ch03_7ed_Nise

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T H R E E 
 
 Modeling 
 in the Time Domain 
 
SOLUTIONS TO CASE STUDIES CHALLENGES 
Antenna Control: State-Space Representation 
For the power amplifier,
 
a
p
E (s)
V (s) 
= 

150
s 150
. Taking the inverse Laplace transform,
 a
e +150ea = 
150vp. Thus, the state equation is 
  a a pe 150e 150v 
For the motor and load, define the state variables as x1 = m and x2 = θ m. Therefore, 
•
x 1 = x2 (1) 
Using the transfer function of the motor, cross multiplying, taking the inverse Laplace transform, 
and using the definitions for the state variables, 
 
•
x 2 = - 
t a
m
m a
K K1
(D )
J R
 x2 + 
t
a m
K
R J
 ea (2) 
Using the gear ratio, the output equation is 
 
y = 0.2x1 (3) 
Also, Jm = Ja+5(
1
5
)2 = 0.05+0.2 = 0.25, Dm = Da+3(
1
5
)2 = 0.01+0.12 = 0.13, t
a m
K
R J
= 
1
(5)(0.25)
 
= 0.8, and  t am
m a
K K1
(D )
J R
= 1.32. Using Eqs. (1), (2), and (3) along with the previous values, the 
state and output equations are, 
3-2 Chapter 3: Modeling in the Time Domain 
 a
0 1 0
e ;y 0.2 0
0 -1.32 0.8
   
     
   
x x x
 
 
Aquifer: State-Space Representation 
C1
1dh
dt
= qi1-qo1+q2-q1+q21 = qi1-0+G2(h2-h1)-G1h1+G21(H1-h1) = 
 -(G2+G1+G21)h1+G2h2+qi1+G21H1 
C2
2dh
dt
= qi2-q02+q3-q2+q32 = qi2-qo2+G3(h3-h2)-G2(h2-h1)+0 = G2h1-[G2+G3]h2+G3h3+qi2-qo2 
C3
3dh
dt
 = qi3-qo3+q4-q3+q43 = qi3-qo3+0-G3(h3-h2)+0 = G3h2-G3h3+qi3-qo3 
Dividing each equation by Ci and defining the state vector as x = [h1 h2 h3]
T 
 
      
   
   
     
    
   
    
   
      
 
 

 
  
1 2 3 1 21 12
1 1 1
.
2 3 3 2 22
2 2 2 2
3 3 3 3
3 3 3
( )
0
( )
( )
0
1 0 0
0 1 0
0 0 1
i
i o
i o
G G G q G HG
C C C
G G G q qG
u t
C C C C
G G q q
C C C
x x
y x
 
where u(t) = unit step function. 
 
ANSWERS TO REVIEW QUESTIONS 
1. (1) Can model systems other than linear, constant coefficients; (2) Used for digital simulation 
2. Yields qualitative insight 
3. That smallest set of variables that completely describe the system 
4. The value of the state variables 
5. The vector whose components are the state variables 
6. The n-dimensional space whose bases are the state variables 
7. State equations, an output equation, and an initial state vector (initial conditions) 
8. Eight 
9. Forms linear combinations of the state variables and the input to form the desired output 
10. No variable in the set can be written as a linear sum of the other variables in the set. 
 Solutions to Problems 3-3 
 
11. (1) They must be linearly independent; (2) The number of state variables must agree with the order of 
the differential equation describing the system; (3) The degree of difficulty in obtaining the state equations 
for a given set of state variables. 
12. The variables that are being differentiated in each of the linearly independent energy storage elements 
13. Yes, depending upon the choice of circuit variables and technique used to write the system equations. 
For example, a three -loop problem with three energy storage elements could yield three simultaneous 
second-order differential equations which would then be described by six, first-order differential equations. 
This exact situation arose when we wrote the differential equations for mechanical systems and then 
proceeded to find the state equations. 
14. The state variables are successive derivatives. 
 
SOLUTIONS TO PROBLEMS 
1. 
 
The state variables are 
2
i , 
4
i and 
o
v . 
We have that 2
1
2
di
v
dt
 , 4
2
di
v
dt
 and 0
5
0.5
dv
i
dt
 . 
Applying KVL around the external loop one gets 
1 3 5
3 0
i o
v i i i v     
And in each one of the nodes 
3 1 2
i i i  and 
5 3 4
i i i  
Substituting 
   1 1 2 3 43 0i ov i i i i i v       
 
or 
    1 1 2 1 2 43 0i ov i i i i i i v        
 
Solving for i1 one gets 
3-4 Chapter 3: Modeling in the Time Domain 
   1 2 4
4 1 1 1
5 5 5 5
i oi i i v v 
Thus 
      1 1 2 4
4 1 4 1
5 5 5 5
i i ov v i i i v v 
     2 1 2 4
2 1 2 1
2 5 10 5 10
i o
di v
i i v v
dt
 
 
Also 
3 1 2 2 4
1 1 1 1
5 5 5 5
i o
i i i i i v v       
and 
      5 3 4 2 4
1 4 1 1
5 5 5 5
i oi i i i i v v 
     5 2 4
2 8 2 2
2
5 5 5 5
o
i o
dv
i i i v v
dt
 
Finally 
       4 2 5 2 4
1 4 1 4
5 5 5 5
o i o
di
v i v i i v v
dt
 
 
In matrix form we have 
     
      
      
         
      
       
       
         
2
2
4
4
2 1 1 2
5 10 10 5
1 4 4 1
5 5 5 5
2 8 2 2
5 5 5 5
o
o
o
di
dt i
di
i v
dt
v
dv
dt
 
 
2
4
0 0 1
o
o
i
y v i
v
 
 
 
 
  
 
 
2. 
Add branch currents and node voltages to the schematic and obtain, 
 
 
 
Write the differential equation for each energy storage element. 
 Solutions to Problems 3-5 
 
 
1
2
3
1
3
1
2
L
dv
i
dt
di
v
dt

 
Therefore the state vector is 
1
3
v
i
 
  
 
x 
Now obtain 
L
v and 
2
i , in terms of the state variables, 
1 2 1 1 3 1 1 3
3 3( 4 ) 11 3
L R
v v v v i v i v v i          
 
2 1 3 1 3 1 3
1 1 1
( )
3 3 3
i i
i i i v v i v i v         
 
Also, the output is 
1 3
4
R
y i v i   
 
Hence, 
 
1 1
1
9 3
9
11 3
2 2
4 1
i
v
y

 
    
   
       

x x
0
x
 
 
 
 
3. 
 
Let C1 be the grounded capacitor and C 2 be the other. Now, writing the equations 
for the energy storage components yields, 
 
1
1
2
1 2
2 3
L
i C
C
C
di
v v
dt
dv
i i
dt
dv
i i
dt
 
 
 
 (1) 
 
Thus the state vector is 
1
2
L
C
C
i
v
v
 
 
  
 
  
x . 
3-6 Chapter 3: Modeling in the Time Domain 
Now, find the three loop currents in terms of the state variables and the input. 
 
Writing KVL around Loop 2 yields 
1 2 2
2
C C
v v i  . 
Or, 
1 22
0.5 0.5
C C
i v v  
 
Writing KVL around the outer loop yields 
3 2
2 2
i
i i v  
Or, 
1 23 2
0.5 0.5 0.5 0.5
i i C C
i v i v v v     
 
Also, 
1 3 L
i i i  . Hence,  
1 21 3
0.5
L L i C C
i i i i v v v      
 
 
 
Substituting the loop currents in equations (1) yields the results in vector-matrix 
form, 
 
1
1
2
2
0 1 0 1
1 1 1 0.5
0 1 1 0.5
L
L
C
C i
C
C
di
dt
i
dv
v v
dt
vdv
dt
 
 
      
      
        
             
 
 
 
 
Since 
1 22o C C
v i v v   , the output equation is 
 
 
1
2
0 1 1
L
C
C
i
y v
v
 
 
   
 
  
 
4. 
Equations of motion in Laplace: 
 
 
2
1 2 3
2
1 2 3
2
1 2 3
(2 3 2) ( ) ( 2) ( ) ( ) 0
( 2) ( ) ( 2 2) ( ) ( ) ( )
( ) ( ) ( 3 ) ( ) 0
s s X s s X s sX s
s X s s s X s sX s F s
sX s sX s s s X s
     
      
    
 
Equations of motion in the time domain: 
 Solutions to Problems 3-7 
 
2
31 1 2
1 22
2
31 2 2
1 22
2
3 31 2
2
2 3 2 2 0
2 2 2 ( )
3 0
dxd x dx dx
x x
dt dt dt dt
dxdx d x dx
x x f t
dt dt dt dt
d x dxdx dx
dt dt dt dt
     
      
    
 
Define state variables: 
 
1 1 1 1
1 1
2 2
3 2 2 3
 or (1)
 or (2)
 or 
z x x z
dx dx
z z
dt dt
z x x z
 
 
 
2 2
4 4
5 3 3 5
3 3
6 6
 (3)
 or(4)
 or (5)
 or 
dx dx
z z
dt dt
z x x z
dx dx
z z
dt dt
 
 
  (6)
 
 
Substituting Eq. (1) in (2), (3) in (4), and (5) in (6), we obtain, respectively: 
 
1
2
3
4
5
6
 (7)
 (8)
 
dz
z
dt
dz
z
dt
dz
z
dt


 (9)
 
 
Substituting Eqs. (1) through (6) into the equations of motion in the time domain and solving for the 
derivatives of the state variables and using Eqs. (7) through (9) yields the state equations: 
 
 
3-8 Chapter 3: Modeling in the Time Domain 
1
2
2
1 2 3 4 6
3
4
4
1 2 3 4 6
5
6
6
2 4 6
3 1 1
2 2 2
2 2 2 ( )
3
dz
z
dt
dz
z z z z z
dt
dz
z
dt
dz
z z z z z f t
dt
dz
z
dt
dz
z z z
dt

     

     

  
 
The output is x3 = z5. 
 
In vector-matrix form: 
 
 
0 1 0 0 0 0 0
1 1.5 1 0.5 0 0.5 0
0 0 0 1 0 0 0
( )
2 1 2 2 0 1 1
0 0 0 0 0 1 0
0 1 0 1 0 3 0
0 0 0 0 1 0
f t

   
   
 
   
   
    
    
   
   
   

Z Z
y Z
 
5. 
 
The impedance equations are: 
 
         2 1 2 32 1 1 0s s X s sX s s X s      
          21 2 32 2 1 1 0sX s s s X s s X s       
             21 2 31 1 2 2s X s s X s s s X s f t        
 
Taking the inverse Laplace transform 
1 1 2 3 31
2 0x x x x xx       
321 2 2 3
2 2 0x x x x xx       
 1 1 2 2 3 3 32 2x x x x x x x f t        
 
1 1 31 2 3
2x x x x xx       
 Solutions to Problems 3-9 
 
2 2 32 1 3
1 1 1 1
0
2 2 2 2
x x x x x x      
 1 1 33 2 2 32 2x x x x x x x f t       
 
Define the state variables 
1 1
z x ; 
2 1
z x ; 
3 2
z x ; 
4 2
z x ; 
5 3
z x ; 
6 3
z x 
The equations are rewritten as 
1 21
x zz   
1 2 1 4 52 6
2x z z z z zz        
2 43
x zz   
     3 6 512 44
2 22 2
z z zz
x zz 
  65 3x zz 
3 26 1 4 3 6 5
2 2x z z z z zz z       
In matrix form 
0 1 0 0 0 0
0
1 2 0 1 1 1
0
0 0 0 1 0 0
0
( )1 1 1 1
00 1
2 2 2 2
0
0 0 0 0 0 1
1
1 1 1 1 2 2
f t
 
  
    
  
          
  
  
    
z z 
 0 0 0 0 1 0y  z 
 
6. 
Drawing the equivalent network, 
 
 
 
Writing the equations of motion, 
3-10 Chapter 3: Modeling in the Time Domain 
 
2
2 3
2
2 3
(555.56 100) 100 3.33
100 (100 100 100) 0
s T
s s
 
 
  
    
 
Taking the inverse Laplace transform and simplifying, 
 
2 2 3
2 3 3 3
0.18 0.18 0.006
0
T  
   

 
  
    
 
Defining the state variables as 
 
1 2 2 2 3 3 4 3
, , , x x x x   
 
    
Writing the state equations using the equations of motion and the definitions of the state variables 
 
 
1 2
2 2 2 3 1 3
3 4
4 3 2 3 3 1 3 4
2 1
0.18 0.18 0.006 0.18 0.18 0.006
3.33 3.33
x x
x T x x T
x x
x x x x
y x
  
   


 

  

        

      
 
, 
In vector-matrix form, 
 
 
0 1 0 0 0
0.18 0 0.18 0 0.006
0 0 0 1 0
1 0 1 1 0
3.33 0 0 0
T
y

   
   

    
   
   
    

x x
x
 
7. 
Drawing the equivalent circuit, 
 
 Solutions to Problems 3-11 
 
 
Writing the equations of motion, 
 
2 3
2 3 4
3 4
12 ( ) 2 ( ) 10 ( )
2 ( ) (3 2) ( ) 3 ( ) 0
3 ( ) 5 ( ) 0
s s T s
s s s s s
s s s s
 
  
 
 
    
  
 
 
Taking the inverse Laplace transform, 
 
2 3
12 ( ) 2 ( ) 10 ( )t t T t   (1) 
• •
2 3 3 4
2 ( ) 3 ( ) 2 3 ( ) 0t t t        (2) 
• •
3 4
3 ( ) 5 ( ) 0t t    (3) 
From (3), 
• •
3 4
5
( ) ( )
3
t t  and 
3 4
5
( ) ( )
3
t t  (4) 
assuming zero initial conditions. 
From (1) 
2 3 4
1 5 5 5
( ) ( ) ( ) ( ) ( )
6 6 18 6
t t T t t T t      (5) 
 
Substituting (4) and (5) into (2) yields the state equation (notice there is only one equation), 
 
•
4 4
25 5
( ) ( ) ( )
18 6
t t T t    
The output equation is given by, 
 
4
1
( ) ( )
10
L
t t  
8. 
Solving Eqs. (3.44) and (3.45) in the text for the transfer functions 1
( )
( )
X s
F s
 and 2
( )
( )
X s
F s
: 
3-12 Chapter 3: Modeling in the Time Domain 
2
2
1 2
1
2
2
0
( )
K
F M s K
X s
M s D s K K
K M s K
 
 
 

   
 
  
 and 
  
 
 

   
 
  
2
1
2 2
1
2
2
0
( )
M s D s K
K F
X s
M s D s K K
K M s K
 
Thus, 
 
1
4 3 2 2
2 1 2 2 1
( )
( )
X s K
F s M M s D M s K M s K M s D K s

   
 
and 
 
2
2 1
4 3 2 2
2 1 2 2 1
( )
( )
X s M s D s K
F s M M s D M s K M s K M s D K s
 

   
 
 
Multiplying each of the above transfer functions by s to find velocity yields pole/zero cancellation at 
the origin and a resulting transfer function that is third order. 
 
9. 
a. Using the standard form derived in the textbook, 
 
 
   
   
    
   
   
         

0 1 0 0 0
0 0 1 0 0
( )
0 0 0 1 0
100 7 10 20 1
100 0 0 0
r t
c
x x
x
 
 
b. Using the standard form derived in the textbook, 
 
 
   
   
   
    
   
   
          

0 1 0 0 0 0
0 0 1 0 0 0
( )0 0 0 1 0 0
0 0 0 0 1 0
30 1 6 9 8 1
30 0 0 0 0
r t
c
x x
x
 
10. 
Program: 
'a' 
num=100; 
den=[1 20 10 7 100]; 
G=tf(num,den) 
[Acc,Bcc,Ccc,Dcc]=tf2ss(num,den); 
Af=flipud(Acc); 
A=fliplr(Af) 
B=flipud(Bcc) 
C=fliplr(Ccc) 
'b' 
num=30; 
 Solutions to Problems 3-13 
 
den=[1 8 9 6 1 30]; 
G=tf(num,den) 
[Acc,Bcc,Ccc,Dcc]=tf2ss(num,den); 
Af=flipud(Acc); 
A=fliplr(Af) 
B=flipud(Bcc) 
C=fliplr(Ccc) 
 
Computer response: 
ans = 
 
 
a 
 
Transfer function: 
 100 
--------------------------------- 
s^4 + 20 s^3 + 10 s^2 + 7 s + 100 
 
 
 
 
 
 
A = 
 
 0 1 0 0 
 0 0 1 0 
 0 0 0 1 
 -100 -7 -10 -20 
 
 
B = 
 
 0 
 0 
 0 
 1 
 
 
C = 
 
 100 0 0 0 
 
ans = 
 
 
b 
 
Transfer function: 
 30 
------------------------------------ 
s^5 + 8 s^4 + 9 s^3 + 6 s^2 + s + 30 
 
 
A = 
 
 0 1 0 0 0 
 0 0 1 0 0 
 0 0 0 1 0 
 0 0 0 0 1 
 -30 -1 -6 -9 -8 
 
3-14 Chapter 3: Modeling in the Time Domain 
B = 
 
 0 
 0 
 0 
 0 
 1 
 
C = 
 
 30 0 0 0 0 
 
 
11. 
a. Using the standard form derived in the textbook, 
 
 
•
0 1 0 0 0
0 0 1 0 0
( )
0 0 0 1 0
13 5 1 5 1
10 8 0 0 0
r t
c
   
   
    
   
   
      

x x
x
 
b. Using the standard form derived in the textbook, 
 
 
 
•
0 1 0 0 0 0
0 0 1 0 0 0
( )0 0 0 1 0 0
0 0 0 0 1 0
0 0 8 13 9 1
6 7 12 2 1
r t
c
   
   
   
    
   
   
        

x x
x
 
 
12. 
Program: 
'a' 
num=[8 10]; 
den=[1 5 1 5 13] 
G=tf(num,den) 
[Acc,Bcc,Ccc,Dcc]=tf2ss(num,den); 
Af=flipud(Acc); 
A=fliplr(Af) 
B=flipud(Bcc) 
C=fliplr(Ccc) 
'b' 
num=[1 2 12 7 6]; 
den=[1 9 13 8 0 0] 
G=tf(num,den) 
[Acc,Bcc,Ccc,Dcc]=tf2ss(num,den); 
Af=flipud(Acc); 
 Solutions to Problems 3-15 
 
A=fliplr(Af) 
B=flipud(Bcc) 
C=fliplr(Ccc) 
 
 
Computer response: 
ans = 
 
ans = 
 
a 
 
 
den = 
 
 1 5 1 5 13 
 
 
 
Transferfunction: 
 8 s + 10 
---------------------------- 
s^4 + 5 s^3 + s^2 + 5 s + 13 
 
 
 
 
A = 
 
 0 1 0 0 
 0 0 1 0 
 0 0 0 1 
 -13 -5 -1 -5 
 
 
 
 
 
B = 
 
 0 
 0 
 0 
 1 
 
 
 
C = 
 
 10 8 0 0 
 
 
ans = 
 
b 
 
 
den = 
 
 1 9 13 8 0 0 
 
 
3-16 Chapter 3: Modeling in the Time Domain 
Transfer function: 
s^4 + 2 s^3 + 12 s^2 + 7 s + 6 
------------------------------ 
 s^5 + 9 s^4 + 13 s^3 + 8 s^2 
 
 
A = 
 
 0 1 0 0 0 
 0 0 1 0 0 
 0 0 0 1 0 
 0 0 0 0 1 
  3 2
1
6 10 5s s s
 0 0 -8 -13 -9 
 
 
B = 
 
 0 
 0 
 0 
 0 
 1 
 
 
C = 
 
 6 7 12 2 1 
 
 
13. 
 
The transfer function can be represented as a block diagram as follows: 
 
 
 
The differential equation of the first box is 
 5 7 5x x x x r t    
 
Define the state variables 
1 ;x x 2 ;x x 3x x 
 Solutions to Problems 3-17 
 
so 
1 2
x x 
2 3
x x 
   3 3 2 15 7 5 5 7 5x x x x r t x x x r t          
From the second box 
3 2
2 2y x x x x    
In vector-matrix form 
 
0 1 0 0
0 0 1 0
5 7 5 1
r t
   
   
 
   
        
x x 
 0 2 1y  x 
 
14. 
a. G(s)=C(sI-A)-1B 
 
 
   
   
  
   
      
0 1 0 0
0 0 1 ; 0 ; 1 0 0
-3 -2 -5 10
A B C 
 

   
 
    
 
 
2
1
3 2
2
s 5s 2 s 5 1
1
(s - ) -3 s(s 5) s
s 5s 2s 3
-3s -2s-3 s
I A 
 
Therefore, G(s) = 
3 2
10
s 5s 2s 3  
. 
b. G(s)=C(sI-A)-1B 
2 3 8
0 5 3
3 5 4
 
 
  
    
A ; 
1
4
6
 
 
  
 
 
B ;  1, 3, 6C 
 
2
1 2
3 2
2
s s 5 3 s 52 8 s 49
1
(s ) 9 s 2 s 32 3 s 6
s 3 s 27 s 157
3 s 15 5 s 1 s 7 s 10

     
 
        
       
I A
 
 
Therefore, 
2
3 2
49 s 349 s 452
G(s)
s 3 s 27 s 157
 

  
 . 
c. G(s)=C(sI-A)-1B 
 
3-18 Chapter 3: Modeling in the Time Domain 
 
3 5 2 5
1 8 7 ; 3 ; 1 4 3
3 6 2 2
   
   
     
   
       
A B C 
 

     
 
        
       
2
1 2
3 2
2
(s 6s 26) (5s 2) (2s 19)
1
( ) (s 23) (s 5s 12) (7s 19)
s 3s 19s 133
(3s 30) (6s 33) (s 5s 19)
sI A 
 
Therefore, G(s) = 
2
3 2
23s 48s 7
s 3s 19s 133
 
  
. 
 
15. 
Program: 
 
'a' 
A=[0 1 5 0;0 0 1 0;0 0 0 1;-7 -9 -2 -3]; 
B=[0;5;8;2]; 
C=[1 3 6 6]; 
D=0; 
statespace=ss(A,B,C,D) 
[num,den]=ss2tf(A,B,C,D); 
G=tf(num,den) 
'b' 
A=[3 1 0 4 -2;-3 5 -5 2 -1;0 1 -1 2 8;-7 6 -3 -4 0;-6 0 4 -3 1]; 
B=[2;7;8;5;4]; 
C=[1 -2 -9 7 6]; 
D=0; 
statespace=ss(A,B,C,D) 
[num,den]=ss2tf(A,B,C,D); 
G=tf(num,den) 
 
 
 
Computer response: 
 
ans = 
 
a 
 
 
a = 
 x1 x2 x3 x4 
 x1 0 1 5 0 
 x2 0 0 1 0 
 x3 0 0 0 1 
 x4 -7 -9 -2 -3 
 
b = 
 u1 
 x1 0 
 x2 5 
 x3 8 
 x4 2 
 
 Solutions to Problems 3-19 
 
c = 
 x1 x2 x3 x4 
 y1 1 3 6 6 
 
d = 
 u1 
 y1 0 
 
Continuous-time model. 
 
Transfer function: 
75 s^3 - 96 s^2 - 2331 s - 210 
------------------------------ 
s^4 + 3 s^3 + 2 s^2 + 44 s + 7 
 
 
ans = 
 
 
b 
 
 
a = 
 x1 x2 x3 x4 x5 
 x1 3 1 0 4 -2 
 x2 -3 5 -5 2 -1 
 x3 0 1 -1 2 8 
 x4 -7 6 -3 -4 0 
 x5 -6 0 4 -3 1 
 
 
b = 
 u1 
 x1 2 
 x2 7 
 x3 8 
 x4 5 
 x5 4 
 
 
 
c = 
 x1 x2 x3 x4 x5 
 y1 1 -2 -9 7 6 
 
 
 
d = 
 u1 
 y1 0 
 
 
 
 
Continuous-time model. 
 
Transfer function: 
-25 s^4 - 292 s^3 + 1680 s^2 + 1.628e004 s + 3.188e004 
------------------------------------------------------ 
 s^5 - 4 s^4 - 32 s^3 + 148 s^2 - 1153 s - 4480 
 
 
16. 
3-20 Chapter 3: Modeling in the Time Domain 
Program: 
syms s 
'a' 
A=[0 1 5 0 
0 0 1 0 
0 0 0 1 
-7 -9 -2 -3]; 
B=[0;5;8;2]; 
C=[1 3 4 6]; 
D=0; 
I=[1 0 0 0 
0 1 0 0 
0 0 1 0 
0 0 0 1]; 
'T(s)' 
T=C*((s*I-A)^-1)*B+D; 
T=simple(T); 
pretty(T) 
'b' 
A=[3 1 0 4 -2 
-3 5 -5 2 -1 
0 1 -1 2 8 
-7 6 -3 -4 0 
-6 0 4 -3 1]; 
B=[2;7;6;5;4]; 
C=[1 -2 -9 7 6]; 
D=0; 
I=[1 0 0 0 0 
0 1 0 0 0 
0 0 1 0 0 
0 0 0 1 0 
0 0 0 0 1]; 
'T(s)' 
T=C*((s*I-A)^-1)*B+D; 
T=simple(T); 
pretty(T) 
 
Computer response: 
 
ans = 
 
a 
 
 
ans = 
 
T(s) 
 
 
 3 2 
 59 s - 148 s - 2241 s - 140 
 ----------------------------- 
 4 3 2 
 s + 3 s + 2 s + 44 s + 7 
 
 Solutions to Problems 3-21 
 
ans = 
 
b 
 
 
ans = 
 
T(s) 
 
 
 4 3 2 
 (- 7 s - 408 s + 1708 s + 14582 s + 
 
 5 4 3 2 
 27665) / (s - 4 s - 32 s + 148 s - 
 
 1153 s - 4480) 
>> 
 
 
17. 
The equivalent cascade transfer function is as shown below. 
 
 
 
δ   0 a2 1
3 3 3 3
K KK K
For the first box, x x x x (t).
K K K K
 
1 2 3
Selecting the phase variables as the state variables: x x, x x, x x.   
Writing the state and output equations: 
•
x 1 = x2 
•
x 2 = x3 
•
x 3 = - 
0
3
K
K
 x1-
1
3
K
K
 x2- 
2
3
K
K
 x3+ 
a
3
K
K
 (t) 
y = (t) = 
•
x + b
a
K
K
 x = 
b
a
K
K
 x1+x2 
In vector-matrix form, 
3-22 Chapter 3: Modeling in the Time Domain 
δ
   
   
     
       
     
   
      
b
a
0 a1 2
3 3 3 3
0 1 0 0
K
0 0 1 0 (t); y 1 0
K
K KK K
- - -
K K K K
x x x 
18. 
Since Tm = Jeq 
ωmd
dt
+ Deqm, and Tm = Kt ia, 
Jeq 
ωmd
dt
+ Deqm = Kt ia (1) 
Or, 
ωmd
dt
 = - 
eq
eq
D
J
 m + 
t
eq
K
J
 ia 
But, m = 
2
1
N
N
 L. 
Substituting in (1) and simplifying yields the first state equation, 
 
ω
L
d
dt
 = -
eq
eq
D
J
 L + 
t 1
eq 2
K N
J N
 ia 
The second state equation is: 
θLd
dt
 = L 
Since 
ea = Raia+La 
adi
dt
 +Kbm = Raia+La 
adi
dt
 +Kb
2
1
N
N
L, 
the third state equation is found by solving for 
adi
dt
. Hence, 
 
a
di
dt
 = - 
b 2
a 1
K N
L N
 L -
a
a
R
L
 ia+ 
a
1
L
ea 
Thus the state variables are: x1 = L, x2 = L , and x3 = ia. 
 
Finally, the output is y = m = 
2
1
N
N
L . 
In vector-matrix form, 
 
   
   
     
           
   
      
eq t 1
eq eq 2
2
a
1
b a2
aa 1 a
D K N
- 0
J J N 0
N
1 0 0 0 e ; y 0 0
N
K R 1N
- 0 -
LL N L
x x x 
 Solutions to Problems 3-23 
 
where, 
ω
θ
L
L
a
i
 
 

 
  
x 
19. 
 
Writing the differential equations, 
 
2
21 1 2
12
2 2 2 0
d x dx dx
x
dt dt dt
    
 
2
2 2 1
2
2 2 2 ( )
d x dx dx
f t
dt dt dt
   
 
Defining the state variables to be x1, v1, x2, v2, where v's are velocities, 
•
x 1 = v1 
•
x 2 = v2 
•
v 1 = -2v1-2x12+2v2 
•
v 2 = v1-v2+0.5f(t) 
Around x1 = 1, x1 = 1+x1, and 
•
x 1 = 
•
x 1 . Also, 
 
δ δ δ2 2 1
1 1 1 1 1 1
x 1 x 1 x 1
dx
x x | | x 1 2x | x 1 2 x
dt  
      
 
Therefore, the state and output equations are, 

•
x 1 = v1 
•
x 2 = v2 
•
v 1 = -2v1-2(1+2x1)+2v2 
•
v 2 = v1-v2+0.5f(t) 
 
3-24 Chapter 3: Modeling in the Time Domain 
y = x2 
In vector-matrix form, 
 
1
1
2 2
1
1
2
2
0 0 1 0 0 0
0 0 0 10 0 1
4 0 2 2 2 2 ( )
0 0 1 1 0 0.5
.
.
.
.
v
x
x
xx
f t
v v
v


 
  

; 
1
2
1
2
0 1 0 0
v
x
x
y
v

 
where f(t) = 2 + f(t), since force in the nonlinear spring is 2 N and must be balanced by 2 N force on 
the damper. 
 
 20. 
Controller: 
The transfer function can be represented as a block diagram as follows: 
 
 
 
Writing the differential equation for the first box, 
 
1
K
s b
 
and solving for 
c
x

, 
1
( )
c c c
x bx K r t

   
From the second box, 
1
1
( )
( ) ( )
c c c c c c
c c
y x ax bx K r t ax
a b x K r t

     
  
 
Wheels: 
The transfer function can be represented as a block diagram as follows: 
 
 Solutions to Problems 3-25 
 
 
Writing the differential equation for the block of the form, 
c
s c
 
and solving for 
w
x

 , 
( )
w w w
x cx cr t

   
The output equation is, 
yw = xw 
Vehicle: 
The transfer function can be represented as a block diagram as follows: 
 
 
 
Writing the differential equation for the block, 
1
s
 
and solving for 
v
x

, 
( )
v v
x r t

 
The output equation is 
 yv = xv 
21. 
Adding displacements to the figure, 
 
 
3-26 Chapter 3: Modeling in the Time Domain 
Writing the differential equations for noncontact, 
 
2
2
2
2
2 2 ( )
0
sr r
r s
s sr
r s
dxd x dx
x x u t
dt dt dt
d x dxdx
x x
dt dt dt
    
     
 
Define the state variables as, 
• •
1 2 3 4
; ; ; 
r r s s
x x x x x x x x    
Writing the state equations, using the differential equations and the definition of the state variables, 
we get, 
 
• •
1 2
• ••
2 1 2 3 4
• •
3 4
• ••
4 1 2 3 4
2 2 ( )
r
r
s
s
x x x
x x x x x x u t
x x x
x x x x x x
 
      
 
    
 
Assuming the output to be xs, the output equation is, 
3
y x 
In vector-matrix form, 
 
 
•
0 1 0 0 0
2 2 1 1 1
( )
0 0 0 1 0
1 1 1 1 0
0 0 1 0
u t
y
   
   
 
    
   
   
    

x x
x
 
Writing the differential equations for contact, 
 
2
2
2
2
2
2
2 2 ( )
0
0
2 2 0
sr r
r s
s sr
r s e
e
s
e e
s e
dxd x dx
x x u t
dt dt dt
d x dxdx
x x z x
dt dt dt
dxdz
x z
dt dt
d x dxdz
x x
dt dt dt
    
       
    
     
 
 Solutions to Problems 3-27 
 
Defining the state variables, 
• • • •
1 2 3 4 5 6 7 8
; ; ; ; ; ; ; 
r r s s e e
x x x x x x x x x z x z x x x x        
Using the differential equations and the definitions of the state variables, we write the state equations. 
•
1 2
•
2 1 2 3 4
•
3 4
•
4 1 2 3 4 5 7
•
5 6
2 ( )
x x
x x x x x u t
x x
x x x x x x x
x x

     

     

 
Differentiating the third differential equation and solving for d
2
z/dt
2
 we obtain, 
 
22•
6 2 2
s e
dx d xd z dz
x
dt dt dt dt
    
But, from the fourth differential equation, 
 
2
3 6 8 72
2 2 2 2e e
s e
d x dxdz
x x x x x x
dt dt dt
        
Substituting this expression back into
•
6
x along with the other definitions and then simplifying yields, 
•
6 4 3 8 7
2 2x x x x x    
Continuing, 
•
7 8
•
8 3 6 7 8
2 2
x x
x x x x x

   
 
Assuming the output is xs, 
s
y x 
Hence, the solution in vector-matrix form is 
 
3-28 Chapter 3: Modeling in the Time Domain 
 
•
0 1 0 0 0 0 0 0 0
1 2 1 1 0 0 0 0 1
0 0 0 1 0 0 0 0 0
1 1 1 1 1 0 1 0 0
( )
0 0 0 0 0 1 0 0 0
0 0 1 1 0 0 2 2 0
0 0 0 0 0 0 0 1 0
0 0 1 0 0 1 2 2 0
0 0 1 0 0 0 0 0
u t
y
   
   
 
   
   
   
     
   
   
    
   
   
       

x x
x
 
 
22. 
 
 a. We begin by calculating 
0.435 0.209 0.02
0.268 0.394 0
0.227 0 0.02
s
s
s
   
 
    
   
sI A 
and 
 
 
0.394 0 0.268 0 0.268 0.394
det( ) ( 0.435) 0.209 0.02
0 0.02 0.227 0.02 0.227 0
s s
s
s s
   
    
   
sI A
 
( 0.435)( 0.394)( 0.02) 0.209( 0.268)( 0.02) 0.02(0.227)( 0.394)s s s s s         
3 20.849 0.188 0.0034 0.056 0.00112 0.00454 0.00179s s s s s        
3 20.849 0.1278 0.00049 ( 0.66)( 0.19)( 0.004)s s s s s s        
 
11 21 31
12 22 32
13 23 33
( )
c c c
Adj c c c
c c c
 
 
   
 
 
sI A 
where 
 
 Solutions to Problems 3-29 
 
11
0.394 0
( 0.394)( 0.02)
0 0.02
s
c s s
s

   

 
12
0.268 0
0.268( 0.02)
0.227 0.02
c s
s

   
 
 
13
0.268 0.394
0.227( 0.394)
0.227 0
s
c s
 
   

 
21
0.209 0.02
0.209( 0.02)
0 0.02
c s
s
 
   

 
2
22
0.435 0.02
( 0.435)( 0.02) 0.00454 0.455 0.00416
0.227 0.02
s
c s s s s
s
 
       
 
 
23
0.435 0.209
0.047443
0.227 0
s
c
 
  

 
31
0.209 0.02
0.02( 0.394)
0.394 0
c s
s
 
  

 
32
0.435 0.02
0.00536
0.268 0
s
c
 
  

 
2
33
0.435 0.209
( 0.435)( 0.394) 0.268(0.209) 0.829 0.1154
0.268 0.394
s
c s s s s
s
 
       
 
 
 
2
2
( 0.394)( 0.02) 0.209( 0.02) 0.02( 0.394)
0.268( 0.02) 0.455 0.0042 0.0054
0.227( 0.394) 0.0474 0.829 0.1154( )
( )
det( ) ( 0.004)( 0.19)( 0.66)
s s s s
s s s
s s sAdj
s s s

     
 
    
 
         
   
1 sI A
sI A
sI A
 
 
 ( 0.394)( 0.02) 0.209( 0.02) 0.02( 0.394)
( )
3333.33( 0.004)( 0.19)( 0.66)
s s s s
s s s

    
 
  
1
C sI A 
( ) ( 0.02)( 0.394)
( )
( ) 3333.33( 0.004)( 0.19)( 0.66)
Y s s s
U s s s s
    
  
1
C sI A B 
 
b. 
>> A=[-0.435 0.209 0.02; 0.268 -0.394 0; 0.227 0 -0.02] 
 
3-30 Chapter 3: Modeling in the Time Domain 
A = 
 
 -0.4350 0.2090 0.0200 
 0.2680 -0.3940 0 
 0.2270 0 -0.0200 
 
>> B = [1;0;0] 
 
B = 
 
 1 
 0 
 0 
 
>> C = [0.0003 0 0] 
 
C = 
 
 1.0e-003 * 
 
 0.3000 0 0 
 
>> [n,d]=ss2tf(A,B,C,0) 
 
n = 
 
 1.0e-003 * 
 Solutions to Problems 3-31 
 
 
 0 0.3000 0.1242 0.0024 
 
 
d = 
 
 1.0000 0.8490 0.1274 0.0005 
 
>> roots(n) 
 
ans = 
 
 -0.3940 
 -0.0200 
 
>> roots(d) 
 
ans = 
 
 -0.6560 
 -0.1889 
 -0.0042 
 
23. By direct observation 
 
3-32 Chapter 3: Modeling in the Time Domain 
0 00 02 0
1 10 11 12 1
02 20 21 22 23 24 2
3 32 33 3
4 42 44 4
0 0 0 1
0 0 0
0
0 0 0 0
0 0 0 0
x a a x
x a a a x
dx a a a a a x
x a a x
x a a x
       
       
       
        
       
       
             
 
 
24. 
a. 
>> A=[-0.038 0.896 0 0.0015; 0.0017 -0.092 0 -0.0056; 1 0 0 -3.086; 0 1 0 0] 
 
A = 
 
 -0.0380 0.8960 0 0.0015 
 0.0017 -0.0920 0 -0.0056 
 1.0000 0 0 -3.0860 
 0 1.0000 0 0 
 
>> B = [-0.0075 -0.023; 0.0017 -0.0022; 0 0; 0 0] 
 
B = 
 
 -0.0075 -0.0230 
 0.0017 -0.0022 
 0 0 
 0 0 
 
>> C = [0 0 1 0; 0 0 0 1] 
 
 Solutions to Problems 3-33 
 
C = 
 
 0 0 1 0 
 0 0 0 1 
 
>> [num,den] = ss2tf(A,B,C,zeros(2),1) 
 
num = 
 
 0 0.0000 -0.0075 -0.0044 -0.0002 
 0 0 0.0017 0.0001 0 
 
 
den = 
 
 1.0000 0.1300 0.0076 0.0002 0 
 
>> [num,den] = ss2tf(A,B,C,zeros(2),2) 
 
num = 
 
 0 -0.0000 -0.0230 0.0027 0.0002 
 0 -0.0000 -0.0022 -0.0001 0 
 
 
den = 
 
3-34 Chapter 3: Modeling in the Time Domain 
 1.0000 0.1300 0.0076 0.0002 0 
 
b. 
 From the MATLAB results 
2
3 2
(0.0075 0.0044 0.0002)
( )
( 0.13 0.0076 0.0002)
B
z s s
s
s s s s
  

  
 
3 2
0.0017 0.0001
( )
0.13 0.0076 0.0002
B
s
s
s s s




  
 
2
3 2
0.023 0.0027 0.0002
( )
( 0.13 0.0076 0.0002)
Sz s s
s
s s s s
  

  
 
3 2
(0.0022 0.0001)
( )
0.13 0.0076 0.0002
S
s
s
s s s


 

  
 
 
25. 
The transfer function is divided into two parts: 
 
 
So we have 
3 2
( ) 1
( ) ( ) ( )
W s
R s s a d s b ad s bd

    
 and 
( )
( )
Y s
s c
W s
  
 
In time domain 
 
( ) ( )w a d w b ad w bdw r      and w cw y  
 
Define the state variables as 
 
 Solutions to Problems 3-35 
 
1
2 1
3 2
x w
x w x
x w x

 
 
 
 
So we can write 
 
3 1 2 3
( ) ( )w x r bdx b ad x a d x       and 
1 2
y cx x  
 
In matrix form these equations are: 
 
1 1
2 2
3 3
0 1 0 0
0 0 1 0
( ) ( ) 1
x x
x x r
x bd b ad a d x
       
       
 
       
                  
 
 
 
1
2
3
1 0
x
y c x
x
 
 

 
  
 
 
26. 
 
a. 
1 21( ) 7( 5) 3
( 5)
G s s
s
     

1
C(sI A) B 
 
b. 
   
1
1
0
5 0 3 35
( ) 7 0 7 0
0 1 1 1 1
0
1
s s
G s
s
s


 
      
          
      
  
1C(sI A) B 
37 21
0
5 1 5s s
  
       
 
 
 
3-36 Chapter 3: Modeling in the Time Domain 
c)    
1
1
0
5 0 3 35
( ) 7 3 7 3
0 1 0 1 0
0
1
s s
G s
s
s


 
      
          
      
  
1C(sI A) B 
 
3
21
7 0 5
5
0
s
s
 
  
  
 
 
27. 
a. 
       1 2 3 4
SO
O A O O SO O IDO
dm
k m t k k m t k m t
dt
    
   3 4
IDO
O SO O IDO
dm
k m t k m t
dt
  
     1 2 3
V
L A L L V
dm
k m t k k m t
dt
   
   3 4
S
L V L S
dm
k m t k m t
dt
  
 
 
 
b. 
 
01 1 02 2 4
02 02 03 04
02 04
1 2 3
2 4
( ) 0 1
( ) 0 0 0
( )0 0 0 0
0 0 ( ) 0 0
0 0 0 0
A
AL L L
SO
SO
EIDO
IDO
VL L L
V
SL L
S
m
mk k k k k
m mk k k k
u tmk km
mk k k
m
mk k
m





 
 
       
       
     
      
             
          
 
 
 
 1 0 0 0 0
A
SO
IDO
V
S
m
m
m
m
m
 
 
 
 
 
 
 
 
y 
 Solutions to Problems 3-37 
 
28. 
The armature loop equation is 
 
   
  
a m
a a a a b
di t d t
e t R i L K
dt dt
 
The motor load equation is 
 
  
  
2
2
m m
m m
d t d
T t J D K
dtdt
 
Substituting the defined state variables one gets 
    3 3 2a a a be t R x L x K x 
Or 
   3 2 3
b a
a a a
K R u
x x x
L L L
 
1 2
x x 
And 
  3 2 2 1mK x Jx Dx Kx 
Or 
   2 1 2 3
mKK Dx x x x
J J J
 
Putting these equations in matrix form: 
 
 
  
     
     
         
           
     
 
1 1
2 2
3 3
0 1 0
0
0
1
0
m
b a
a
a a
x x
KK D
x x u
J J J
x x
K R
L
L L
 
 
 
 
 

 
  
1
2
3
1 0 0
x
y x
x
 
29. 
Let  2 1/q hK A K . Cross multiplying the transfer function, 
   3 2 2( 2 ) ( ) ( )h h p vs s s X s KX s 
Taking the inverse Laplace transform, 
22
p h p h p v
x x x Kx    
3-38 Chapter 3: Modeling in the Time Domain 
 
Defining the state variables, 
1
2
3
p
p
p
x x
x x
x x



 
Writing the state equations, 
1 2
2 3
2
3 2 3
2
h h v
x x
x x
x x x Kx 


   
 
In vector-matrix form, 
 
 
1 1
2 2
2
33
1
2
3
0 1 0 0
0 0 1 0
0 2
1 0 0
v
h h
x x
x x x
x Kx
x
y x
x
 
       
       
        
              
 
 

 
  
 
 
30. 
a. The output equation is 
 



 
 
    
 
 
 
0 1 0y l x l
x
x
Cx 
b. Substituting the given values into the A matrix and B and C vectors yields, 
 
 
 
 
 
 
 
0 1 0 0
29.8615 0 0 0
0 0 0 1
0.9401 0 0 0
A ; 
 
 

 
 
 
 
0
1.1574
0
0.4167
B 
  0 36 0 1 0.C ; D = 0 
 Solutions to Problems 3-39 
 
 
Using the following MATLAB M-file: 
 
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]; 
D=0; 
[numg, deng]=ss2tf(A, B, C, D, 1); 
display G(s)=Y(s)/U(s); 
G=tf(numg, deng) 
The computer output (in the “Command” window) is: 
 
G(s)=Y(s)/U(s) 
G = 
 3.6e-05 s^2 - 2.378e-15 s - 11.36 
 --------------------------------- 
 s^4 - 29.86 s^2 
Continuous-time transfer function. 
 
Note: As will be noted in chapter 6, a system with missing powers of ‘s’ and alternating signs 
of its denominator coefficients is unstable and needs to be stabilized. 
 
31. 
a. 
1 1
(1 )f s dT u Tv    
*
2 1
(1 )f u Tv T    
*
3 2
(1 )f u kT cv   
 
1
1 0 10 0
0
(1 ) | (1 )
f
d u v d u v
T
 

       

 
3-40 Chapter 3: Modeling in the Time Domain 
1
*
0
0
f
T



 
1
1 0 10 0
0
(1 ) | (1 )
f
u v u v
v
 

   

 
2
1 0 10 0
0
(1 ) | (1 )
f
u v u v
T
 

   

 
2
*
0
f
T


 

 
2
1 0 10 0
0
(1 ) | (1 )
f
u T u T
v
 

   

 
3
0
0
f
T



 
3
20*
0
(1 )
f
u k
T

 

 
3
0
f
c
v

 

 
1
0 0
1 0
f
T v
u




 
1
2 0
0
f
u



 
2
0 0
1 0
f
T v
u


 

 
2
2 0
0
f
u



 
3
1 0
0
f
u



 
*3
0
2 0
f
kT
u

 

 
 
 
 
 Solutions to Problems 3-41 
 
Then just by direct substitution. 
b. 
Substituting values one gets: 
 
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
 
 

 
  
 
32. 
a. The following basic equations characterize the relationships between the state, input, and output 
variables for the HEV common forward path of the figure: 
( ) ( )
a A c
u t K u t  
( ) ( ) ( ) ( ) ( )
A a a a a b A c b
L I R I t u t e t K u t k t        (1) 
( ) ( ) ( )
tot f c
J T t T t T t    , where 
tot m veh w
J J J J   , 
( ) ( )
t a
T t k I t  , ( ) ( )
f f
T t k t  
b. Given that the state variables are the motor armature current, Ia(t), and angular speed,  (t), we re-
write the above equations as: 
 
( ) ( ) ( )a b A
a a c
a a a
R k K
I I t t u t
L L L
      (2) 
 
1
( ) ( ) ( )t f
a c
tot tot tot
k k
I t t T t
J J J
    (3) 
In matrix form, the resulting state-space equations are: 
 
0
1
0
a b A
a a a caa
ct f
tottot tot
R k K
L L L uII
Tk k
JJ J

   
    
                       
  
 (4) 
3-42 Chapter 3: Modeling in the Time Domain 
 
1 0
0 1
a a
I I
 
     
     
     
 (5) 
 
33. 
The corresponding differential equation is: 
   6 60.0224 196 10 137.2 10 39( ) ( ) h th qt t th        
 
We select the state variables as 
1
x h 
2
x h 
So we can write 
 
1 2
x x 
  6 62 2 10.0024 196 10 132.2 10 39x x x q t
        
 In matrix form: 
 
 1 1 66
2 2
0 1 0
137.2 10 39
196 10 0.0024 1
x x
q t
x x


      
         
        
 
    1
2
1 0
x
h t y
x
 
   
 
 
Alternatively this can also be written as: 
 
 1 1 66
2 2
0 1 0
137.2 10
196 10 0.0024 1
x x
q t
x x


      
        
        
 
      1
2
39 39 1 0
x
h t y t
x
 
     
 
 
 
 
ONLINEFFIRS 11/25/2014 13:29:37 Page 1
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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
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