Text Material Preview
S E V E N Steady-State Errors SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Steady-State Error Design via Gain a. G(s) = 76.39K s(s 150)(s 1.32) . System is Type 1. Step input: e() = 0; Ramp input: e() = V 1 K = 1 76.39K 150 1.32 = 2.59 K ; Parabolic input: e() = . b. V 1 K = 2.59 K = 0.2. Therefore, K = 12.95. Now test the closed-loop transfer function, T(s) = 3 2 989.25 s 151.32s 198s 989.25 , for stability. Using Routh-Hurwitz, the system is stable. s3 1 198 s2 151.32 989.25 s1 191.46253 0 s0 989.25 0 Video Laser Disc Recorder: Steady-State Error Design via Gain a. The input, 15t2, transforms into 30/s3. e() = 30/Ka = 0.005. Ka = 0.2*600 20000 * K1K2K3 = 6x10 -3 K1K2K3. Therefore: e() = 30/Ka = 3 1 2 3 30 6x10 K K K = 5x10 -3 . Therefore K1K2K3 = 10 6 . b. Using K1K2K3 = 10 6 , G(s) = 5 2 4 2x10 (s 600) s (s 2x10 ) . Therefore, T(s) = 5 3 4 2 5 8 2x10 (s 600) s 2x10 s 2x10 s 1.2x10 . Making a Routh table, 7-2 Chapter 7: Steady-State Errors s3 1 2x10 5 s2 2x10 4 1.2x10 8 s1 194000 0 s0 120000000 0 we see that the system is stable. c. Program: numg=200000*[1 600]; deng=poly([0 0 -20000]); G=tf(numg,deng); 'T(s)' T=feedback(G,1) poles=pole(T) Computer response: ans = T(s) Transfer function: 200000 s + 1.2e008 ------------------------------------ s^3 + 20000 s^2 + 200000 s + 1.2e008 poles = 1.0e+004 * -1.9990 -0.0005 + 0.0077i -0.0005 - 0.0077I ANSWERS TO REVIEW QUESTIONS 1. Nonlinear, system configuration 2. Infinite 3. Step(position), ramp(velocity), parabola(acceleration) 4. Step(position)-1, ramp(velocity)-2, parabola(acceleration)-3 5. Decreases the steady-state error 6. Static error coefficient is much greater than unity. 7. They are exact reciprocals. 8. A test input of a step is used; the system has no integrations in the forward path; the error for a step input is 1/10001. Solutions to Problems 7-3 9. The number of pure integrations in the forward path 10. Type 0 since there are no poles at the origin 11. Minimizes their effect 12. If each transfer function has no pure integrations, then the disturbance is minimized by decreasing the plant gain and increasing the controller gain. If any function has an integration then there is no control over its effect through gain adjustment. 13. No 14. A unity feedback is created by subtracting one from H(s). G(s) with H(s)-1 as feedback form an equivalent forward path transfer function with unity feedback. 15. The fractional change in a function caused by a fractional change in a parameter 16. Final value theorem and input substitution methods SOLUTIONS TO PROBLEMS 1. s 0 s 0 s R(s) e( ) lim sE(s) lim 1 G(s) where 2 450(s 12)(s 8)(s 15) G(s) s(s 38)(s 2s 28) . For step, e () = 0. For 37tu(t), 2 37 R(s) s . Thus, e () = 6.075x10-2. For parabolic input, e() = . 2. a. From the figure 6 4 2 ss ss ss e r c . b. Since the system is linear, and because the original input was 2r t tu t , the new steady state error is 2 1 2 sse . 3. s 0 s 0 s R(s) e( ) lim sE(s) lim 1 G(s) = 3 s 0 2 s(160/s ) lim 2.83 60(s 3)(s 4)(s 8) 1 s (s 6)(s 17) 7-4 Chapter 7: Steady-State Errors 4. Reduce the system to an equivalent unit feedback system by first moving 1/s to the left past the summing junction. This move creates a forward path consisting of a parallel pair 1 1 s in cascade with a feedback loop consisting of 3 4 G s s and 2H s . Thus, 3 / 41 1 24 / 4 e ss G s s s Hence the system is Type 1, and the steady-state errors are as follows: Steady state error for 10 0u t Steady state error for 10 10 10 93.33 3 / 28v tu t K Steady state error for 210t u t 5. System is type 0. Kp = 1.4881. For 20 u(t), 20 ( ) 8.04 1 P e K For 60 t u(t), ( )e For 81t 2 u(t), ( )e 6. 4 3 ( ) 150 / ( ) 210( 4)( 6)( 11)( 13)1 ( ) 1 ( 7)( 14)( 19) R s S E s S S S SG s S S S S Thus, 0 150 ( ) lim ( ) 0.3875 (210)(4)(6)(11)(13) (7)(14)(19) s e sE s 7. de s E (s) dt Solutions to Design Problems 7-5 Therefore, e( ) = 0 lim s s2E(s) = 0 lim s s2 R(s) 1 G(s) = 2 4 0 2 6 lim 100( 1)( 2) 1 ( 10)( 3) s s s s s s s s = 9 10 . 8. 15 1020(13)(26)(33) ( ) ; 25.65 1 (65)(75)(91) p p e K K . Therefore, e() = 0.563. 9. For 70u(t), ess = 70 70 14 1 5 p K ; For 70tu(t), ess = , since the system is Type 0. 10. a. The closed-loop transfer function is, 2 5000 ( ) 75 5000 T s s s from which, n = 5000 and 2n = 75. Thus, = 0.53 and 2/ 1 %OS=e x100 = 14.01%. b. Ts = 4 n = 4 75 / 2 = 0.107 second. c. Since system is Type 1, ess for 5u(t) is zero. d. Since Kv is 5000 75 = 66.67, ess = 5 v K = 0.075. e. ess = , since system is Type 1. 11. 0 500000 7 20 45 lim 35000 30 50s K sG s v Thus 60 . 12. 2 0 2 4 6 lim ( ) 10,000 5 7 a s Kx x x K s G s x . Therefore, K = 7291.667. 7-6 Chapter 7: Steady-State Errors 13. a. Ge(s) = 5 s(s 1)(s 2) 5(s+3) 1 s(s 1)(s 2) = 3 2 5 s 3s 7s 15 Therefore, Kp = 1/3; Kv = 0; and Ka = 0. b. For 50u(t), e() = p 50 1 K = 37.5; For 50tu(t), e() = ; For 50t2u(t), e() = c. Type 0 14. ( ) ( ) 1 ( ) R s E s G s . Thus, 0 ( ) lim ( ) s e sE s = 4 2 20 3 60 lim 1030( 8 23)( 21 18) 1 ( 6)( 13) s s s s s s s s s s = 0.0110. 15. Collapsing the inner loop and multiplying by 1000/s yields the equivalent forward-path transfer function as, 5 2 10 ( 2) ( ) ( 1005 2000) e s G s s s s Hence, the system is Type 1. 16. The transfer function from command input to error signal can be found using Mason’s rule or any other method: 2 2 1 1 20 1 ( 3) 20( ) ( 3) 20( ) ( 3) 20 1 ( 3) G s s GE s s s R s s s G G s s Letting 1 ( )R s s and by the final value theorem: 2 0 0 1 ( ) ( ) ( ) ss s s G s e LimsE s Lim G s Solutions to Design Problems 7-7 a. If 1 G is type 0, it is required that 2 ( ) 0G s b. If 1 G is type 1, it is required that 2 ( )G s must be type 0 c. If 1 G is type 2, it is required that 2 ( )G s must be type 1 17. 2 2 0 0 ( ) ( )= lims E(s) = lims 1 ( )s s R s e G s . For Type 0, step input: R(s) = 1 s , and 0 ( )= lim = 0 1 ( )s s e G s For Type 0, ramp input: R(s) = 2 1 s , and 0 0 1 1 1 ( )= lim = = 1 ( ) 1 lim ( ) 1s p s e G s G s K For Type 0, parabolic input: R(s) = 3 1 s , and 0 1 ( )= lim = ( )s e s sG s For Type 1, step input: R(s) = 1 s , and 0 ( )= lim = 0 1 ( )s s e G s For Type 1, ramp input: R(s) = 2 1 s , and 0 1 ( )= lim = 0 1 ( )s e G s For Type 1, parabolic input: R(s) = 3 1 s , and 0 v 1 1 ( )= lim = ( ) Ks e s sG s For Type 2, step input: R(s) = 1 s , and 0 ( )= lim = 0 1 ( )s s e G s For Type 2, ramp input: R(s) = 2 1 s , and 0 1 ( )= lim = 0 1 ( )s e G s For Type 2, parabolic input: R(s) = 3 1 s , and 0 1 ( )= lim = ( )s e s sG s 0 7-8 Chapter 7: Steady-State Errors 18. a. 1/10 7 e( )= 0.01; where 10. 5 8 12 v v K K K x x Thus, 685.71K . b. Kv = 10. c. The minimum error will occur for the maximum gain before instability. Using the Routh-Hurwitz Criterion along with 4 3 2 ( 7) ( ) 25 196 480 7 K s T s s s s K s K : s 4 1 196 7K For Stability s 3 25 480+K s 2 4420-K 175K K < 4420 s 1 2 435 2121600K K -1690.2 < K < 1255.2 s 0 175K K > 0 Thus, for stability and minimum error K = 1255.2. Thus, 7 18.3 5 8 12 v K K x x and 1/10 1/10 e( )= 0.0055 18.3 v K . 19. 30 30 900 ( ) 0.005 / 30 v e K Ka Ka . Hence, Ka = 180000. Solutions to Design Problems 7-9 20. Find the equivalent G s for a unit feedback system. 2 5 7 1 2 K s s K G s s s s . Thus 100 100 0.01 / 7 v e K K ; from which 70000K . 21. 3 20 ; ( ) 0.061. Hence, 765.03. 7 a a K K e K K 22. a. e() = V 10 K = 1 6000 . But, Kv = 30K 5 = 60,000. Hence, K = 10,000. For finite error for a ramp input, n = 1. b. 2 0 0 10000( 3 30) lim ( ) lim ( 5) p s s s s K G s s s 2 0 0 10000( 3 30) lim ( ) lim 60000 ( 5) v s s s s K sG s s s s 2 2 2 0 0 10000( 3 30) lim ( ) lim 0 ( 5) a s s s s K s G s s s s 23. a. Type 0 b. E(s) = R(s) 1 G(s) . Thus, 20 0 2 12 / 12 ( ) lim ( ) lim 1 0.08( 6 6) 1 ( 5) ( 3) s s s e sE s s KK s s s s . c. e() = , since the system is Type 0. 24. 27 27 e( ) = = = 0.4. Thus, K = 325 247 /1188vK K . 25. e() = 1 1 pK = 1 6K 1 58 = 0.08. Thus, K = 111. 26. The system is stable for 0 < K < 2000. Since the maximum Kv is Kv = K 320 = 2000 320 = 6.25, the minimum steady-state error is V 1 K = 1 6.25 = 0.16. 27. To meet steady-state error characteristics: 7-10 Chapter 7: Steady-State Errors 2 2 2 2 R(s) 1 E(s) 1 G(s) K(s ) s 1 (s ) 1 e(t) sE(s) 0.1 1t s 0 K K Therefore, K = 92. To meet the transient requirement: Since T(s) = α β β α2 2 K(s ) s (K 2 )s ( K ) , n 2 = 10 = 2 + K ; 2n = 10 = K+2Solving for , = ±1. For = +1, K = 1.16 and = 7.76. An alternate solution is = -1, K = 5.16, and = 1.74. 28. a. System Type = 1 b. Assume G(s) = α K s(s ) . Therefore, e() = V 1 K = α 1 K/ = 0.01, or α K = 100. But, T(s) = G(s) 1 G(s) = α 2 K s s K . Since n = 10, K = 100, and = 1. Hence, G(s) = 100 s(s+1) . c. 2n = = 1. Thus, = 1 20 . 29. a. Since the steady-state output of the system can follow a ramp input with a finite error, then the system is type-1; b. The velocity error constant is given by: lim ( ) 0 v K K sG s s . Given that 1 ( ) 0.01 v e K , we have: 100 v K K . But, 2 ( ) ( ) 1 ( ) ( ) G s K K T s G s s s K s s K , where 2 25K n . Thus, 0.25 100 K . c. But 0.25 2 0.25 0.025 2 5 n Solutions to Design Problems 7-11 d. When K = 4 and = 0.4, 2 ( ) 4 ( ) ( ) 0.4 4 C s T s R s s s , then: 4 1 10 ( ) 0.1 0.4 v v K K e K ; 4 2n rad/sec; 0.4 2 0.4 0.1 2 2 n . 30. T(s) = G(s) 1 G(s) = α β α2 K(s ) s (K ) s K . Hence, K+ = 2, K = n 2 = (12+12) = 2. Also, e() = V 1 K = β β α α α 0.1. Therefore, 0.1K 0.2, K 1.8, and 1.111. K 31. System Type = 1. T(s) = G(s) 1 G(s) = 2 K s as K . From G(s), Kv = K a = 110. For 12% overshoot, = 0.56. Therefore, 2n = a, and n 2 = K. Hence, a = 1.12 K . Also, a = 110 K . Solving simultaneously, K = 1.52 x 104, and a = 1.38 x 102. 32. a. For 20% overshoot, = 0.456. Also, Kv = 1000 = K a . Since T(s) = 2 K s as K , 2n = a, and n = K . Hence, a = 0.912 K . Solving for a and K, K = 831,744, and a = 831.744. b. For 10% overshoot, = 0.591. Also, V 1 K = 0.01. Thus, Kv = 100 = K a . Since T(s) = 2 K s as K , 2n = a, and n = K . Hence, a = 1.182 K . Solving for a and K, K = 13971 and a = 139.71. 33. a. For the inner loop: G1(s) = 2 4 3 3 1 ss (s 1) 1 s s 1 1 s (s 1) Ge(s) = 2 1 s (s 3) G1(s) = 5 4 3 1 s(s 4s 3s s 3) 7-12 Chapter 7: Steady-State Errors T(s) = e e G (s) 1 G (s) = 6 5 4 2 1 s 4s 3s s 3s 1 b. From Ge(s), system is Type 1. c. Since system is Type 1, ess = 0 d. ; From Ge(s), Kv = 0 lim ( )e s sG s = 1 3 . Therefore, ess = V 5 K = 15. e. Poles of T(s) = -3.0190, -1.3166, 0.3426 ± j0.7762, -0.3495. Therefore, system is unstable and results of (c) and (d) are meaningless 34. a. For the inner loop: G1(s) = 10 s(s+1)(s+3)(s+4) 20 1 (s+1)(s+3)(s+4) = 3 2 10 s(s +8s 19s 32) Ge(s) = 3 2 20 s(s +8s 19s 32) T(s) = e e G (s) 1 G (s) = 4 3 2 20 s 8s 19s 32s 20 b. From Ge(s), system is Type 1. c. Since system is Type 1, ess = 0 d. From Ge(s), Kv = 0 lim ( )e s sG s = 20 32 = 5 8 . Therefore, ess = V 5 K = 8. e. Poles of T(s) = -5.4755, -0.7622 ± j1.7526, -1. Therefore, system is stable and results of parts c and d are valid. 35. Program: numg1=[1 9];deng1=poly([0 -6 -12 -14]); 'G1(s)=' G1=tf(numg1,deng1) numg2=6*poly([-9 -17]);deng2=poly([-12 -32 -68]); 'G2(s)=' G2=tf(numg2,deng2) numh1=13;denh1=1; 'H1(s)=' H1=tf(numh1,denh1) numh2=1;denh2=[1 7]; 'H2(s)=' H2=tf(numh2,denh2) %Close loop with H1 and form G3 'G3(s)=G2(s)/(1+G2(s)H1(s)' G3=feedback(G2,H1) %Form G4=G1G3 'G4(s)=G1(s)G3(s)' G4=series(G1,G3) %Form Ge=G4/1+G4H2 Solutions to Design Problems 7-13 'Ge(s)=G4(s)/(1+G4(s)H2(s))' Ge=feedback(G4,H2) %Form T(s)=Ge(s)/(1+Ge(s)) to test stability 'T(s)=Ge(s)/(1+Ge(s))' T=feedback(Ge,1) 'Poles of T(s)' pole(T) %Computer response shows that system is stable. Now find error specs. Kp=dcgain(Ge) 'sGe(s)=' sGe=tf([1 0],1)*Ge; 'sGe(s)' sGe=minreal(sGe) Kv=dcgain(sGe) 's^2Ge(s)=' s2Ge=tf([1 0],1)*sGe; 's^2Ge(s)' s2Ge=minreal(s2Ge) Ka=dcgain(s2Ge) essstep=100/(1+Kp) essramp=100/Kv essparabola=200/Ka Computer response: ans = G1(s)= Transfer function: s + 9 ------------------------------- s^4 + 32 s^3 + 324 s^2 + 1008 s ans = G2(s)= Transfer function: 6 s^2 + 156 s + 918 ------------------------------ s^3 + 112 s^2 + 3376 s + 26112 ans = H1(s)= Transfer function: 13 ans = H2(s)= Transfer function: 1 ----- s + 7 ans = G3(s)=G2(s)/(1+G2(s)H1(s) Transfer function: 6 s^2 + 156 s + 918 ------------------------------ s^3 + 190 s^2 + 5404 s + 38046 Solutions to Problems 7-13 ans = G4(s)=G1(s)G3(s) Transfer function: 6 s^3 + 210 s^2 + 2322 s + 8262 ------------------------------------------------------ s^7 + 222 s^6 + 11808 s^5 + 273542 s^4 + 3.16e006 s^3 + 1.777e007 s^2 + 3.835e007 s ans = 7-14 Chapter 7: Steady-State Errors Ge(s)=G4(s)/(1+G4(s)H2(s)) Transfer function: 6 s^4 + 252 s^3 + 3792 s^2 + 24516 s + 57834 ------------------------------------------------------- s^8 + 229 s^7 + 13362 s^6 + 356198 s^5 + 5.075e006 s^4 + 3.989e007 s^3 + 1.628e008 s^2 + 2.685e008 s + 8262 ans = T(s)=Ge(s)/(1+Ge(s)) Transfer function: 6 s^4 + 252 s^3 + 3792 s^2 + 24516 s + 57834 ------------------------------------------------------- s^8 + 229 s^7 + 13362 s^6 + 356198 s^5 + 5.075e006 s^4 + 3.989e007 s^3 + 1.628e008 s^2 + 2.685e008 s + 66096 ans = Poles of T(s) ans = -157.1538 -21.6791-14.0006 -11.9987 -11.1678 -7.0001 -5.9997 -0.0002 Kp = 7 ans = sGe(s)= ans = sGe(s) Transfer function: 6 s^5 + 252 s^4 + 3792 s^3 + 2.452e004 s^2 + 5.783e004 s -------------------------------------------------------- s^8 + 229 s^7 + 1.336e004 s^6 + 3.562e005 s^5 + 5.075e006 s^4 + 3.989e007 s^3 + 1.628e008 s^2 + 2.685e008 s + 8262 Kv = 0 ans = s^2Ge(s)= ans = s^2Ge(s) Transfer function: 6 s^6 + 252 s^5 + 3792 s^4 + 2.452e004 s^3 + 5.783e004 s^2 -------------------------------------------------------- s^8 + 229 s^7 + 1.336e004 s^6 + 3.562e005 s^5 + 5.075e006 s^4 + 3.989e007 s^3 + 1.628e008 s^2 + 2.685e008 s + 8262 Solutions to Problems 7-15 Ka = 0 essstep = 12.5000 essramp = Inf essparabola = Inf 36. Solutions to Design Problems 7-15 The equivalent forward transfer function is 15 2 5 f K G s s s K . Also 1 2 1 5 1 2 5 5f G s K T s G s s K s K . From the problem statement 15 20 2 5 v f K K K and 2 2 5n fK where 15n K . Solving simultaneously for 1K and fK we get 1 156.8K , 7.44fK . 37. We calculate the Velocity Error Constant, 0 0 3 2 2 ( 34.16 144.4 7047 557.2) 0.00842( 7.895)( 0.108 0.3393) 25 4 3 2 ( 0.07895)( 4 8)13.18 95.93 14.61 31.94 ( ) ( ) v s s s s s s s s s s s ss s s s s K Lim sG s P s Lim 557.2 0.0357 0.623 31.94 For a unit ramp input the steady state error is 1 1.605ss v e K . The input slope is 25 15.6 1.605 38. e() = s 0 lim 2 1 2 sR(s)-sD(s)G (s) 1 G (s)G (s) , where G1(s) = 1 s 5 and G2 = 100 s 2 . From the problem statement, R(s) = D(s) = 1 s . Hence, e() = s 0 100 1 s 2lim 1 100 1 s 5 s 2 = - 49 11 . 39. Error due only to disturbance: Rearranging the block diagram to show D(s) as the input, Therefore, 7-16 Chapter 7: Steady-State Errors -E(s) = D(s) 2 1 2 K s(s 4) K K (s 2) 1 s(s 3)(s 4) = D(s) 2 1 2 K (s 3) s(s 3)(s 4) K K (s 2) For D(s) = 1 s , e D () = 0 1 3 lim ( ) 2s sE s K . Error due only to input: e R () = V 1 K = 1 2 1 K K 6 = 1 2 6 K K . Design: e D () = - 0.00001 = - 1 3 2K , or K 1 = 150,000. e R () = 0.002 = 1 2 6 K K , or K 2 = 0.02 40. a. The open loop transmission is 35 ( ) ( ) 2 G s P s s , so 0 35 ( ) ( ) 2 P s K Lim G s P s . For a unit step input 1 0.0541 1 ssr P e K . Since the input is threefold that we have that 3(0.0541) 0.1622 ssr e b. Solutions to Design Problems 7-17 7-18 Chapter 7: Steady-State Errors b. The transfer function from disturbance to error signal is 7 ( ) 72 7( ) 37 1 5 2 E s s D s s s Using the final value theorem 0 0 7 1 7 ( ) 0.1892 37 37 ssd s s e Lim sE s Lim s s s c. Solutions to Design Problems 7-19 e. 0.1622 0.1892 0.351 tot ssr ssd e e e f. 7-20 Chapter 7: Steady-State Errors 41. a11 2 1 2 1 2 1 1 1 0 2 1 E ( )( ) ( ) ( )( ) ; ( ) 1 ( ) ( ) ( ) 1 ( ) ( ) ( ) ( ) ( ) lim 1 ( ) ( ) a s sG s G s G sC s R s G s H s R s G s H s sR s G s e G s H s 42. System 1: Forming a unity-feedback path, the equivalent unity feedback system has a forward transfer function of 2 10( 10) 10( 10)( 2) ( ) 10( 10)( 3) 11 132 300 1 ( 2) e s ss s G s s s s s s s a. Type 0 System; b. Kp = 0 lim ( ) 1/ 3 p e s K G s ; c. step input; d. e() = 1 1 p K = 3/4; e. 0 0 1 ( ) ( ) lim lim 0 10( 10)( 4)1 ( ) ( ) 1 ( 2) a step s s s sR s s e s sG s H s s s . System 2: Solutions to Design Problems 7-21 Forming a unity-feedback path, the equivalent unity feedback system has a forward transfer function of 10( 10) 10( 10)( 2) ( ) 10( 10) (11 102) 1 ( 2) e s ss s G s s s s s s s a. Type 1 System; b. 0 lim ( ) 0.98v e s K sG s ; c. ramp input; d. 1 ( ) 1.02 v e K ; e. 2 0 0 1 ( ) 1 ( ) lim lim 10( 10)( 1)1 ( ) ( ) 50 1 ( 2) a ramp s s s sR s s e s sG s H s s s . 43. System 1. Push 5 to the right past the summing junction: Produce a unity-feedback system: Thus, 2 5( 4) 5( 4)( 3)( 7) ( ) 5( 4) 15 41 1 ( 3)( 7) e s ss s G s s s s s s . Kp = 20 41 . estep = p 1 1+K = 0.67, eramp = , eparabola = . Checking for stability, from first block diagram above, T(s) = 2 5( 4) 20 61 s s s . The system is stable. 7-22 Chapter 7: Steady-State Errors System 2. Push 20 to the right past the summing junction and push 10 to the left past the pickoff point: Produce a unity-feedback system: Thus, 2 200( 4) 200( 4)( 3)( 7) ( ) 200( 4) 39 185 759 1 ( 3)( 7) 40 e s ss s G s s s s s s . Kp = 200(4) 1.05 759 . estep = p 1 1 K = -20, eramp = , eparabola = . Checking for stability, from first block diagram above, 2 ( ) 200( 4) ( ) 1 ( ) 15 41 e e G s s T s G s s s . Therefore, system is stable and steady-state error calculations are valid. ( ) 1 ( ) e e G s G s 44. Produce a unity-feedback system: Solutions to Design Problems 7-23 Thus, Ge(s) = 2 2 (s 1) s (s 2) (s 1)(K-1) 1 s (s 2) = 3 2 s 1 s 2s (K-1)s (K-1) . Error = 0.001 = p 1 1 K . Therefore, Kp = 999 = 1 K-1 . Hence, K = 1.001001. Check stability: Using original block diagram, T(s) = 2 2 (s 1) s (s 2) K(s 1) 1 s (s 2) = 3 2 s 1 s 2s Ks K . Making a Routh table: s3 1 K s2 2 K s1 K 2 0 s0 K 0 Therefore, system is stable and steady-state error calculations are valid. 45. a. Produce a unity-feedback system: 1 5 3 ( ) 1 2 2 s H s s s 7-24 Chapter 7: Steady-State Errors Thus, 2 4 3 2 2 ( 1) ( 1)( 2)( 4) ( ) 3 ( 1) 6 8 3 3 1 ( 4)( 2) e K s K s ss s G s K s s s s Ks K s s s . System is Type 0. b. Since Type 0, appropriate static error constant is Kp. c. 2 2 3 3 P K K K d. 1 1 3 0.6 21 5 1 3 step P e K Check stability: Using original block diagram, 2 4 3 2 2 ( 1) ( 1)( 2)( 3) ( ) ( 1)( 4) 5 ( 6) 5 4 1 ( 3)( 2) K s K s ss s T s K s s s s K s Ks K s s s . 2 4 3 2 2 ( 1) ( 1)( 2)( 4) ( ) ( 1)( 5) 6 ( 8) 6 5 1 ( 4)( 2) K s K s ss s T s K s s s s K s Ks K s s s Making a Routh table: Solutions to Design Problems 7-25 s4 1 K+8 5K s3 6 6K 0 s2 8 5K 0 s1 9 4 K 0 0 s0 5K 0 0 Therefore, system is stable for K > 0 and steady-state error calculations are valid. 46. Program: K=10 numg1=K*poly([-1 -2]);deng1=poly([0 0 -4 -5 -6]); 'G1(s)=' G1=tf(numg1,deng1) numh1=[1 6];denh1=poly([-8 -9]); 'H1(s)=' H1=tf(numh1,denh1) 'H2(s)=H1-1' H2=H1-1 %Form Ge(s)=G1(s)/(1+G1(s)H2(s) 'Ge(s)=G1(s)/(1+G1(s)H2(s))' Ge=feedback(G1,H2) %Test system stability 'T(s)=Ge(s)/(1+Ge(s))' T=feedback(Ge,1) pole(T) Kp=dcgain(Ge) 'sGe(s)' sGe=tf([1 0],1)*Ge; sGe=minreal(sGe) Kv=dcgain(sGe) 's^2Ge(s)' s2Ge=tf([1 0],1)*sGe; s2Ge=minreal(s2Ge) Ka=dcgain(s2Ge) essstep=30/(1+Kp) essramp=30/Kv essparabola=60/Ka K=1E6 numg1=K*poly([-1 -2]);deng1=poly([0 0 -4 -5 -6]); 'G1(s)=' G1=tf(numg1,deng1) 7-26 Chapter 7: Steady-State Errors numh1=[1 6];denh1=poly([-8-9]); 'H1(s)=' H1=tf(numh1,denh1) 'H2(s)=H1-1' H2=H1-1 %Form Ge(s)=G1(s)/(1+G1(s)H2(s) 'Ge(s)=G1(s)/(1+G1(s)H2(s))' Ge=feedback(G1,H2) %Test system stability 'T(s)=Ge(s)/(1+Ge(s))' T=feedback(Ge,1) pole(T) Kp=dcgain(Ge) 'sGe(s)' sGe=tf([1 0],1)*Ge; sGe=minreal(sGe) Kv=dcgain(sGe) 's^2Ge(s)' s2Ge=tf([1 0],1)*sGe; s2Ge=minreal(s2Ge) Ka=dcgain(s2Ge) essstep=30/(1+Kp) essramp=30/Kv essparabola=60/Ka Computer response: K = 10 ans = G1(s)= Transfer function: 10 s^2 + 30 s + 20 ------------------------------- s^5 + 15 s^4 + 74 s^3 + 120 s^2 ans = H1(s)= Transfer function: s + 6 --------------- s^2 + 17 s + 72 ans = H2(s)=H1-1 Solutions to Design Problems 7-27 Transfer function: -s^2 - 16 s - 66 ---------------- s^2 + 17 s + 72 ans = Ge(s)=G1(s)/(1+G1(s)H2(s)) Transfer function: 10 s^4 + 200 s^3 + 1250 s^2 + 2500 s + 1440 ----------------------------------------------------------------------- s^7 + 32 s^6 + 401 s^5 + 2448 s^4 + 7178 s^3 + 7480 s^2 - 2300 s - 1320 ans = T(s)=Ge(s)/(1+Ge(s)) Transfer function: 10 s^4 + 200 s^3 + 1250 s^2 + 2500 s + 1440 --------------------------------------------------------------------- s^7 + 32 s^6 + 401 s^5 + 2458 s^4 + 7378 s^3 + 8730 s^2 + 200 s + 120 ans = -8.5901 + 0.3993i -8.5901 - 0.3993i -6.0000 -4.4042 + 0.1165i -4.4042 - 0.1165i -0.0057 + 0.1179i -0.0057 - 0.1179i Kp = -1.0909 ans = sGe(s) 7-28 Chapter 7: Steady-State Errors Transfer function: 10 s^5 + 200 s^4 + 1250 s^3 + 2500 s^2 + 1440 s ----------------------------------------------------------------------- s^7 + 32 s^6 + 401 s^5 + 2448 s^4 + 7178 s^3 + 7480 s^2 - 2300 s - 1320 Kv = 0 ans = s^2Ge(s) Transfer function: 10 s^6 + 200 s^5 + 1250 s^4 + 2500 s^3 + 1440 s^2 ----------------------------------------------------------------------- s^7 + 32 s^6 + 401 s^5 + 2448 s^4 + 7178 s^3 + 7480 s^2 - 2300 s - 1320 Ka = 0 essstep = -330.0000 essramp = Inf essparabola = Inf K = 1000000 ans = G1(s)= Solutions to Design Problems 7-29 Transfer function: 1e006 s^2 + 3e006 s + 2e006 ------------------------------- s^5 + 15 s^4 + 74 s^3 + 120 s^2 ans = H1(s)= Transfer function: s + 6 --------------- s^2 + 17 s + 72 ans = H2(s)=H1-1 Transfer function: -s^2 - 16 s - 66 ---------------- s^2 + 17 s + 72 ans = Ge(s)=G1(s)/(1+G1(s)H2(s)) Transfer function: 1e006 s^4 + 2e007 s^3 + 1.25e008 s^2 + 2.5e008 s + 1.44e008 -------------------------------------------------------------------------------- s^7 + 32 s^6 + 401 s^5 - 997542 s^4 - 1.899e007 s^3 - 1.16e008 s^2 - 2.3e008 s - 1.32e008 ans = T(s)=Ge(s)/(1+Ge(s)) Transfer function: 1e006 s^4 + 2e007 s^3 + 1.25e008 s^2 + 2.5e008 s + 1.44e008 -------------------------------------------------------------------------------- 7-30 Chapter 7: Steady-State Errors s^7 + 32 s^6 + 401 s^5 + 2458 s^4 + 1.007e006 s^3 + 9.009e006 s^2 + 2e007 s + 1.2e007 ans = -28.2460 +22.2384i -28.2460 -22.2384i 16.7458 +22.2084i 16.7458 -22.2084i -6.0000 -1.9990 -1.0007 Kp = -1.0909 ans = sGe(s) Transfer function: 1e006 s^5 + 2e007 s^4 + 1.25e008 s^3 + 2.5e008 s^2 + 1.44e008 s -------------------------------------------------------------------------------- s^7 + 32 s^6 + 401 s^5 - 9.975e005 s^4 - 1.899e007 s^3 - 1.16e008 s^2 - 2.3e008 s - 1.32e008 Kv = 0 ans = s^2Ge(s) Transfer function: 1e006 s^6 + 2e007 s^5 + 1.25e008 s^4 + 2.5e008 s^3 + 1.44e008 s^2 -------------------------------------------------------------------------------- s^7 + 32 s^6 + 401 s^5 - 9.975e005 s^4 - 1.899e007 s^3 - 1.16e008 s^2 - 2.3e008 s - 1.32e008 Solutions to Design Problems 7-31 Ka = 0 essstep = -330.0000 essramp = Inf essparabola = Inf 47. a. Mason’s rule can be used to find the open loop transfer from input to output: Only one forward path, 1 2 1 1 ( ) T v m T K K K s LCs Three touching loops, 1 m K L Ls , 2 2 1 L LCs , 3 1 L L Z Cs 2 1 1 1 m L K Ls LCs Z Cs ; 1 1 2 1 1 2 1 ( ) ( ) 1 1 1 T m v m L K K K T LCs sG s K Ls LCs Z Cs . Letting 1 L L Z sC 2 2 2 1 ( ) ( 1) ( ) 1 ( ) 1 T m v T m v m L L m K K K K K K ssLCsG s K C s L C C s K C Ls CLCs Since the system is not unity feedback, we calculate 7-32 Chapter 7: Steady-State Errors 2 2 2 ( 1) ( )( ) ( 1) ( 1)1 ( ) ( ) 1 ( ) ( ) T m v L m T m v T m v L m L m K K K s s L C C s K CG s K K K s K K K sGH s G s s L C C s K C s L C C s K C 2 ( 1) ( ) ( 1) ( 1) T m v T m L m v K K K s K K s L C C s K C K s The system is type 0. b. For a step input we calculate 0 ( ) 1 1 ( ) ( ) 1 P s G s K Lim GH s G s Then 1 1 1 1 ss P e K 48. Y(s) = R(s) 1 2 1 2 G (s)G (s) 1 G (s)G (s)H(s) + 2 1 2 D(s)G (s) 1 G (s)G (s)H(s) E(s) = R(s) - Y(s) = R(s) - 1 2 1 2 G (s)G (s) 1 G (s)G (s)H(s) R(s) - 2 1 2 D(s)G (s) 1 G (s)G (s)H(s) = 1 2 1 2 G (s)G (s) 1 1 G (s)G (s)H(s) R(s) - 2 1 2 G (s) 1 G (s)G (s)H(s) D(s) Thus, e() = 1 2 2 s 0 s 0 1 2 1 2 G (s)G (s) G (s) lim sE(s) lim 1- R(s)- D(s) 1 G (s)G (s)H(s) 1 G (s)G (s)H(s) 49. a. E(s) = R(s) - C(s). But, C(s) = [R(s) - C(s)H(s)]G1(s)G2(s) + D(s). Solving for C(s), C(s) = 1 2 1 2 R(s)G (s)G (s) 1 G (s)G (s)H(s) + 1 2 D(s) 1 G (s)G (s)H(s) Substituting into E(s), E(s) = 1 2 1 2 G (s)G (s) 1- 1 G (s)G (s)H(s) R(s) - 1 2 1 1 G (s)G (s)H(s) D(s) b. For R(s) = D(s) = 1 s , Solutions to Design Problems 7-33 e() = s 0 lim sE(s) = 1 - 1 2 s 0 1 2 s 0 lim G (s)G (s) 1 lim G (s)G (s)H(s) - 1 2 s 0 1 1 lim G (s)G (s)H(s) c. Zero error if G1(s) and/or G2(s) is Type 1. Also, H(s) is Type 0 with unity dc gain. 50. First find the forward transfer function of an equivalent unity feedback system. 3 2 ( 2)( 5) ( ) ( 1) 7 ( 10) ( 1)1 ( 2)( 5) e K Ks s s G s K s a s s K s K a s s s Thus, 1 1 1 ( ) 1 1 ( 1) p a e KK a K a Finding the sensitivity of e(), we get: Se:a = a e e a = 2 ( 1) 1 a a a a a a = 1 1a . The following MATLAB M-file was written and used to plot the sensitivity, e, as a function of the parameter a. The graph obtained is shown below. a = 0 : 0.1 : 10; e = (a-1).^-1; plot (a, e,'LineWidth',2) grid title 'Sensitivity to Parameter a' xlabel 'a' ylabel 'Sensitivity of errror, e, to a' 7-34 Chapter 7: Steady-State Errors 51 a. : 2 1 1 1 T P L s F s G s H s F s L s G s H sP sP T S F s L sT P H s L s H s L s 1 1 1 P s F s G s H s F s L s L s L s b. : 1 1 1 1 T P T s H s L s S s F s L s L s . 52. a. 2 : 2 1 2 1 2 20 T P s s S G s P s s s so 20 2 G s P s s s and 20 2 4 5 2 s s G s s s . Also 2 20 2 20 201 2 20 1 2 G s P s s s G s P s s s s s , so 2 2 200 1 3 2 20 10 20 1 3 2 201 K s s s sT s K F s G s P s s s s sG s P s . Solutions to Design Problems 7-35 b. The system is type 1, so for 0sse it is required that 0 10 lim 1 3s F s K . So 3 10 K . 53. From Eq. (7.70), e() = 1 - 1 2 s 0 1 2 K K (s 2) lim K K (s 1) 1 (s 2) - 2 s 0 1 2 K (s 2) lim K K (s 1) 1 (s 2) = 2 1 2 2 K 2 K K Sensitivity to K1: Se:K1 = δ δ 1 1 K e e K = - 1 2 1 2 K K 2 K K = - (100)(0.1) 2 (100)(0.1) = - 0.833 Sensitivity to K2: Se:K2 = δ δ 2 2 K e e K = 2 1 2 1 2 2K (1 K ) (K -2)(2 K K ) = 2(0.1)(1 100) (0.1-2)(2 (100)(0.1)) = - 0.89 54. a. Using Mason’s rule: 1 1T ; Loops 2 0 1 2 2 2 2 0 0 1t r r US K L m m M s ss s and 2 rmL s , no non-touching loops. 1 1 rm s 1 1 2 0 2 2 0 1 1 r r r m TE s R m m s ss b. For a unit step input, 20 0 2 2 0 1 11 1 1 1 r r ss r s r r r r m mse Lims m s m m m m s ss c. For a unit ramp input, 2 20 0 2 2 0 1 1 1 r ss s r r m se Lim s s m m s ss d. The system is type 0. 55. a. Using Eq. (7.89) with 7-36 Chapter 7: Steady-State Errors 2 -1 2 3 2 2 s 15s 50 -(4s 22) -(2s 20) 1 (s - ) -(3s 15) s 10s 23 6 s 20s 111s 164 -(s 13) s 9 s 15s 38 I A yields e() = 1.09756 for a step input and e() = for a ramp input. The same results are obtained using 1 50 22 20 1 15 23 6 164 13 9 38 A and Eq. (7.96) for a step input and Eq. (7.103) for a ramp input. b. Using Eq. (7.89) with 2 1 2 3 2 2 s 9s s 7 1 (s - ) -(5s 7) s 7s s 9s 5s 7 -(s 9) -1 s 9s 5 I A yields e() = 0 for a step input and e() = 5 7 for a ramp input. The same results are obtained using 1 0 0 7 1 7 0 0 7 9 1 5 A and Eq. (12.123) for a step input and Eq. (12.130) for a ramp input. c. Using Eq. (7.89) with 2 1 2 3 2 2 15 4 5 23) 10) 1 (s ) 11 14 42 2 19 14 43 17 3 2) 2 3 9 5 s s s s s s s s s s s s s s s I A yields e() = 6 for a step input and e() = for a ramp input. The same results are obtained using 1 4 23 10 1 11 42 19 17 2 3 5 A and Eq. (7.96) for a step input and Eq. (7.103) for a ramp input. 56. a. Following Figure P7.26, the transfer function from f to e is given by: * 1 f f r r Ge KG Solutions to Design Problems 7-37 For 1 f s we have that in steady state 2 2 4 2 0.8 1 0 b a e b K a It can be seen from this expression that if K s is type 1 or larger 0e . b. From Figure P7.26: * 1 f r f r G r KGr KG The error is now defined as * 1 1 1 f r f r G r KGr KG = *1 1 r f r r KG G r KG KG In steady state this expression becomes: 4 2 4 4 2 2 2 2 2 2 4 4 2 2 1 0 0.8 0 1 0.2 0 1 0 1 0 1 0f b b b b b K K K a a a a ar b b K K a a It can be seen in this equation that the steady state error cannot be made zero. 57. a. Letting 0x , 0y , we get two equations: 0k x a g h y 1 1q k x qa g q qh y Solving the first equation for x , substituting into the second and simplifying we get 1 y a h b. The systems characteristic equation is calculated from 2 0 1 1 s k a g h s s k g q a h g s k h a q k s qa g q qh I A The Routh array is 2s 1 k h a s k g q a h g 1 k h a For closed loop stability the second row requires 0k g q a h g or 1k g q h a q . 7-38 Chapter 7: Steady-State Errors The third row requires 0h a . The intersection of both requirements is 1 0 k g q h a q c. From part a. zero steady state error requires a h , however this value is limited by stability requirements as shown in part b. Therefore zero steady state error is not possible. 58. a. The system’s type is 0 because the system shows a nonzero steady state error for a step input b. 1 0.37 0.63sse c. Since the system is type 0, for a ramp input ss e . 59. a. We will calculate the steady state error from e t r t c t so we start by calculating the system’s closed loop transfer function. From Mason’s Rule: 1 2 3 4 2 3 4 1 2 3 4 2 3 4 2 3 4 3 2 3 4 2 3 4 3 4 (1 )(1 )(1 )( ) ( ) 1 (1 )(1 )(1 ) (1 )(1 )(1 ) (1 )(1 ) K K K K sT sT sTC s T K K K K K K K sv K K svR s sT sT sT sT sT sT sT sT 1 2 3 4 2 3 4 1 2 3 4 2 3 4 2 3 4 3 21 1 1 1 K K K K sT sT sT K K K K K K K sv K K sv sT E s R s C s 1 2 3 4 2 3 4 1 2 3 4 2 3 4 2 3 4 3 2 1 1 1 1 1 1 K K K K s sT sT sT K K K K K K K sv K K sv sT 2 3 4 1 2 3 4 2 3 4 2 3 4 3 2 1 2 3 4 2 3 4 1 2 3 4 2 3 4 2 3 4 3 2 (1 )(1 )(1 ) (1 )1 (1 )(1 )(1 ) (1 ) sT sT sT K K K K K K K sv K K sv sT K K K K s sT sT sT K K K K K K K sv K K sv sT Applying the Final Value Theorem 2 3 4 1 2 3 4 2 3 4 2 3 4 3 2 1 2 3 4 0 0 2 3 4 1 2 3 4 2 3 4 2 3 4 3 2 1 2 3 4 1 2 3 4 1 2 3 4 (1 )(1 )(1 ) (1 ) lim ( ) lim (1 )(1 )(1 ) (1 ) 1 1 ss s s e sT sT sT K K K K K K K sv K K sv sT K K K K sE s sT sT sT K K K K K K K sv K K sv sT K K K K K K K K K K K K b. Since the steady state error to a step input is non-zero the system is type 0 60. The Simulink model of this system and its step response (displayed in blue) from t = 0 to 300 seconds are shown below. Here, the reference input, r(t), is a unit step, u(t), applied at t = 0 (displayed in green), and the disturbance is d(t) = 0.25 u(t) applied at t = 150 seconds (displayed in yellow). As Solutions to Design Problems 7-39could be seen from this plot, the steady-state errors due to a step reference input, r(t) = u(t), and due to a disturbance, d(t) = 0.25 u(t), are both equal to zero. It should be noted, however, that the relative stability of this Type 3 system is poor, since it exhibits a high percent overshoot, % OS 80 %, due to the unit-step reference input. 61. The modified Simulink models of this system and their responses (displayed in blue) from t = 0 to 100 seconds are shown below. In the top model, the reference input, r(t), is a unit ramp, t u(t), applied at t = 7-40 Chapter 7: Steady-State Errors 0 (displayed in green), and the disturbance is d(t) = 0.25 t u(t) applied at t = 50 sec (displayed in yellow). As could be seen from this plot, the steady-state position errors due to the unit-ramp reference input and the 0.25 t u(t) disturbance ramp are equal to zero. Solutions to Design Problems 7-41 7-42 Chapter 7: Steady-State Errors SOLUTIONS TO DESIGN PROBLEMS 62. The force error is the actuating signal. The equivalent forward-path transfer function is 1 1 2 ( ) ( ) e K G s s s K K . The feedback is ( ) e e H s D s K . Using Eq. (7.72) ( ) ( ) 1 ( ) ( ) a e R s E s G s H s . Applying the final value theorem, 2 2 _ 0 1 1 2 1 ( ) lim 0.1 ( ) 1 ( ) a ramp s e e e s Ks e K D s K K s s K K . Thus, K2 < 0.1Ke. Since the closed-loop system is second-order with positive coefficients, the system is always stable. 63. a. The minimum steady-state error occurs for a maximum setting of gain, K. The maximum K possible is determined by the maximum gain for stability. The block diagram for the system is shown below. Pushing the input transducer to the right past the summing junction and finding the closed-loop transfer function, we get 2 3 2 2 3 3( 10)( 4 10) ( ) 3 14 50 (3 100) 1 ( 10)( 4 10) K Ks s s T s K s s s K s s s Solutions to Design Problems 7-43 Forming a Routh table, s 3 1 50 s 2 14 3K+100 s 1 3 600 14 K 0 s 0 3K+100 0 The s 1 row says - < K < 200. The s0 row says 100 3 < K. Thus for stability, 100 3 < K < 200. Hence, the maximum value of K is 200. b. 3 6 100 p K K . Hence, 1 1 ( ) 1 7 step p e K . c. Step input. 64. Substituting values we have 0.1140625 ( ) ( 2.67)( 10) se P s s s , 0.005 ( ) 0.005 L G s s The proportional error constant 0.1 0 0 0.005 140625 ( ) ( ) 5273.44 0.005 ( 2.67)( 10) s P s s L e K Lim G s P s Lim L s s s 1 1 0.1 1 1 5273.44 ss P e K L which gives 31.71 10L . 65. a. The open loop transmission is 2 48500 ( ) 2.89 P I K s K GP s s s . The system is type 2. b. The Transfer function from disturbance to error signal is 2 3 2 2 48500 ( ) 485002.89 48500( ) 2.89 48500( ) 1 2.89 P I P I E s ss s K s KD s s s K s K s s s Using the Final value theorem 3 20 0 48500 1 ( ) 0 2.89 48500( ) ss s s P I s e Lim sE s Lim s s s K s K s 7-44 Chapter 7: Steady-State Errors c. We calculate 2 0 48500 ( ) ( ) 2.89 I a s K K Lim s G s P s so 1 2.89 0.05 48500 SS a I e K K . So we get 0.0120 I K d. The system’s characteristic equation is 2 0.01248500 1 0 2.89 P K s s s s or 3 22.89 48500 584.021 0 P s s K s . The Routh array is: 3s 1 48500 PK 2s 2.89 584.021 s 140165 584.21 2.89 P K 1 48500 PK The dominant requirement is given by the third row 0.00417 P K 66. a. A bode plot of the open loop transmission ( ) ( ) c G s P s shows that the open loop transfer function has a crossover frequency of 0.04c rad day . A convenient range for sampling periods is 0.15 0.5 3.75 12.5 c c day T day . T=8 days fall within range. b. We substitute 1 1 4 1 z s z into ( ) c G s we get 4 2 2 2 10 (1.145 1.71 0.8489) ( ) 1.852 0.8519 c z z G z z z Solutions to Design Problems 7-45 c. 67. a. When the speed controller is configured as a proportional controller, the forward-path transfer function of this system is: SC 0.11 ( 0.6) ( ) ( 0.5173) 5 (s 0.6) (s 0.01908) Ps K G s s s (1) For the steady-state error for a unit-step input, r(t) = u(t), to be equal to 1%: 7-46 Chapter 7: Steady-State Errors SC step 0 0 1 1 ( ) 0.01 0.11 ( 0.6)1 lim ( ) 1 lim ( 0.5173) 5 (s 0.6) (s 0.01908) P s s e s KG s s s (2) From equation (2), we get: SC 1 0.01 0.11 0.6 1 0 5 0.6 0.01908 PK , which yields: KP SC = 85.9. b. When the speed controller is configured as a proportional plus integral controller, the forward-path transfer function of the system becomes: SC 0.11 ( 0.6) (100 ) ( ) ( 0.5173) 5 (s 0.6) (s 0.01908) Is s K G s s s s (3) For the steady-state error for a unit-ramp input, r(t) = t u(t), to be equal to 2.5%: SC ramp 0 0 1 1 ( ) 0.025 lim ( ) 0.11 ( 0.6) (100 ) lim ( 0.5173) 5 (s 0.6) (s 0.01908) I s s e sG s s s K s s s s (4) From equation (4), we get: SC 1 0.025 0.11 0.6 0 5 0.6 0.01908 IK , which yields: KI SC = 34.7. c. We’ll start by finding G1(s), the equivalent transfer function of the parallel combination, representing the torque and speed controllers, shown in Figure P7.35: 2 1 13.53 3 ( 0.6) 100 40 313.53 300 72 ( ) ( 0.5) ( 0.5) ( 0.5) s s s s s G s s s s s s (5) Given that the equivalent transfer function of the car is: 3 2 6.13 10 ( ) 0.01908 G s s , we apply equation 7.62 * of the text taking into consideration that the disturbance here is a step with a magnitude equal to 83.7: 1 0 0 2 83.7 83.7 ( ) 0 1 3.11 lim lim ( ) ( )s s e G s G s Solutions to Design Problems 7-47 68. a. The system is type 0 so 0lim 0.7p sK KP s K . It follows that 1 1 0.03 1 1 0.7 ss p e K K . So 46.19K . From Problem 76, Chapter 6, the system is closed-loop stable for K < 9.63, so this steady-state error is not achievable. b. Due to stability constraints, the minimum steady state error for a unit step input is achievable when K = 9.63. 1 0.1292 1 9.63 0.7 sse or 12.92% c. For zero steady-state error to a step input, the system must be augmented to Type 1. The simplest compensator that can be used to achieve this is an integrator, namely C K G s . ONLINEFFIRS 11/25/2014 13:29:37 Page 1 Copyright 2015 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201)748-6011, fax (201)748-6008, website http://www.wiley.com/go/permissions. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Ourcompany is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. The software programs and experiments available with this book have been included for their instructional value. They have been tested with care but are not guaranteed for any particular purpose. The publisher and author do not offer any warranties or restrictions, nor do they accept any liabilities with respect to the programs and experiments. AMTRAK is a registered trademark of National Railroad Passenger Corporation. Adobe and Acrobat are trademarks of Adobe Systems, Inc. which may be registered in some jurisdictions. FANUC is a registered trademark of FANUC, Ltd. Microsoft, Visual Basic, and PowerPoint are registered trademarks of Microsoft Corporation. QuickBasic is a trademark of Microsoft Corporation. MATLAB and SIMULINK are registered trademarks of The MathWorks, Inc. The Control System Toolbox, LTI Viewer, Root Locus Design GUI, Symbolic Math Toolbox, Simulink Control Design, and MathWorks are trademarks of The MathWorks, Inc. LabVIEW is a registered trademark of National Instruments Corporation. Segway is a registered trademark of Segway, Inc. in the United States and/or other countries. Chevrolet Volt is a trademark of General Motors LLC. Virtual plant simulations pictured and referred to herein are trademarks or registered trademarks of Quanser Inc. and/or its affiliates. 2010 Quanser Inc. All rights reserved. Quanser virtual plant simulations pictured and referred to herein may be subject to change without notice. ASIMO is a registered trademark of Honda. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel. Outside of the United States, please contact your local representative. 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 http://www.wiley.com/go/permissions http://www.wiley.com/go/citizenship http://www.wiley.com/go/returnlabel
