Solution Manual For Feedback Control Systems, 5th Edition

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CHAPTER 2
2.1 (a) An ideal voltage source model implies constant voltage regardless of the external connections. This
could result in current approaching infinity in the case of a short circuit and therefore would require infinite power
to supplied by the source. Similarly, an ideal current source implies constant current regardless of external
connections. This cannot possibly be true in the case of an open circuit.
(b) An appropriately sized resistor in series with the ideal voltage source, or in parallel with the ideal
current source, will provide a more accurate model of a physical power supply.
2.2 The operational amplifiers must be operating in their linear amplification range. This means that the output
voltage can never exceed the power supply voltage. Usually, the ideal op-amp assumptions are used to develop a
linear model.
2.3 (a)1 2 2 1 1
2 3 3
2 2 3 1 2
( ) ( ) , ( ) ( )
( ) 0
R R R I s V s V s R I s
R R R L s I s
   
  
     
     
    
 
1 2
12
2
1 2 2 1 2 1 1 2 1 3 2 3
2 2 3 1
2 2 3
1 1 2 1 1 2 1 3 2 3
( )
( )0
( )
( )
( )
R R V s
R V sR
I s R R R R R L s R R R R R R
R R R L s
V s R R
V s R R L s R R R R R R


 
     
  
    
(b) Replace R3 with1 2 3 2
3 2
2 2 3 2
2
1 1 2 1 2 3 1 2 1 2 1 2 2 1 2 3
( )
( )
( ) ( ) [ ( )( ) ]
R R R L s
R L s
V s R R L s
V s L L R R s R L L R R R R L s R R R


      
(c) 
1
2 3 2 3
2 0
1 2 1 1 2 1 3 2 3 1 2 1 3 2 3
10( ) ,
10( ) 10
limss s
V s s
s R R R Rsv R R L s R R R R R R R R R R R R

 
     

2.4 (a)   2 2 2 2
2 1
1
( ) ( ) ( ) ( ) ( ) ( ) 11 1 10, ( ) ( )
10 10 5 10 1010 10 5 ( ) 3
i i
V s V s V s V s V s V s ss sV s V s V s s
s
  
         
(b)   2 2 2 2 2
2 1
1
( ) ( ) ( ) ( ) ( ) ( ) ( ) 11 1 10, ( ) ( )
10 10 5 10 1010 10 5 ( ) 6 322
i i
V s V s V s V s V s V s V s ss sV s V s V s s
s
s
s
  
         
 
(c)2 2 2 2
0 0
10( 1) 10( 1)10 10( ) ( ) , lim ( ) ( ) ( ) , lim ( )
3 3( 3) (6 3)s s
s s
a V s sV s b V s sV s
s s s s 
 
   
 
2.5( ) ( )
( ) ( ) ( ) 1. ( ) ( ) 1 ( ) 1
( ) ( )
( ) 1
( ) ( ) ( ) ( ) ( )
f fo o
o i o i
i i i i
t
i
o f o i
R RV s V s
a v t v t b v t v t
V s R V s R
dv t
c v t R C d v t v d
dt RC
 

 
        
 

   
2.6 From the solution to Problem 2.5(a), The gain of the first op-amp stage is one. For the second stage,2
10 10 ( 12) 10 2.4
1 50
o s s
K K
v e e
K K

     
. For the third stage,2
2 20 4.8
1
o o s
K
v v e
K

   .
2.7 (a)10, 10 , let 10 , then 100 .f
f i i f
i
Z Z Z R K R K
Z
      
(b)1
10 , let 10 , then 10 .f
i
Z sC R K C F
Z s R

 
   
(c)

Let R3 = R4 = 100K. Gain = -1 for second stage. For first stage,2
1 12
1 1 1 2 1
1 1 1 1
1 2 1
10
( ) 10 , 10
1 1 10
Let 10 , then 100 , and 10 .
R s
RR s
G s R C R R
sR s
sC R C
R K R K C F


         
 
  
2.7 (d)
Let R3 = = R4 =100K. Gain = -1 for second stage. For first stage,1 2 12
1 1 1 2
1 1 1 1
1 1 2
11 10
( ) 10 ,
1 1 10
Let 10 , then 10 .
R CsC
G s R C R C
sR s
sC R C
R K C C F


          
 
  
(e)2 1 1 2
2 1 1 2 1 1
1 1
(1 ) (10 1) 10, 1, let 100 , 100 .f f
f
i i
Z RR R C s R R C s s R C R R K C F
Z R R R

   
             
 

2.7 (f)1 1
1 2 1 2 1
2 1 2 2
1 0.1
10 10, 10 Let 1 , then 10 , and 100 .f
i
Z C C R C R M C F C F
Z C R C s s C


    
               
  
2.8 
1
a 2 2 , 8 18, 2 2, and 8 18
1 2
8 1
1 2
1 2 2 2 28 1
( ) 1 28 1 18 18 2
8 1
2 2 1 2
18 1 8 1834 34
( ) ,
1 2 1 217 17
8 1 8 1
( ) [1, 2; 8,1], [2;18]
A B B A A B A B
A
B
A
b B
c A B
d M V

         
          
 
                                


   
 
   
 ( ) *AB inv M V 
2.9 (a) (b)1 2 2
1 2 1
Figure P2.9(a):
Figure P2.9(b):
and
a a b
a b
C G G E G HF
C G E G G F
HG G G G G
 
 
  1 2 2
Figure P2.9(c):
and
c d
c d
C G E G F
G G G G G H
 
  
2.10 (a) (b)1 2 1
1 2 2
Figure P2.10(a): ,
Figure P2.10(b): ,
and
a a b
a b
C G G E F G HE
C G E F G G E
HG G G G G
 
 
  2
1
Figure P2.10(c): ,
and
c d c
c d
C G G E F G E
G
G G H G H
 
  

2.11  4 2 3 , 3 , 1A B C B A C A B      a  (4)(1) ( 3)(1)(1)
b 1 ( 6 9 3) 1, 1
1 1
(3)(1)(4) (1)( 3)(3) (4)(1)(1) (4)(1)(3)( 1) (1)(1)[1 ( 6)]
3, 1
1 1 1 1 1
A
B C

         
   
       
(c)
>> M=[1 2 3;3 -1 0;1 -1 -1], V=[4;0;-1]
2.12 (a) Δ = 1 – (-G1-G2-G4-G3 G4 -G1G2G3 G4) + (G1G2+G1 G4+ G2 G4) – (-G1 G2 G4)
Δ = 1+ G1+G2+G4+G3 G4 + G1G2+G1 G4+ G2 G4 +G1 G2 G4+G1G2G3 G4
M1 = G1G2G3 G4, Δ1 = 1.
M2 = G3 G4, Δ2 = 1+ G1
M3 = G1, , Δ3 = 1+ G4
M4 = -1, Δ4 = 1+ G1+ G4+ G1G4
M5 = -G3G4G1, Δ5 = 11 2 3 4 3 4 4 1G G G G G G GD
R
  
 
(b)2 1
3 2 3 4
4
B R G B G A
C G R G G B G C
D B G C
   
  
 
2.13. (a)1 2 3 2 3 1 1 2 3 2 3
1 2 3 1 2 3 1 2 1 3 1 2 3 1 2 3 1 2 1 3
2 (1 ) 3 2( )
( ) 1 ( ) 1
G G G G G G G G G G GC s
R s G G G G G G G G G G G G G G G G G G G G
  
 
            
(b)1 2 1 3, (2 )

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2.151 1 1 3 2 2 1 2 3
3 2 1
1 1 2 2
3 2
1 [ 2 3 4 ] 2 6 1 6 9 4
2 , 1; , 1 3
5
( ) 6 9 4
s s s s s s s s s
M s M s s
s
T s s s s
        
  
            
      

   
2.16 (a)2 1 1 2 2 1 1 3 1 2 1 2 1 2 2
1 2 1 3 1 2 3 1
2 1 1 2 2
1 [ ]; , 1 ; , 1
( ) 1
G H G G H M G G G H M G G
G G G G G G G H
T s G H G G H
           
 
  
(b)2
3 2
6( 2) 2 4
8 ( 1) 24 84 72
( ) 9 20 246( 2) 2
1 8 ( 1)
s
s s s s s s
T s s s ss
s s s
   
          
   
   
(c)2
24 36
( ) 8 12
s
T s s s

  
2.17 (a)1 1 2 2 1 2( ); 0 ( )Mx B x x B x x Kx      
(b)2 1 1 2 2 1 1 2( ) 0; ( ) 0Mx K x B x x B x x K x      
(c)1 1 1 1 1 2 2 1 1 1 1 1( ) 0; ( ) ( ); ( )M x B x K x x K x x f t M x B x f t        
(d)1 1 1 1 2 1 1 2 1 2 2 2 1 2 1 2 2 1( ) ( ); ( ) ( )M x b x x k x k x x M x b x x k x x

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2
2 1 2 1 1 2
1 1 1
1 3 22
1 1 2 1 22
1
( ) ( ) ( ) ( ); ( ) ( ) ( ) 0
( )
0 ( )
( ) ; ( ) ( )
Ms Bs K X s BsX s F s BsX s Bs K X s
F s Bs
Bs K X s Bs K
X s F s MBs MK s B K K s K KMs Bs K Bs
Bs Bs K
       

 
       
 (c)2 1
1 1 1 2
1 1
( ) 1 2
( ) ( ) ( ); ( ) ( 4)
X s
M s X s cX s F s F s M s B s s s
   
 
(d)2
1 1 1 2 1 1 2 2
2
1 2 1 2 1 2 2
1 2
2
2 1 2
1 2
1 1 1 2 1 2
2
1 2 2 1 2
1 1 2
4 3
1 2 1 1 2 1 2 2
( ) ( ) ( ) ( ) 0
( ) ( ) ( ) ( ) ( );
0 ( )
( ) ( )
( ) ;
( ) ( )
( ) ( )
( )
( ) ( ) [
m s b s k k X s b s k X s
b s k X s m s b s k X s F s
b s k
F s m s b s k
X s m s b s k k b s k
b s k m s b s k
X s b s k
F s m m s b m m s k m

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(c)2
2
2
2 4 3 22
2
4 2 7 0
2 ( ) ( ) 4 2 7
( ) ; ( ) 24 24 84 39 594 2 7 2
2 6 3 9
s s
F s X s s s
X s F s s s s ss s
s s
 
  
       
  
2.22 (a)1 1 2 1 1 1 2 2 2 1 2 1( ) ( ); 0 ( )t J B B J B
    
        
(b)2
11 1 2 1
2
21 2 1
2
1 1 2
1 2 1
2 3 22
1 2 1 1 2 1 2 2 1 21 1 2 1
2
1 2 1
( ) ( )( )
( ) 0
( ) ( )
0 ( )
( ) ; ( ) ( )( )
s sJ s B B s B s
sB s J s B s
J s B B s s
B s s B
s s J J s J B J B J B s B B sJ s B B s B s
B s J s B s
        
       

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(c)1 2
1 2 2 3
( )
( ) 1
m
a
s G G
E s G G H KG G
   
(d)1 2 3
1 2 2 3
( )
( ) 1
L
a
G G Gs
E s G G H KG G
   
(e) From (c) and (d), or by inspection,1
3
mL L
m a a
G
E E

  
  
  
2.26 From Example 2.11 write analogous differential equation models.1
1 1 1 2 1 1 2 1 1 2 1 2
1
2
2 2 2 1 1 2 1 2 2 1 2 1 2
1 2
1
( ) ( ) 0 ( ) 0
1 1
( ) ( ) ( )
di
M x B x x K x x L R i i i dt i dt
dt C
di
M x B x x K x x f L R i i i dt i dt i dt v
dt C C
            
             
 
  
2.27 (a)1 2
1 2 1 1 2 2
( ) . Let and .
1( ) , let , , , and = .
My B B y Ky f i y v f
di
M B B i K idt v L M R B R B K
Cdt
     
      
(b)1 1 2 2 1 1 2( ) 0; ( ) 0Mx B x x B x x K x     1 1 2 2 1Let , , , , and .i x i x K L M R B
C
    
(c)

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1
1 1 2 2 1 1 2 1
2
2 1 1 2 2 1 2
1
1
( ) 0 ( ) 0
1
( ) 0 ( ) 0
di
Mx B x x K x L R i i i dt
dt C
B x x K x R i i i dt
C
        
      

2.28 (a) From Problem 2.17(c) solution,1 1 1 1 1 2 2 1 1 1 1 1( ) 0; ( ) ( ); ( )M x B x K x x K x x f t M x B x f t        
.1 1 1 1 1, , , ( ), and LL M R B i x v f t v M x    
(b) From Problem 2.17(d) solution, let1 1 2 2, , .i x i x v F   1
1 1 1 1 2 1 2 1 1 2 2 2 1 1 1 2 1 1 2 1 2
2
2 2 1 2 1 2 2 1 2 1 2 1 2 2 1
( ) ( ) 0 ( ) 0
( ) ( ) ( )
di
M x b x x k k x b x k x M b i i k i dt k i i dt
dt
di
M x b x x k x x F M b i i k i dt i dt v
dt
             
            
 
 
Let1 1 2 2 1 1 1 2 2, , , 1 , 1L M L M b R C k C k    
2.29 (a)

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(b)' ' ' '
1 1 1 3 1 3 1 3 1 3 1
2 2 2
'1 1 1 2
1 1 1 3 3
2 2 2 1
( ); 0 ( )
; ; ;
K J B K
r r r r
K K J J B B
r r r r
  
   
 
     
     
         
     
2.30 22
1 2 1 0.01 100 10z n z z K     
2.31 (a)1 100 (rpm)
100
g mr r  . The encoder produces 100 pulses per revolution of the gearbox
output shaft. Therefore, the resolution is360 3.6
100  per pulse.
(b) 22 6 2
0.01 (Kg ) 0.01 10 (Kg )L m LJ m J J m
    
2.32 (a)  
  
2
38
2 21 2 1( ) 1 1 38 0.475
19( ) 20 20 4 44 40 ( 1)( 10)1 2 21 2 1
L
a
s ss
E s s s s s s s s
s s
 
              
   
(b)3 2
0.475
20( )
( ) 9.5( )( 1)( 10)
20( )0.475( ) 11 (10 9.5 ) 9.51 ( 1)( 10)
P D
L P D
P DC P D
K s K
s K s Ks s s
K s Ks s s K s K
s s s

  
 
      
(c)3 2
( ) 20( ) 20( ) ( 1)( 10)
20( )0.475( ) 11 (10 9.5 ) 9.51 ( 1)( 10)
a P D P D
P DC P D
E s K s K K s K s s s
K s Ks s s K s K
s s s
   
 
      
(d)( )( ) ( ) 0.475
( ) ( ) ( ) ( 1)(

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8../. lc) F~):: ~:li-z.5")e-stdk : J:'-$~ = e-si/O::l
_ e-2.~5 -Z.S - s 2. ~
--5
8.2. (b) Fb) = -5:s lC) FlS): 6,5
~,+ ) t~2.+'1)l
.) iFt.s ) = 7l ~ + t:J. S-) = 11:5+3.~
(:Sf~.5)~f-l3)~ c jf- + ~ -f. 9.zs
I
llii) ::5 ~~ UI t +3/JD) = 9..
1
.33cs« J.f.:l-2.54 iM.,.Ij;t
FL~) s: ~.33.5 2.5'lJJ) ~ ~.33~ -J~
;52.+ lip. :52+11, :5'+11.
if) jU) ::: (, e-lZ ~ t.t-Jf51J) =~lJle-~ht. t-,.lfe-~ 1
.4. rL~) = .LJ.1.1I LJ) _ Jj.z'1 1.~fj2) = -"1.21/:5 - ~.2,LJ
{~tl.l)l.~1 (~.,..~)LflJ ~l. + J.j5 +5

Loading page 13...

(b Fe.•..)- J J -I ..L-
v - st lS+l) = :52. + T + s.rl
~ l;6} == ;t -/ + e-:C v
tc: J:I.-j):: Z~+J :. Z~+, =- Z{~rJ) 4- - J
jl.fls-l-IO l~+Il2.+3~ (.'S+IJ4-31. (~+j)~+j'~
~l.i) =. ;, e-C e.&.d 3;/; - ~ e-~~ 3,t
8.3. (A) Ft..~)=: 5 :. 2,5' .•.. -5 .f- z.s"
j(~H)(~+,) ~ ~ s+a
~(1):: z.S- - 5 e-z T l,S e-2~
I
8 .LJ F~) = s+5 = k., + A~.-,. c ) I. .l:st2)1.+ £3)2- :s.J.. z.-j 5 ~+l. TJ'.3
kJ= -2+i3r5 =~_ ,_I :: (J,7tJ7/9Sb
-2t'3+Z+j3 1- J 1-
:. Pl.:tJ: ;.7tJ7e-j~e.l-2.:lj3)i+ tJ.7tJ7eilfe..t-Z:;;T3)£
2:1:[ &.i-lf) . l3i -1r')]
( ) ~!:t) = CJ, 7~ 7 e- e.J f e.-J 7i"
= /.LJJ~ e.-2.Z elM (3~ -.lf5°) I
) F~l)h1. la.).J k 1= o. TIJ1 1_~5°
.; +' it) ;:2. t.~.71)7) ~ f3i _~5D) = /, .l.JJ~M:J. (3-i-4!t'j
-1-

Loading page 14...

8. ,. to: tLi) =.If e-u;i;'3) ; dfje:Ji = .Jfe-lli--8)t.-l.J= _2e-lLi-3)
:.ge tf. e-2~ : - y t.lJ1J3.J.j) e-2 t ::-3.) 2.27/-/ e -2t
.: J'[dr/dt].: - 32.21
~+2..
(b d[~~/di] = ~ r{:5)- ~~t1+) = 5J'[~e(,e-2t] -.lJe~
= 1.;4.)3 -0"13: - ~l2'7
j+2 ~+2.
(C +' it-) :: -<Ie - 2li -~)uti-3)
alo/di = _ge-2.'i-.3)U.~-3) + .1Je-2ti:.-3)}~Li-3)
1~:'3
- 1: [d~/d:t] ~ _9e:3~ r#e-3$ :: 216 e-3s
j+~ ~+2
(ell ;[[df/dt] ::::sFl:5) -t~) =~rl:5) :: J.j~e.-3~
;:5+2.

Loading page 15...

B.7,
B, g " - rUL ~ H~ ~ tb)j Hr.t- Juu",. -'/::,'
~fk.f.J~~~&~
A r +it.. t t.tJ ~ lhL
1~).LtM- t'U} = ~ sF'ts) = ~
.t, -';let> ~-YO
B,tt. to: S2.A'lS) T 55 XlS) T l.JXl~) = J~/S
X l6) = 10 == 2'.5 + -1b/3 +- )M2
~(5+J)&t!l) S s+1 :5+4
. 'X-l:t):: 2.5 -~ e-t T..2 e-llt J t ~~. ~ ~
lb) l.A'l~) - S X~)- "itA) r 5[~X~) -~ltJ)] + J..j -= )&/5
• X is) = 10 +- Sr- L~) '*" "X u>} + 5XlO)
s z s T I}l.S+lf) cs + 1) (~+Li)
_ s 2-J- 1,6+/0 _ Z.J5 -513 y~
- ~ (~+n(s+1.J) - s-+ ~t-I t- ~""LJ
" 'XU)= 2.5 --!e-.i:+ 1 e-~.t
1t {k)} = 2.5"- 5"13 +- Y" =Lt=a
1l Lt:1}t:tf [~ e-t: - ~e-JJ*J-t:.;1-

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Solution Manual For Feedback Control Systems, 5th Edition

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