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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 1 2 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 s R 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 R 3 with 1 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 lim ss s V s s s R R R R s v R R L s R R R R R R R R R R R R          

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    2.4 (a)     2 2 2 2 2 1 1 ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 0, ( ) ( ) 10 10 5 10 10 10 10 5 ( ) 3 i i V s V s V s V s V s V s s s s V s V s V s s s              (b)     2 2 2 2 2 2 1 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 0, ( ) ( ) 10 10 5 10 10 10 10 5 ( ) 6 3 2 2 i i V s V s V s V s V s V s V s s s s V 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 f o o o i o i i i i i t i o f o i R R V 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)

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    Let R 3 = R 4 = 100K. Gain = -1 for second stage. For first stage, 2 1 1 2 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 R R s G s R C R R s R s sC R C R K R K C F                      2.7 (d) Let R 3 = = R 4 = 100K. Gain = -1 for second stage. For first stage, 1 2 1 2 1 1 1 2 1 1 1 1 1 1 2 1 1 10 ( ) 10 , 1 1 10 Let 10 , then 10 . R C sC G s R C R C s R 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 R R R C s R R C s s R C R R K C F Z R R R                       

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    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 2 8 1 ( ) 1 2 8 1 18 18 2 8 1 2 2 1 2 18 1 8 18 34 34 ( ) , 1 2 1 2 17 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 H G 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 H G 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     

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    2.11   4 2 3 , 3 , 1 A 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 – (-G 1 -G 2 -G 4 -G 3 G 4 - G 1 G 2 G 3 G 4 ) + ( G 1 G 2 + G 1 G 4 + G 2 G 4 ) – (- G 1 G 2 G 4 ) Δ = 1+ G 1 +G 2 +G 4 +G 3 G 4 + G 1 G 2 + G 1 G 4 + G 2 G 4 + G 1 G 2 G 4 + G 1 G 2 G 3 G 4 M 1 = G 1 G 2 G 3 G 4, Δ 1 = 1. M 2 = G 3 G 4, Δ 2 = 1+ G 1 M 3 = G 1, , Δ 3 = 1+ G 4 M 4 = -1 , Δ 4 = 1+ G 1 + G 4 + G 1 G 4 M 5 = -G 3 G 4 G 1 , Δ 5 = 1 1 2 3 4 3 4 4 1 G G G G G G G D 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 G C 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 ) , A R G A C B G R G A B C C G B         1 1 2 2 2 3 1 2 3 2 3 1 1 2 3 1 2 3 1 2 1 3 1 2 2 3 1 0 1 1 2 0 0 3 2 1 0 1 1 1 1 0 1 G G G G G G G G G G G C G G G G G G G G G G G G G G G                  2.14 (a) 1 1 1 2 1 1 2 1 1 1 1 2 2 1 1 2 2 1 [1 2 8 3] [2 6 ] 4 17 2 3, 1 2 8 1 3(1 2 ) 8 3 14 ( ) 4 17 2 s s s s s s s M s M s s s s s T s s s                                     (b) 1 1 1 ( ) ( ) 2 ( ) ( ) ; ( ) ; ( ) 4 ( ) 3 ( ) 1 1 R s C s A s A s B s C s s B s A s s s          

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    2.15 1 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 24 6( 2) 2 1 8 ( 1) s s s s s s s T s s s s s 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; ( ) 0 Mx 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 F           2.18 (a) 2 1 2 1 2 2 1 1 2 2 1 3 2 ( ) ( ) ( ) ( ); ( ) ( ) ( ) 0 ( ) ( ) ( ) 0 ; ( ) ( ) 0 ( ) ( ) Ms Bs X s BsX s F s BsX s Bs K X s F s Bs X s F s Bs K Ms Bs Bs X s X s Bs Bs K Ms Bs Bs Bs Bs K X s Bs K F s MBs MKs BKs                                          (b)

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    2 2 1 2 1 1 2 1 1 1 1 3 2 2 1 1 2 1 2 2 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 K Ms 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 k                                 2 2 1 2 1 1 2 ( )] m m s b k s k    2.19 (a) 1 2 ( ) My B B y Ky f     (b) 2 1 2 2 1 2 ( ) 1 [ ( ) ] ( ) ( ); ( ) ( ) Y s Ms B B s K Y s F s F s Ms B B s K         2.20 (a) 2 1 1 2 1 5 1 1 2 2 1 2 1 2 3 4 2 1 1 ( ) ( ) ( ) 1 ( ) ( ) ( ) ( ) ( ) M s Bs K X s X s G X s Bs K M s Bs K K X s F s X s G F s G X s Bs K Bs K                (b) See Example 2.11. 2.21 (a) 1 1 1 1 2 2 2 2 2 1 4 2 5 2( ) 0; 6 3 7 2( ) x x x x x x x x x x f           (b)     2 2 1 2 2 1 2 1 1 4 3 2 2 2 4 2 7 ( ) 2 ( ) 0; 6 3 9 ( ) 2 ( ) ( ) 0 2 ( ) 6 3 9 ( ) 2 ( ) ; ( ) 24 24 84 39 59 4 2 7 2 2 6 3 9 s s X s X s s s X s X s F s F s s s X s X s F s s s s s s s s s                       

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    (c) 2 2 2 2 4 3 2 2 2 4 2 7 0 2 ( ) ( ) 4 2 7 ( ) ; ( ) 24 24 84 39 59 4 2 7 2 2 6 3 9 s s F s X s s s X s F s s s s s s 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 1 1 1 2 1 2 2 1 2 1 2 1 1 2 1 2 1 2 3 2 2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2 1 2 1 ( ) ( ) ( ) ( ) 0 ( ) ( ) 0 ( ) ( ) ; ( ) ( ) ( ) s s J s B B s B s s B 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 s J s B B s B s B s J s B s                                              2.23 (a) 1 2 ( ) J B B       (b) 2 1 2 ( ) 1 ( ) ( ) s s Js B B s      2.24 (a) From (2-51): Note that ( ) ( ). s s s    1 4 7 4 2 2 ( ) 0.005 714285714.2857 ( ) 7 10 + 1.289 (s+4.99 10 ) (s+369) 3.519 10 10 a s G E s s s              (b) 4 0 6 714285714.2857 ( ) ( ); lim ( ) 232.7535(rad/s) 369(4.99 10 ) ss s s G s s s s                 2.25 (a) 2 2 2 2 ; ; ; ( ); 0 ( ) m a m m a m a m m a m m L L m m m L L L m d di e K e R i L e K i dt dt d d d d J B K J B K dt dt dt dt                          (b) 2 2 2 2 ( ) ( ); ( ) ( ) ( ) ( ) ( ) ( ) ( ) ; ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ( )) ( ) ( ) 0 ( ) ( ) ( ( ) ( )) ( ) m m m a m a m a m a m a m m a L m m m m m L m m m m L L L L m L L E s K s s E s E s E s R I s L sI s E s I s L s R s K I s s K s s J s s B s s K s s s J s B s K K s J s s Bs s K s s s J s Bs                                           K 

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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 G s E s G G H KG G     (e) From (c) and (d), or by inspection, 1 3 m L 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 C dt               (b) 1 1 2 2 1 1 2 ( ) 0; ( ) 0 Mx B x x B x x K x       1 1 2 2 1 Let , , , , 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 L L M R B i x v f t v M x      (b) From Problem 2.17(d) solution, let 1 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                                  Let 1 1 2 2 1 1 1 2 2 , , , 1 , 1 L M L M b R C k C k      2.29 (a) 2 2 3 3 2 3 3 1 2 2 1 2 1 2 1 2 2 1 1 2 1 1 3 1 1 3 1 2 2 1 1 3 1 3 3 2 ( ); 0 ( ) , 0 ( ) K K B J r r r r r r r r K K r r r r r K B J r                                                        

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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   2 2 1 2 1 0.01 100 10 z n z z K       2.31 (a) 1 100 (rpm) 100 g m r r   . The encoder produces 100 pulses per revolution of the gearbox output shaft. Therefore, the resolution is 360 3.6 100  per pulse. (b)   2 2 6 2 0.01 (Kg ) 0.01 10 (Kg ) L m L J 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 s s 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 ) 9.5 1 ( 1)( 10) P D L P D P D C P D K s K s K s K s s s K s K s s s K s K s s s                 (c) 3 2 ( ) 20( ) 20( ) ( 1)( 10) 20( )0.475 ( ) 11 (10 9.5 ) 9.5 1 ( 1)( 10) a P D P D P D C P D E s K s K K s K s s s K s K s s s K s K s s s                (d) ( ) ( ) ( ) 0.475 ( ) ( ) ( ) ( 1)( 10) C L L a C a s s s E s s E s s s s        

  13. Page 13
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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 .) iF t.s ) = 7l ~ + t:J. S-) = 11:5 + 3.~ (:Sf~.5)~f-l3)~ c jf- + ~ -f. 9.zs I llii) :: 5 ~~ UI t +3/J D ) = 9.. 1 .33 cs« J.f.:l-2.54 iM.,.Ij;t FL~) s: ~.33.5 2.5'lJJ) ~ ~.33~ -J~ ;52.+ lip. :5 2 + 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

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    (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 k J = -2+i 3r5 =~_ ,_I :: (J,7tJ7/9S b -2t'3+Z+j3 1- J 1- :. Pl.:tJ: ;.7tJ7e-j~e.l-2.:lj3)i+ tJ.7tJ7e ilf e.. 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. TIJ 1 1_~5° .; +' it) ;: 2. t.~. 71) 7) ~ f3i _~5D) = /, .l.JJ~ M:J. (3-i-4!t'j -1-

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    8. ,. to: tLi) =.If e-u ;i;'3) ; dfje:Ji = .Jfe- lli --8)t.-l.J= _2e- lLi - 3 ) :.g e 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.

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    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" = L t=a 1l Lt:1}t:tf [~ e-t: - ~e-JJ*J-t:.; 1-

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