CH 02
CH 02
CH 02
Modeling in the
Frequency Domain
SOLUTIONS TO CASE STUDIES CHALLENGES
d(i 0 + i)
Writing the differential equation, + 2(i0 + i) 2 − 5 = v(t) . Linearizing i2 about i0,
dt
2 2
(i +i) - i = 2i i = 2i i. Thus, (i +i)2 = i2 + 2i i.
0 0 0 0 0 0
i=i
0
2-2 Chapter 2: Modeling in the Frequency Domain
di
Substituting into the differential equation yields, + 2i02 + 4i0i - 5 = v(t). But, the
dt
resistor voltage equals the battery voltage at equilibrium when the supply voltage is zero since
the voltage across the inductor is zero at dc. Hence, 2i02 = 5, or i0 = 1.58. Substituting into the linearized
di i(s) 1
differential equation, + 6.32i = v(t). Converting to a transfer function, = . Using the
dt V(s) s+6.32
linearized i about i0, and the fact that vr(t) is 5 volts at equilibrium, the linearized v r(t) is vr(t) = 2i2 =
2(i0+i)2 = 2(i02+2i0i) = 5+6.32i. For excursions away from equilibrium, v r(t) - 5 = 6.32i = vr(t).
Vr(s) 6.32
Therefore, multiplying the transfer function by 6.32, yields, = as the transfer function about
V(s) s+6.32
v(t) = 0.
SOLUTIONS TO PROBLEMS
1.
1 −st 1
0
− st
a. F(s) = e dt = − e =
s 0 s
e −st −(st + 1)
b. F(s) = te dt = 2 (−st − 1) 0 =
− st
0
s s2 e st 0
Solutions to Problems 2-3
−s 1
F(s) t → = = 0. Therefore, F(s) = .
s 3e st t → s2
e − st
c. F(s) = sin t e dt = 2
− st
2 (−s sin t − cos t) = 2
0
s + 0
s +2
e − st s
d. F(s) = cost e dt = 2
− st
2 (−s cos t + sin t) = 2
0
s + 0
s +2
2.
a. Using the frequency shift theorem and the Laplace transform of sin t, F(s) = .
(s+a)2+2
(s+a)
b. Using the frequency shift theorem and the Laplace transform of cos t, F(s) = .
(s+a)2+2
t2
c. Using the integration theorem, and successively integrating u(t) three times,
dt = t;
tdt = ;
2
3.
a. Taking the sum of the voltages around the loop and assuming zero initial conditions yields:
t
di(t ) 1
Ri(t ) + L + i ( )d = v(t )
dt C0
I ( s) 1 1
= =
V ( s) Ls + R + 1 R 1
L( s + + )
Cs L LCs
I ( s) 2 2s
= = 2
V ( s) ( s + 2 + 16 ) s + 2s + 16
s
2-4 Chapter 2: Modeling in the Frequency Domain
2
I ( s) =
s + 2s + 16
2
Observing that the denominator has complex roots, we re-write the above equation as:
2
I ( s) =
( s + 1) + ( 15 ) 2
2
Applying the frequency shift theorem to the Laplace transform of sin t u(t), we find that the
− at
transform for f (t ) = e sin(t ) is F ( s) = .
(s + a)2 + 2
Comparing F(s) to I(s), we conclude that in the latter: a = 1 and = 15 . Thus, the current, i(t),
2
i (t ) = 15 e −t sin( 15 t )
15
c.
0.5
0.4
0.3
Current, i(t), A
0.2
0.1
-0.1
-0.2
0 0.5 1 1.5 2 2.5 3 3.5 4
Time, sec
Solutions to Problems 2-5
4.
a.
The Laplace transform of the differential equation, assuming zero initial conditions, is,
2𝑠
(𝑠 + 5)𝑋(𝑠) = 2
(𝑠 + 32 )
Multiplying by the lowest common denominator and equating the same powers of s on both
sides,
𝐴 + 𝐵 = 0, 5𝐵 + 𝐶 = 2, 9𝐴 + 5𝐶 = 0
Combining equations,
5 5 9
𝐴=− , 𝐵= , 𝐶= ,
17 17 17
Thus,
5 5 9 1
− 17 𝑠 .3.3
𝑋(𝑠) = 17
+ 2 + 17
𝑠 + 5 𝑠 + 32 𝑠2 + 9
Taking the inverse Laplace transform,
5 5 3
𝑥(𝑡) = − 𝑒 −5𝑡 + cos 3𝑡 + sin 3𝑡
17 17 17
b.
The Laplace transform of the differential equation, assuming zero initial conditions, is,
2
(𝑠 2 + 4𝑠 + 2)𝑋(𝑠) = 2
𝑠 +1
Solving for 𝑋(𝑠) and expanding by partial fractions using the two real roots of the quadratic,
2 𝐴 𝐵 𝐶𝑠 + 𝐷
𝑋(𝑠) = 2 = + +
(𝑠 + 4𝑠 + 2)(𝑠 2 + 1) 𝑠 + 3.414 𝑠 + 0.586 𝑠 2 + 1
Multiplying by the lowest common denominator and equating the same powers of s on both
sides,
𝐴 + 𝐵 + 𝐶 = 0, 0.586𝐴 + 3.414𝐵 + 4𝐶 + 𝐷 = 0,
𝐴 + 𝐵 + 2𝐶 + 4𝐷 = 0, 0.586A+3.414B+2D=2
2-6 Chapter 2: Modeling in the Frequency Domain
Combining equations,
Therefore,
8 2
2 −0.56 0.527 𝑠
X(𝑠) = (𝑠2 +4𝑠+2)(𝑠2+1) = 𝑠+3.414 + 𝑠+0.586 − 𝑠17
2 +1
+ 17
𝑠2 +1
c.
The Laplace transform of the differential equation, assuming zero initial conditions, is
5
((𝑠 2 + 6𝑠 + 20)𝑋(𝑠) =
𝑠
Solving for 𝑋(𝑠) and expanding by partial fractions,
5 𝐴 𝐵𝑠 + 𝐶
𝑋(𝑠) = 2
= + 2
𝑠(𝑠 + 6𝑠 + 20) 𝑠 𝑠 + 6𝑠 + 20
Multiplying by the lowest common denominator and equating the same powers of s on both
sides,
𝐴 + 𝐵 = 0, 6𝐴 + 𝐶 = 0, 20𝐴 = 5
Combining equations,
1 1 3
𝐴= , 𝐵=− , 𝐶=−
4 4 2
Thus,
1 1 3
𝑠+2
4
𝑋(𝑠) = − 2 4
𝑠 𝑠 + 6𝑠 + 20
1 1 (𝑠 + 3) + 3 √11
4 4√11
𝑋(𝑠) = 4 −
𝑠 (𝑠 + 3)2 + 11
Solutions to Problems 2-7
5.
a.
Obtaining the Laplace transform on both sides of the equation one gets
2
𝑠 2 𝑋(𝑠) − 2𝑠 + 2 + 2(𝑠𝑋(𝑠) − 2) + 2𝑋(𝑠) = 2
𝑠 +4
From which
𝑠 3 + 𝑠 2 + 4𝑠 + 5 𝐴𝑠 + 𝐵 𝐶𝑠 + 𝐷
X(s)=2 2 2
= 2 + 2
(𝑠 + 4)(𝑠 + 2𝑠 + 2) (𝑠 + 4) (𝑠 + 2𝑠 + 2)
with 𝐴 = −0.2, 𝐵 = −0.2, 𝐶 = 2.2, 𝐷 = 2.6. So, the latter expression can be written as
−0.2𝑠 2 2.2(𝑠 + 1) 0.4
X(s)= 2 − 0.1 2 + +
(𝑠 + 4) (𝑠 + 4) (𝑠 + 1) + 1 (𝑠 + 1)2 + 1
2
𝑠3 + 2 𝐴 𝐵 𝐶 𝐷𝑠 + 𝐸
𝑋(𝑠) = 3 2
= 3+ 2+ + 2
𝑠 (𝑠 + 4) 𝑠 𝑠 𝑠 𝑠 +4
1 1 1
The constants are found to be 𝐴 = 2, 𝐵 = 0, 𝐶 = − 8 , 𝐷 = 8, 𝐸 = 1. So,
1 2! 1 1 1 𝑠 1 2
𝑋(𝑠) = − + +
4 𝑠3 8 𝑠 8 𝑠2 + 4 2 𝑠2 + 4
Computer response:
ans =
2
(s + 5) (s + 3 s + 10)
--------------------------------
2
(s + 3) (s + 4) (s + 2 s + 100)
/ 1/2 1/2 \
| 1/2 11 sin(3 11 t) |
5203 exp(-t) | cos(3 11 t) - -------------------- |
20 exp(-3 t) 7 exp(-4 t) \ 57233 /
------------ - ----------- + ------------------------------------------------------
103 54 5562
ans =
3 2
s + 4 s + 2 s + 6
-------------------------------------
2 2
(s + 8) (s + 8 s + 3) (s + 5 s + 7)
Solutions to Problems 2-9
/ 1/2 1/2 \
| 1/2 4262 13 sinh(13 t) |
1199 exp(-4 t) | cosh(13 t) - ------------------------ |
\ 15587 /
----------------------------------------------------------- -
417
/ / 1/2 \ \
| 1/2 | 3 t | |
| / 1/2 \ 131 3 sin| ------ | |
/ 5 t \ | | 3 t | \ 2 / |
65 exp| - --- | | cos| ------ | + ---------------------- |
\ 2 / \ \ 2 / 15 / 266 exp(-8 t)
---------------------------------------------------------- - -------------
4309 93
7.
The Laplace transform of the differential equation, assuming zero initial conditions, is,
(s3+3s2+5s+1)Y(s) = (s3+4s2+6s+8)X(s).
Y(s) s3 + 4s2 + 6s + 8
Solving for the transfer function, = .
X(s) s3 + 3s 2 + 5s + 1
8.
(𝑠 2 + 7𝑠 + 80)𝑋(𝑠) = 10𝐹(𝑠)
2
𝑠 𝑋(𝑠) + 7𝑠𝑋(𝑠) + 80𝑋(𝑠) = 10𝐹(𝑠)
Then obtain the inverse Laplace transform on both sides with zero initial conditions
𝑑2 𝑥(𝑡) 𝑑𝑥(𝑡)
2
+7 + 80𝑥(𝑡) = 10𝑓(𝑡)
𝑑𝑡 𝑑𝑡
c. Cross-multiplying
9.
C(s) s5 + 2s 4 + 4s3 + s 2 + 4
The transfer function is = .
R(s) s 6 + 7 s 5 + 3s 4 + 2 s 3 + s 2 + 5
d 6 c d 5 c d 4c d 3 c d 2 c d 5r d 4r d 3r d 2r
+7 5 +3 4 +2 3 + + 5c = +2 4 +4 3 + + 4 r.
dt 6 dt dt dt dt 2 dt 5 dt dt dt 2
10.
The block diagram represents the transfer function
𝐶(𝑠) 𝑠 4 + 2𝑠 3 + 3𝑠 2 + 𝑠 + 1
= 5
𝑅(𝑠) 𝑠 + 5𝑠 4 + 8𝑠 3 + 2𝑠 2 + 3𝑠 + 4
Cross-multiplying
(𝑠 5 + 5𝑠 4 + 8𝑠 3 + 2𝑠 2 + 3𝑠 + 4)𝐶(𝑠) = (𝑠 4 + 2𝑠 3 + 3𝑠 2 + 𝑠 + 1)𝑅(𝑠)
11.
s 2 X ( s ) − s + 1 + 4 sX ( s ) − 4 + 5 X ( s ) = R ( s )
Collecting terms: ( s + 4 s + 5) X ( s ) = R ( s ) + s + 3
2
2-12 Chapter 2: Modeling in the Frequency Domain
R( s ) s+3
Solving for X(s), X ( s ) = + 2
s + 4s + 5 s + 4s + 5
2
R(s) X(s)
+ X(s)
R(s)
+ +
+
s +3
12.
Program:
'Factored'
Gzpk=zpk([-15 -26 -72],[0 -55 roots([1 5 30])' roots([1 27 52])'],5)
'Polynomial'
Gp=tf(Gzpk)
Computer response:
ans =
Factored
Zero/pole/gain:
5 (s+15) (s+26) (s+72)
--------------------------------------------
s (s+55) (s+24.91) (s+2.087) (s^2 + 5s + 30)
ans =
Polynomial
Transfer function:
5 s^3 + 565 s^2 + 16710 s + 140400
--------------------------------------------------------------------
s^6 + 87 s^5 + 1977 s^4 + 1.301e004 s^3 + 6.041e004 s^2 + 8.58e004 s
13.
Program:
numg=[-5 -70];
deng=[0 -45 -55 (roots([1 7 110]))' (roots([1 6 95]))'];
[numg,deng]=zp2tf(numg',deng',1e4);
Solutions to Problems 2-13
Gtf=tf(numg,deng)
G=zpk(Gtf)
[r,p,k]=residue(numg,deng)
Computer response:
Transfer function:
10000 s^2 + 750000 s + 3.5e006
-------------------------------------------------------------------------------
s^7 + 113 s^6 + 4022 s^5 + 58200 s^4 + 754275 s^3 + 4.324e006 s^2 + 2.586e007 s
Zero/pole/gain:
10000 (s+70) (s+5)
------------------------------------------------
s (s+55) (s+45) (s^2 + 6s + 95) (s^2 + 7s + 110)
r =
-0.0018
0.0066
0.9513 + 0.0896i
0.9513 - 0.0896i
-1.0213 - 0.1349i
-1.0213 + 0.1349i
0.1353
p =
-55.0000
-45.0000
-3.5000 + 9.8869i
-3.5000 - 9.8869i
-3.0000 + 9.2736i
-3.0000 - 9.2736i
0
k =
[]
14.
a. The circuit elements are converted into their Laplace transform equivalents. The
2𝑠
equivalent parallel of the rightmost inductor in parallel with the resistor is 𝑍 = 𝑠||2 = .
𝑠+2
Applying the voltage divider rule one gets
2𝑠
𝑉𝑜 𝑠 +2 = 𝑠
=
𝑉𝑖 2 + 2𝑠 2(𝑠 + 1)
𝑠+2
b. The circuit elements are converted into their Laplace transform equivalents. The
rightmost resistor in parallel with the branch with an inductor and a capacitor in series is
1
1 2𝑠+ 2𝑠2 +1
𝑠
𝑍 = 1|| (2𝑠 + 𝑠
) = 1 = 2𝑠2 +𝑠+1. Applying the voltage divider rule
1+2𝑠+
𝑠
2-14 Chapter 2: Modeling in the Frequency Domain
2𝑠 2 + 1 1
𝑉𝑜 2𝑠 2+𝑠+1 4
= 2 =
𝑉𝑖 2𝑠 + 1 𝑠 1
1+ 2 𝑠2 + 4 + 2
2𝑠 + 𝑠 + 1
15.
a.
s s
(s + 2 )I1 (s) − 2 I (s) − sI3 (s) = Vi (s)
s +1 s +1 2
s s 1
− 2 I1 (s) + ( 2 + 1 + )I2 (s) − I3 (s) = 0
s +1 s +1 s
−sI1(s) − I2 (s) +(2s +1)I3 (s) = 0
Solving for I2(s),
s(s2 + 2s + 2)
I2 (s) = 4 V (s)
s + 2s 3 + 3s2 + 3s + 2 i
I2(s) (s 2 + 2s + 2)
But, Vo(s) =
s
= V (s) . Therefore,
s 4 + 2s3 + 3s2 + 3s + 2 i
Vo (s) s2 + 2s + 2
=
Vi (s) s4 + 2s3 + 3s2 + 3s + 2
16.
a. Writing the nodal equations yields,
6s + 5 1 1
VR (s) − VC (s) = Vi (s)
6s 3s 2s
1 3s2 + 2
− VR (s) + V (s) = 0
3s 6s C
1 1 6s + 5 1
V (s) − V (s)
2s i 3s 6s 2s i
3s2 + 2 1
0 − 0
VR (s) = 6s ; V (s) = 3s
6s + 5 1 C
6s + 5 1
− −
6s 3s 6s 3s
1 3s2 + 2 1 3s 2 + 2
− −
3s 6s 3s 6s
2 1
(s + + 1)V1 (s) − Vo (s) = Vi (s)
s s
1 1
V1 (s) + (s + + 1)Vo (s) = Vi (s)
s s
(s 2 + 2s + 2)
Vo(s) = V (s) .
s 4 + 2s3 + 3s2 + 3s + 2 i
Hence,
Vo (s) (s2 + 2s + 2)
= 4
Vi (s) s + 2s3 + 3s2 + 3s + 2
17.
𝑉𝑜 𝑍𝑓 (𝑠)
a. The amplifier is in an inverting amplifier configuration. Therefore 𝐺(𝑠) = (𝑠) = −
𝑉𝑖 𝑍𝑖 (𝑠)
1 1
where 𝑍𝑓 (𝑠) = 200𝑘 + and 𝑍𝑖 (𝑠) = 500𝑘 + . The transfer function is:
2𝜇𝑠 2𝜇𝑠
Solutions to Problems 2-17
1
200𝑘 + 𝑠 + 2.5
2𝜇𝑠
𝐺(𝑠) = − = −0.4
1 𝑠+1
500𝑘 +
2𝜇𝑠
𝑉𝑜 𝑍𝑓 (𝑠)
b. The amplifier is in an inverting amplifier configuration. Therefore 𝐺(𝑠) = (𝑠) = −
𝑉𝑖 𝑍𝑖 (𝑠)
1 1
where 𝑍𝑓 (𝑠) = 100𝑘 + 200𝑘|| and 𝑍𝑖 (𝑠) = 100𝑘 + . The transfer function is:
2𝜇𝑠 1𝜇𝑠
1 2 × 10 5
100𝑘 + 200𝑘|| 100𝑘 + 𝑠(𝑠 + 7.5)
2𝜇𝑠 0.4𝑠 + 1 = −
𝐺(𝑠) = − =−
1 1 (𝑠 + 2.5)(𝑠 + 10)
100𝑘 + 100𝑘 +
1𝜇𝑠 1𝜇𝑠
18.
a.
1
Z1 ( s ) = 4 x105 +
4 x10−6 s
1
Z 2 ( s ) = 1.1x105 +
4 x10−6 s
Therefore,
Z1 ( s) + Z 2 ( s) ( s + 0.98)
G ( s) = = 1.275
Z1 ( s) ( s + 0.625)
b.
1011
Z1 ( s) = 4 x105 + s
0.25 x106
4 x105 +
s
109
27.5
Z 2 ( s) = 6 x105 + s
0.25 x106
110 x103 +
s
Therefore,
19.
The system has two independent translational displacements: 𝑥1 (𝑡), shown in the figure, and a
displacement 𝑥2 (𝑡) on the right-hand side of the spring where the force is applied. We can write
20.
Writing the equations of motion,
(s 2 + s + 1)X1 (s) − (s + 1)X2 (s) = F(s)
−(s + 1)X1 (s) + (s2 + s + 1)X 2 (s) = 0
From which,
X2 (s) (s + 1)
= 2 2 .
F(s) s (s + 2s + 2)
21.
The system has two independent translational displacements, so we can write the following two
equations:
𝑋1 : (𝑠 2 + 2𝑠 + 7)𝑋1 (𝑠) − (𝑠 + 5)𝑋2 (𝑠) = 0
𝑋2 : − (𝑠 + 5)𝑋1 (𝑠) + (2𝑠 2 + 3𝑠 + 5)𝑋2 (𝑠) = 𝐹(𝑠)
Solving we get:
𝑠 2 + 2𝑠 + 7 0
| | (𝑠 2 + 2𝑠 + 7)𝐹(𝑠)
−(𝑠 + 5) 𝐹(𝑠)
𝑋2 (𝑠) = 2 = 2
𝑠 + 2𝑠 + 7 −(𝑠 + 5) (𝑠 + 2𝑠 + 7)(2𝑠 2 + 3𝑠 + 5) − (𝑠 + 5)2
| |
−(𝑠 + 5) 2𝑠 2 + 3𝑠 + 5
(𝑠 2 + 2𝑠 + 7)𝐹(𝑠)
= 4
2𝑠 + 7𝑠 3 + 24𝑠 2 + 21𝑠 + 10
𝑋2 (𝑠) 1 𝑠 2 +2𝑠+7
The resulting transfer function can be written as = .
𝐹(𝑠) 2 𝑠 4 +3.5𝑠 3 +12𝑠 2 +10.5𝑠+5
Solutions to Problems 2-19
22.
a.
X3 (s) 13s + 20
=
F(s) 4s(4s + 25s 2 + 43s + 15)
3
b.
or
23.
Writing the equations of motion,
(4 s 2 + 4 s + 8) X 1 ( s) − 4 X 2 ( s) − 2 sX 3 ( s) = 0
−4 X 1 ( s ) + (5s 2 + 3s + 4) X 2 ( s) − 3sX 3 ( s ) = F ( s)
−2sX 1 ( s ) − 3sX 2 ( s ) + (5s 2 + 5s + 5) = 0
24.
a.
x = 0 is at equilibrium.
𝑑2 𝑥
𝑀 𝑑𝑡 2 + 𝐾𝑥 = 0
b.
c.
𝐾 𝑀 𝐾
𝑥(𝑡)= 𝑥0 𝑐𝑜𝑠√𝑀 𝑡 + √ 𝐾 𝑥1 𝑠𝑖𝑛√𝑀 𝑡
d.
𝐾
𝜔=√𝑀 rad/sec,
Thus
Solutions to Problems 2-21
1 𝐾
𝑓 = 2𝜋 √𝑀 𝐻𝑧
25.
a.
Defining
1 (s) = rotation of J1
2 (s) = rotation between K1 and D1
3 (s) = rotation of J 3
4 (s) = rotation of right - hand side of K2
26.
This system has two independent rotations. One, shown in the figure 𝜃2 , and 𝜃1 associated with
the inertia where the input torque is applied. The two impedance equations that describe the
system are:
𝜃1 : (2𝑠 2 + 3𝑠 + 1)𝜃1 (𝑠) − (2𝑠 + 1)𝜃2 (𝑠) = 𝑇(𝑠)
𝜃2 : − (2𝑠 + 1)𝜃1 (𝑠) + (3𝑠 + 1)𝜃2 (𝑠) = 0
27.
Thus,
N 4N 2
3 (s) N3 N1
=
T (s) Jeq s2 + Deq s
where
N 2 2
Jeq = J4+J5+(J2+J3) 4 + J1 N4 N2 , and
N3 N3 N1
N4 2 NN
Deq = (D4 + D5 ) + (D2 + D3 )( ) + D1 ( 4 2 ) 2
N3 N 3 N1
Solutions to Problems 2-23
28.
Reflecting all impedances to 2(s),
2 2 2 2 2
{[J2+J1(NN21 ) +J3 (N3
N4
) ] s2 + [f2+f1(
N2
N1
) +f3(
N3
N4 ) ] s + [K (
N3
N4
) ]} 2 (s) = T(s)
N2
N1
Substituting values,
2 2 2
{[1+2(3)2+16(14 ) ]s2 + [2+1(3)2+32(14 ) ]s + 64(14 ) }2(s) = T(s)(3)
Thus,
2(s) 3
=
T(s) 20s2+13s+4
29.
Reflecting impedances across gears from the right hand side to the left hand side one gets:
5 2 5 2
𝐽𝑒𝑞 = 2 + 100 ( ) + 150 ( ) = 7.5
25 50
5 2
𝐷𝑒𝑞 = 300 ( ) = 12
25
5 2
𝐾𝑒𝑞 = 2 + 400 ( ) = 6
50
𝜃 𝑁
So (7.5𝑠 2 + 12𝑠 + 6)𝜃(𝑠) = 𝑇(𝑠). Since 𝜃 = 𝑁2 = 10, (7.5𝑠 2 + 12𝑠 + 6) 10 𝜃2 (𝑠) =
2 1
𝑇(𝑠)
𝜃2 (𝑠) 1 0.0133
= 2
= 2
𝑇(𝑠) 75𝑠 + 120𝑠 + 60 𝑠 + 1.6𝑠 + 0.8
2-24 Chapter 2: Modeling in the Frequency Domain
30.
Reflecting impedances and applied torque to respective sides of the spring yields the
following
equivalent system:
-22(s) + (2.7777s+2)3(s) = 0
3 ( s ) 0.72 ( s) 0.24
Hence, = . But, 4 (s) = 0.33333 ( s) . Thus, 4 =
T (s) s T (s) s
Solutions to Problems 2-25
31.
Reflecting the 0.02 Nm/rad damper towards the left we get
2
T1 1 kg-m 2 Nm/rad
2 2 Nm/rad
0.32 Nm/rad
𝜃1 : (𝑠 2 + 2𝑠)𝜃1 − 2𝑠𝜃2 = 𝑇1
Solving:
2
|𝑠 + 2𝑠 𝑇1 | 2𝑠𝑇1
𝜃2 = 2 −2𝑠 0 = 2
𝑠 + 2𝑠 −2𝑠 (𝑠 + 2𝑠)(2.32𝑠 + 2) − 4𝑠 2
| |
−2𝑠 2.32𝑠 + 2
2𝑠𝑇1 2𝑇1
= 3 2 2 2
= 2
2.32𝑠 + 2𝑠 + 4.64𝑠 + 4𝑠 − 4𝑠 2.32𝑠 + 2.64𝑠 + 4
So
𝜃2 2
= 2
𝑇1 2.32𝑠 + 2.64𝑠 + 4
𝜃𝐿
𝑇 5 1 𝜃2 10 1 𝜃𝐿 4 1 𝜃𝐿
Using the gear ratios we get = = and = = . It follows that = = .
𝑇1 20 4 𝜃𝐿 40 4 𝑇1 4𝑇 16 𝑇
Finally
𝜃𝐿 32 13.8
= =
𝑇 2.32𝑠 2 + 2.64𝑠 + 4 𝑠 2 + 1.14𝑠 + 1.72
2-26 Chapter 2: Modeling in the Frequency Domain
32.
Reflect impedances to the left of J5 to J5 and obtain the following equivalent circuit:
[Jeqs2+(Deq+D)s+(K2+Keq)]5(s) -[Ds+K2]6(s) = 0
2 2 2
where Jeq = J1[ (NN43NN21 ) + (J2+J3) (NN43 ) ]
+ (J4+J5) , Keq = K1
N4
( )
N3
, and
2 2
Deq = D [(NN34NN12 ) + (NN43 ) ]
+1 .
Solutions to Problems 2-27
33.
Draw a freebody diagram of the translational system and the rotating member connected to the
translational system.
2 3 2
From the freebody diagram of the mass, F(s) = (2s2+2s+3)X(s). Summing torques on the rotating
member,
(Jeqs2 +Deqs)(s) + F(s)2 = Teq(s). Substituting F(s) above, (Jeqs2 +Deqs)(s) + (4s2+4s+6)X(s) =
X(s)
Teq(s). However, (s) = . Substituting and simplifying,
2
Jeq Deq
Teq = [(2
+4 s2 + ) (
2
+4 s+6 X(s) ) ]
But, Jeq = 3+3(4)2 = 51, Deq = 1(2)2 +1 = 5, and Teq(s) = 4T(s). Therefore,
34.
Reflecting through gears the inertia and damping from the load side to motor shaft one
gets,
50 2 50 2
𝐽𝑚 = 6 + 24 (150) = 8.667 and 𝐷𝑚 = 50 + 36 (150) = 54
𝐾𝑡 𝑇𝑠𝑡𝑎𝑙𝑙 150 𝑒𝑎 60 3
Note from the motor load curve that = = = 2.5 and 𝐾𝑏 = = = .
𝑅𝑎 𝑒𝑎 60 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 100 5
Substituting all of the above, one gets
𝐾𝑡
𝜃𝑚 𝑅𝑎 𝐽𝑚 0.2885
= =
𝐸𝑎 1 𝐾𝑡 𝐾𝑏 𝑠(𝑠 + 6.4036)
𝑠 (𝑠 + 𝐽 (𝐷𝑚 + 𝑅 ))
𝑚 𝑎
𝜃𝑚 𝑁
Noting that 𝜃𝐿
= 𝑁2 = 3
1
𝜃𝐿 0.09615
=
𝐸𝑎 𝑠(𝑠 + 6.4036)
2-28 Chapter 2: Modeling in the Frequency Domain
35.
Kt Ts 5 E 5 1
= = = 1 ; Kb = a = = ;
Ra Ea 5 600 2 1 4
60
2 2 2
1 1 1
J m = 18 + 4 + 1 = 3.125 ; Dm = 36 = 2.25
4 2 4
Thus,
1
m ( s) 3.125 0.32
= =
Ea ( s ) s(s +
1 1
(2.25 + (1)( ))) s ( s + 0.8)
3.125 4
1
Since: 2 ( s ) = m ( s ) ; then:
4
2 ( s) 0.08
=
Ea ( s) s( s + 0.8)
36.
From Eqs. (2.45) and (2.46),
Also,
Tm(s) = KtIa(s) = (Jms2+Dms)(s). Solving for (s) and substituting into Eq. (1), and simplifying
yields
Dm
(s + )
Ia (s) 1 Jm
= (2)
Ea (s) Ra s + Ra Dm + K b Kt
Ra J m
37.
For the rotating load, assuming all inertia and damping has been reflected to the load,
(JeqLs2+DeqLs)L(s) + F(s)r = Teq(s), where F(s) is the force from the translational system, r=2 is
the radius of the rotational member, JeqL is the equivalent inertia at the load of the rotational load and
the armature, and DeqL is the equivalent damping at the load of the rotational load and the armature.
Since JeqL = 1(2)2 +1 = 5, and DeqL = 1(2)2 +1 = 5, the equation of motion becomes, (5s2+5s)L(s) +
F(s)r = Teq(s). For the translational system, (s2+s)X(s) = F(s). Since X(s) = 2L(s), F(s) =
(s2+s)2L(s). Substituting F(s) into the rotational equation, (9s2+9s) L(s) = Teq(s). Thus, the
equivalent inertia at the load is 9, and the equivalent damping at the load is 9. Reflecting these back to
9 9 Kt
the armature, yields an equivalent inertia of and an equivalent damping of . Finally, = 1; Kb
4 4 Ra
4 4 2
m(s) 9 9 1 L (s) 9
= 1. Hence, = = . Since L(s) = m(s), = . But X(s) =
Ea(s) 49 13 2 Ea(s) 13
s(s+ ( +1)) s(s+ ) s(s+ )
94 9 9
4
X(s) 9
rL(s) = 2L(s). therefore, = .
E (s) s(s+13)
a 9
2-30 Chapter 2: Modeling in the Frequency Domain
38.
The equations of motion in terms of velocity are:
K1 K2 K
[M1s + ( fv1 + fv 3 ) + + ]V1 (s) − 2 V2 (s) − fv 3V3 (s) = 0
s s s
K K
− 2 V1 (s) + [M2 s + ( fv 2 + f v 4 ) + 2 ]V2 (s) − f v4 V3 (s) = F(s)
s s
− f v3 V1 (s) − f v4 V2 (s) + [M3 s + fV3 + fv 4 ]V3 (S) = 0
For the series analogy, treating the equations of motion as mesh equations yields
In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.
For the parallel analogy, treating the equations of motion as nodal equations yields
In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.
Solutions to Problems 2-31
39.
Writing the equations of motion in terms of angular velocity, (s) yields
K1 K
(J1s + D1 + )1 (s) − (D1 + 1 )2 (s) = T(s)
s s
K (K + K2 )
−(D1 + 1 ) 1 (s) + (J 2 s + D1 + 1 ) 2 (s) = 0
s s
K K
− 2 2 (s) − D2 3 (s) + (D2 + 2 ) 4 (s) = 0
s s
K
(J3 s + D2 + 3 ) 3 (s) − D2 4 (s) = 0
s
For the series analogy, treating the equations of motion as mesh equations yields
In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.
For the parallel analogy, treating the equations of motion as nodal equations yields
In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.
40.
An input r1 yields c1 = 5r1+7. An input r2 yields c2 = 5r2 +7. An input r1 +r2 yields, 5(r1+r2)+7 =
5r1+7+5r2 = c1+c2-7. Therefore, not additive. What about homogeneity? An input of Kr1 yields c =
41.
The truncated Taylor series expansion of 𝑓(𝑥) = 3𝑒 −5𝑥 ≈ 𝑓(0) + 𝑓 ′ (0)𝑥 = 3 − 15𝑥
Letting 𝑥 = 𝛿𝑥 and substituting for 𝑓(𝑥) one gets
𝑑3 𝛿𝑥 𝑑2 𝛿𝑥 𝑑𝛿𝑥
3
+ 10 2
+ 20 + 15𝛿𝑥 = 3 − 15𝛿𝑥
𝑑𝑡 𝑑𝑡 𝑑𝑡
Simplifying
𝑑3 𝛿𝑥 𝑑2 𝛿𝑥 𝑑𝛿𝑥
3
+ 10 2
+ 20 + 30𝛿𝑥 = 3
𝑑𝑡 𝑑𝑡 𝑑𝑡
42.
The relationship between the nonlinear spring’s displacement, xs(t) and its force, fs(t) is
xs (t) = 1 − e− fs (t)
d 2 x(t ) dx(t )
2 + 2 − ln(1 − x(t )) = f (t ) (2)
dt 2 dt
1 1
ln(1− x) = ln(1 − x0 ) − x = ln(1− x 0 ) − x (3)
1 − x x = x0 1− x 0
1
ln(1- x) = ln(1 – 0.6321) - x = -1 - 2.718x
1- 0.6321
Solutions to Problems 2-33
Placing this value into Eq. (2) along with x(t) = x0 + x and f(t) = 1 + f , yields the linearized
d x
2
dx
differential equation, 2 + 2 + 1 + 2.718x = 1 + f
dt 2 dt
d 2x dx
or 2 + 2 + 2.718x = f
dt 2 dt
Taking the Laplace transform and rearranging yields the transfer function,
x( s) 1
= 2
f ( s) 2s + 2s + 2.718
43.
a. The three equations are transformed into the Laplace domain:
~
Ss − S 0 = k K S C − k S
~
Cs = k ( S − K M C )
Ps = k 2 C
~
S0 k K S
S= + C
s + k s + k
Sk
C= ~
s + k K M
k2
P= C
s
~
( s + k K M )
S= 2 ~ ~ ~ S0
s + k (1 + K M ) s + k 2 ( K M − K S )
k
C= ~ ~ ~ S0
( s 2 + k (1 + K M ) s + k 2 ( K M − K S ))
k 2 k
P= ~ ~ ~ S0
s( s 2 + k (1 + K M ) s + k 2 ( K M − K S ))
b.
2-34 Chapter 2: Modeling in the Frequency Domain
S () = Lim sS ( s ) = 0
s →0
C () = Lim sC ( s ) = 0
s →0
k 2 k S 0 k2 S0
P() = Lim sP ( s) = 2 ~ ~ = = S0
s →0 k ( K M − K S ) k ( K
~ k2 ~
S + − KS )
k
44.
Obtaining Laplace transform on both sides of this equation and eliminating terms one gets that:
1
𝛩 𝑠
𝐽
(𝑠) = 3 2
𝑇𝑑 𝑠 + 𝜂𝑠 + 𝑘𝑠 − 𝜌
45.
𝑇 𝜆 2𝑎𝜋𝑓 𝑠
(𝑠) = +
𝑈 𝑠+𝜆 𝑇𝑟𝑒𝑓 𝑠 + 4𝜋 2 𝑓 2
2
Solutions to Problems 2-35
46.
dV (t ) (1− e − at )
a. By direct differentiation = V0 (e −t )e = e −tV (t )
dt
(1− e − t )
b. V () = Lim V (t ) = Lim V0 e
= V0 e
t ⎯⎯→ t ⎯⎯→
c.
Lambda = 2.5;
alpha = 0.1;
V0=50;
t=linspace(0,100);
V=V0.*exp(Lambda.*(1-exp(-alpha.*t))/alpha);
plot(t,V)
grid
xlabel('t (days)')
ylabel('mm^3 X 10^-3')
12
x 10
4
3.5
2.5
mm3 X 10-3
1.5
0.5
0
0 10 20 30 40 50 60 70 80 90 100
t (days)
47.
Using the impedance method the two equations are:
𝒙𝟏 : (𝒎𝒔𝟐 + 𝒌)𝒙𝟏 − 𝒙𝒎 𝒌 = 𝑭𝟏
𝒙𝒎 : −𝒙𝟏 𝒌 + (𝑩𝒔 + 𝒌)𝒙𝒎 = 𝑭𝒊𝒔𝒐
𝐹 −𝑘
|𝐹 1 𝐵𝑠 + 𝑘
| 𝐹1 (𝐵𝑠 + 𝑘) + 𝐹𝑖𝑠𝑜 𝑘 𝐹1 𝐵𝑠 + 𝑘(𝐹1 + 𝐹𝑖𝑠𝑜 )
𝑖𝑠𝑜
𝑥1 = = =
𝒎𝒔𝟐 +𝒌 −𝑘 2
𝑠(𝑚𝐵𝑠3 + 𝑘𝑚𝑠 + 𝑘𝐵)
| (𝑚𝑠 + 𝑘)(𝐵𝑠 + 𝑘) − 𝑘
2
|
−𝑘 𝐵𝑠 + 𝑘
48.
Opening the current source, we find the contribution of the voltage source, Va(s), to the ac current,
IacF (s).
1
Va ( s) Va ( s) Cs
I acF ( s) = = = Va ( s)
1 Z ( s) Ls + R + 1 LCs + RCs + 1
2
Cs
Short-circuiting the voltage source, Va(s), we find the contribution of the current source, IacR(s), to
the ac current, IacF (s).
2
1
R+
Cs RCs + 1
I acF ( s ) = I acR ( s ) = I acR ( s )
2
Ls + R +
1 LCs 2 + RCs + 1
Cs
Thus, the total current, IacF (s), is given by:
1 + RCs Cs
I acF ( s) = I acF ( s) + I acF 1 ( s) = I acR ( s) + Va ( s)
2 LCs + RCs + 1
2
LCs + RCs + 1
2
Solutions to Problems 2-37
49.
Writing the loop equation around the armature circuit for the motor in Figure 2.35:
𝑑𝑖𝑎 𝑑𝜃𝑚
𝑒𝑎 (𝑡) = 𝑅𝑎 𝑖𝑎 + 𝐿𝑎 + 𝐾𝑏
𝑑𝑡 𝑑𝑡
d 2 m (t ) d (t )
Tm (t ) = J m + Dm m + K m m (t )
dt dt
Thus
m ( s) Kt
=
Ea ( s) J m La s + ( J m Ra + Dm La ) s + ( Dm Ra + K m La + K t K b ) s + K m Ra
3 2
50.
Ae
a. Expressing 2 gh as a Taylor series around h0i
A
Ae A A A Ae g
2 gh e 2 gh0 + e 2 gh h = e 2 gh0 + h (1)
A A h A h0 A A 2 gh0
Also,
2-38 Chapter 2: Modeling in the Frequency Domain
h = h0 + h (2)
and
q = q0 + q (3)
Substituting (1), (2), and (3) into the given nonlinear equation and eliminating
d h Ae g q
+ h =
dt A 2 gh0 A
H (s) 1/ A
=
Q( s ) A g
s + e
A 2 gh0
d
b. Substituting qe = Ae 2 ghav into e1q − eqe = ( Aehav )
dt
de
e1q − e Ae 2 ghav = Ahav
dt
Rearranging
de
Ahav + e Ae 2 ghav = e1q
dt
Simplifying,
de A
+ e 2 ghav e = e1q
dt Ahav
From which,
E ( s) e1
=
Q( s) ( s + Ae 2 gh )
av
Ahav
Solutions to Problems 2-39
51.
a. The first two equations are nonlinear because of the Tv products on their right hand side.
Otherwise the equations are linear.
dT dT * dv
b. To find the equilibria let = = =0
dt dt dt
Leading to
s − dT − T = 0
Tv − T * = 0
kT * − cv = 0
The first equilibrium is found by direct substitution. For the second equilibrium, solve the last two
equations for T*
Tv cv c
T* = and T* = . Equating we get that T =
k k
Substituting the latter into the first equation after some algebraic manipulations we get that
ks d cv s cd
v= − . It follows that T * = = − .
c k k
52.
F − Fw
a. From a= , we have: F = Fw + km • m • a = FRO + FL + FSt + +km • m • a (1)
km • m
Substituting for the motive force, F, and the resistances FRo, FL, and Fst using the equations given in
P • tot 2
F= = f • m • g • cos + m • g • sin + 0.5 • • Cw • A • v + vhw + k m • m • a (2)
v
b. Noting that constant acceleration is assumed, the average values for speed and acceleration are:
aav = 20 (km/h)/ 4 s = 5 km/h.s = 5x1000/3600 m/s2 = 1.389 m/s2
The motive force, F (in N), and power, P (in kW) can be found from eq. 2:
Fav = 0.011 x 1590 x 9.8 + 0.5 x 1.2 x 0.3 x 2 x 13.89 2 + 1.2 x 1590 x 1.389 = 2891 N
Pav = Fav. v / η tot = 2891 x 13.89 / 0.9 = 44, 617 N.m/s = 44.62 kW
o
To maintain a speed of 60 km/h while climbing a hill with a gradient α = 5 , the car engine or motor
Thus, the additional power, Padd, the car needs after reaching 60 km/h to maintain its speed while
o
climbing a hill with a gradient α = 5 is:
To linearize this equation about vo = 50 km/h = 13.89 m/s, we use the truncated taylor series:
d (v 2 )
v − vo
2 2
(v − vo) = 2vo • (v − vo) (4), from which we obtain:
dv
v = vo
Fo = 69.46 N
Excess
Motive Motive Car Speed,
Force, + Force, v(t)
F (t) + Fe(t)
Gv
_
FRo = 171.4 N
d. Taking the Laplace transform of the left and right-hand sides of equation (6) gives,
Thus the transfer function, Gv(s), relating car speed, V(s) to the excess motive force, Fe(s), when the
car travels on a level road at speeds around vo = 50 km/h = 13.89 m/s under windless conditions is:
V (s) 1
Gv ( s ) = = (8)
Fe (s) 10 + 1908 s
53.
a.
Since the system’s transfer function exhibits a pure time delay of T seconds, the
unit step response of the system is the unit step response of a first order system
delayed T seconds, namely
𝑡−𝑇
ℎ(𝑡) = 𝐾 (1 − 𝑒 − 𝜏 ) 𝑢(𝑡 − 𝑇)
b.
h(t)
0.63K
t(sec)
T
T+ɀ 4
Ts=T+ɀ
2-42 Chapter 2: Modeling in the Frequency Domain
c.
𝐻(𝑠) 𝐾
= 𝑒 −𝑠𝑇
𝑄(𝑠) (1 + 𝜏𝑠)
Then cross-multiplying
𝐻(𝑠)(1 + 𝜏𝑠)𝑒 𝑠𝑇 = 𝐾𝑄(𝑠)