ch04_7ed_Nise

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F O U R 
 
 Time Response 
 
SOLUTIONS TO CASE STUDIES CHALLENGES 
 
Antenna Control: Open-Loop Response 
The forward transfer function for angular velocity is, 
 
G(s) = 
0
P
(s)
V (s)
 = 
 
24
(s 150)(s 1.32)
 
a. 0(t) = A + Be
-150t + Ce-1.32t 
b. G(s) = 
 2
24
s 151.32s 198
. Therefore, 2n =151.32, n = 14.07, and  = 5.38. 
c. 0(s) = 
 2
24
s(s 151.32s 198)
 = 
  
   
24 1 1 1
0.12121 0.0010761 0.12229
( 150)( 1.32) 150 1.32s s s s s s
 
Therefore, 0(t) = 0.12121 + .0010761 e
-150t - 0.12229e-1.32t. 
d. Using G(s), 
0 0 0151.32 198 24 ( )pv t     
Defining, 
x1  0
x2  0
• 
Thus, the state equations are, 
 
•
1 2
•
2 1 2
1
198 151.32 24 ( )p
x x
x x x v t
y x

   

 
In vector-matrix form, 
 
 
• 0 1 0
( ); 1 0
198 151.32 24
pv t y
   
     
    
x x x 
4-2 Chapter 4: Time Response 
 
e. 
Program: 
'Case Study 1 Challenge (e)' 
num=24; 
den=poly([-150 -1.32]); 
G=tf(num,den) 
step(G) 
 
Computer response: 
ans = 
 
Case Study 1 Challenge (e) 
 
 
Transfer function: 
 24 
------------------- 
s^2 + 151.3 s + 198 
 
 
 
 
Ship at Sea: Open-Loop Response 
a. Assuming a second-order approximation: n
2 = 2.25, 2n = 0.5. Therefore  = 0.167, n = 1.5. 
Ts = 
n
4

 = 16; TP = 
ωn 1



 = 2.12 ; 
%OS = e- / 1  x 100 = 58.8%; nTr = 1.169 therefore, Tr = 0.77. 
b. 
 
 
 2
2.25
( )
0.5 2.25
s
s s s
 = 
2
1 0.5
0.5 2.25
s
s s s


 
 
 = 
2
0.25
( 0.25) 2.1875
1 2.1875
( 0.25) 2.1875
s
s s
 

 
 
 Answers to Review Questions 4-3 
 
 = 
2
1 ( 0.25) 0.16903 1.479
( 0.25) 2.1875
s
s s
  

 
 
Taking the inverse Laplace transform, 

(t) = 1 - e-0.25t (cos1.479t +0.16903 sin1.479t) 
c. 
Program: 
'Case Study 2 Challenge (C)' 
'(a)' 
numg=2.25; 
deng=[1 0.5 2.25]; 
G=tf(numg,deng) 
omegan=sqrt(deng(3)) 
zeta=deng(2)/(2*omegan) 
Ts=4/(zeta*omegan) 
Tp=pi/(omegan*sqrt(1-zeta^2)) 
pos=exp(-zeta*pi/sqrt(1-zeta^2))*100 
t=0:.1:2; 
[y,t]=step(G,t); 
Tlow=interp1(y,t,.1); 
Thi=interp1(y,t,.9); 
Tr=Thi-Tlow 
'(b)' 
numc=2.25*[1 2]; 
denc=conv(poly([0 -3.57]),[1 2 2.25]); 
[K,p,k]=residue(numc,denc) 
'(c)' 
[y,t]=step(G); 
plot(t,y) 
title('Roll Angle Response') 
xlabel('Time(seconds)') 
ylabel('Roll Angle(radians)') 
 
Computer response: 
ans = 
 
Case Study 2 Challenge (C) 
 
 
ans = 
 
(a) 
 
 
Transfer function: 
 2.25 
------------------ 
s^2 + 0.5 s + 2.25 
 
 
 
 
4-4 Chapter 4: Time Response 
 
 
omegan = 
 
 1.5000 
 
 
zeta = 
 
 0.1667 
 
 
Ts = 
 
 16 
 
 
Tp = 
 
 
 2.1241 
 
 
pos = 
 
 
 58.8001 
 
 
Tr = 
 
 0.7801 
 
 
 
ans = 
 
 
(b) 
 
 
K = 
 
 0.1260 
 -0.3431 + 0.1058i 
 -0.3431 - 0.1058i 
 0.5602 
 
 
 
p = 
 
 -3.5700 
 -1.0000 + 1.1180i 
 -1.0000 - 1.1180i 
 0 
 
 
 
 Answers to Review Questions 4-5 
 
k = 
 
 [] 
 
 
ans = 
 
(c) 
 
 
 
 
ANSWERS TO REVIEW QUESTIONS 
 
1. Time constant 
2. The time for the step response to reach 67% of its final value 
3. The input pole 
4. The system poles 
5. The radian frequency of a sinusoidal response 
6. The time constant of an exponential response 
7. Natural frequency is the frequency of the system with all damping removed; the damped frequency of 
oscillation is the frequency of oscillation with damping in the system. 
8. Their damped frequency of oscillation will be the same. 
9. They will all exist under the same exponential decay envelop. 
10. They will all have the same percent overshoot and the same shape although differently scaled in time. 
11. , n, TP, %OS, Ts 
12. Only two since a second-order system is completely defined by two component parameters 
13. (1) Complex, (2) Real, (3) Multiple real 
14. Pole's real part is large compared to the dominant poles, (2) Pole is near a zero 
15. If the residue at that pole is much smaller than the residues at other poles 
16. No; one must then use the output equation 
17. The Laplace transform of the state transition matrix is (sI -A)-1 
4-6 Chapter 4: Time Response 
18. Computer simulation 
19. Pole-zero concepts give one an intuitive feel for the problem. 
20. State equations, output equations, and initial value for the state-vector 
21. Det(sI-A) = 0 
 
SOLUTIONS TO PROBLEMS 
1. 
a. Overdamped Case: 
C(s) = 
2
9
s(s 9s 9) 
 
 
Expanding into partial fractions, 
 
9 1 0.171 1.171
C(s)
s(s 7.854)(s 1.146) s (s 7.854) (s 1.146)
   
   
 
Taking the inverse Laplace transform, 
c(t) = 1 + 0.171 e-7.854t - 1.171 e-1.146t 
b. Underdamped Case: 
2 31
2 2
K s KK9
C(s)
s(s 3s 9) s (s 3s 9)

  
   
 
 
 
 K2 and K3 can be found by clearing fractions with K1 replaced by its value. Thus, 
 9 = (s2 + 3s + 9) + (K2s + K3)s 
 or 
 9 = s2 + 3s +9 + K2s
2 + K3s 
 Hence K2 = -1 and K3 = -3. Thus, 
2
1 s 3
C(s)
s (s 3s 9)

 
 
 
 
 
2
1 s 3 27
C(s)
3s 4
(s )
2

  

 
 Solutions to Problems 4-7 
 
 
2 2
3 3s
1 2 2C(s)
s 3 27 3 27
s s
2 4 2 4
 
 
 
  
   
      
   
 
 
2 2
3 3 27
s
1 2 427
C(s)
s 3 27 3 27
s s
2 4 2 4
 
 
 
  
   
      
   
 
 
 
  
3 3
t t
2 2
27 3 27
c(t) 1 e cos t e sin t
4 427
 
 
c(t) = 1 - 
2
3
 e-3t/2 cos (
27
4
 t -  
 = 1 - 1.155 e -1.5t cos (2.598t - ) 
 where 
 = arctan (
3
27
) = 30o 
 c. Oscillatory Case: 
2
9
C(s)
s(s 9)


 
 
2 31
2 2
K s KK9
C(s)
ss(s 9) (s 9)

  
 
 
 The evaluation of the constants in the numerator are found the same way as they were for the 
underdamped case. The results are K2 = -1 and K3 = 0. Hence, 
2
1 s
C(s)
s (s 9)
 

 
 Therefore, 
c(t) = 1 - cos 3t 
 d. Critically Damped 
2
9
C(s)
s(s 6s 9)

 
 
 
31 2
2 2
KK K9
C(s)
s (s 3)s(s 6s 9) (s 3)
   
  
 
4-8 Chapter 4: Time Response 
 The constants are then evaluated as 
 
 
 
Now, the transform of the response is 
   
  2 2
9 1 3 1
C(s)
s (s 3)s(s 6s 9) (s 3)
 
 
c(t) = 1 - 3t e-3t - e-3t 
2. 
a. C(s) = 

5
s(s 5)
 = 
1
s
 -

1
 
s 5
 . Therefore, c(t) = 1 - e-5t. 
Also, T = 
1
5
 , Tr = 
2.2
a
 = 
2.2
5
 = 0.44, Ts = 
4
a
 = 
4
5
 = 0.8. 
 b. C(s) = 

20
s(s 20)
 = 
1
s
 - 

1
s 20
 . Therefore, c(t) = 1 - e-20t. Also, T = 
1
20
 , 
Tr = 
2.2
a
 = 
2.2
20
 = 0.11, Ts = 
4
a
 = 
4
20
 = 0.2. 
 
3. 
Program: 
'(a)' 
num=5; 
den=[1 5]; 
Ga=tf(num,den) 
subplot(1,2,1) 
step(Ga) 
title('(a)') 
'(b)' 
num=20; 
den=[1 20]; 
Gb=tf(num,den) 
subplot(1,2,2) 
step(Gb) 
title('(b)') 
 
Computer response: 
ans = 
 
(a) 
 
Transfer function: 
 5 
----- 
s + 5 
 
 Solutions to Problems 4-9 
 
ans = 
 
(b) 
 
Transfer function: 
 20 
------ 
s + 20 
 
 
 
 
 
 
 
4. 
Using voltage division, 
1
( ) 1
1( ) 1
C
i
V s RC
V s s
s
RC
 


. Since 
5
( )
i
V s
s
 
 
5 1 5 5
( )
1 1
C
V s
s s s s
 
   
  
. 
 
Therefore: ( ) 5 5 t
c
v t e  . 
 
Also, 
1 2.2 4
1sec; 2.2sec; 4sec
1 1 1
r s
T T T      . 
 
5. 
4-10 Chapter 4: Time Response 
Program: clf 
num=1; 
den=[1 1]; 
G=tf(num,den); 
G; 
step(5*G) 
 
 
Computer response: 
 
G = 
 1 
 ----- 
 s + 1 
Continuous-time transfer function. 
 
 
 
6. 
Writing the equation of motion, 
2( 6 ) ( ) ( )Ms s X s F s  
Thus, the transfer function is, 
2
( ) 1
( ) 6
X s
F s Ms s


 
 
Differentiating to yield the transfer function in terms of velocity, 
 
( ) 1 1/
6( ) 6
sX s M
F s Ms
s
M 


 
 
 Solutions to Problems 4-11 
 
 Thus, the settling time, Ts, and the rise time, Tr, are given by 
 
 
4 2 2.2 1.1
0.667 ; 0.367
6 / 3 6 / 3
s r
T M M T M M
M M
      
 
7. 
Program: 
Clf 
M=1 
num=1/M; 
den=[1 6/M]; 
G=tf(num,den) 
step(G) 
pause 
M=2 
num=1/M; 
den=[1 6/M]; 
G=tf(num,den) 
step(G) 
 
Computer response: 
 
M = 
 
 1 
 
 
Transfer function: 
 1 
----- 
s + 6 
 
 
M = 
 
 2 
 
 
Transfer function: 
 0.5 
----- 
s + 3 
 
 
 
 
 
 
 
 
 
 
 
 
 
4-12 Chapter 4: Time Response 
 
From plot, time constant =.0.16 s. 
 
 
 
 Solutions to Problems 4-13 
 
 
 
From plot, time constant = 0.33 s. 
8. 
a. Pole: -2; c(t) = A + Be-2t ; first-order response. 
b. Poles: -3, -6; c(t) = A + Be-3t + Ce-6t; overdamped response. 
c. Poles: -10, -20; Zero: -7; c(t) = A + Be-10t + Ce-20t; overdamped response. 
d. Poles: (-3+j3 15 ), (-3-j3 15 ) ; c(t) = A + Be-3t cos (3 15 t + ); underdamped. 
e. Poles: j3, -j3; Zero: -2; c(t) = A + B cos (3t + ); undamped. 
f. Poles: -10, -10; Zero: -5; c(t) = A + Be-10t + Cte-10t; critically damped. 
 
9. 
Program: 
p=roots([1 6 4 7 2]) 
 
Computer response: 
p = 
 
4-14 Chapter 4: Time Response 
 -5.4917 
 -0.0955 + 1.0671i 
 -0.0955 - 1.0671i 
 -0.3173 
 
 
 
 
 
10. 
 
3 4 2
2 1
4 7 5
s
s s
s
  
 
  
 
    
I A 
 
2
2
2
1
5 7 4 6 2 4
2 6 2 23 7
4 14 7 37 3 8
s s s s
s s s s
s s s s
s

     
 
     
     
 

I A 
Where 
3 22 38 25s s s     
   

   
 
      
   

2
1 2
2
9 7
1
2 31
3 27 64
s s
s s s
s s
X s I A B 
 1 7 1y  X 
So 
 
2
3 2
12 11 288
2 38 25
s s
G s
s s s
  

  
 
The poles are found by obtaining the roots in the denominator of G(s) and are 0.692, 4.8144 
and -7.5062 
 
11. 
>> A=[3 -4 2;-2 0 1;4 7 -5]; 
>> B=[-1;-2;3]; 
>> C=[1 7 1]; 
>> D=0; 
>> G=ss2tf(A,B,C,D,1); 
>> [nG,dG]=ss2tf(A,B,C,D,1); 
>> G=tf(nG,dG) 
 
G = 
 
 -12 s^2 - 11 s + 288 
 ----------------------- 
 s^3 + 2 s^2 - 38 s + 25 
 
Continuous-time transfer function. 
 
 
 Solutions to Problems 4-15 
 
>> poles=roots(dG) 
poles = 
 
 -7.5062 
 4.8144 
 0.6918 
 
12. 
Writing the node equation at the capacitor, VC(s) (
2
1
R
 + 
1
Ls
 + Cs) + C
1
V (s)-V(s)
R
 = 0. 
Hence, C
V (s)
V(s)
 = 
  
1
1 2
1
1 1 1
Cs
R R Ls
R
 =
2
10s
s 20s 500 
 . The step response is 
2
10
s 20s 500 
 .The poles are at 
-10 ± j20. Therefore, vC(t) = Ae
-10t cos (20t + ). 
13. 
Program: 
num=[10 0]; 
den=[1 20 500]; 
G=tf(num,den) 
step(G) 
 
Computer response: 
 
Transfer function: 
 10 s 
---------------- 
s^2 + 20 s + 500 
 
 
4-16 Chapter 4: Time Response 
 
 
14. 
The equation of motion is: (Ms2+fvs+Ks)X(s) = F(s). 
 
Hence, 
X(s)
F(s)
 = 
2
v s
1
Ms f s K 
 = 
2 2
1 0.5
2 2 6 3s s s s

   
. 
The step response is now evaluated: 
 
X(s) = 
2
0.5
( 3)s s s 
 = 
2
1 1
1
6 6 6
1 11
( )
2 4
s
s
s


 
 = 
2
1 1 1 11
( )
1 / 6 6 2 26 11
1 11
( )
2 4
s
s
s
 

 
. 
 
Taking the inverse Laplace transform, 
 
0.5 11 1 11
2 211
1 1
( ) cos sin
6 6
t
t tx t e
 
  
 
  = 
 
0.51 111 1.091 cos 16.78
6 2
t oe t
  
    
   
. 
 
15. 
 Solutions to Problems 4-17 
 
 C(s) = 


  
2
n
22
n ns(s 2 s )
 = 
1
s
 - 
 
  
n
22
n n
s 2
s 2 s
 = 
1
s
 - 
 
    
n
2 22 2
n n n
s 2
(s ) -
 
= 
1
s
 - 
   
    
n n
2 2 2
n n
(s )
(s ) ( 1- )
 = 
1
s
 - 

    
 
    
2n
n n2
n
2 2 2
n n
(s ) 1-
1-
(s ) ( 1- )
 
 Hence, 

   


 
 
 
n 2 2
n n
2
c(t) = 1-e cos 1- t + sin 1- t
1
t 
 = 1 - e-n
t 

2
2
1
1-
 cos (n 
21- t - ) = 1 - e-n
t 
2
1
1-
 cos (n 
21- t - ), 
where  = tan-1 

21-
 
16. %OS = e- / 21- x 100. Dividing by 100 and taking the natural log of both sides, 
ln (
%OS
100
) = - 

21-
 . Squaring both sides and solving for 2, 2 =
 
2
2 2
%OS
ln ( )
100
%OS
ln ( )
100
 . Taking the 
negative square root,  = 
 2 2
%OS
-ln( )
100
%OS
ln ( )
100
 . 
17. 
 a. 
 

2
C( )
( 2)
s
s s
 
 
1 1
C( )
2
s
s s
 

 
 
2( ) 1 tc t e  
 b. 
 
5
C( )
( 3)( 6)
s
s s s

 
 
 
5 1 5 1 5 1
C( )
18 9 3 18 6
s
s s s
  
 
 
 3 6
5 5 5
( )
18 9 18
t tc t e e    
 c. 
 
10( +7)
C( )=
( +10)( +20)
s
s
s s s
 
 
7 1 3 1 13 1
C( )= +
20 10 +10 20 +20
s
s s s
 
  10 20
7 3 13
( )= +
20 10 20
t tc t e e 
 d. 
 
2
20
C( )=
( +6 +144)
s
s s s
 
4-18 Chapter 4: Time Response 
 
2
5 1 5 +6
( )=
36 36 6 +144
s
C s
s s s
 
 
2
3
( +3)+ 135
5 1 5 135
C( )=
36 36 ( +3) +135
s
s
s s
 
 
           
35 5 3( )= cos 135 + sin 135
36 36 135
tc t e t t 
 e. 
 
2
+2
C( )=
( +9)
s
s
s s
 
 

2
2 1 1 2 +9
C( )= +
9 9 +9
s
s
s s
 
 
2
2 1 1 2 3.3
C( )
9 9 9
s
s
s s
 
 

 
 
2 2 1
( ) cos.3 sin.3
9 9 3
c t t t
 
   
 
 
 f. 
 
2
5
C( )
( 10)
s
s
s s



 
 
 2
1 1 1 1 1 1
C( )= +
20 20 10 2 ( +10)
s
s s s
 
 10 10
1 1 1
( ) te
20 20 2
t tc t e    
18. 
 a. N/A 
 b. s2+9s+18, n
2 = 18, 2n = 9, Therefore  = 1.06, n = 4.24, overdamped. 
 c. s2+30s+200, n
2 = 200, 2n = 30, Therefore  = 1.06, n = 14.14, overdamped. 
 d. s2+6s+144, n
2 = 144, 2n = 6, Therefore  = 0.25, n = 12, underdamped. 
 e. s2+9, n
2 = 9, 2n = 0, Therefore  = 0, n = 3, undamped. 
 f. s2+20s+100, n
2 = 100, 2n = 20, Therefore  = 1, n = 10, critically damped. 
19. 
 
       2 2 22
400 1 6 1 3 3 391
6 400 3913 391 3 391 3 391
s s
X s
s ss s s s s s
 
     
       
 
  3 3 11 (cos19.77 0.152sin19.77 ) 1 1.048 (cos19.77 tan 0.314)t tx t e t t e t        
20. 
 Solutions to Problems 4-19 
 
 a. n
2 = 16 r/s, 2n = 3. Therefore  = 0.375, n = 4. Ts = 
n
4

 = 2.667 s; TP = 

 2n 1- .
 = 
0.8472 s; %OS = e- /  21- x 100 = 28.06 %; nTr = (1.76
3 
- 0.417
2 
+ 1.039+ 1) = 1.4238; 
therefore, Tr = 0.356 s. 
 b. n
2 = 0.04 r/s, 2n = 0.02. Therefore  = 0.05, n = 0.2. Ts = 
n
4

 = 400 s; TP = 

 2n 1- .
 
= 15.73 s; %OS = e- /  21- x 100 = 85.45 %; nTr = (1.76
3 
- 0.417
2 
+ 1.039+ 1); therefore, 
Tr = 5.26 s. 
 c. n
2 = 1.05 x 10
7
 r/s, 2n = 1.6 x 10
3
. Therefore  = 0.247, n = 3240. Ts = 
n
4

 = 0.005 s; TP 
= 
2
n 1-

 
 = 0.001 s; %OS = e- / 21- x 100 = 44.92 %; nTr = (1.76
3 
- 0.417
2 
+ 1.039+ 
1); therefore, Tr = 3.88x10
-4 s. 
 
21. 
Program: 
'(a)' 
clf 
numa=16; 
dena=[1 3 16]; 
Ta=tf(numa,dena) 
omegana=sqrt(dena(3)) 
zetaa=dena(2)/(2*omegana) 
Tsa=4/(zetaa*omegana) 
Tpa=pi/(omegana*sqrt(1-zetaa^2)) 
Tra=(1.76*zetaa^3 - 0.417*zetaa^2 + 1.039*zetaa + 1)/omegana 
percenta=exp(-zetaa*pi/sqrt(1-zetaa^2))*100 
subplot(221) 
step(Ta) 
title('(a)') 
'(b)' 
numb=0.04; 
denb=[1 0.02 0.04]; 
Tb=tf(numb,denb) 
omeganb=sqrt(denb(3)) 
zetab=denb(2)/(2*omeganb) 
Tsb=4/(zetab*omeganb) 
Tpb=pi/(omeganb*sqrt(1-zetab^2)) 
Trb=(1.76*zetab^3 - 0.417*zetab^2 + 1.039*zetab + 1)/omeganb 
percentb=exp(-zetab*pi/sqrt(1-zetab^2))*100 
subplot(222) 
step(Tb) 
title('(b)') 
'(c)' 
numc=1.05E7; 
denc=[1 1.6E3 1.05E7]; 
Tc=tf(numc,denc) 
 
4-20 Chapter 4: Time Response 
omeganc=sqrt(denc(3)) 
zetac=denc(2)/(2*omeganc) 
Tsc=4/(zetac*omeganc) 
Tpc=pi/(omeganc*sqrt(1-zetac^2)) 
Trc=(1.76*zetac^3 - 0.417*zetac^2 + 1.039*zetac + 1)/omeganc 
percentc=exp(-zetac*pi/sqrt(1-zetac^2))*100 
subplot(223) 
step(Tc) 
title('(c)') 
 
Computer response:ans = 
 
(a) 
 
 
Transfer function: 
 16 
-------------- 
s^2 + 3 s + 16 
 
omegana = 
 
 4 
 
zetaa = 
 
 0.3750 
 
 
Tsa = 
 
 2.6667 
 
 
Tpa = 
 
 0.8472 
 
 
Tra = 
 
 0.3559 
 
 
percenta = 
 
 28.0597 
 
 
ans = 
 
(b) 
 
 
Transfer function: 
 0.04 
------------------- 
s^2 + 0.02 s + 0.04 
 
 
 
omeganb = 
 Solutions to Problems 4-21 
 
 
 0.2000 
 
 
zetab = 
 
 0.0500 
 
 
Tsb = 
 
 400 
 
 
Tpb = 
 
 15.7276 
 
 
Trb = 
 
 5.2556 
 
 
percentb = 
 
 85.4468 
 
 
ans = 
 
 
(c) 
 
 
Transfer function: 
 1.05e007 
----------------------- 
s^2 + 1600 s + 1.05e007 
 
 
omeganc = 
 
 3.2404e+003 
 
 
zetac = 
 
 0.2469 
 
 
Tsc = 
 
 0.0050 
 
 
Tpc = 
 
 0.0010 
 
 
Trc = 
 
 3.8810e-004 
4-22 Chapter 4: Time Response 
 
 
percentc = 
 
 44.9154 
 
 
 
 
 
 
 
 
 
 
 
22. 
Program: 
T1=tf(16,[1 3 16]) 
T2=tf(0.04,[1 0.02 0.04]) 
T3=tf(1.05e7,[1 1.6e3 1.05e7]) 
ltiview 
 
Computer response: 
Transfer function: 
 16 
-------------- 
s^2 + 3 s + 16 
 
Transfer function: 
 0.04 
------------------- 
s^2 + 0.02 s + 0.04 
 
 
Transfer function: 
 1.05e007 
----------------------- 
s^2 + 1600 s + 1.05e007 
 
 Solutions to Problems 4-23 
 
 
 
 
 
 
 
4-24 Chapter 4: Time Response 
 
 
 
 
23. 
 a.  = 
2 2
%OS
- ln ( )
100
%OS
+ ln ( )
100
 = 0.56, n = 
s
4
T
 = 11.92. Therefore, poles = -n ± jn 
21- 
 = -6.67 ± j9.88. 
 b.  = 
2 2
%OS
- ln ( )
100
%OS
+ ln ( )
100
 = 0.591, n = 
π
2PT 1-
 = 0.779. 
 Therefore, poles = -n ± jn 
21- = -0.4605 ± j0.6283. 
 c. n = 
s
4
T
 = 0.571, n 
21- = 
p


 = 1.047. Therefore, poles = -0.571 ± j1.047. 
24. 
The corresponding damping factor is 

 
2 2
(- ln(0.15)
0.517
(0.15)ln
. The settling 
time is 
4
0.7
s
n
T

  sec, so 11.053
n
  . The transfer function is 
2
2 2 2
122.164
( )
2 11.43 122.164
n
n n
G s
s s s s

 
 
   
. 
 
 Solutions to Problems 4-25 
 
 
 
25. 
a. Adding impedances 2(5 2 20) ( ) ( )s s X s F s   . So the transfer function is 
 
  2 2
1 0.2
5 2 20 0.4 4
X s
F s s s s s
 
   
. 
b. Follows that 2 4n  , or 2n  . 2 0.4n  , so 0.1.  %OS = 
21
100 72.93%e




 . 
4
20s
n
T

  
sec. 
2
1.58
1
p
n
T

 
 

 sec. To obtain the rise time Figure 4.16 is used 1.104n rT  , or 
0.552rT  sec. The dc gain of the system is 
0.2
0.05
4
maxc   . 
 
26. 
a. The impedance equations are: 
 2 1 22s s s T    
 1 21 0s s     
Solving for 
2 
2
2 22
2
0
2 2 12
1
s s T
s T
s ss s s
s s



 
  
 
 
So 
 
 
2
2
1
2
1
2
s
T s
s s


 
 
b. 
1
2
n  ,  2 1n or 
1
2
  . 
4
8s
n
T

  sec. 
2
4.44
1
p
n
T

 
 

 sec and %OS = 
21
100 4.32%e




 . 
 
 
27. 
a. Applying the input to the transfer function, and using a partial fraction expansion 
 
   
2
2 2
a A B C
C s
s s as s a s a
   
 
 
 
2
02
| 1s
a
A
s a
 

 
2
|s a
a
C a
s
   
Substituting we have 
 
   
2
2 2
1a B a
C s
s s as s a s a
   
 
 
To equate coefficients the above equation is multiplied on both sides by  
2
s s a so 
4-26 Chapter 4: Time Response 
   2 2 21 2a B s a Ba a s a      from which it is readily apparent that 1B   . So 
 
   
2
2 2
1 1a a
C s
s s as s a s a
   
 
 
The inverse Laplace transform of this equation is 
   1 1 1at at atc t e ate e at        
 
b. The settling time will be achieved when   0.98sc T  or  1 1 0.98s
aT
s
e aT

   or 
 1 0.02.saT se aT

  
c. 
>> x=linspace(0,7,2000); 
>> y=exp(-x).*(1+x); 
>> plot(x,y) 
 
 
 
So, 5.834
s
x T a  . We have that 
5.834
s
T
a
 . 
 
 Solutions to Problems 4-27 
 
 
28. 
Note that 
 
22
53.176 11.56
4.6 31.281 53.176 3.4ref s s s


 
  
 
 
So the system is critically damped. Therefore %OS=0, 
P
T does not exist, and 
5.834 5.834
1.72
3.4
s
T
a
   sec. For a unit step input 1finalc  . So the output for an 
input with a magnitude of 3 will look as follows: 
 
 
29. 
The peak times are visually identified as 0.15, 0.2, 0.25, 0.35 and 0.45 sec. A %OS = 20% corresponds to a 
damping factor of =0.456. Using 

 
 21
n
pT
 the corresponding natural frequencies are 25.53, 
17.65 14.12, 10.09 and 7.84 rad/sec. The real part of the poles is  n = -10.7311 -8.0483 -6.4386 - 
4.5990 -3.5770. The imaginary part of the poles is  21n = 20.9440 15.7080 12.5664 8.9760 
6.9813. The corresponding plot is: 
 
4-28 Chapter 4: Time Response 
 
30. 
 a. 
2
24.542
s(s 4s 24.542) 
 = 
1
s
 - 
2
s 4
(s 2) 20.542

 
 = 
1
s
 - 
2
2
(s 2) 4.532
4.532
(s 2) 20.542
 
 
. 
Thus c(t) = 1 - e-2t (cos4.532t+0.441 sin 4.532t) = 1-1.09e-2t cos(4.532t -23.80). 
 b. 
 
2 2
245.42 1 1 0.70971 5.7418
0.29029
( 10)( 4 24.542) 10 4 24.542
s
s s s s s s s s

  
     
 
 
2 2
245.42 1 1 0.70971 5.7418
0.29029
( 10)( 4 24.542) 10 ( 2) 20.542
s
s s s s s s s

  
     
 
 
 
  
     2 2
4.3223
0.70971( 2) 20.542
245.42 1 1 20.542
0.29029
( 10)( 4 24.542) 10 ( 2) 20.542
s
s s s s s s s
 
 Solutions to Problems 4-29 
 
 
 
  
     2 2
245.42 1 1 0.70971( 2) 0.95367 20.542
0.29029
( 10)( 4 24.542) 10 ( 2) 20.542
s
s s s s s s s
 
 Therefore, c(t) = 1 - 0.29e-10t - e-2t(0.71 cos 4.532t + 0.954 sin 4.532t) 
= 1 - 0.29e-10t - 1.189 cos(4.532t - 53.34o). 
 c. 
 
2 2
73.626 1 1 0.13926 2.8607
1.1393
( 3)( 4 24.542) 3 4 24.542
s
s s s s s s s s

  
     
 
 
 
  
     2 2
3.1393
0.13926( 2) 20.542
73.626 1 1 20.542
1.1393
( 3)( 4 24.542) 3 ( 2) 20.542
s
s s s s s s s
 
 
 
  
     2 2
73.626 1 1 0.13926( 2) 0.69264 20.542
1.1393
( 3)( 4 24.542) 3 ( 2) 20.542
s
s s s s s s s
 
 Therefore, c(t) = 1 - 1.14e-3t + e-2t (0.14 cos 4.532t - 0.69 sin 4.532t) 
= 1 - 1.14e-3t + 0.704 cos(4.532t +78.53o). 
31. 
Since the third pole is more than five times the real part of the dominant pole, 
s2+0.842s+2.829 determines the transient response. 
Since 2n = 0.842, and n = 93.2 = 1.71   = 0.246, 
2/ 1
%OS x100 44.4%e
  
  , Ts = 
4
n

 = 9.50 sec, 
Tp = 
21
n



 = 1.895 sec; 
nTr = (1.76
3 
- 0.417
2 
+ 1.039+ 1) = 1.257, therefore, Tr = 0.735. 
32. 
a. Measuring the time constant from the graph, T = 0.0244 seconds. 
 
 
4-30 Chapter 4: Time Response 
 
Estimating a first-order system, G(s) = 
K
s a
. But, a = 1/T = 40.984, and 
K
a
 = 2. Hence, K = 
81.967. Thus, 
G(s) = 
81.967
s 40.984
 
b. Measuring the percent overshoot and settling time from the graph: %OS = (13.82-11.03)/11.03 = 
25.3%, 
 
 
 Solutions to Problems 4-31 
 
and Ts = 2.62 seconds. Estimating a second-order system, we use Eq. (4.39) to find  = 0.4 , and Eq. 
(4.42) to find n = 3.82. Thus, G(s) = 2
n n
K
s 2 s 2  
 . Since Cfinal = 11.03, 2
n
K

 = 11.03. 
Hence, K = 160.95. Substituting all values, 
G(s) = 
2
160.95
s 3.056s 14.59 
 
c. From the graph, %OS = 40%. Using Eq. (4.39),  = 0.28. Also from the graph, 
2
4.
1
p
n
T

 
 

 Substituting  = 0.28, we find n = 0.818. 
Thus, 
G(s) = 
2 2
n n
K
s 2 s  
 = 
2
0.669
s 0.458s 0.669 
. 
 
 
 
 
33. 
a. 
 
22
4 0.2 0.125 ( 1.5) 0.1302 7.75
0.075
( 2)( 3 10) 2 1.5 7.75
s s
s s s s s s s
  
  
     
 
 Since the amplitudes of the sinusoids are of the same order of magnitude as the 
residue of the pole at -2, pole-zero cancellationcannot be assumed. 
 b. 
 
2 2
2.5 1 1 1 1 1 3 14
( 2)( 4 20) 16 64 2 64 4 20
s s
s s s s s s s s
 
  
     
 
 
2 2
2.5 1 1 1 1 1 3( 2) 2 16
( 2)( 4 20) 16 64 2 64 ( 2) 16
s s
s s s s s s s
  
  
     
 
 Since the amplitude of the sinusoids are of the same order of magnitude as the 
residue of the pole at -2, pole-zero cancellation cannot be assumed. 
 c. 
2 2
( 2.2) 0.22 0.0143 0.2057 0.2629
( )
( 2)( 5) 2 5
s s
C s
s s s s s s s s
 
   
     
 
22
1 19
0.2057 0 07344
( 2.2) 0.22 0.0143 2 4
( )
( 2)( 5) 2 1 19
2 4
s .
s
C s
s s s s s s
s
 
       
     
  
 
 
4-32 Chapter 4: Time Response 
 
Since the amplitudes of the sinusoids are of 1.5 orders of magnitude larger than 
the residue of the pole at -2, pole-zero cancellation can be assumed. Since 2n 
= 1, and n = 5 = 2.236,  = 0.224, 
  
 
2/ 1
%OS x100 48.64%e , Ts = 
n
4

 = 8 sec, Tp = 

 2n 1-
 = 1.44 
sec; nTr = 1.23, therefore, Tr = 0.55. 
 d. 
 
22
2.01 1 1 0.049893 0.25018
0.05025 0.00035714
( 2)( 5 20) 2 5 55
2 4
s s
s s s s s s
s
 
  
     
  
 
 
 
22
5 55
0.049893 0.03383
2.01 1 1 2 4
0.05025 0.00035714
( 2)( 5 20) 2 5 55
2 4
s
s
s s s s s s
s
 
      
     
  
 
 
Since the amplitude of the sinusoids are of two orders of magnitude larger than the residue of the 
pole at -2, pole-zero cancellation can be assumed. Since 2n = 5, and n = 20 = 4.472,  = 
0.559, 
2/ 1
%OS x100 12.03%e
  
  , Ts = 
n
4

 = 1.6 sec, Tp = 

 2n 1-
 = 0.847 sec; 
nTr = 1.852, therefore, Tr = 0.414. 
 
34. 
Program: 
%Form sC(s) to get transfer function 
clf 
num=[1 3]; 
den=conv([1 3 10],[1 2]); 
T=tf(num,den) 
step(T) 
 
Computer response: 
Transfer function: 
 s + 3 
----------------------- 
s^3 + 5 s^2 + 16 s + 20 
 
 
 Solutions to Problems 4-33 
 
 
 
%OS = 
(0.163-0.15)
0.15
 = 8.67% 
 35. 
Only part c can be approximated as a second-order system. From the exponentially decaying cosine 
the poles are located at 
1,2
2 9.796s j   . Thus, 
4 4
2 s; 0.3207 s
Re 2 Im 9.796
s p
T T
 
      
Also, 
2 22 9.796 10
n
    and Re 2
n
   . Hence, 0.2  , yielding 52.66 percent 
overshoot. 
36. 
a. 
(1) 

   
       1 2 2 2 2
1
33.75
1 0.17213 33.75 0.17213 5.809533.75
( )
3 36 ( 1.5) 33.75 ( 1.5) 33.75 ( 1.5) 33.75
aC s
s s s s s
 
Taking the inverse Laplace transform 
Ca1(t) = 0.17213 e
-1.5t sin 5.8095t 
(2) 
2 2
2
( )
( 3 36)
aC s s
s s s

 
 = 
2
1 1
1 1 18 6
18 3 36
s
s s s


 
 = 
 
4-34 Chapter 4: Time Response 
2
1 3 0.083333
33.75
1 1 18 2 33.75
18 3
33.75
2
s
s
s
 
  
 
 
  
 
 
 
= 
2
3
0.055556 0.014344 33.75
1 2
0.055556
3
33.75
2
s
s
s
 
  
 
 
  
 
 
Taking the inverse Laplace transform 
Ca2(t) = 0.055556 - e
-1.5t (0.055556 cos 5.809t + 0.014344 sin 5.809t) 
The total response is found as follows: 
Cat(t) = Ca1(t) + Ca2(t) = 0.055556 - e
-1.5t (0.055556 cos 5.809t - 0.157786 sin 5.809t) 
Plotting the total response: 
 
b. 
(1) Same as (1) from part (a), or Cb1(t) = Ca1(t) 
(2) Same as the negative of (2) of part (a), or Cb2 (t) = - Ca2(t) 
The total response is 
Cbt(t) = Cb1(t) + Cb2(t) = Ca1(t)- Ca2(t) = -0.055556 + e
-1.5t (0.055556 cos 5.809t + 0.186474 sin 
5.809t) 
 Solutions to Problems 4-35 
 
 
Notice the nonminimum phase behavior for Cbt(t). 
4-36 Chapter 4: Time Response 
37. 
 
 
 
 
 
 Solutions to Problems 4-37 
 
 
 
38. 
1 3
4 2
s
s
s
  
   
 
I A and      2det 1 2 12 3 14s s s s s       I A . Solving for the roots the 
two poles are -1.5j3.428. 
39. 
 a. 
1 0 0 0 2 3 2 3
0 1 0 0 6 5 0 ( 6) 5
0 0 1 1 4 2 1 4 ( 2)
s
s s s
s
      
     
     
     
            
I A 
 
3 28 11 8s s s s    I A 
4-38 Chapter 4: Time Response 
 b. Factoring yields poles at 9.111, 0.5338, and –1.6448. 
40. 
 x = (sI - A ) -1 (x0 + B u ) 
 
1
2
1 0 1 2 3 1 3
0 1 3 1 1 1 9
s
s

          
            
            
X 
 
3 2
2 2
3 2
2 2
3 5 30 54
( 5)( 9)
10 12 102
( 5)( 9)
s s s
s s
s s s
s s
   
 
  
   
 
   
X 
  ( ) 1 2Y s  X 
 
3 2
2 2
5 15 54 150
( )
( 5)( 9)
s s s
Y s
s s
  

 
 
 
41. 
x = (sI - A ) -1 (x0 + B u ) 
 
1
1 0 0 0 1 0 0 0
1
0 1 0 2 4 1 0 0
1
0 0 1 0 0 6 0 1
s
s

          
          
                                   
x 
2
1
[ 6][ 1][ 0.58579][ 3.4142]
[ 6][ 1][ 0.58579][ 3.4142]
4 2
[ 6][ 1][ 0.58579][ 3.4142]
s s s s
s
s s s s
s s
s s s s
 
 
    
 
  
    
  
  
    
x 
 
( ) [1 0 0]
1
( )
( 6)( 1)( 0.58579)( 3.4142)
Y s
Y s
s s s s


   
X
 
42. 
 


      
          
      
    
           
                  
1
1
3 0 2 2 1
( ) ( ) ( ) ( )
1 1 0 1
2( 1)1 2 2
0
( 3)3
1 1 1 1
( 1)( 3) 1 ( 1)( 3)
s
s s u t
s s
ss
s ss s
s
s s s s s s s
0X I A x B
 
 Solutions to Problems 4-39 
 
   
 
 
  
 
 
2 1
3 2 1 2 1 5 1
1 0
3 3 3 31
1 3
s
s s s
Y s
s s s ss
s s s
 
 
     
   
 
   
 
 
Obtaining the inverse Laplace transform 
  3
2 5
3 3
ty t e  
 
 
43. 
1
1
( ) ( )
1 0 0 3 1 0 0 0
1
0 1 0 0 6 1 0 1
0 0 1 0 0 5 0 1
s u
s
s


  
          
          
             
                    
0x I A x B
x
 
1
( 3)( 5)
1
( 5)
1
( 5)
s s s
s s
s s
 
  
 
 
  
 
 
 
 
x
 
3 5
5
5
1 1 1
15 6 10
1 1
( )
5 5
1 1
5 5
t t
t
t
e e
t e
e
 


 
  
 
  
 
 
 
  
x
 
  5
2 2
( ) 0 1 1
5 5
ty t e  x 
 
44. 
Program: 
A=[-3 1 0;0 -6 1;0 0 -5]; 
B=[0;1;1]; 
C=[0 1 1]; 
D=0; 
S=ss(A,B,C,D) 
step(S) 
 
Computer response: 
a = 
 x1 x2 x3 
 x1 -3 1 0 
 x2 0 -6 1 
4-40 Chapter 4: Time Response 
 x3 0 0 -5 
 
 
 
 
 
b = 
 u1 
 x1 0 
 x2 1 
 x3 1 
 
 
c = 
 x1 x2 x3 
 y1 0 1 1 
 
 
 
d = 
 u1 
 y1 0 
 
Continuous-time model. 
 
45. 
Program: 
syms s %Construct symbolic object for 
 %frequency variable 's'. 
'a' %Display label 
A=[-3 1 0;0 -6 1;0 0 –5] %Create matrix A. 
B=[0;1;1]; %Create vector B. 
C=[0 1 1]; %Create C vector 
X0=[1;1;0] %Create initial condition vector,X(0). 
 
 
 
 Solutions to Problems 4-41 
 
 
U=1/s; %Create U(s). 
I=[1 0 0;0 1 0;0 0 1]; %Create identity matrix. 
X=((s*I-A)^-1)*(X0+B*U); %Find Laplace transform of state vector. 
x1=ilaplace(X(1)) %Solve for X1(t). 
x2=ilaplace(X(2)) %Solve for X2(t). 
x3=ilaplace(X(3)) %Solve for X3(t). 
y=C*[x1;x2;x3] %Solve for output, y(t). 
y=simplify(y) %Simplify y(t). 
'y(t)' %Display label. 
pretty(y) %Pretty print y(t).Computer response: 
 
ans = 
 
a 
 
A = 
 -3 1 0 
 0 -6 1 
 0 0 -5 
 
X0 = 
 1 
 1 
 0 
x1 = 
 
7/6*exp(-3*t)-1/3*exp(-6*t)+1/15+1/10*exp(-5*t) 
 
x2 = 
 
exp(-6*t)+1/5-1/5*exp(-5*t) 
 
x3 = 
 
1/5-1/5*exp(-5*t) 
 
y = 
 
2/5+exp(-6*t)-2/5*exp(-5*t) 
 
y = 
 
2/5+exp(-6*t)-2/5*exp(-5*t) 
 
ans = 
 
y(t) 
 2/5 + exp(-6 t) - 2/5 exp(-5 t) 
 
 
46. 
 |I - A | = 2 + 5 +1 
 |I - A | = ( + 0.20871) ( + 4.7913) 
 
 0.20871P   
 
4-42 Chapter 4: Time Response 
 4.7913Q   
 
0.20871 4.7913 0.20871 4.7913
1 2 5 6
0.20871 4.7913 0.20871 4.7913
3 4 7 8
t t t t
t t t t
A e A e A e A e
A e A e A e A e
   
   
  
  
  
 
 
2 1 6 5
0
4 3 8 7
A A A A
A A A A
  
  
  
 
 
 
0.20871 -4.7913 -0.20871 4.7913
1 2 5 6
0.20871 -4.7913 -0.20871 4.7913
3 4 7 8
0.20871 4.7913 0.20871 4.7913
0.20871 4.7913 0.20871 4.7913
t t t t
t t t t
A e A e A e A e
t A e A e A e A e
 
 
    
  
     
 
 
 
2 1 6 5
0
4 3 8 7
4.7913 0.20871 4.7913 0.20871
4.7913 0.20871 4.7913 0.20871
A A A Ad
A A A At
    
  
     
 
 
Therefore, 
 
2 1 6 5
4 3 8 7
1 0
0 1
A A A A
A A A A
    
   
    
 
 
2 1 6 5
4 3 8 7
4.7913 0.20871 4.7913 0.20871 0 1
4.7913 0.20871 4.7913 0.20871 1 5
A A A A
A A A A
      
   
       
 
 
Solving for Ai's two at a time, and substituting into the state-transition matrix 
 
 
   
   
  
      
0.20871 4.7913 0.20871 4.7913
0.20871 4.7913 0.20871 4.7913
1.0455 0.045545 0.21822 0.21822
0.21822 0.21822 0.045545 1.0455
t t t t
t t t t
e e e e
e e e e
 
To find x(t), 
 
 
0x x 
 
0.20871 4.7913 0.20871 4.7913
0.20871 4.7913 0.20871 4.7913
11.0455 0.045545 0.21822 0.21822
00.21822 0.21822 0.045545 1.0455
t t t t
t t t t
e e e e
x
e e e e
   
   
    
   
      
 
 
 
    
-0.20871 4.7913
-0.20871 -4.7913
1.0455 - 0.045545
-0.21822 0.21822
t t
t t
e e
x
e e
 
 
To find the output, 
 (1,2)y x 
 
 
 
 
    
0.20871 4.7913
0.20871 4.7913
1.0455 0.045545
(1,2)
0.21822 0.21822
t t
t t
e e
y
e e
 
 
 
0.20871 4.7913(0.60911 0.39089 )t ty e e   
47. 
 

  
    
1
1
/ A 
 
 |I - A | = 2 + 1 
 Solutions to Problems 4-43 
 
 
1 2 5 6
3 4 7 8
cos[ ] sin[ ] cos[ ] sin[ ]
cos[ ] sin[ ] cos[ ] sin[ ]
A t A t A t A t
A t A t A t A t
  
  
  
 
 
2 1 6 5
4 3 8 7
cos[ ] sin[ ] cos[ ] sin[ ]
cos[ ] sin[ ] cos[ ] sin[ ]
A t A t A t A td
A t A t A t A tt
  
   
   
 
 
1 5
0
3 7
A A
A A
 
  
 
 
 
2 6
0
4 8
A Ad
A At
 
  
  
 
 
1 5
3 7
1 0
0 1
A A
A A
   
   
  
 
 
2 6
4 8
0 1
1 0
A A
A A
   
   
  
 
Solving for the Ai's and substituting into the state-transition matrix, 
 
 
cos[ ] sin[ ]
sin[ ] cos[ ]
t t
t t
 
   
 
 
To find the state vector, 
      0 [ ] [ ]
t
x t B d 
 
       
             
0
cos( ) sin( ) 0
sin( ) cos( ) 1
t t t
x d
t t
 
 
  
    0
sin[ ]
d
cos[ ]
t t
x
t
 
 t    
 
  
    
0 sin[ ]
cos[ ]t
x d 
 
1 cos[ ]
sin[ ]
t
x
t
 
  
 
 
(3,4)
1 cos[ ]
(3,4)
sin[ ]
( 3cos[ ] 4sin[ ] 3)
y x
t
y
t
y t t

 
   
 
   
 
 
48. 
 |I - A | = ( + 2) ( + 0.5 - 2.3979i) ( + 0.5 + 2.3979i) 
 Let the state-transition matrix be 
 
    
    

    
     
.5 t .5 t 2 t .5 t .5 t
1 2 3 10 11 12
.5 t .5 t 2 t .5 t .5 t
4 5 6 13 14 15
.5 t
7
A e cos(2.3979t) A e sin(2.3979t) A e A e cos(2.3979t) e sin(2.3979t) A
A e cos(2.3979t) A e sin(2.3979t) A e A e cos(2.3979t) A e sin(2.3979t) A
A e cos(2.3
A

   
 
 
 
      
.5 t 2 t .5 t .5 t
8 9 16 17 18979t) A e sin(2.3979t) A e A e cos(2.3979t) A e sin(2.3979t) A
 
4-44 Chapter 4: Time Response 
 Since (0) = I, (0) = A, and (0) = A2, we can evaluate the coefficients, Ai's. Thus, 
 
     
      
   
       
3 1 12 10 21 19
6 4 15 13 24 22
9 7 18 16 27 25
A A A A A A 1 0 0
A A A A A A 0 1 0
A A A A A A 0 0 1
 
 
3 2 1 12 11 10 21 20 19
6 5 4 15 14 13 24 23 22
9 8 7 18 17 16 27 26 25
2A 2.3979A 0.5A 2A 2.3979A 0.5A 2A 2.3979A 0.5A
2A 2.3979A 0.5A 2A 2.3979A 0.5A 2A 2.3979A 0.5A
2A 2.3979A 0.5A 2A 2.3979A 0.5A 2A 2.3979A 0.5A
         
 
          
          
2 1 0
0 0 1
0 6 1
 
 
 
   
 
     
     
     
3 2 1 12 11 10 21 20 19
6 5 4 15 14 13 24 23 22
9 8 7 18 17 16 27 26
4A 2.3979A 5.4999A 4A 2.3979A 5.4999A 4A 2.3979A 5.4999A
4A 2.3979A 5.4999A 4A 2.3979A 5.4999A 4A 2.3979A 5.4999A
4A 2.3979A 5.4999A 4A 2.3979A 5.4999A 4A 2.3979A 5.
   
     
   
    25
4 2 1
0 6 1
4999A 0 6 5
 
 Solving for the Ai's taking three equations at a time, 
 
   
 

  
  
 

 
  
2 0.5 0.5 2
0.5 0.5
0.5
0.5 0.5 2
0.125 cos[2.3979 ] 0.33884 sin[2.3979 ] 0.125
0 cos[2.3979 ] 0.20851 sin[2.3979 ]
0 2.502 sin[2.3979 ]
0.125 cos[2.3979 ] 0.078194 sin[2.3979 ] 0.125
t t t t
t t
t
t t
e e t e t e
e t e t
e t
e t e t e


 
0.5
0.5 0.5
0.41703 sin[2.3979 ]
cos[2.3979 ] 0.20852 sin[2.3979 ]
t
t
t t
e t
e t e t
 
 
      
t
0
Using (t) (t) (0) (t- ) ( ) d , and y [1 0 0] (t),x x Bu x  
2( )
0
t
ty e d    
= 
1
2
 - 
1
2
 e-2t 
49. 
Program: 
syms s t tau %Construct symbolic object for 
 %frequency variable 's', 't', and 'tau. 
'a' %Display label. 
A=[-2 1 0;0 0 1;0 -6 -1] %Create matrix A. 
B=[1;0;0] %Create vector B. 
C=[1 0 0] %Create vector C. 
 
 
X0=[1;1;0] %Create initial condition vector,X(0). 
I=[1 0 0;0 1 0;0 0 1]; %Create identity matrix. 
'E=(s*I-A)^-1' %Display label. 
 Solutions to Problems 4-45 
 
E=((s*I-A)^-1) %Find Laplace transform of state 
 %transition matrix, (sI-A)^-1. 
Fi11=ilaplace(E(1,1)); %Take inverse Laplace transform 
Fi12=ilaplace(E(1,2)); %of each element 
Fi13=ilaplace(E(1,3)); 
Fi21=ilaplace(E(2,1)); 
Fi22=ilaplace(E(2,2)); 
Fi23=ilaplace(E(2,3)); 
Fi31=ilaplace(E(3,1)); 
Fi32=ilaplace(E(3,2)); %to find state transition matrix. 
Fi33=ilaplace(E(3,3)); %of (sI-A)^-1. 
'Fi(t)' %Display label. 
Fi=[Fi11 Fi12 Fi13 %Form Fi(t). 
Fi21 Fi22 Fi23 
Fi31 Fi32 Fi33]; 
pretty(Fi) %Pretty print state transition matrix, Fi. 
Fitmtau=subs(Fi,t,t-tau); %Form Fi(t-tau). 
'Fi(t-tau)' %Display label. 
pretty(Fitmtau) %Pretty print Fi(t-tau). 
x=Fi*X0+int(Fitmtau*B*1,tau,0,t); 
 %Solve for x(t). 
x=simple(x); %Collect terms. 
x=simplify(x); %Simplify x(t). 
x=vpa(x,3); 
'x(t)' %Display label. 
pretty(x) %Pretty print x(t). 
y=C*x; %Find y(t) 
y=simplify(y); 
y=vpa(simple(y),3); 
y=collect(y); 
'y(t)' 
pretty(y) %Pretty print y(t). 
 
 
Computer response: 
 
ans = 
 
a 
 
 
 
 
 
A = 
 
 -2 1 0 
 0 0 1 
 0 -6 -1 
 
 
 
B = 
 
 1 
 0 
 0 
 
C = 
 
 1 0 0 
 
X0 = 
4-46 Chapter 4: Time Response 
 
 1 
 1 
 0 
 
ans = 
 
E=(s*I-A)^-1 
 
E = 
 
[ 1/(s+2), (s+1)/(s+2)/(s^2+s+6), 1/(s+2)/(s^2+s+6)] 
[0, (s+1)/(s^2+s+6), 1/(s^2+s+6)] 
[ 0, -6/(s^2+s+6), s/(s^2+s+6)] 
 
ans = 
 
Fi(t) 
 
 [ 13 
 [exp(-2 t) , - 1/8 exp(-2 t) + 1/8 %1 + --- %2 , 
 [ 184 
 ] 
 1/8 exp(-2 t) - 1/8 %1 + 3/184 %2] 
 ] 
 
 [ 
 [0 , 1/23 %2 + %1 , - 1/23 
 
 1/2 1/2 1/2 
 (-23) (exp((-1/2 + 1/2 (-23) ) t) - exp((-1/2 - 1/2 (-23) ) t)) 
 
 ] 
 ] 
 
 [ 
 [0 , 6/23 
 
 1/2 1/2 1/2 
 (-23) (exp((-1/2 + 1/2 (-23) ) t) - exp((-1/2 - 1/2 (-23) ) t)) 
 
 ] 
 , - 1/23 %2 + %1] 
 
 1/2 
 
 Solutions to Problems 4-47 
 
 %1 := exp(- 1/2 t) cos(1/2 23 t) 
 
 1/2 1/2 
 %2 := exp(- 1/2 t) 23 sin(1/2 23 t) 
 
ans = 
 
Fi(t-tau) 
 
 [ 
 [exp(-2 t + 2 tau) , 
 [ 
 
 13 1/2 
 - 1/8 exp(-2 t + 2 tau) + 1/8 %2 cos(%1) + --- %2 23 sin(%1) , 
 184 
 
 1/2 ] 
 1/8 exp(-2 t + 2 tau) - 1/8 %2 cos(%1) + 3/184 %2 23 sin(%1)] 
 ] 
 
 [ 1/2 1/2 
 [0 , 1/23 %2 23 sin(%1) + %2 cos(%1) , - 1/23 (-23) ( 
 
 1/2 
 exp((-1/2 + 1/2 (-23) ) (t - tau)) 
 
 1/2 ] 
 - exp((-1/2 - 1/2 (-23) ) (t - tau)))] 
 
 [ 1/2 1/2 
 [0 , 6/23 (-23) (exp((-1/2 + 1/2 (-23) ) (t - tau)) 
 
 1/2 
 - exp((-1/2 - 1/2 (-23) ) (t - tau))) , 
 
 1/2 ] 
 - 1/23 %2 23 sin(%1) + %2 cos(%1)] 
 
 1/2 
 %1 := 1/2 23 (t - tau) 
 
 %2 := exp(- 1/2 t + 1/2 tau) 
 
 
ans = 
 
x(t) 
 
 [.375 exp(-2. t) + .125 exp(-.500 t) cos(2.40 t) 
 
 + .339 exp(-.500 t) sin(2.40 t) + .500] 
 
 
 
4-48 Chapter 4: Time Response 
 [.209 exp(-.500 t) sin(2.40 t) + exp(-.500 t) cos(2.40 t)] 
 
 [1.25 i (exp((-.500 + 2.40 i) t) - 1. exp((-.500 - 2.40 i) t))] 
 
ans = 
 
y(t) 
 
 .375 exp(-2. t) + .125 exp(-.500 t) cos(2.40 t) 
 
 + .339 exp(-.500 t) sin(2.40 t) + .500 
 
50. 
 The state-space representation used to obtain the plot is, 
 
  
0 1 0
u(t); y(t) 1 0
-1 -0.8 1
   
     
   
x x x 
 Using the Step Response software, 
 
 
 
Calculating % overshoot, settling time, and peak time, 
 Solutions to Problems 4-49 
 
 
2n = 0.8, n = 1,  = 0.4. Therefore, 
2/ 1
%OS x100 25.38%e
  
  , Ts = 
n
4

 = 10 sec, 
Tp = 

 2n 1-
 = 3.43 sec. 
 
 
 
 
51. 
 
Program: 
A=[0 1 0;-12 -8 1;0 0 -2]; 
B=[0;0;1]; 
C=[1 1 0]; 
D=0; 
S=ss(A,B,C,D); 
stepplot(S) 
 
Computer response: 
S = 
 a = 
 x1 x2 x3 
 x1 0 1 0 
 x2 -12 -8 1 
 x3 0 0 -2 
 
 b = 
 u1 
 x1 0 
 x2 0 
 x3 1 
 c = 
 x1 x2 x3 
 y1 1 1 0 
 
 d = 
 u1 
 y1 0 
 
Continuous-time state-space model. 
4-50 Chapter 4: Time Response 
 
 
 
 
 
52. 
 a. P(s) = 
s 0.5
s(s 2)(s 5)

 
 = 
1 / 20
s
 + 
1 / 4
s 2
 - 
3 /10
s 5
. Therefore, p(t) = 
1
20
+ 
1
4
 e-2t - 
3
10
 e-5t. 
 b. To represent the system in state space, draw the following block diagram. 
 
 
 
 For the first block, 
 y 7y 10y v(t)   
 Let x1 = y, and x2 = y . Therefore, 
x 1 = x2 
x 2 = -10x1 - 7x2 + v(t) 
 Also, 
 p(t) = 0.5y + y = 0.5x1 + x2 
 
 
 Solutions to Problems 4-51 
 
 Thus, 
  
   
     
    
0 1 0
1;p(t) 0.5 1
10 7 1
x x x 
 c. 
Program: 
A=[0 1;-10 -7]; 
B=[0;1]; 
C=[.5 1]; 
D=0; 
S=ss(A,B,C,D) 
step(S) 
 
Computer response: 
a = 
 x1 x2 
 x1 0 1 
 x2 -10 -7 
 
 
b = 
 u1 
 x1 0 
 x2 1 
 
 
c = 
 x1 x2 
 y1 0.5 1 
 
 
d = 
 u1 
 y1 0 
 
Continuous-time model. 
 
 
4-52 Chapter 4: Time Response 
 
 
 
 
53. Consider the un-shifted Laplace transform of the output 
 
2 2 2 2
2.5(1 0.172 )(1 0.008 ) 280.82( 5.814)( 125)
( )
(1 0.07 ) (1 0.05 ) ( 14.286) ( 20)
s s s s
Y s
s s s s s s
   
 
   
 
2 2( 14.286) ( 14.286) ( 20) ( 20)
A B C D E
s s s s s
    
   
 
02 2
280.82( 5.814)( 125)
2.5
( 14.286) ( 20)
s
s s
A
s s

 
 
 
 
14.2862
280.82( 5.814)( 125)
564.7
( 20)
s
s s
B
s s

 
 

 
 
2
14.286 14.2862 3 2
280.82( 5.814)( 125) 280.82 36735.2 204085.94
( 20) 40 400
s s
d s s d s s
C
ds s s ds s s s
 
   
 
  
 

       

 
3 2 2 2
14.2863 2 2
( 40 400 )(561.64 36735.2) (280.82 36735.2 204085.94)(3 80 400)
( 40 400 )
s
s s s s s s s s
s s s
 
219.7  
 
 Solutions to Problems 4-53 
 
202
280.82( 5.814)( 125)
640.57
( 14.286)
s
s s
D
s s

 
 

 
2
20 202 3 2
280.82( 5.814)( 125) 280.82 36735.2 204085.94
( 14.286) 28.572 204.09
s s
d s s d s s
E
ds s s ds s s s
 
   
 
  
 

       

 
3 2 2 2
203 2 2
( 28.572 204.09 )(561.64 36735.2) (280.82 36735.2 204085.94)(3 57.144 204.09)
( 28.572 204.09 )
s
s s s s s s s s
s s s
217.18 
 
thus 
 
2 2
2.5 564.7 219.7 640.57 217.18
( )
( 14.286) ( 14.286) ( 20) ( 20)
Y s
s s s s s
    
   
 
Obtaining the inverse Laplace transform of the latter and delaying the equation in time domain we 
get 
 
14.286( 0.008) 14.286( 0.008) 20( 0.008)
20( 0.008)
( ) [2.5 564.7( 0.008) 219.7 640.57( 0.008)
217.18 ] ( 0.008)
t t t
t
y t t e e t e
e u t
     
 
     
 
 
 
54. 
a. The transfer function can be written as 
 
0.1
2
2.5056( 3.33)
( )
( 1)( 0.72 1.44)
ss e
s
I s s s
 

  
 
 
It has poles at s=-0.36±j1.145 and s=-1. A zero at s=-3.33 
The ‘far away’ pole at -1 is relatively close to the complex conjugate poles as 0.36*5>1 so a 
dominant pole approximation can’t be applied. 
 
b) In time domain the input can be expressed as: 
 
4-54 Chapter 4: Time Response 
( ) 250 ( ( ) ( 0.15))i t A u t u t   
 
 
Obtaining Laplace transforms this can be expressed as 
 
0.151
( ) 250
se
I s
s


 
 
We first obtain the response to an unshifted unit step: 
 
2 2
2.5056( 3.33)
( )
( 1)( 0.72 1.44) 1 0.72 1.44
s A B Cs D
s
s s s s s s s s

 
   
     
 
 
 
02
2.5056( 3.33)
5.8
( 1)( 0.72 1.44)
s
s
A
s s s


 
  
 
12
2.5056( 3.33) 2.5056(2.33)
3.4
( 0.72 1.44) ( 1)(1.72)
s
s
B
s s s


   
  
 
We will get C and D by equating coefficients. Substituting these two values and multiplying both 
sides by the denominator we get. 
 
2 22.5056( 3.33) 5.8( 1)( 0.72 1.44) 3.4 ( 0.72 1.44) ( ) ( 1)s s s s s s s Cs D s s          
 
3 2 3 2 3 22.5056( 3.33) 5.8( 1.72 2.16 1.44) 3.4( 0.72 1.44 ) ( ( ) )s s s s s s s Cs C D s Ds           
 
3 22.5056( 3.33) (2.4 ) (7.528 ) (10.632 ) 8.352s C s C D s D s         
 
We immediately get C=-2.4 and D=-5.128 
 Solutions to Problems 4-55 
 
 
So 
2 2
5.8 3.4 2.4 5.128 5.8 3.4 2.4 5.128
( )
1 0.72 1.44 1 ( 0.36)1.3104
s s
s
s s s s s s s

 
     
     
 
  
      
      2 2 2
5.8 3.4 2.4( 0.15) 4.768 5.8 3.4 2.4( 0.15) 1.145
4.164
1 1( 0.36) 1.3104 ( 0.36) 1.3104 ( 0.36) 1.3104
s s
s s s ss s s
Obtaining inverse Laplace transform we get 
 
0.36 0.36( ) 5.8 3.4 2.4 cos(1.145 ) 4.164 sin(1.145 )t t tt e e t e t       
0.365.8 3.4 2.4 sin(1.145 30 )t te e t     
 
So the actual (shifted) unit step response is given by 
 
( 0.1) 0.36( 0.1)( ) [5.8 3.4 2.4 sin(1.145( 0.1) 30 )] ( 0.1)t tt e e t u t          
 
The response to the pulse is given by: 
 
( 0.1) 0.36( 0.1)( ) [1.45 0.85 0.6 sin(1.145( 0.1) 30 )] ( 0.1)t tt m me me t u t           
( 0.25) 0.36( 0.25)[1.45 0.85 0.6 sin(1.145( 0.25) 30 )] ( 0.25)t tm me me t u t        
 
55. 
At steady state the input is  9V and the output is  6V Thus G(0) = 6/9 = 0.667 
The maximum peak is achieved at  285 with a %OS = (7.5/6-1)*100 = 25% 
This corresponds to a damping factor of 
2 2 2
ln(% /100) 1.3863
0.4
ln (% /100) 1.9218
OS
OS

 

  
 
 
2
12027.2
(285 )(0.9165)1
n
p
T
 


  

 
So the approximated transfer function is 
4-56 Chapter 4: Time Response 
 
2 2 6
2 2 2 2 2 7
0.667*12027.2 96.5*10
( )
2 2*0.4*12027.2 12027.2 9622 14.5*10
n
n n
K
G s
s s s s s s

 
  
     
 
56. 
The oscillation period is 
22 1
n
T

   and from the figure 0.0675 0.0506 0.0169
2
T
s s s   
Thus T=0.0338 sec from which we get 
21 185.8931
n
   
The peaks of the response occur when the ‘cos’ term of the step response is 1 thus from the figure 
we have: 
 
(0.0506)
2
1 1.1492
1
ne



 

 and 
(0.0675)
2
1 0.9215
1
ne



 

 
From which we get 
(0.0506)
(0.0675)
0.1492
1.9006
0.0785
n
n
e
e




  or (0.0169) 1.9006ne   or 38
n
  
Substituting this result we get 
2 2381 1 185.8931
n
  

    
or 
2
2
1444
(1 ) 34556.2284

  or 2 0.0436  or 0.21  
Finally 
38
180.9
n


  
57. 
The step input amplitude is the same for both responses so it will just be assumed to be unitary. 
For the ‘control’ response we have: 
0.018
final
c  , 0.024
pt
M  from which we get 
max 0.024 0.018% 100% 100% 33.33%
0.018
final
final
c c
OS
c
 
     
2 2 2 2
ln(% /100) ln(0.333)
0.33
ln (% /100) ln (0.333)
OS
OS

 
 
  
 
 
 Solutions to Problems 4-57 
 
2 2
33.3
1 0.1 1 0.333
n
p
T
 


  
 
 
Leading a transfer function 
2
1108.9
( )
22 1108.9
c
G s
s s

 
 
Similarly for the ‘hot tail’: 
0.023
final
c  , 0.029ptM  
0.029 0.023
% 100% 26.1%
0.023
OS

   
2 2
ln(0.261)
0.393
ln (0.261)



 

 
2 2
34.17
1 0.1 1 0.261
n
p
T
 


  
 
 
2
1167.6
( )
26.9 1167.6
h
G s
s s

 
 
Using MATLAB: 
>> syms s 
>> s=tf('s') 
 
Transfer function: 
s 
>> Gc = 1108.89/(s^2+22*s+1108.89); 
>> Gh = 1167.6/(s^2+26.9*s+1167.6); 
>> step(Gc,Gh) 
 
4-58 Chapter 4: Time Response 
 
Both responses are equivalent if error tolerances are considered. 
58. 
The original transfer function has zeros at 7200 7400s j   
And poles at 1900 4500s j   ; 120 1520s j   
With (0) 0.1864G  
The dominant poles are those with real parts at -120, so a real pole is added at 
-1200 giving the following approximation: 
3 2 6
6 2 3
(1200)(2324.8 10 ) ( 14400 106.6 10 )
( ) 0.1864
106.6 10 ( 240 2324.8 10 )( 1200)
s s
G s
s s s
   

    
 
2 6
2 3
4.8782( 14400 106.6 10 )
( 240 2324.8 10 )( 1200)
s s
s s s
  

   
 
Using MATLAB: 
>> syms s 
>> s=tf('s'); 
 Solutions to Problems 4-59 
 
>>G=9.7e4*(s^2-14400*s+106.6e6)… 
/(s^2+3800*s+23.86e6)/(s^2+240*s+2324.8e3); 
>> Gdp=4.8782*(s^2-14400*s+106.6e6)/(s^2+240*s+2324.8e3)/(s+1200); 
 >> step(G,Gdp) 
 
 
 
Both responses differ because the original non-dominant poles are very close to the complex pair of 
zeros. 
 
59. 
M(s) requires at least 4 ‘far away’ poles that are added a decade beyond all original poles and zeros. 
This gives 
2 2 2 2
8 4 4
( 0.009) ( 0.018 0.0001) 1028.81( 0.009) ( 0.018 0.0001)
( )
9.72 10 ( 0.0001)(1 / 0.1) ( 0.0001)( 0.1)
s s s s s s
M s
s s s s
     
 
    
 
 
4-60 Chapter 4: Time Response 
60. 
2 2 2 2
ln(% /100) ln(0.30)
0.36
ln (% /100) ln (0.30)
OS
OS

 
 
  
 
 
2 2
0.026
1 127 1 0.30
n
p
T
 


  
 
 
2
2 2 2
0.00067
( )
2 0.0187 0.00067
n
n n
G s
s s s s

 
 
   
 
61. 
 a. Let the impulse response of T(s) be h(t). We have that 
 
205)20)(5(
450
)(






s
B
s
A
ss
sH 
 30
20
450
5 

 s
s
A ; 30
5
450
20 

 s
s
B 
 
20
30
5
30
)(




ss
sH . Obtaining the inverse Laplace transform we get 
 
tt eeth 205 3030)(   
 
b. Let the step response of the system be g(t). We have that 
 
5 20 5 20
0 0
0 0 0
30 30
( ) ( ) 30 30
5 20
t t t
t t t t t tg t h t dt e dt e dt e e        
  
 
5 20 5 206( 1) 1.5( 1) 4.5 6 1.5t t t te e e e           
 
c. 
450
( )
( 5)( 20) 5 20
A B C
G s
s s s s s s
   
   
 
0
450
4.5
( 5)( 20)
s
A
s s

 
 
; 5
450
6
( 20)
s
B
s s

  

; 20
450
1.5
( 5)
s
C
s s

 

 
Leading 
4.5 6 1.5
( )
5 20
G s
s s s
  
 
. After the inverse Laplace we get 
5 20( ) 4.5 6 1.5t tg t e e    
62. 
a. The poles given by 
2 3 38.99 10 3.97 10 0s s      have an 0.063 / sec
n
rad  and 
0.0714  
 Solutions to Problems 4-61 
 
The poles given by 
2 4.21 18.23 0s s   have an 4.27 / sec
n
rad  and 0.493  Thus 
the former represent the Phugoid and the latter the Short Period modes. 
 
b. In the original we have (0) 4.85
e


  so the Phugoid approximation is given by: 
2 3 3
1.965( 0.0098)
( 8.99 10 3.97 10 )
e
s
s s

  

 
   
 
c. 
>> syms s 
>> s=tf('s'); 
>>G=-26.12*(s+0.0098)*(s+1.371)/(s^2+8.99e-3*s+3.97e-3)/(s^2+4.21*s+18.23); 
>> Gphug=-1.965*(s+0.0098)/(s^2+8.99e-3*s+3.97e-3); 
>> step(G,Gphug) 
 
 
 
Both responses are indistinguishable. 
4-62 Chapter 4: Time Response 
 
63. 
 a. 
Program 
numg=[33 202 10061 24332 170704]; 
deng=[1 8 464 2411 52899 167829 913599 1076555]; 
G=tf(numg,deng) 
[K,p,k]=residue(numg,deng) 
 
Computer Response 
K = 
 
 0.0018 + 0.0020i 
 0.0018 - 0.0020i 
 -0.1155 - 0.0062i 
 -0.1155 + 0.0062i 
 0.0077 - 0.0108i 
 0.0077 + 0.0108i 
 0.2119 
 
 
p = 
 
 -1.6971 +16.4799i 
 -1.6971 -16.4799i 
 -0.5992 +12.1443i 
 -0.5992 -12.1443i 
 -1.0117 + 4.2600i 
 -1.0117 - 4.2600i 
 -1.3839 
 
 
k = 
 
 [] 
b. 
 
Therefore, an approximation to G(s)/ is: 
 
0.2119
1.3839
( )
s
G s

 
 
c. 
 
Program 
numg=[33 202 10061 24332 170704]; 
deng=[1 8 464 2411 52899 167829 913599 1076555]; 
 Solutions to Problems 4-63 
 
G=tf(numg,deng); 
numga=0.2119; 
denga=[1 1.3839]; 
Ga=tf(numga,denga); 
step(G) 
hold on 
step(Ga) 
 
Computer Response 
 
 
 
 
Approximation does not show oscillations and is slightly off of final value. 
 
64. 
Computer Response 
Transfer function: 
 
 s^15 + 1775 s^14 + 1.104e006 s^13 + 2.756e008 s^12 + 2.272e010 s^11 
 
 + 7.933e011 s^10 + 1.182e013 s^9 + 6.046e013 s^8 + 1.322e014 s^7 
 
 + 1.238e014 s^6 + 3.977e013 s^5 + 5.448e012 s^4 + 3.165e011 s^3 
 
 + 6.069e009 s^2 + 4.666e007 s + 1.259e005 
 
----------------------------------------------------------------------------31.62 s^17 + 4.397e004 s^16 + 1.929e007 s^15 + 2.941e009 s^14 
 
 + 1.768e011 s^13 + 4.642e012 s^12 + 5.318e013 s^11 + 2.784e014 s^10 
 
 + 7.557e014 s^9 + 1.238e015 s^8 + 1.356e015 s^7 + 8.985e014 s^6 
 
 + 2.523e014 s^5 + 3.179e013 s^4 + 1.732e012 s^3 + 3.225e010 s^2 
 
 + 2.425e008 s + 6.414e005 
 
4-64 Chapter 4: Time Response 
 
 
 
 
65. 
a. To find the step responses for these two processes, ya(t) and yp(t), we consider 
first the un-shifted Laplace transform of their outputs for Xa(s) = Xp(s) = 1/s: 

 

   
    
3
*
4 4
14.49 9.8 10
( )
(1478.26 1) ( 6.77 10 ) ( 6.77 10 )
a
A B
Y s
s s s s s s
 (1), 
where 
3
4
9.8 10
14.49
6.77 10
0
A
s
s



 
 

 and 
3
4
9.8 10
14.49
6.77 10
B
s
s



  
  
 (2) 
Substituting the values of A and B into equation (1) gives: 
*
4 4
1 1
( ) 14.49
( 6.76 10 ) ( 6.76 10 )
a
A B
Y s
s s s s 
 
    
    
 (3) 
Taking the inverse Laplace transform of *( )
a
Y s and delaying the resulting response 
in the time domain by 3.3 seconds, we get: 
46.67 10 ( 3.3)( ) 14.49 1 ( 3.3)t
a
y t e u t
     
 
 (4) 
 Solutions to Problems 4-65 
 
Noting that the denominator of ( )
p
G s can be factored into 
3 3( 0.174 10 )( 6.814 10 )s s     , we have: 

   

   
       
5
*
3 3 3 3
1.716 10
( )
( 0.174 10 )( 6.814 10 ) ( 0.174 10 ) ( 6.814 10 )
p
C D E
Y s
s s s s s s
 (5), 
where: 
5
3 3
1.716 10
14.48
( 0.174 10 )( 6.814 10 )
0
C
s s
s

 

 
   

; 
5
3
3
1.716 10
14.85
( 6.814 10 )
0.174 10
D
s s
s




  
 
  
; 
5
3
3
1.716 10
0.37
( 0.174 10 )
6.814 10
E
s s
s




 
 
  
. (6) 
Substituting the values of C, D and E into equation (5) and simplifying gives: 
   
 
               
*
3 3 3 3
14.48 14.85 0.37 1 1.0256 0.0256
( ) 14.48
( 0.174 10 ) ( 6.814 10 ) ( 0.174 10 ) ( 6.814 10 )
pY s
s ss s s s
 (7) 
Taking the inverse Laplace transform of *( )
p
Y s and delaying the resulting response 
in the time domain by 25 seconds, we get: 
3 30.174 10 ( 25) 6.814 10 ( 25)( ) 14.48 1 1.0256 0.0256 ( 25)t t
p
y t e e u t
          
 
 (8) 
 
b. Using Simulink to model the two processes described above, ya(t) and yp(t) were 
output to the “workspace.” MATLAB plot commands were then utilized to plot 
ya(t) and yp(t) on a single graph. The model and graph are shown below. 
4-66 Chapter 4: Time Response 
 
 
 
 
 
 Solutions to Problems 4-67 
 
 
66. 
a. 
>> A=[-8.792e-3 0.56e-3 -1e-3 -13.79e-3; -0.347e-3 -11.7e-3 -0.347e-3 0; 0.261 -20.8e-3 -96.6e-3 0; 
0 0 1 0] 
 
A = 
 
 -0.0088 0.0006 -0.0010 -0.0138 
 -0.0003 -0.0117 -0.0003 0 
 0.2610 -0.0208 -0.0966 0 
 0 0 1.0000 0 
 
>> eig(A) 
 
ans = 
 
 -0.1947 
 0.0447 + 0.1284i 
 0.0447 - 0.1284i 
 -0.0117 
 
b. 
Given the eigenvalues, the state-transition matrix will be of the form 
 
4-68 Chapter 4: Time Response 
 
 
 
 
 
 
11 12 13 14
21 22 23 24
31 32 33 34
41 42 43 44
( )
K K K K
K K K K
t
K K K K
K K K K
 with 
 
0.1947 0.0117 0.0447 0.0447
1 2 3 4
sin(0.1284 ) cos(0.1284 )t t t t
ij ij ij ij ij
K K e K e K e t K e t       
Thus 64 constants have to be found. 
 
67. 
a. The equations are rewritten as 
1 1L
C s
di d
u E
dt L L

   
1 1C
L C
du d
i u
dt C RC

  
from which we obtain 
 
1
10
1 1
0
L
L
s
CC
di d
idt L
EL
uddu
C RCdt
   
                             
 
 0 1 L
C
i
y
i
 
  
 
 
 
b. To obtain the transfer function we first calculate 
 
1
1
2
1 1
1 1
( )
1 1 1 (1 )
( )
d
s
RC L
d d
s s
L C
s
d d
s ss
RC LCC RC


 
  
 
   
       
     
  
I A 
So 
 
 Solutions to Problems 4-69 
 
 1 2
1 1
1
1
( ) ( ) 0 1
1 (1 )
0( )
d
s
RC L
d
s
C
G s s L
d
s s
RC LC

 
  
 
   
      
 
   
C I A B 
2 2
2 2
1 11
1 (1 ) 1 (1 )
0
d ds
C LC
L
d d
s s s s
RC LC RC LC
  
  
    
  
    
 
68. 
a. We have 
8.34 2.26
( )
1
s
s
s
 
   
 
I A and 
1
2
2.26 2.26
1 8.34 1 8.34
( )
8.34 2.26 ( 0.28)( 8.06)
s s
s s
s
s s s s

    
   
      
   
I A 
We first find 
 
 
  
-1 1L
( 0.28)( 8.06)s s
 
 
 
1 21
( 0.28)( 8.06) 0.28 8.06
K K
s s s s
 
   
 
1 0.28
1
0.129
8.06
s
K
s

 

; 
2 8.06
1
0.129
0.28
s
K
s

  

 so 
     
  
-1 0.28 8.061L 0.129 0.129
( 0.28)( 8.06)
t te e
s s
 
Follows that 
 
           
     
-1 -1 0.28 8.062.26 1L 2.26L 0.292 0.292
( 0.28)( 8.06) ( 0.28)( 8.06)
t te e
s s s s
 
4-70 Chapter 4: Time Response 
          
     
-1 -1 0.28 8.061L L 0.036 1.04
( 0.28)( 8.06) ( 0.28)( 8.06)
t ts d e e
s s dt s s
 
And 
 
     
      
         
-1 -1 -18.34 18.34
( 0.28)( 8.06) ( 0.28)( 8.06) ( 0.28)( 8.06)
s s
L L L
s s s s s s
0.28 8.06 0.28 8.06 0.28 8.060.036 1.04 1.076 1.076 1.04 0.036t t t t t te e e e e e            
Finally the state transition matrix is given by: 
 
   
   
    
  
  
0.28 8.06 0.28 8.06
0.28 8.06 0.28 8.06
0.036 1.04 0.292 0.292
( )
0.129 0.129 1.04 0.036
t t t t
t t t t
e e e e
t
e e e e
 
b. 
 
 
  
  
 
0.28 8.06
0.28 8.06
0.036 1.04
( )
0.129 0.129
t t
t t
e e
t
e e
B 
 
            0.28 8.06 0.28 8.06 0.28 8.06( ) 0.451 13.04 0.292 0.292 0.159 12.748t t t t t tt e e e e e eC B
 
Since ( ) 1u t  
            
0.28( ) 8.06( )
0 0
( ) ( ) [ 0.159 12.748 ]
t t
t ty t t d e e dC B 
0.28 0.28 8.06 8.06
0 0
0.159 12.748
t t
t te e d e e d       
0.28 0.28 8.06 8.06
0
0.159 12.748
0.28 8.06
t t te e e e  

  
0.28 8.06 0.28 8.060.568[1 ] 1.582[1 ] 1.014 0.568 1.582t t t te e e e           
c. 
>> A=[-8.34 -2.26; 1 0]; 
>> B = [1; 0]; 
>> C = [12.54 2.26]; 
 Solutions to Problems 4-71 
 
>> D = 0; 
>> t = linspace(0,15,1000); 
>> y1 = step(A,B,C,D,1,t); 
>> y2 = 1.014+0.568*exp(-0.28.*t)-1.582*exp(-8.06.*t); 
>> plot(t,y1,t,y2) 
 
 
 
69. 
a. Assuming that the input step amplitude is 560 (N-m) and that the settling time is 0.4 sec the 
transfer function is: 
 
40
0.714560
0.1 1 10
G s
s s
 
 
 
 
b.        10 10* 1 560*0.0714 1 40 1at t tBRKT t M K e e e        where M is the 
input step amplitude. 
4-72 Chapter 4: Time Response 
c.     10 0.20.2 40 1 36.6BRKT e   (Bar) 
 
70. 
From the problem statement 
 
( ) ( ) ( ) ( ) ( )
c e a m cam
s G s E s G s T s   (1) 
45
( )
( ) 0.2 1
e
m m
K
G s
R Js D K K s


 
  
 (2) 
5
( )
( ) 0.2 1
m
m
m m
R k
G s
R Js D K K s
 
 
  
 (3) 
Substituting from equations (2) and (3) into (1) gives: 
 
45 5
( ) ( ) ( )
0.2 1 0.2 1
c a cam
s E s T s
s s
  
 
 (4) 
 
The Laplace transforms of the inputs are: 
100
( )
a
E s
s
 and 
0.410
( )
s
cam
e
T s
s

 (5) 
First, we find the response to ( ) 100 ( )
a
e t u t , c1(t): 
 
1
4500 4500 900
( )
(0.2 1) (0.2 1)
c
s
s s s s
   
 
 (5)  
5
1
( ) 4500(1 ) ( )t
c
t e u t  (6) 
Next we find the response the un-shifted step in torque: ( ) 10 ( )
cam
T t u t , c2(t): 
 
2
50 50 10
( )
(0.2 1) (0.2 1)
c
s
s s s s
   
 
 (7)  
5
2
( ) 50(1 ) ( )t
c
t e u t   (8) 
So the actual response to a shifted unit step in torque is given by: 
5( 0.4)
2
( 0.4) 50(1 ) ( 0.4)t
c
t e u t      (9) 
Therefore, the final analytical expression for c(t) is: 
 Solutions to Design Problems 4-73 
 
5 5( 0.4)
1 2
( ) ( ) ( 0.4) 4500(1 ) ( ) 50(1 ) ( 0.4)t t
c c c
t t t e u t e u t             (10) 
71. 
From the problem statement 
2
2
1
( ) 1
( )
( ) ( )
1
( )
L
T T
M Lem M L
T M L T
J D
s s
C Cs
G s
J J DT s s J J
s s
C J J C

 

 

 

 (1) 
Substituting the parameters given in the problem into equation (1) yields: 
 
 
2
2
2
0.1287 0.15
1
( ) 1 1700 1700( )
1 0.15( ) 0.1625
1
17002 40.1
em
s s
s
G s
T s s
s s


 

 
 

 (2) 
 
or 
2
2
( ) 29.578 ( 1.1655 13209)
( )
( ) ( 5.6013 63481.7)
em
s s s
G s
T s s s s
  
 
 
 (3) 
 
Since Tem(t) = 50 u(t), the Laplace transform of the shaft speed at no-load is: 
 
2
2
1478.88 ( 1.1655 13209)
( ) ( ) ( ) ( )
( 5.6013 63481.7)
nl em
s s
s T s G s U s
s s s
 
  
 
 (4) 
 
With a load torque, TL = 0.2  (t), N.m., the speed, L(s), is given by: 
 
 ( ) ( ) ( ) ( ) ( ) ( ) 0.2 ( ) ( )L em L em Ls T s T s G s T s G s s G s      (5) 
 
Solving equation (5) for L(s), substituting for G(s) from (3), and simplifying, we have: 
 
4-74 Chapter 4: Time Response 
2
3 2
( ) 1478.88 ( 1.1655 13209)
( ) ( ) ( )
1 0.2 ( ) ( 11.52 63489 78139)
L em
G s s s
s T s U s
G s s s s
 
  
   
 (6) 
 
The MATLAB M-file is: 
 
 num1=1478.88; 
 num2=[1 1.1655 13209]; 
 den1=[1 0]; 
 den2=[1 5.6013 63481.7]; 
 num=conv(num1, num2); 
 den=conv(den1, den2); 
 Omega_nl=tf(num, den); 
 step(Omega_nl, 0:0.01:3); 
 hold on; 
 numL=num; 
 denL=[1 11.52 63489 78139]; 
 Omega_L=tf(numL, denL); 
 step(Omega_L, 0:0.01:3); 
 hold off; 
 grid; 
 
The MATLAB figure, shown below, illustrates the two step-responses obtained: the blue curve 
corresponds to the n.l., while the green one shows that the angular speed drops markedly when a 
load torque, TL = 0.2  (t), N.m., is applied (The steady-state value of Ls.s. 
= 250 rad/second.). 
 
 Solutions to Design Problems 4-75 
 
 
 
72. 
From the problem in chapter 3 
 
 
 
 
 
 
0 1 0 0
29.8615 0 0 0
0 0 0 1
0.9401 0 0 0
A ;
 
 

 
 
 
 
0
1.1574
B
0
0.4167
 
 C 0 36 0 1 0. ; D = 0. 
The MATLAB M-file is: 
 
A=[0 1 0 0;29.8615 0 0 0;0 0 0 1;-0.9401 0 0 0]; 
B=[0;-1.1574;0; 0.4167]; 
C=[0.36 0 1 0]; 
4-76 Chapter 4: Time Response 
D=0; 
S=ss(A, B, C, D); 
impulseplot(S,0:0.1:11.0); 
 
The impulse response, xG(t), is shown below: 
 
 
Note: As would be clear in chapter 6, such a response to a unit impulse indicates that this 
system is unstable and needs to be stabilized. 
 
SOLUTIONS TO DESIGN PROBLEMS 
73. 
 Writing the equation of motion, ( 2) ( ) ( )
v
f s X s F s  . Thus, the transfer function is
1 /( )
2( )
v
v
fX s
F s
s
f


. Hence, 
4 4
2
2s v
v
T f
a
f
   , or 
2
s
v
T
f  . 
 
 
 
 Solutions to Design Problems 4-77 
 
74. 
 The transfer function is, 
2
1 /
( )
1
M
F s
K
s s
M M

 
. Now,
 
4 4
4 8
1Re
2
s
T M
M
    . Thus, 
1
2
M  . Substituting the value of M in the denominator of the transfer function yields, 
2 2 2s s K  . Identify the roots 1,2 1 2 1s j K    . Using the imaginary part and substituting 
into the peak time equation yields 1
Im 2 1
p
T
K
 
  

, from which 5.43K  . 
 
75. 
 Writing the equation of motion, 2( 1) ( ) ( )vMs f s X s F s   . Thus, the transfer function is 
 
2
( ) 1/
1( ) v
X s M
fF s
s s
M M

 
. Since 
4
10 , 0.4s n
n
T 

   . But, 2 0.8.v n
f
M
  Also, from Eq. 
(4.39) 17% overshoot implies  = 0.491. Hence, n = 0.815. Now, 1/M = n
2
 = 0.664. Therefore, M 
1.51. Since 2 0.8, 1.21v n v
f
f
M
   . 
76. 
 Writing the equation of motion: (Js2+s+K)(s) = T(s). Therefore the transfer function is 
 
(s)
T(s)

 = 
2
1
J
1 K
s s
J J
 
 . 
 = 
 2 2
%OS
- ln( )
100
%OS
ln ( )
100
 = 0.358. 
 
Ts = 
n
4

 = 
4
1
2J
 = 8J = 3. 
 
Therefore J = 
3
8
. Also, Ts = 3 = 
n
4

 = 
n
4
(0.358)
. Hence, n = 3.724. Now, 
K
J
 = n2 = 13.868. 
Finally, K = 5.2. 
 
 
77. 
 Writing the equation of motion 
[s2+D(5)2s+
21 (10)
4
](s) = T(s) 
The transfer function is 
(s)
T(s)

 = 
2
1
s 25Ds 25 
 
4-78 Chapter 4: Time Response 
 
Also, 
 = 
π 2 2
%OS
- ln( )
100
%OS
ln ( )
100
 = 0.358 
and 
2n = 2(0.358)(5) = 25D 
Therefore D = 0.14. 
 78. 
 The equivalent circuit is: 
 
 
 where Jeq = 1+(
1
2
N
N
)2 ; Deq = (
1
2
N
N
 )2; Keq = (
1
2
N
N
)2. Thus, 
 
 1
(s)
T(s)

 = 
2
eq eq eq
1
J s D s K 
 . Letting 1
2
N
N
 = n and substituting the above values into the transfer 
function, 
 
 1
(s)
T(s)

 =
2
2 2
2
2 2
1
1 n
n n
s s
1 n 1 n

 
 
 . Therefore, n = 
2
2
n
2(1 n )
 . Finally, Ts = 
n
4

 = 
2
2
8(1 n )
n

 = 
16. Thus n = 1. 
 79. 
Let the rotation of the shaft with gear N2 be L(s). Assuming that all rotating load 
has been reflected to the N2 shaft,  2 ( ) ( ) ( )eqL eqL L eqJ s D s K s F s r T 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 N2 shaft, and DeqL is the equivalent damping at 
the N2 shaft. Since JeqL = 1(2)
2
 + 1 = 5 and DeqL = 1(2)
2
 = 4, the equation of motion 
becomes,  25 4 ( ) 2 ( ) ( )L eqs s K s F s T s    . For the translational system 
2( ) ( ) ( )Ms s X s F s  . Substituting F(s) into the rotational equation of motion, 
   2 25 4 ( ) 2 ( ) ( )L eqs s K s Ms s X s T s     . But,
 
 Solutions to Design Problems 4-79 
 
( ) ( )
( ) and ( ) 2 ( ).
2
L eq
X s X s
s T s T s
r
    Substituting these quantities in the equation 
above yields  2
( )
(5 4 ) 8 ( )
4
X s
M s s K T s    . 
Thus, the transfer function is 
2
( ) 4 /(5 4 )
8( )
(5 4 ) (5 4 )
X s M
KT s
s s
M M


 
 
. 
Now, 
4 4
20 (5 4 )
8
2(5 4 )
s
e
T M
R
M
    

. Hence, M = 15/4. 
For 16% overshoot,  = 0.504 from Eq. (4.39). 
Therefore, 
8
2 0.4
(5 4 )
n
M
  

. Solving for n yields n = 0.3968. 
 But, 0.3968
(5 4 ) 20
n
K K
M
   

. Thus, K = 3.15. 
 
 
 80. 
The transfer function for the capacitor voltage is C
V (s)
V(s)
 = 
1
Cs
1
R Ls
Cs
 
 = 
6
2 6
10
s Rs 10 
 . 
 For 20% overshoot,  = 
 2 2
%OS
- ln( )
100
%OS
ln ( )
100
 = 0.456. Therefore, 2n = R = 2(0.456)(103) = 912. 
81. 
Solving for the capacitor voltage using voltage division, 
1/ ( )
( ) ( )
1C i
CS
V s V s
R LS
CS

 
. Thus, the 
transfer function is 
2
( ) 1/( )
1( )
C
i
V s LC
RV s
s s
L LC

 
. Since 3
4
10 , Re 4000
Re 2
s
R
T
L
    . Thus 
8 KR   . Also, since 20% overshoot implies a damping ratio of 0.46 and 
n
1
2 8000, 8695.65n
LC
    . Hence, 0.013 FC  . 
82. 
 Using voltage division the transfer function is, 
 
2
1 1
( )
1 1( )
C
i
V s Cs LC
RV s
R Ls s s
Cs L LC
 
   
 
4-80 Chapter 4: Time Response 
 Also,
 
3 4 4 87 10
Re
2
s
L
T x
R R
L
    . Thus, 1143
R
L
 . Using Eq. (4.39) with 15% overshoot, 
 = 0.5169. But, 2n = R/L. Thus, 5
1 1
1105.63
(10 )
n
LC L


   . Therefore, L = 81.8 mH 
and R = 98.5 . 
 
83. 
 For the circuit shown below 
 
write the loop equations as 
1 1 1 2
( ) ( ) ( ) ( )
i
R Ls I s RI s V s   
1 1 1 2 2
1
( ) ( ) 0R I s R R I s
Cs
 
     
 
 
Solving for I2(s) 
1
1
2
1 1
1 1 2
( )
0
( )
1
R Ls Vi s
R
I s
R Ls R
R R R
Cs
 
 
 

  
 
   
 
 
But, 
2
1
( ) ( )
o
V s I s
Cs
 . Thus, 
1
2
2 1 2 1 1
( )
( ) ( ) ( )
o
i
V s R
V s R R CLs CR R L s R

   
 
Substituting component values, 
2
2 2
2 2
1
( ) ( 1000000)
1000000
(1000000 1) 1( )
1000000
( 1000000) ( 1000000)
o
i
V s R C
CRV s
s s
R C R C



 
 
 
 
For 8% overshoot,  = 0.6266. For Ts = 0.001, n = 
4
0.001
 = 4000. Hence, n = 6383.66. Thus, 
 Solutions to Design Problems 4-81 
 
 
2
2
1
1000000 6383.66
( 1000000)R C


 
or, 
2
1
0.0245
1000000
C
R


 (1) 
Also, 
 2
2
1000000 1
8000
( 1000000)
CR
R C



 (2) 
Solving (1) and (2) simultaneously, 
2
8023R  , and C = 2.4305 x 10-2 F. 
 
 
 
84. 
 
a. In Problem 3.31 we had 
 
0 0 0 0
1* *
0 0 0 0
2*
0
( ) 0 0
0
0 0
T d v T T T v
u
T v T T T v
u
v k c v kT
  
   
         
        
            
               
 
   *0 0 1
T
y T
v
 
 

 
  
 
 When 
2
0u  the equations are equivalent to 
 
0 0 0 0
* *
0 0 0 0 1
( ) 0
0 0
T d v T T T v
T v T T T v u
v k c v
  
   
         
       
          
             
 
  *0 0 1
T
y T
v
 
 

 
  
 
Substituting parameter values one gets 
 
4-82 Chapter 4: Time Response 
 
* *
1
0.04167 0 0.0058 5.2
0.0217 0.24 0.0058 5.2
0 100 2.4 0
T T
T T u
v v
        
       
          
             
 
  *0 0 1
T
y T
v
 
 

 
  
 
b. 
1
1
0.04167 0 0.0058
( )
( ) 0.0217 0.24 0.0058
det( )
0 100 2.4
s
Adj s
s s
s
s


 
 
     
  
   
I A
I A
I A
 
 
3 2
2
0.24 0.0058 0.0217 0.24
det( ) ( 0.04167) 0.0058
100 2.4 0 100
( 0.04167) ( 0.24)( 2.4) 0.58 (0.0058)(2.17)
2.6817 0.11 0.0126
( 2.6419)( 0.0398 0.0048)
s s
s s
s
s s s
s s s
s s s
   
   
  
     
   
   
I A
 
To obtain the adjoint matrix we calculate the cofactors: 
 
11
0.24 0.0058
( 2.64)
100 2.4
s
C s s
s
 
  
 
 
12
0.0217 0.0058
0.0217( 2.4)
0 2.4
C s
s
 
   

 
13
0.0217 0.24
2.17
0 100
s
C
 
 

 
21
0 0.0058
0.58
100 2.4
C
s
 
 
 
22
0.04167 0.0058
( 0.04167)( 2.4)
0 2.4
s
C s s
s

   

 
23
0.04167 0
100( 0.04167)
0 100
s
C s

   

 
31
0 0.0058
0.0058( 0.24)
0.24 0.0058
C s
s
   
 
 
2
32
0.04167 0.0058
2.4117 0.1001
0.0217 2.4
s
C s s
s

   
 
 
 Solutions to Design Problems 4-83 
 
2
33
0.04167 0
0.2817 0.01
0.0217 0.24
s
C s s
s

   
 
 
Then we have 
2
2
( 2.64) 0.58 0.0058( 0.24)
( ) 0.0217( 2.4) ( 0.04167)( 2.4) ( 2.4417 0.1101)
2.17 100( 0.04167) 0.2817 0.01
s s s
Adj s s s s s s
s s s
    
 
       
 
    
I A 
Finally 
2
1
2
1
2 2
5.2
2.17 100( 0.04167) 0.28171 0.01
( ) ( ) 5.2
( 2.6419)( 0.0398 0.0048)
0
520 10.3844 0.02
520
( 2.6419)( 0.0398 0.0048) ( 2.6419)( 0.0398 0.0048)
s s sY
s s
U s s s
s s
s s s s s s

 
         
   
  
  
  
     
C I A B
 
 
c. 100% effectiveness means that 
1
1u  or 
1
1
( )U s
s
 , so by the final value theorem 
20 0
520( 0.02) 1
( ) ( ) 820.1168
( 2.6419)( 0.0398 0.0048)s s
s
y Lim sY s Lim s
s s s s 

     
  
 (virus 
copies per mL of plasma) 
The closest poles to the imaginary axis are 0.0199 0.0661j  so the approximate settling time 
will be 
4
210
0.0199
s
T   days. 
85. 
a. 
Substituting 
2650
 F(s) 
s
  into the transfer function and solving for V(s) gives: 
F(s) 2650
V(s) 
1908 (1908 10) (1908 10)
A B
s s s s s

    
    
 
 
Here: 
2650
265
(1908 10) 0
A
s s
 
  
 and 
2650
505,620
1
190.8
B
s s
  
 
 
Substituting we have: 
4-84 Chapter 4: Time Response 
3
265 505620 1 1
V(s) 265
(1908 10) ( 5.24 10 )s s s s 
 
     
    
 
Taking the inverse Laplace transform, we have: 
35.24 10( ) 265(1 ) ( ), in m/s tv t e u t
     
b. 
>> s=tf('s'); 
>> G=1/(1908*s+10); 
>> t=0:0.1:1000; 
>> y1=2650*step(G,t); 
>> y2=265*(1-exp(-5.24e-3.*t)); 
>> plot(t,y1,t,y2) 
>> xlabel('sec') 
>> ylabel('m/s') 
 
 
 Solutions to Design Problems 4-85 
 
Both plots are identical. 
86. 
Since no overshoot is observed in the response, for simplicity, we postulate a first order system with 
some delay, namely  
 1
sTKG s e
s


. 
The steady state change in temperature is of 8C, so the transfer function’s dc gain is -8 since the step 
is negative. The time delay observed is approximately 0.017h=60sec. To find the systems time 
constant we find the point at which the temperature reaches 0.63(8C)=5.04C. 
This happens approximately 0.02hours = 75 seconds after the observed response delay. So the 
approximate transfer function is  
 
60 608 0.107
1 75 0.0133
s sG s e e
s s
   
 
. 
 
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Library of Congress Cataloging-in-Publication Data
Nise, Norman S.
Control systems engineering / Norman S. Nise, California State Polytechnic University, Pomona. — Seventh edition.
1 online resource.
Includes bibliographical references and index.
Description based on print version record and CIP data provided by publisher; resource not viewed.
ISBN 978-1-118-80082-9 (pdf) — ISBN 978-1-118-17051-9 (cloth : alk. paper)
1. Automatic control–Textbooks. 2. Systems engineering–Textbooks. I. Title.
TJ213
629.8–dc23
2014037468
Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
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