S E V E N. Steady-State Errors SOLUTIONS TO CASE STUDIES CHALLENGES

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1 S E V E N Steady-State Error SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Steady-State Error Deign via Gain a. G( (50)(.3). Syte i Type. Step input: e( ) 0; Rap input: e( ) v ; Parabolic input: e( ). 50 x.3.59 v 0.. Therefore,.95. Now tet the cloed-loop tranfer function, b T( , for tability. Uing Routh-Hurwitz, the yte i table Video Laer Dic Recorder: Steady-State Error Deign via Gain a. The input, 5t, tranfor into 30/ 3. e( ) 30/ a a 0.* * 3 6x Therefore: e( ) 30/ a 5x0-3. Therefore b. Uing 3 0 6, G( x0 5 ( 600) ( x0 4. Therefore, T( ) x0 5 ( 600) 3 x0 4 x0 5.x0 8. Making a Routh table, 30 6x0 3 3 Copyright 0 by John Wiley & Son, Inc.

2 7- Chapter 7: Steady-State Error we ee that the yte i table. c. Progra: nug00000*[ 600]; dengpoly([ ]); Gtf(nug,deng); 'T(' Tfeedback(G,) polepole(t) Coputer repone: an T( 3 x0 5 x0 4.x Tranfer function: e ^ ^ e008 pole.0e004 * i I ANSWERS TO REVIEW QUESTIONS. Nonlinear, yte configuration. Infinite 3. Step(poition), rap(velocity), parabola(acceleration) 4. Step(poition)-, rap(velocity)-, parabola(acceleration)-3 5. Decreae the teady-tate error 6. Static error coefficient i uch greater than unity. 7. They are exact reciprocal. 8. A tet input of a tep i ued; the yte ha no integration in the forward path; the error for a tep input i / The nuber of pure integration in the forward path Copyright 0 by John Wiley & Son, Inc.

3 Solution to Proble Type 0 ince there are no pole at the origin. Miniize their effect. If each tranfer function ha no pure integration, then the diturbance i iniized by decreaing the plant gain and increaing the controller gain. If any function ha an integration then there i no control over it effect through gain adjutent. 3. No 4. A unity feedback i created by ubtracting one fro H(. G( with H(- a feedback for an equivalent forward path tranfer function with unity feedback. 5. The fractional change in a function caued by a fractional change in a paraeter 6. Final value theore and input ubtitution ethod SOLUTIONS TO PROBLEMS. where e ( ) li E( ) 0 R( li 0 G( ) 450( )( 8)( 5) G( ( 38)(. 8) For tep, e ( ) 0. For 37tu(t), R( 37. Thu, e ( ) 6.075x0-. For parabolic input, e( ).. a. Fro the figure e r c 5 3 b. Since the yte i linear, and becaue the original input wa r ( t).5tu( t), the new teady.5 tate error i e e ( ) li E( ) 0 R( li 0 G( ) 3 (60/ ) li ( 3)( 4)( 8) ( 6)( 7) Copyright 0 by John Wiley & Son, Inc.

4 7-4 Chapter 7: Steady-State Error 4. Reduce the yte to an equivalent unity feedback yte by firt oving / to the left pat the uing junction. Thi ove create a forward path coniting of a parallel pair, with a feedback loop coniting of G( and H( 7. Thu, 3 G e ( ( /( 3) 4 /( 3) Hence, the yte i Type and the teady-tate error are a follow: ( ) ( 7) in cacade Steady-tate error for 5u(t) 0. Steady-tate error for 5tu(t) 5 v 5 / Steady-tate error for 5t u(t) Syte i type 0. p For 30u(t), e( ). For 70tu(t), e( ) For 8t u(t), e( ) 4 R ( ) 50 / S E () G ( ) 0( S 4)( S 6)( S )( S 3) 3 S ( S 7)( S 4)( S 9) p Thu, 50 e( ) li E( (0)(4)(6)()(3) (7)(4)(9) 7. Therefore, e. ( ) li 0 E( li 0 de dt R( G( E ( 6 4 li 0 00( )( ) ( 0)( 3) 9 0. Copyright 0 by John Wiley & Son, Inc.

5 Solution to Proble (3)(6)(33) e( ) ; p Therefore, e( ) (65)(75)(9) p For 70u(t), e 4 p 5 ; For 70tu(t), e, ince the yte i Type a. The cloed-loop tranfer function i, 5000 T( fro which, ω n 5000 and ζω n 75. Thu, ζ 0.53 and %OS e ζπ / ζ x00 4.0%. b. T 4 ζω n econd. 75 / c. Since yte i Type, e for 5u(t) i zero. d. Since v i , e e. e, ince yte i Type. 5 v v 00500(5)(4)(3) li G( (7)( α)(33) Thu, α xx4x6 li G( 0, x7 a. Therefore, ( )( ) a. G e ( 5( 3) ( )( ) Therefore, p /3; v 0; and a 0. Copyright 0 by John Wiley & Son, Inc.

6 7-6 Chapter 7: Steady-State Error b. For 50u(t), e( ) c. Type 0 50 p 37.5; For 50tu(t), e( ) ; For 50t u(t), e( ) R( E(. Thu, e( ) li G( E( 4 li ( 8 3)( 8) 3 ( 6)( 3) Collaping the inner loop and ultiplying by 000/ yield the equivalent forward-path tranfer function a, Hence, the yte i Type. G e ( 0 5 ( ) ( ) 6. The tranfer function fro coand input to error ignal can be found uing Maon rule or any other ethod: 0 G E( ( 3) ( 3) 0G R( 0 ( 3) 0G G ( 3) Letting R( and by the final value theore: G ( e Li E( Li 0 0 G ( a. If G i type 0, it i required that G ( 0 b. If G i type, it i required that G ( ut be type 0 c. If G i type, it i required that G ( ut be type 7. e( ) li E( li 0 0 G ( ) For Type 0, tep input: R( R()., and e( ) li 0 0 G ( ) Copyright 0 by John Wiley & Son, Inc.

7 Solution to Proble 7-7 For Type 0, rap input: R(, and e( ) li 0 G ( ) li G ( ) p 0 For Type 0, parabolic input: R( 3, and e( )li 0 G() For Type, tep input: R(, and e( ) li 0 0 G ( ) For Type, rap input: R(, and e( )li 0 0 G ( ) For Type, parabolic input: R( 3, and e( )li 0 G() v For Type, tep input: R(, and e( ) li 0 0 G ( ) For Type, rap input: R(, and e( )li 0 0 G ( ) For Type, parabolic input: R( 3, and e( )li 0 0 G() 8. a. /0 7 e( ) 0.0; where v x x Thu, v b. v 0. c. The iniu error will occur for the axiu gain before intability. Uing the Routh-Hurwitz ( 7) Criterion along with T() : ( ) Copyright 0 by John Wiley & Son, Inc.

8 7-8 Chapter 7: Steady-State Error For Stability < < < > Thu, for tability and iniu error 55.. Thu, /0 /0 e( ) v e( ) Hence, a a/6 a v Find the equivalent G( for a unity feedback yte. G( 00 v 00 / 0.0; fro which 0, a ; e( ) Hence, a v and 58 x x () 0 (). Thu, e( ). a. e( ) 0 v But, v ,000. Hence, 0,000. For finite error for a rap input, n. 0000( 3 30) b. p li G( li 0 0 ( 5) 0000( 3 30) v li G( li ( 5) 0000( 3 30) a li G( li ( 5) Copyright 0 by John Wiley & Son, Inc.

9 Solution to Proble a. Type 0 R( b. E( G( / e( ) li E( li 0 0 ( 6 6) Thu, ( 5) ( 3). c. e( ), ince the yte i Type e( ) 0.4. Thu, /88. e( ) v 6 p Thu,. The yte i table for 0 < < 000. Since the axiu v i v iniu teady-tate error i v To eet teady-tate error characteritic: , the 30 Therefore, α 9β. (α) To eet the tranient requireent: Since T( (β) (β α), ω n 0 β α ; ζω n 0 β. Solving for β, β ±. For β,.6 and α An alternate olution i β -, 5.6, and α.74. a. Syte Type b. Aue G( G( But, T( G( (α). Therefore, e( ) v α. Since ω n 0, 00, and α. Hence, G( 00 (). /α 0.0, or α 00. Copyright 0 by John Wiley & Son, Inc.

10 7-0 Chapter 7: Steady-State Error 9. c. ζω n α. Thu, ζ 0. G( T( G( Alo, e( ) v (α) (β) α. Hence, β, α ω n ( ). β 0.. Therefore, β 0.α 0.,.8, and α.. α 30. G( Syte Type. T( G( a. Fro G(, v a 0. For % overhoot, ζ Therefore, ζω n a, and ω n. Hence, a.. Alo, a. Solving iultaneouly, 0.5 x 0 4, and a.38 x a. For 0% overhoot, ζ Alo, v 000 a. Since T( a, ζω n a, and ω n. Hence, a 0.9. Solving for a and, 83,744, and a b. For 0% overhoot, ζ Alo, v 0.0. Thu, v 00 a. Since T( a, ζω n a, and ω n. Hence, a.8. Solving for a and, 397 and a a. For the inner loop: () G ( () G e ( (3) G ( ( ) G e ( T( G e ( b. Fro G e (, yte i Type. c. Since yte i Type, e 0 d. ; Fro G e (, v li 0 G e ( 3. Therefore, e 5 v 5. Copyright 0 by John Wiley & Son, Inc.

11 Solution to Proble 7- e. Pole of T( , -.366, ± j0.776, Therefore, yte i untable and 33. reult of (c) and (d) are eaningle a. For the inner loop: 0 ()(3)(4) G ( 0 ()(3)(4) 0 ( ) G e ( 0 ( ) 34. G e ( T( G e ( b. Fro G e (, yte i Type. c. Since yte i Type, e 0 d. Fro G e (, v li G e ( Therefore, e 5 8. v e. Pole of T( , ± j.756, -. Therefore, yte i table and reult of part c and d are valid. Progra: nug[ 9];dengpoly([ ]); 'G(' Gtf(nug,deng) nug6*poly([-9-7]);dengpoly([ ]); 'G(' Gtf(nug,deng) nuh3;denh; 'H(' Htf(nuh,denh) nuh;denh[ 7]; 'H(' Htf(nuh,denh) %Cloe loop with H and for G3 'G3(G(/(G(H(' G3feedback(G,H) %For G4GG3 'G4(G(G3(' G4erie(G,G3) %For GeG4/G4H 'Ge(G4(/(G4(H()' Gefeedback(G4,H) %For T(Ge(/(Ge() to tet tability 'T(Ge(/(Ge()' Tfeedback(Ge,) 'Pole of T(' pole(t) %Coputer repone how that yte i table. Now find error pec. pdcgain(ge) 'Ge(' Getf([ 0],)*Ge; 'Ge(' Copyright 0 by John Wiley & Son, Inc.

12 7- Chapter 7: Steady-State Error Geinreal(Ge) vdcgain(ge) '^Ge(' Getf([ 0],)*Ge; '^Ge(' Geinreal(Ge) adcgain(ge) etep00/(p) erap00/v eparabola00/a Coputer repone: an G( Tranfer function: ^4 3 ^3 34 ^ 008 an G( Tranfer function: 6 ^ ^3 ^ an H( Tranfer function: 3 an H( Tranfer function: an G3(G(/(G(H( Tranfer function: 6 ^ ^3 90 ^ Solution to Proble 7-3 an G4(G(G3( Tranfer function: 6 ^3 0 ^ ^7 ^6 808 ^ ^4 3.6e006 ^3.777e007 ^ 3.835e007 an Ge(G4(/(G4(H() Tranfer function: 6 ^4 5 ^3 379 ^ ^8 9 ^7 336 ^ ^ e006 ^ e007 ^3.68e008 ^.685e an T(Ge(/(Ge() Tranfer function: 6 ^4 5 ^3 379 ^ ^8 9 ^7 336 ^ ^ e006 ^ e007 ^3.68e008 ^.685e an Pole of T( an Copyright 0 by John Wiley & Son, Inc.

13 Solution to Proble p 7 an Ge( an Ge( Tranfer function: 6 ^5 5 ^4 379 ^3.45e004 ^ 5.783e ^8 9 ^7.336e004 ^6 3.56e005 ^ e006 ^ e007 ^3.68e008 ^.685e v 0 an ^Ge( an ^Ge( Tranfer function: 6 ^6 5 ^5 379 ^4.45e004 ^ e004 ^ ^8 9 ^7.336e004 ^6 3.56e005 ^ e006 ^ e007 ^3.68e008 ^.685e Solution to Proble 7-5 a 0 etep.5000 erap Inf eparabola Inf The equivalent forward tranfer function i, G( (0 f ). Alo, T( G( G( 0 (0 f )0. Fro the proble tateent, v 0 0 f 0. Alo, ζω n 0 f (0.5) 0 0. Solving for and f iultaneouly, 0 and f We calculate the Velocity Error Contant, Copyright 0 by John Wiley & Son, Inc.

14 7-4 Chapter 7: Steady-State Error v 3 ( ) ( 7.895)( ) Li G( P( Li ( )( 4 8) For a unit rap input the teady tate error i e.. The input lope i v 37. a. Fro the point of view of e(t) the above block diagra i equivalent to the original. In thi unity α feedback block diagra the open loop traniion i G( α, the yte i β β type 0. b. The poition error contant i P β e. α P α β β G(0) α β. The teady tate error i Copyright 0 by John Wiley & Son, Inc.

15 Solution to Proble R( D(G ( e( ) li 0 G (G ( R( D(. Hence, 0, where G ( 5 and G e( ) li Fro the proble tateent, 39. Error due only to diturbance: Rearranging the block diagra to how D( a the input, Therefore, -E( D( (4) () (3)(4) D( (3) (3)(4) () For D(, e 3 D( ) li E( -. 0 Error due only to input: e R ( ) v 6 6. Deign: 3 e D ( ) , or 5, e R ( ) 0.003, or 0.06 Copyright 0 by John Wiley & Son, Inc.

16 7-6 Chapter 7: Steady-State Error 40. a. The open loop traniion i 35 G ( P(, o P LiG( P( 0 tep input e Since the input i threefold that we have that r e b. r 3 (0.054) P For a unit Copyright 0 by John Wiley & Son, Inc.

17 Solution to Proble E( 7 c. The tranfer function fro diturbance to error ignal i D( Uing the final value theore e ( ) d Li E Li d. Copyright 0 by John Wiley & Son, Inc.

18 7-8 Chapter 7: Steady-State Error e. e e e f. tot r d Copyright 0 by John Wiley & Son, Inc.

19 Solution to Proble C( R( G (G ( G (H ( ; E ( a R( R(G e a ( ) li ( 0 G (H ( G ( G (H ( Syte : Foring a unity-feedback path, the equivalent unity feedback yte ha a forward tranfer function of G e ( 0( 0) ( ) 0( 0)( 3) ( ) 0( 0) a. Type 0 Syte; b. p p li 0 G e ( /3; c. tep input; d. e( ) p 3/4; R( e. e a tep ( ) li 0 G(H( li 0 0( 0)( 4) 0. ( ) Syte : Foring a unity-feedback path, the equivalent unity feedback yte ha a forward tranfer function of Copyright 0 by John Wiley & Son, Inc.

20 7-0 Chapter 7: Steady-State Error 43. 0( 0) ( ) G e ( 0( 0) ( ) 0( 0) ( 0) a. Type Syte; b. v li 0 G e ( 0.98 ; c. rap input; d. e( ) v.0; e. e a rap ( ) li 0 R( G(H( li 0 Syte. Puh 5 to the right pat the uing junction: 0( 0)( ) ( ) 50. R( - 5(4) ( 3)( 7) ( 5)( 8) C( Produce a unity-feedback yte: R( - - 5( 5(4) ( 3)( 7) ( 5)( 8) C( 5( 4) ( 3)( 7) 5( 4) G (). 5( 4) p e tep ( 3)( 7) Thu, e e parabola. Checking for tability, fro firt block diagra above, T( 0.67, e rap, p 5( 4). The yte i table. 0 6 Syte. Puh 0 to the right pat the uing junction and puh 0 to the left pat the pickoff point: Copyright 0 by John Wiley & Son, Inc.

21 Solution to Proble 7- R( - 00( 00(4) 4) ( 3)( 7) (5)(8) C( 40 Produce a unity-feedback yte: R( (4) ( 3)( 7) (5)(8) C( ( 4) ( 3)( 7) 00( 4) Thu, Ge () 00( 4) ( 3)( 7) 40 e tep -0, e rap, e parabola. p Checking for tability, fro firt block diagra above, Therefore, yte i table and teady-tate error calculation are valid.. p 00(4) Ge ( 00( 4) T() G ( 5 4 e. 44. a. Produce a unity-feedback yte: Copyright 0 by John Wiley & Son, Inc.

22 7- Chapter 7: Steady-State Error R( () C( - - () - Thu, G e ( () () ()(-) () 3 (-)(-). Syte i Type 0. b. Since the yte i Type 0, the appropriate tatic error contant i p. Thu, etep ( ) 0.00 p Therefore, p Hence, Check tability: Uing original block diagra, T( Making a Routh table: () () () () Therefore, yte i table and teady-tate error calculation are valid. 45. a. Produce a unity-feedback yte: 4 H ( - Copyright 0 by John Wiley & Son, Inc.

23 Solution to Proble 7-3 R( - - () ( ) () ( 3) C( 3 ( ) ( 3) ( )( ) Thu, Ge(). Syte i Type ( ) 5 6 ( 3)( ) b. Since Type 0, appropriate tatic error contant i p. c. p d. e tep p Check tability: Uing original block diagra, ( ) ( 3) ( )( ) T(). 4 3 ( )( 4) 5 ( 6) 5 4 ( 3)( ) Making a Routh table: Copyright 0 by John Wiley & Son, Inc.

24 7-4 Chapter 7: Steady-State Error Therefore, yte i table for 0 < and teady-tate error calculation are valid. 46. Progra: 0 nug*poly([- -]);dengpoly([ ]); 'G(' Gtf(nug,deng) nuh[ 6];denhpoly([-8-9]); 'H(' Htf(nuh,denh) 'H(H-' HH- %For Ge(G(/(G(H( 'Ge(G(/(G(H()' Gefeedback(G,H) %Tet yte tability 'T(Ge(/(Ge()' Tfeedback(Ge,) pole(t) pdcgain(ge) 'Ge(' Getf([ 0],)*Ge; Geinreal(Ge) vdcgain(ge) '^Ge(' Getf([ 0],)*Ge; Geinreal(Ge) adcgain(ge) etep30/(p) erap30/v eparabola60/a E6 nug*poly([- -]);dengpoly([ ]); 'G(' Gtf(nug,deng) nuh[ 6];denhpoly([-8-9]); 'H(' Htf(nuh,denh) 'H(H-' HH- %For Ge(G(/(G(H( 'Ge(G(/(G(H()' Gefeedback(G,H) %Tet yte tability 'T(Ge(/(Ge()' Tfeedback(Ge,) pole(t) pdcgain(ge) 'Ge(' Getf([ 0],)*Ge; Geinreal(Ge) Copyright 0 by John Wiley & Son, Inc.

25 Solution to Proble 7-5 vdcgain(ge) '^Ge(' Getf([ 0],)*Ge; Geinreal(Ge) adcgain(ge) etep30/(p) erap30/v eparabola60/a Coputer repone: 0 an G( Tranfer function: 0 ^ ^5 5 ^4 74 ^3 0 ^ an H( Tranfer function: ^ 7 7 an H(H- Tranfer function: -^ ^ 7 7 an Ge(G(/(G(H() Tranfer function: 0 ^4 00 ^3 50 ^ ^7 3 ^6 40 ^5 448 ^4 778 ^ ^ Copyright 0 by John Wiley & Son, Inc.

26 7-6 Chapter 7: Steady-State Error an T(Ge(/(Ge() Tranfer function: 0 ^4 00 ^3 50 ^ ^7 3 ^6 40 ^5 458 ^ ^ ^ 00 0 an i i i i i i p an Ge( Tranfer function: 0 ^5 00 ^4 50 ^3 500 ^ ^7 3 ^6 40 ^5 448 ^4 778 ^ ^ v 0 an ^Ge( Tranfer function: 0 ^6 00 ^5 50 ^4 500 ^3 440 ^ ^7 3 ^6 40 ^5 448 ^4 778 ^ ^ a 0 etep Copyright 0 by John Wiley & Son, Inc.

27 Solution to Proble 7-7 erap Inf eparabola Inf an G( Tranfer function: e006 ^ 3e006 e ^5 5 ^4 74 ^3 0 ^ an H( Tranfer function: ^ 7 7 an H(H- Tranfer function: -^ ^ 7 7 an Ge(G(/(G(H() Tranfer function: e006 ^4 e007 ^3.5e008 ^.5e008.44e ^7 3 ^6 40 ^ ^ e007 ^3 -.6e008 ^ -.3e e008 Copyright 0 by John Wiley & Son, Inc.

28 7-8 Chapter 7: Steady-State Error an T(Ge(/(Ge() Tranfer function: e006 ^4 e007 ^3.5e008 ^.5e008.44e ^7 3 ^6 40 ^5 458 ^4.007e006 ^ e006 ^ e007.e007 an i i i i p an Ge( Tranfer function: e006 ^5 e007 ^4.5e008 ^3.5e008 ^.44e ^7 3 ^6 40 ^ e005 ^ e007 ^3 -.6e008 ^ -.3e e008 v 0 an ^Ge( Tranfer function: e006 ^6 e007 ^5.5e008 ^4.5e008 ^3.44e008 ^ ^7 3 ^6 40 ^ e005 ^ e007 ^3 -.6e008 ^ Copyright 0 by John Wiley & Son, Inc.

29 Solution to Proble 7-9 Copyright 0 by John Wiley & Son, Inc. -.3e e008 a 0 etep erap Inf eparabola Inf 47. a. Maon rule can be ued to find the open loop tranfer fro input to output: Only one forward path, ) ( LC T v T τ Three touching loop, L L α, LC L, C Z L L 3 C Z LC L L Δ α ; Δ C Z LC L LC T G L v T ) ( ) ( Δ Δ α τ. Letting L L C Z [ ] C C C L C C LC L LC G L v T L v T α τ τ α τ ) ( ) ( ) ( ) ( Since the yte i not unity feedback, we calculate [ ] [ ] [ ] C C C L C C C L C C C L G GH G L v T L v T L v T α τ τ α τ τ β α τ τ ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( [ ] ) ( ) ( ) ( ) ( C C C L v T L v T τ τ β α τ τ

30 7-30 Chapter 7: Steady-State Error The yte i type 0. b. For a tep input we calculate Then e P β P G( Li 0 GH ( G( β 48. G (G ( Y( R( G (G (H( D(G ( G (G (H( G (G ( E( R( - Y( R( - G (G (H( R( - D(G ( G (G (H( Thu, G (G ( - G (G (H( G ( R( - G (G (H( D( G (G ( G ( e( ) lie( li R( D( 0 0 G (G (H( G (G (H( 49. a. E( R( - C(. But, C( [R( - C(H(]G (G ( D(. Solving for C(, Subtituting into E(, R(G (G ( C( G (G (H( D( G (G (H( E( G (G ( - G (G (H( R( - G (G (H( D( b. For R( D(, li G (G ( 0 e( ) lie( 0 lig (G (H( lig (G (H( c. Zero error if G( and/or G( i Type. Alo, H( i Type 0 with unity dc gain. Firt find the forward tranfer function of an equivalent unity feedback yte. Copyright 0 by John Wiley & Son, Inc.

31 Solution to Proble 7-3 ( )( 4) G e ( ( a ) ( )( 4) Thu, e( ) e( ) p (a ) 3 5 ( 4) (a ) a a Finding the enitivity of e( ), S e:a a δe e δa a a a a (a ) a a a. 5. a. P δt P( ( L( )( F( G( H ( ) F( L( ( G( H ( ) L( ) F( L( S T : P T δp L( ( P( F( G( H ( F( L( ( L( ) L( T ( L( b. S T : P F( L( L( Copyright 0 by John Wiley & Son, Inc.

32 7-3 Chapter 7: Steady-State Error P δt 5 4 a. S T : P o GP( and T δp 5 4 GP( ( 5) ( 5) ( 5) 7 GP( ( 5) 4 G (. Alo, o P( 5 GP( ( 5) 4 T ( ( )( )( 5 4) 4 F ( GP( 4 ( )( ) GP( 5 4 b. The yte i type, o for e 0 it i required that Li F( 7. So. 0 7 Fro Eq. (7.70), ( ) ( ) - e( ) li li ( ) ( ) Senitivity to : S e: δe (00)(0.) e - δ - (00)(0.) Senitivity to : S e: δe ( ) (0.)(00) e δ ( -)( ) (0.-)((00)(0.)) ( ) ( ) 54. a. Uing Maon rule: t T ; Loop L ε ω0 ε r r ω M ε ω ε non-touching loop. Δ E R TΔ Δ b. For a unit tep input, 0 US ε r ε εr ε ε ω0 r ε ω 0 ε ε r 0 and εr L, no ε Copyright 0 by John Wiley & Son, Inc.

33 Solution to Proble 7-33 εr ε r e Li 0 ε ω0 ε r r r r ε ω0 ε c. For a unit rap input, e Li 0 d. The yte i type 0. r εr ε ε ω0 ε ω 0 ε ε r r 55. a. Uing Eq. (7.89) with (I - A ) (4) - (0) - (3 5) ( 3) yield e( ) for a tep input and e( ) for a rap input. The ae reult are obtained uing and Eq. (7.96) for a tep input and Eq. (7.03) for a rap input. b. Uing Eq. (7.89) with (I - A ) (5 7) 7 - ( 9) - 95 yield e( ) 0 for a tep input and e( ) 5 7 for a rap input. The ae reult are obtained uing and Eq. (.3) for a tep input and Eq. (.30) for a rap input. c. Uing Eq. (7.89) with Copyright 0 by John Wiley & Son, Inc.

34 7-34 Chapter 7: Steady-State Error yield e( ) 6 for a tep input and e( ) for a rap input. The ae reult are obtained uing and Eq. (7.96) for a tep input and Eq. (7.03) for a rap input. 56. Find G 4 (: Since 00 i/hr ft/ec, the velocity repone of G 4 ( to a tep diplaceent of the accelerator i v(t) 46.67( - e -αt ). Since 60 i/hr 88 ft/ec, the velocity equation at 0 econd 56becoe ( - e -α0 ). Solving for α yield α Thu, G 4 ( But, fro the velocity equation, the dc value of G 4 ( i Solving for, G 4 ( Find error: The forward tranfer function of the velocity control loop i G 3 (G 4 ( 3.49 ( )( 0.09). Therefore, v e( ) v 6.8 x Firt, reduce the yte to an equivalent unity feedback yte. Puh to the right pat the uing junction. Convert to a unity feedback yte by adding a unity feedback path and ubtracting unity fro 3. The equivalent forward tranfer function i, J D G e ( J D 3 - J D ( 3 - ) Copyright 0 by John Wiley & Son, Inc.

35 Solution to Proble 7-35 The yte i Type 0 with p ( ) 3. Auing the input concentration i R o, e( ) R o R o 3. The error can be reduced if 3. p a. For the inner loop, G e ( (0.0) (0.0) For G e ( G e ( (0.0) (0.0) ( 0.0). Syte i Type. Therefore, e tep 0, b. e rap 0, c. e parabola a G e ( d. T( G e ( (0.0) (0.0) 0.0, where c J < < < < Thu, for tability c J > a. Following Figure P7.9, the tranfer function fro δ f to e i given by: e δ f * r Gf G r For δ f we have that in teady tate Copyright 0 by John Wiley & Son, Inc.

36 7-36 Chapter 7: Steady-State Error b 0.8 a e( ) b (0) a It can be een fro thi expreion that if () i type or larger e( ) 0. b. Fro Figure P7.9: r δ f Gf G 4 * r Gr r The error i now defined a r G r G G G r G δ G G * * f r r f r f r r In teady tate thi expreion becoe: b b b b b (0) 0.8 (0) 0. (0) r a a a a a (0) δ b f 4 b4 (0) (0) a a It can be een in thi equation that the teady tate error cannot be ade zero. SOLUTIONS TO DESIGN PROBLEMS 60. Pot gain: 3π π 3; Aplifier gain: ; Motor tranfer function: Since tie contant 0.5, α. Alo, α () 0. Hence, 0. The otor tranfer function i now coputed a C( E a (. The following block diagra reult after puhing the potentioeter to the right pat the uing junction: Copyright 0 by John Wiley & Son, Inc.

37 Solution to Proble Finally, ince v 0 60, fro which 3. Firt find v: Circuference π nautical ile. Therefore, boat ake revolution in π hr. rev π rad Angular velocity i thu, 0.34 hr 3600 x 0.34 ec 5.56 x 0-3 rad ec. For 0. o error, e( ) /0o 5.56 x o x π rad. Thu v 3.9 v 4 ; fro which, a. Perforing block diagra reduction: R( (0.5)( ) C( Copyright 0 by John Wiley & Son, Inc.

38 7-38 Chapter 7: Steady-State Error R( - 00 ( 0. ) ( 0.5)( ) - C( R( - - Ge( C( G e ( Syte i unity feedback with a forward tranfer function, G t (, where G t ( Thu, yte i Type 0. b. Fro G t (, p Thu, c. T G t G t For 3650, T Becaue of the negative coefficient in the denoinator the yte i untable and the pilot would not be hired. Copyright 0 by John Wiley & Son, Inc.

39 Solution to Proble The force error i the actuating ignal. The equivalent forward-path tranfer function i G e ( ( ). The feedback i H( D e e. Uing Eq. (7.7) R( E a ( G e (H( e a _ rap ( ) li 0. Applying the final value theore, (D e e ) ( ) i econd-order with poitive coefficient, the yte i alway table. e < 0.. Thu, < 0. e. Since the cloed-loop yte a. The iniu teady-tate error occur for a axiu etting of gain,. The axiu poible i deterined by the axiu gain for tability. The block diagra for the yte i hown below. ω i _deired ( 3 V i ( ( 0)( 4 0) - ω o ( 3 Puhing the input tranducer to the right pat the uing junction and finding the cloed-loop tranfer function, we get Foring a Routh table, 3 ( 0)( 4 0) T( 3 ( 0)( 4 0) (3 00) Copyright 0 by John Wiley & Son, Inc.

40 7-40 Chapter 7: Steady-State Error The row ay - < < 00. The 0 row ay 00 <. Thu for tability, 3 00 < < 00. Hence, the axiu value of i b. p Hence, e ( ) tep p 7. c. Step input. 65. Subtituting value we have P( e 0.005L, G ( (.67)( 0) The proportional error contant L 4065e LiG( P( Li 573. L P (.67)( 0) e 0. which give L L P 66. P I a. The open loop traniion i GP(. The yte i type..89 b. The Tranfer function fro diturbance to error ignal i E( D( P I ( ) P I.89 Uing the Final value theore e ( ) ( ) Li E Li P I I.89 c. We calculate a Li G( P( o e SS So we get I P 0.0 d. The yte characteritic equation i 0 or P The Routh array i: a I Copyright 0 by John Wiley & Son, Inc.

41 Solution to Proble P P P The doinant requireent i given by the third row > P Deired force a. Input tranducer Input voltage F up ( 53.85) ( )( ) Controller Actuator Pantograph dynaic Y h -Y cat Spring diplaceent 8300 Spring F out 00 Senor Y h -Y cat Spring diplaceent Deired force F up ( 53.85) ( )( ) Controller Actuator Pantograph dynaic 8300 Spring F out b. G( Y h ( Y cat ( F up ( ( 53.85) ( )( ) Ge( (/00)*(/000)*G(*8.3e3 G e ( (53.85) ( ) ( ) p *53.85/[(376.3)(983)] *.000E-5 Copyright 0 by John Wiley & Son, Inc.

42 7-4 Chapter 7: Steady-State Error Maxiu iniize the teady-tate error. Maxiu poible i that which yield tability. Fro Chapter 6 axiu for tability i x 0 5. Therefore, p c. e /( p ) a. The yte i Type 0 ince there are no open-loop pole at the origin. b. The open loop tranfer function i: G() P() b So that LiG( P( 84.6 e p p which reult in ( ) The Routh array i 69. a. When the peed controller i configured a a proportional controller, the forward-path tranfer function of thi yte i: 0. ( 0.6) P SC G( ( 0.573) 5 ( 0.6) ( ) () e For the teady-tate error for a unit-tep input, r(t) u(t), to be equal to %: ( ) li G( 0. ( 0.6) P 0 SC li 0 ( 0.573) 5 ( 0.6) ( ) tep Fro equation (), we get: 0. 0, which yield: PSC P SC () b. When the peed controller i configured a a proportional plu integral controller, the forward-path tranfer function of the yte becoe: Copyright 0 by John Wiley & Son, Inc.

43 Solution to Proble ( 0.6) (00 I ) SC G( [ ( 0.573) 5 ( 0.6) ( ) ] (3) For the teady-tate error for a unit-rap input, r(t) t u(t), to be equal to.5%: e ( ) li G( 0 li 0 0. ( 0.6) (00 rap I ) [ ( 0.573) 5 ( 0.6) ( ) ] SC 0.05 (4) Fro equation (4), we get: 0. 05, which yield: ISC I SC c. We ll tart by finding G (, the equivalent tranfer function of the parallel cobination, repreenting the torque and peed controller, hown in Figure P7.35: ( 0.6) G ( (5) ( 0.5) ( 0.5) ( 0.5) Given that the equivalent tranfer function of the car i: G (, we apply equation * of the text taking into conideration that the diturbance here i a tep with a agnitude equal to 83.7: e( ) 0 li li G 3. ( G () 0 0 Copyright 0 by John Wiley & Son, Inc.

44 7-44 Chapter 7: Steady-State Error Founded in 807, John Wiley & Son, Inc. ha been a valued ource of knowledge and undertanding for ore than 00 year, helping people around the world eet their need and fulfill their apiration. Our copany i built on a foundation of principle that include reponibility to the counitie we erve and where we live and work. In 008, we launched a Corporate Citizenhip Initiative, a global effort to addre the environental, ocial, econoic, and ethical challenge we face in our buine. Aong the iue we are addreing are carbon ipact, paper pecification and procureent, ethical conduct within our buine and aong our vendor, and counity and charitable upport. For ore inforation, pleae viit our webite: Copyright 0 by John Wiley & Son, Inc. No part of thi publication ay be reproduced, tored in a retrieval yte or tranitted in any for or by any ean, electronic, echanical, photocopying recording, canning or otherwie, except a peritted under Section 07 or 08 of the 976 United State Copyright Act, without either the prior written periion of the Publiher or authorization through payent of the appropriate per-copy fee to the Copyright Clearance Center, Roewood Drive, Danver, MA 093, (978) , fax (978) Requet to the Publiher for periion hould be addreed to the Periion Departent, John Wiley & Son, Inc., River Street, Hoboken, NJ , (0) , fax (0) or online at Evaluation copie are provided to qualified acadeic and profeional for review purpoe only, for ue in their coure during the next acadeic year. Thee copie are licened and ay not be old or tranferred to a third party. Upon copletion of the review period, pleae return the evaluation copy to Wiley. Return intruction and a free of charge return hipping label are available at Outide of the United State, pleae contact your local repreentative. ISBN Copyright 0 by John Wiley & Son, Inc.

Control Systems Engineering ( Chapter 7. Steady-State Errors ) Prof. Kwang-Chun Ho Tel: Fax:

Control Systems Engineering ( Chapter 7. Steady-State Errors ) Prof. Kwang-Chun Ho Tel: Fax: Control Sytem Engineering ( Chapter 7. Steady-State Error Prof. Kwang-Chun Ho kwangho@hanung.ac.kr Tel: 0-760-453 Fax:0-760-4435 Introduction In thi leon, you will learn the following : How to find the