
x=0
x
|V
x|
2
= max
x=0,y=V
y
|y|
2
= max
x=0,y=V
x
_
2
1
[y
1
[
2
+
2
2
[y
2
[
2
+. . . +
2
p
[y
p
[
2
|y|
2
max
x=0,y=V
x
_
2
1
[y
1
[
2
+
2
1
[y
2
[
2
+. . . +
2
1
[y
p
[
2
|y|
2

1
max
x=0,y=V
x
_
[y
1
[
2
+[y
2
[
2
+. . . +[y
p
[
2
|y|
2

1
.
Let
x = V e
1
where e
1
= (1, 0, 0 . . . , 0)
T
. Then it follows that
max
x=0
|Ax|
2
|x|
2
A x
2
x
2
= |UV
x|
2
= |UV
V e
1
|
2
= |Ue
1
|
2
=
1
.
Therefore
|A|
2ind
=
1
.
Also the notation (A) =
1
is used to denote the maximum singular value of
A and (A) =
p
is utilized to denote the smallest singular value of A.
It follows from UV
= A that
Av
i
=
i
u
i
A
u
i
=
i
v
i
Thus
A
Av
i
=
i
A
u
i
=
2
i
v
i
AA
u
i
=
i
Av
i
=
2
i
u
i
Thus
2
i
are the eigenvalues of A
A and AA
.
Lemma 21. 1. (A) = max
x=0
|Ax|
2
|x|
2
.
2. = min
x=0
|Ax|
2
|x|
2
.
3. [(A+ ) (A)[ ().
4. (A) (A)()
5. (A
1
) =
1
(A)
.
The L
and the H
spaces
L
(jR) is the Banach space of matrix valued functions that are essentially
bounded on the imaginary axis with the norm
|F|
:= ess sup
R
[F(j)].
H
is the Banach space of matrix valued functions that are essentially

bounded on the imaginary axis and analytic in the closed right half plane
s : Re(s) 0 with the norm
|F|
:= sup
s:Re(s)0
[F(s)] = ess sup
R
[F(j)].
1H
is the subspace of H
that consist of elements that are real and

rational functions of the complex variable s.
The H
norm is the two induced norm

Suppose G L
be a p q transfer matrix. Consider the multiplication

operator induced by G on
L
2
= f : |f|
2
:=
_

[f(j)[
2
d < |
dened by
M
G
: L
2
H
2
; M
G
f = Gf.
Theorem 37. Let G L
be a p q transfer matrix. Then

|M
G
| := |M
G
|
2ind
= |G|
.
Proof: Note that
|M
G
| = sup|Gf|
2
: |f|
2
1.
Now
|Gf|
2
2
=
_
(j)G
(j)G(j)F(j)d
=
_
|G(j)f(j)|
2
2

2
[G(j)]|f(j)|
2
2
|G|
|f(j)|
2
2
Thus
|M
G
| = sup|Gf|
2
: |f|
2
1 |G|
.
This proves one side of the theorem. That |M
G
| |G|
is left as an
exercise.
Consider a p q MIMO transfer matrix G. Let y = G(s)u with
y(t) =
_
_
_
_
y
1
sin(
0
t +
1
)
y
2
sin(
0
t +
2
)
.
.
.
y
p
sin(
0
t +
p
)
_
_
_
_
and u(t) =
_
_
_
_
u
1
sin(
0
t +
1
)
u
2
sin(
0
t +
2
)
.
.
.
u
q
sin(
0
t +
q
)
_
_
_
_
sup
i
,
0
, u
2
| y|
2
= |G|
where
y =
_
_
_
_
y
1
y
2
.
.
.
y
p
_
_
_
_
and u =
_
_
_
_
u
1
u
2
.
.
.
u
q
_
_
_
_
Performance Specications
Figure 9:
L
i
= KP is the input loop transfer matrix
L
o
= PK is the output loop transfer matrix
S
i
= (I +L
i
)
1
is the transfer matrix from d
i
to u
p
is the input sensitivity
matrix.
S
o
= (I +L
o
)
1
is the output sensitivity matrix.
T
i
= L
i
(I +L
i
)
1
is the input complimentary sensitivity matrix
T
o
= L
o
(I +L
o
)
1
is the output complimentary sensitivity matrix.
Figure 10:
Loop equations are given by
y = T
o
(r n) +S
o
Pd
i
+S
o
d
r y = S
o
(r d) +T
o
n S
o
Pd
i
u = KS
o
(r n) KS
o
d T
i
d
i
u
p
= KS
o
(r n) KS
o
d +S
i
d
i
Disturbance Rejection
y = T
o
(r n) +S
o
Pd
i
+S
o
d
r y = S
o
(r d) +T
o
n S
o
Pd
i
u = KS
o
(r n) KS
o
d T
i
d
i
u
p
= KS
o
(r n) KS
o
d +S
i
d
i
Good output disturbance rejection at the output y would require small
(S
o
) = [(I +PK)
1
] =
1
(I +PK)
Good input disturbance rejection at the output y would require small
(S
o
P) = [(I +PK)
1
P] = (PS
i
)
Good input disturbance rejection at the plant input u
p
would require small
(S
i
) = [(I +KP)
1
] =
1
(I +KP)
Good output disturbance rejection at the plant input u
p
would require small
(S
i
K) = [(I +KP)
1
K] = (KS
o
)
Note that
(A) 1 (I +A) (A) + 1.
It follows that
1
(PK)+1

1
(PK+I)
= (S
o
)
1
(PK)1
if (PK) > 1
1
(KP)+1

1
(KP+I)
= (S
i
)
1
(KP)1
if (KP) > 1
It follows that (S
0
) and (S
i
) are small if and only if (PK) and (KP) are
respectively large.
Thus for good output disturbance rejection at the output one needs
(PK) >> 1
Thus for good input disturbance rejection at the plant input one needs
(KP) >> 1
Now if its assumed that P and K are invertible and that (PK) >> 1 then it
follows that
(S
o
P) = ((I +PK)
1
P)) = ((I +PK)
1
PKK
1
))
= ((I +PK)
1
PK)) (K
1
)
(K
1
)
=
1
(K)
.
Now (S
o
P) has to be small for rejection of input disturbance at the output.
Thus
good input and output rejection at the output (PK) >> 1 and
(K) >> 1 in the appropriate frequency range.
Similarly if its assumed that P and K are invertible and that (KP) >> 1 then
it follows that
(S
i
K) = ((I +KP)
1
K)) = ((I +KP)
1
KPP
1
))
= ((I +KP)
1
KP)) (P
1
)
(P
1
)
=
1
(P)
.
Now (S
i
K) has to be small for rejection of output disturbance at the input.
Thus
good input and output rejection at the input (KP) >> 1 and
(P) >> 1 in the appropriate frequency range.
The condition that (P) >> 1 is a fundamental limitation in the sense that no
controller can alleviate the situation if its not met.
Noise Rejection
y = T
o
(r n) +S
o
Pd
i
+S
o
d
r y = S
o
(r d) +T
o
n S
o
Pd
i
u = KS
o
(r n) KS
o
d T
i
d
i
u
p
= KS
o
(r n) KS
o
d +S
i
d
i
Good noise rejection at the output requires (T
o
) = (L
o
(I +L
o
)
1
) to be
small. This implies that (PK) << 1 in the frequency range where the
noise effects are predominant.
Thus a tradeoff has to be struck between good noise rejection and good
disturbance rejection. Also note that if (L
o
) << 1 then S
o
I and KS
o
K.
Now the effect of noise on the control output u is given by
u = KS
o
n.
Thus
To prevent the noise from saturating the controller (K) M in the
frequency range where the loop gain is small.
LFT 342
Performance Specications Summarized
In a frequency range [0,
] that characterizes the frequency content of the

disturbances and tracking needs
(PK) >> 1, (KP) >> 1, (K) >> 1.
In a frequency range [
u
, ) that characterizes the frequency content of the
noise and saturation effects
(PK) << 1, (KP) << 1, (K) M.
LFT 343
Linear Fractional Transformations
LFT 344
Lower Linear Fractional Transformations
Figure 11:
Suppose
_
z
y
_
=
_
M
11
M
12
M
21
M
22
__
w
u
_
and suppose
u = Ky.
LFT 345
Then the map from w z is given by
T
(M, K) = M
11
+M
12
K(I M
22
K)
1
M
21
called the lower fractional transformation of M and K.
LFT 346
Upper Linear Fractional Transformations
Figure 12:
Suppose
_
s
z
_
=
_
M
11
M
12
M
21
M
22
__
v
w
_
and suppose
v = s.
LFT 347
Then the map from w z is given by
T
u
(M, ) = M
22
+M
21
(I M
11
)
1
M
12
called the upper fractional transformation of M and .
LFT 348
GK Framework
Figure 13:
Note in the above map
z = T
u
(T
(G, K), )w.

LFT 349
The following lemma follows from simple algebra
Lemma 22. Suppose C is invertible. Then
(A+BQ)(C +DQ)
1
= T
(M, Q)
(C +DQ)
1
(A+QB)
1
= T
u
(N, Q)
where
M =
_
AC
1
B AC
1
D
C
1
C
1
D
_
, N =
_
C
1
A C
1
B DC
1
A DC
1
_
.
LFT 350
Lemma 23. Let M =
_
M
11
M
12
M
21
M
22
_
and M
22
be invertible. Then
(T
u
(M, ))
1
= T
(N, )
where N is given by
N =
_
M
11
M
12
M
1
22
M
21
M
12
M
1
22
M
1
22
M
21
M
1
22
_
.
LFT 351
Parametric Uncertainty: An Example
Figure 14:
Consider a spring-mass-damper system where the spring constant is k, the
mass m, and the damping factor is c. The dynamical equation is given by
x +
c
m
x +
k
m
x =
F
m
as describe by Figure 14.
Suppose k, m and c are each uncertain by 1% of their nominal values

k, m,
LFT 352
and c. Thus
k =

k(1 + 0.1
k
),
1
m
=
1
m(1 + 0.1
m
)
and c = c(1 + 0.1
c
).
Note that
1
m(1+0.1
m
)
= T
(M
1
,
m
) where
M =
_
1
m

0.1
m
1 0.1
_
.
The block diagram in terms of the uncertainties
k
,
m
, and
c
is given in
Figure 15.
LFT 353
Figure 15:
It can be veried that
_
x
1
x
2
_
= T
(M, )
_
_
x
1
x
2
F
_
_
LFT 354
where
M =
_
_
_
_
_
_
0 1 0 0 0 0
k
m

c
m
1
m

1
m

1
m

0.1
m
0.1
k 0 0 0 0 0
0 0.1 c 0 0 0 0
k c 1 1 1 0.1
_
_
_
_
_
_
=
_
_
k
0 0
0
c
0
0 0
m
_
_
.
LFT 355
Polynomial dependence on uncertain parameters
Figure 16:
LFT 356
Polynomial dependence on uncertain parameters
Figure 17:
LFT 357
Rational dependence on uncertain parameters
Figure 18:
LFT 358
A General Uncertainty Description
Figure 19:
Note in the above map
z = T
u
(T
(G, K), )w.

LFT 359
We have seen that a general description of the uncertainty is well captured by
diag[
1
I
r
1
,
2
I
r
2
, . . . ,
s
I
r
s
,
1
,
2
, . . . ,
F
] :
i
1,
i
1H
.
The allowable class of uncertainty is
LTI
= diag[
1
I
r
1
,
2
I
r
2
, . . . ,
s
I
r
s
,
1
,
2
, . . . ,
F
] :
i
R,
i
1H
.
Associated with the above class of allowable perturbations we also dene
B
LTI
=
LTI
: ||
1.
Note that for any B
LTI
with
= diag[
1
I
r
1
,
2
I
r
2
, . . . ,
s
I
r
s
,
1
,
2
, . . . ,
F
] the following conditions are
equivalent
||
1
LFT 360
[
i
[ 1 for all i = 1, . . . , s and |
i
(s)|
1 for all i = 1, . . . , F.
We also dene the following sets of constant matrices
= diag[
1
I
r
1
,
2
I
r
2
, . . . ,
s
I
r
s
,
1
,
2
, . . . ,
F
] :
i
R,
i
c
m
j
m
j
B = : () 1.
Lemma 24. Given a constant matrix and R there exists a transfer
matrix
(s) B
LTI
such that
=
(j).
MIMO ROBUST STABILITY 361
Robust Stability of MIMO Systems
Nominal Stability
Figure 20:
Denition 51. The GK interconnection in Figure 20(a) is Nominally
stable (NS) if the GK interconnection in Figure 20(b) is internally stable.
Note that the interconnection in Figure 20(b) can be internally stabilized if G
can be stabilized through the control input u. In other words if
_
A B
C D
_
is
a minimal realization of G then the inherited realization of G
33
has to be
stabilizable and detectable. Otherwise there will be no controller that can
internally stabilize the interconnection and thus no controller can achieve
nominal stability.
Figure 21:
Suppose the interconnection in Figure 21(a) is internally stable. That is we
have nominal stability. Then
N = F
(G, K) = G
11
+G
12
K(I P
22
K)
1
P
21
will be stable.
The N Interconnection
Figure 22:
Suppose the GK interconnection is internally stable. Then it follows that the
GK interconnection is nominally stable and N is stable with
N = T
(G, K).
Now consider the N interconnection shown in Figure 22(a). It is evident
that the N interconnection is stable if and only if N
11
interconnection
in Figure 22(b) is stable. This follows as N is stable and therefore N trivially
stabilizable through v.
The Robust Stability
Figure 23:
Denition 52. The GK interconnection is said to be robustly stable
for all B
LTI
if the interconnection is internally stable for all B
LTI
.
From the discussion above it follows that the following two statements are
equivalent
The GK interconnection is robustly stable
The GK interconnection is internally stable (Nominal Stability) and the
M interconnection is internally stable for all B
LTI
with M = N
11
.
A Robust Stability Theorem
Theorem 38. Assume that M is a stable transfer matrix. The following
statements are equivalent.
The M interconnection is robustly stable with respect to
LTI
(that is
the M interconnection is internally stable for all B
LTI
).
det(I M(j)(j) ,= 0 for all B
LTI
and for all R.
Proof: For B
LTI
, is stable. Note that as M and both are stable we
have from the Nyquist criterion for MIMO systems the M interconnection
is stable if and only if the Nyquist contour of det(I M(j)(j)) does not
encircle the origin and does not touch the origin.
(1 2) This follows easily from the Nyquist criterion. Note that the Nyquist
criterion states that the contour of det(I M(j)(j)) should not touch the
origin for stability. From (1) we have that the M interconnection is stable
for all in B
LTI
it follows that
det(I M(j)(j)) ,= 0 for all B
LTI
and for all R.
(2 1) Suppose there exists a B
LTI
such that the M
interconnection is not internally stable. From the Nyquist criterion atleast one
of the following conditions have to be violated
det(I M(j)(j)) = 0 for some R.
The Nyquist contour of det(I M(j)(j)) encircles the origin at least
once.
If the rst condition holds then the statement is proven. Suppose not. Then
the Nyquist contour of det(I M(j)(j)) encircles the origin at least once.
Note that if f(, s) := det(I M(s)(s) then the Nyquist contour of f(, s)
changes continuously with respect to . For = 1 the Nyquist contour of f
encircles 0. For = 0 the Nyquist contour is a single point 1. Thus for some
[0, 1] and some
R, f(
) = 0 that is det(I M(j
(j
)) = 0.
Its evident that
(s) :=
(s) B
LTI
. Thus there exists a
B
LTI
such that for some
, det(I M(j
(j
)) = 0.
The following Corollary follows from the theorem above and Lemma 24
Corollary 4. Assume that M is a stable transfer matrix. The following
statements are equivalent.
The M interconnection is robustly stable with respect to
LTI
.
det(I M(j)) ,= 0 for all Band for all R.
A Robust Stability for Unstructured Uncertainty
Consider the uncertainty class
1H
[||
1.
There is no structure to the class above. It is relatively easy to obtain
necessary and sufcient conditions for robust stability with respect to the
above class.
Lemma 25. Let A be a complex matrix. Then it follows that
max
(B)1
(AB) max
(B)1
(AB) = (A).
Proof: Note that as () is an induced norm it follows that
(AB) (AB)
and thus
max
(B)1
(AB) max
(B)1
(AB).
Now suppose the singular value decomposition of A is given by UV
. Let
B
:= V U
. Then it follows that

(B
) (V ) (U
) = 1.
Furthermore we have that
(AB
) = (UV
V U
) = (UV
) = () =:
1
.
Now
AB
= UU

(AB
) =
1
= (A).
Thus we have constructed a matrix B
with bar(B
) 1 and
(AB
) = (AB
). Thus it follows that

max
(B)1
(AB) (A).
This proves the lemma.
Theorem 39. Let M(s) be a stable transfer matrix. The M
interconnection is internally stable for all
1H
: ||
1 if and
only if |M(s)|
< 1.
Proof: () Suppose |M|
< 1. Let
1H
: ||
1. Let
R. Then
(M(j)(j)) (M(j)(j)) (M(j)) ((j)) < 1.
det(I M(j)(j)) ,= 0 for all R.
Thus from Theorem 38 that the M interconnection is robustly stable with
respect to the class
1H
: ||
1.
() Suppose |M|
1 and suppose is such that (M(j)) 1. It follows

from Lemma 25
max
()1
(M(j)) = (M(j)) 1.
Thus there exists a constant matrix with () 1 such that
M(j)x = x
with x ,= 0 and [[ 1. Let
=
1
.
Note that (
) 1 Then it follows that M(j)
has an eigenvalue at 1 and

thus det(I M(j)
) = 0. One can construct a (s) 1H
with the
property that (j) =
and ||
1. Thus we have constructed a (s)

with ||
1 with
det(I M(j)(j)) = det(I M(j)
) = 0.
From Theorem 38 it follows that the M interconnection is not robustly
stable.
A General Uncertainty Description
Let the structure of the uncertainty structure be captured by the class .
Dene
: c
nn
: R by
(M) :=
1
min () : det(I M) = 0 with
.
The following theorem elucidates the signicance of this denition.
Theorem 40. Suppose the class of allowable uncertainty is given by .
Then the M interconnection is stable for all (s) B
LTI
if and only if
(M(j)) < 1 for all .

Proof: From Corollary 4 that the M interconnection is stable if and only if
for any R
det(I M(j)) ,= 0 for all B
[det(I M(j)) = 0 for any ] () > 1
min
() : det(I M(j)) = 0 > 1
1
min
{ ():det(IM(j))=0 }
< 1

(M(j)) < 1.
.
Properties of .
1. For any uncertainty structure and scalar , (M) = [[(M).
2. For any uncertainty structure
(M) (M).
3. If the uncertainty structure is such that it consists only of full complex
blocks = c
nn
then
(M) = (M).
4. Let D be a set of matrices that commute with the matrices in (that is if
D then D = D for all ) then for any D D
(M) = (DMD
1
).
5. Let D be a set of matrices that commute with the matrices in (that is if
D then D = D for all ) then for any D D
(M) (DMD
1
).
6. If the uncertainty structure consists only of complex blocks then
(M) = max
B
(M).
7. For any uncertainty structure and for any unitary matrix U
(MU) = (M) = (UM).
8. Let consist only of complex blocks then
(M) = max
UU
(MU)
where
U = U : U
U = I and U .
Note that from Theorem 40 we have that the M interconnection is
robustly stable with respect to if and only if
(M(j)) 1 for all R.

This condition can be replaced by
sup
R
[
(M(j))] < 1.
Note M = N
11
where N = T
(G, K). Thus robust stability is guaranteed if one

can nd a controller K that internally stabilizes the GK interconnection and
sup
R
[
(M(j))] < 1.
Thus the synthesizing the optimal controller will be obtained by solving the
following problem:
inf
K stabilizing
sup
R
[
(M(K)(j))].
Note that computing
(M(K)(j)) is not easy and thus we replace it with its

upper bound (M(K)(j)). This bound can be improved by using the fact that
(M(K)(j)) =
(D()M(K)(j)D
1
())] [D()M(K)(j)D
1
()]
for all D() D where D is the set of complex matrices that commute with
matrices in . Thus we can use the following bound
M(K)(j) inf
D()D
(D()M(K)(j)D
1
).
Thus we have the following problem to solve
inf
K stabilizing
sup
inf
D()D
(D()M(K)(j)D
1
()).
Let

D be the set of maps from R c
nn
such that every element

D

D
satises

D() D. Thus the problem is
inf
K stabilizing
sup
inf
D
(D()M(K)(j)D
1
()).
It is true that
sup
inf
D
(D()M(K)(j)D
1
()) = inf
D
sup
(D()M(K)(j)D
1
()).
Thus the problem becomes
inf
K stabilizing
inf
D
sup
(D()M(K)(j)D
1
()).
We will use another upper bound. Let
D
s
:= D(s) 1H
nn
: D
1
(s) 1H
nn
and D(j) commutes with .

Note that
inf
K stabilizing
inf
D
sup
(D()M(K)(j)D
1
())
inf
K stabilizing
inf
D
s
D
s
sup
(D
s
()M(K)(j)D
1
s
()).
We will use the greater upper bound as it is easier to solve.
inf
K stabilizing
inf
D
s
D
s
sup
(D(j)M(K)(j)D
1
(j))
= inf
K stabilizing
inf
D
s
D
s
|D
s
M(K)D
1
s
|
Things to note
Suppose K is a stabilizing controller. Then
inf
D
s
D
s
|D
s
M(K)D
1
s
|
(28)
can be solved.
Suppose D
s
1H
nn
is a xed transfer matrix. Then

inf
K stabilizing
|D
s
M(K)D
1
s
|
(29)
MIMO ROBUST PERFORMANCE 386
is a standard H
problem and can be solved.

The joint problem
inf
K stabilizing, D
s
D
s
|D
s
M(K)D
1
s
|
is hard to solve.
The D K iteration scheme operates by rst assuming D
s
= I. Compute K
1
that solves (28). Then with K = K
1
solve the problem (29). Let D
1
(s) be the
solution. Solve (28) with D
s
= D
1
to obtain K
2
. Iterate to get a satisfactory
result. Note that there is no guarantee of convergence for this problem.
Multiple Input Multiple Output
Interconnections: Robust Performance
Robust Performance
Figure 24:
Denition 53. The GK interconnection achieves robust performance
if the GK is robustly stable with respect to
LTI
and
|T
u
(T
(G, K), )|
< 1 for all B

LTI
.
Theorem 41. The GK interconnection achieves robust performance if
and only if the GK interconnection is nominally stable and
sup
P
[T
(G, K)(j)] < 1

where
P
:= diag(
p
, ) :
LTI
,
p
1H
n
w
n
z
.
Proof: () Suppose the GK framework achieves robust performance.
Then it follows that GK interconnection is internally stable (Nominal
Stability) and the N (with N = T
(G, K)) interconnection is stable for all

B
LTI
. Let B
LTI
and let M := T
u
(N, ). The N
interconnection is stable and |T
u
(N, )|
< 1. Thus from Theorem 39 (small

gain theorem) it follows that the the M
p
interconnection is stable for any
p

s
1H
n
w
n
z
: |
s
|
1.
This proves that the M
p
interconnection is internally stable. This implies
that the interconnection of T
(G, K) and any

p

P
with |
p
|
1 is
stable. That is T
(G, K) achieves robust stability with respect to

P
. Thus
from Theorem 40
sup
P
[T
(G, K)(j)] < 1.

() Suppose
sup
P
[T
(G, K)(j)] < 1

and GK interconnection is internally stable (Nominal stability). Then
T
(G, K) is stable and T
(G, K)
P
interconnection is internally stable for
any
P
B
P
.
Let
LTI
. Then the M
p
interconnection is stable with
M := T
u
(N, ) and
p

s
1H
n
w
n
z
: |
s
|
1.
From the small gain theorem on unstructured uncertainty it follows that
|M|
< 1.
Thus
|T
u
(N, )|
< 1 for all B

LTI
.
H
LOOP SHAPING 392

H
Loop Shaping
H
LOOP SHAPING 393

McFarlane Glover Design
H
LOOP SHAPING 394

McFarlane Glover Design
+
+
N
N
K
+
+
M
M
+
+
e
2
e
1+
+
Figure 25:
The nominal plant is given in the coprime factor form as G =

M
1

N.
The perturbed plant is given by
G
= G = (

M +
M
)
1
(

N +
N
)
H
LOOP SHAPING 395

where
M
and
N
are stable unknown transfer functions.
The robust design objective is to stabilize not only the nominal model but
the family of the perturbed plants given by
(
= (

M +
M
)
1
(

N +
N
) : |[
M
,
N
]|
H
<
with a controller K as shown in Figure 25.
H
LOOP SHAPING 396

Advantages of Coprime Factor Perturbation Form
In the other forms of perturbations (e.g. additive uncertainty, multiplicative
uncertainty and inverse multiplicative forms) the number of unstable poles
of the nominal and the perturbed plants has to be the same. In the coprime
perturbation form the number of unstable poles for the perturbed form can
be different than the number of unstable poles of the nominal plant.
The solution in this case is particularly elegant.
It can be used for robustyfying any existing closed loop design.
H
LOOP SHAPING 397

Preliminaries
Denition 54. The feedback system of Figure 25 denoted by (

M,

N, K, ) is
robustly stable if and only if the interconnection (G
, K) is internally stable for

all G
.
H
LOOP SHAPING 398

M Structure
P
K
z w
u
y
v s
Figure 26:
We will rst cast the coprime perturbation robust stability problem into the
H
LOOP SHAPING 399

standard framework shown in Figure 26 with := [
M
,
N
]. Note that in this
case the signals w and z are absent and v in Figure 26 corresponds to v in
Figure 25. Also, s in Figure 26 is given by the vector (s
M
, s
N
)
T
.
H
LOOP SHAPING 400

+
+
N
K
M
y
u
v
s
M
s
N
Figure 27:
Note that
_
_
s
M
s
N
y
_
_
=
_
_
M
1
M
1
N
0 I
M
1
M
1
N
_
_
. .
P
_
v
u
_
u = Ky
.
Note that y = M
1
v +M
1
Nu = M
1
v +Gu = M
1
v +GKy. Thus
H
LOOP SHAPING 401

y = (I GK)
1
M
1
v. This implies that u = K(I GK)
1
M
1
v. Thus we
obtain
_
s
M
s
N
_
=
_
y
u
_
=
_
(I GK)
1
M
1
K(I GK)
1
M
1
_
. .
M
v.
H
LOOP SHAPING 402

Small Gain Theorem
From the small gain theorem it follows that the interconnection is stable for
all (
if and only if
|M|
H
=
_
_
_
_
_
K
I
_
(I GK)
1
M
1
_
_
_
_
H
.
Thus the robust stability problem in the coprime perturbation setting can be
solved by an equivalent H
problem.
The solution is even more elegant; there is no need for iterations to obtain
the optimal controller which achieves the largest .
H
LOOP SHAPING 403

Glover McFarland Design
Theorem 42. A controller K is stabilizing and satises
_
_
_
_
_
K
I
_
(I GK)
1
M
1
_
_
_
_
H

if and only if K has a rcf K = UV
1
for some U, V RH
satisfying
_
_
_
_
_
N
_
+
_
U
V
__
_
_
_
H
(1
2
)
1
2
.
H
LOOP SHAPING 404

Theorem 43. 1. Optimal solutions to the normalized lcf robust stabilization
problem gives
inf
K stabilizing
_
_
_
_
_
K
I
_
(I GK)
1
M
1
_
_
_
_
H
= 1 |[N, M|
2
H
1
2
.
2. The maximum robust stability margin is
max
= 1 |[N, M]|
2
H
1
2
3. All optimal controllers are given by K = UV
1
where U, V RH
satisfy
_
_
_
_
_
N
_
+
_
U
V
__
_
_
_
H
= |[N, M|
H
.
H
LOOP SHAPING 405

McFarland Glover Controller
Theorem 44. The controller K (a positive feedback controller) that
guarantees
_
_
_
_
_
K
I
_
(I GK)
1
M
1
_
_
_
_
H

for a specied >
min
is given by
_
A+BF +
2
(L
T
)
1
ZC
T
(C +DF)
2
(L
T
)
1
ZC
T
B
T
X D
T
_
where
min
=
1
max
= (1 +(XZ))
1
2
, F = S
1
(D
T
C +B
T
X), L = (1
2
)I +XZ,
H
LOOP SHAPING 406

and X and Z are the solutions to
(ABS
1
D
T
C)
T
X +X(ABS
1
D
T
C) XBS
1
B
T
X +C
T
R
1
C = 0 and
(30)
(ABS
1
D
T
C)Z +Z(ABS
1
D
T
C)
T
ZC
T
R
1
CZ +BS
1
B
T
= 0 (31)
with R = I +DD
T
and S = I +D
T
D.
SISO H
2
Problem 407
SISO H
2
Problem
SISO H
2
Problem 408
Consider a discrete time generalized system G(z) and suppose that we
denote by := z
1
. Then G(z) = G() is stable if and only if all poles are
outside the unit disc. The unstable poles and zeros are the ones that are
inside the closed unit disc. Suppose the set of closed loop maps achieveable
via stabilizing controllers is given by
H() U()Q() : Q stable
where h, u are stable transfer functions.
Suppose we denote the closed-loop map by (q) := H UQ. We denote the
impulse response of the H, U, Q and by h, u, q and respectively. As all
the transfer functions are stable we have that h, u, q and
1
.
Suppose the input to this system is white with variance
2
. Then the output
variance is given by
2
[
k=0
(k)
2
] =:
2
||
2
2
.
SISO H
2
Problem 409
Thus if denotes a system transfer function between a white noise input with
unit variance then the output variance is given by ||
2
. The following problem
is of relevance
= min
K stabilizing
|(K)|
2
2
= min
Q stable
|(K)|
2
2
: = h uq
Suppose the unstable zeros of u() are given by z
1
, z
2
, . . . , z
n
that are all
SISO H
2
Problem 410
distinct and real. Note that
() = H() U()q() for some Q stable
stable and
H()()
U()
is stable
stable and H(z
i
) (z
i
) = 0 for all i = 1, . . . , n
stable and

k=0
(k)z
k
i
=

k=0
h(k)z
k
i
=: b
i
for all i = 1, . . . , n
stable and

k=0
(k)z
k
i
=

k=0
h(k)z
k
i
=: b
i
for all i = 1, . . . , n

1
and
_
_
_
_
1 z
1
z
2
1
. . .
1 z
2
z
2
2
. . .
.
.
.
.
.
.
.
.
.
1 z
n
z
2
n
. . .
_
_
_
_
. .
A
_
_
_
_
(0)
(1)
(2)
.
.
.
_
_
_
_
. .
=
_
_
_
_
b
1
b
2
.
.
.
b
n
_
_
_
_
. .
b

1
and A = b.
SISO H
2
Problem 411
Thus the problem becomes
= inf
1
,A=b
k=0
[(k)[
2
.
Using Lagrange multipliers the problem is equivalent to solving
= max
yR
n inf
k=0
[(k)[
2
+y
[A b]
= max
yR
n inf
k=0
[(k)[
2
+
y b
y]
Note that
v := A
y =
_
v
1
v
2
. . . v
n
_
. .
A
_
_
_
_
y
1
y
2
. . .
y
n
_
_
_
_
SISO H
2
Problem 412
Thus v =

n
i=1
y
i
v
i
. Thus
= max
yR
n inf
k=0
[(k)[
2
+
y b
y]
= max
yR
n inf
k=0
[(k)[
2
+
v b
y]
= max
yR
n inf
k=0
[[(k)[
2
+(k)v(k)] b
y
Consider
inf
k=0
[[(k)[
2
+(k)v(k)] b
y.
The solution can be obtained by minimizing over each (k) the expression
[[(k)[
2
+(k)v(k)]
which is minimized by
(k) =
v(k)
2
.
SISO H
2
Problem 413
Thus
= max
yR
n inf
k=0
[[(k)[
2
+(k)v(k)] b
y
= max
yR
n
k=0
1
4
v(k)
2
1
2
v(k)
2
b
y
= max
yR
n
k=0
1
4
v(k)
2
b
y
= min
yR
n
k=0
1
4
v(k)
2
+b
y
Thus the problem reduces to solving the problem
min
yR
n
1
4
y
AA
y +b
y.
The solution to the above problem is given by
y = 2(AA
)
1
b.
Thus the optimal v is given by
v
o
= A
y = 2A
(AA
)
1
b
SISO H
2
Problem 414
and the optimal is given by
=
1
2
v = A
(AA
)
1
b
and the minimum value = (b
(AA
)
1
b)
1
.

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Robust Control: EE 575

takes large values near =

for all <

1 and thus does not conict with the sensitivity

software solves. Instead the problem of nding a

norm for a vector valued transfer function f : C C

paradigm these concerns of sensor noise rejection are reected

optimal controllers are not unique.

1 for all relevant i and j

to be achieved the following conditions have to be satised

< 1 is a necessary condition.

< 1 is a necessary condition.

Theorem 15. Consider the interconnection depicted in the Figure above

< 1 then 1 [M(j)[ [(j))[ > 0 for any ||

1 then we can construct a such that is a stable

1 and there exists an where

< 1, stable with ||

< 1 is satised. Let be xed

SISO CONTROLLER SYNTHESIS 206

denote the truncation operator on L

T(s) is a bi-proper transfer matrix.

P() = P(0) +P(1) +. . . +P(k)

M is called the Smith-Mcmillan form of

MIMO INTERCONNECTIONS 272

sense if and only if

stable systems is utilized.

is the Banach space of matrix valued functions that are essentially

that consist of elements that are real and

norm is the two induced norm

be a p q transfer matrix. Consider the multiplication

be a p q transfer matrix. Then

] that characterizes the frequency content of the

(G, K), )w.

(G, K), )w.

[0, 1] and some

) = 0 that is det(I M(j

. Then it follows that

Thus it follows that

). Thus it follows that

1 and suppose is such that (M(j)) 1. It follows

) 1 Then it follows that M(j)

has an eigenvalue at 1 and

) = 0. One can construct a (s) 1H

1. Thus we have constructed a (s)

(M(j)) < 1 for all .

() : det(I M(j)) = 0 > 1

(M(j)) 1 for all R.

(G, K). Thus robust stability is guaranteed if one

(M(K)(j)) is not easy and thus we replace it with its

and D(j) commutes with .

is a xed transfer matrix. Then

problem and can be solved.

< 1 for all B

(G, K)(j)] < 1

(G, K)) interconnection is stable for all

< 1. Thus from Theorem 39 (small

(G, K) and any

(G, K) achieves robust stability with respect to

(G, K)(j)] < 1.