# Chapter9 - CHAPTER 9 9.1 9.2 9.3 9.4 FEEDBACK SYSTEM...

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CHAPTER 9 FEEDBACK SYSTEM PERFORMANCE BASED ON THE FREQUENCY RESPONSE 9.1 For derivation of the result, refer Section 9.3; Eqn. (9.9). 9.2 For derivation of the result, refer Section 9.3; Eqn. (9.18). 9.3 For derivation of the result, refer Section 9.3; Eqns (9.15) – (9.16). 9.4 The relations z 4 z 2 + 1 4 2 M r = 0 ; w r = w z n 1 2 2 - yield z = 0.6 and w n = 21.8 Note that M r z relation gives two values of z for M r = 1.8; z = 0.6 and z = 0.8. We select z = 0.6 as damping ratio larger than 0.707 yields no peak above zero frequency. The characteristic equation of the given system is s 2 + as + K = s 2 + 2 zw n s + w n 2 = 0 This gives K = 475 and a = 26.2 t s = 4 zw n = 0.305 sec w b = w n 12 24 4 22 4 -+ zz z / = 25.1 rad/sec 9.5 From the response curve, we find M p = 0.135, t p = 0.185 sec z 2 = () ln ln M M p p 2 + p gives z = 0.535 for M p = 0.135 p wz n 1 2 - = 0.185 gives w n = 20 The corresponding frequency response performance indices are: M r = 1 21 2 - = 1.11 ; w r = n 2 - = 13.25

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90 CONTROL SYSTEMS: PRINCIPLES AND DESIGN w b = wz z z n 12 24 4 22 4 -+ / = 24.6 9.6 z » 0.01 f m The characteristic equation of the system is t s 2 + (1 + KK t ) s + K = 0 This gives w n = / K t ; z = 1 2 + KK K t t = 0.01 f m K = 1 2 10 2 10 10 42 2 K KK t mt m ·- · - -- - ft f t 9.7 The polar plot of G ( j w ) crosses the real axis at w = 1/ t t 1 2 . The magnitude | G ( j w )| at this frequency is given by | G ( j w )| = K tt + Therefore G m × K + = 1 This gives K = 11 1 G m + F H G I K J 9.8 (a) It can be determined from the Bode plot that w g = 1 rad/sec and F M = 59.2º (b) F M = tan –1 2 41 2 z zz +- F M = 59.2º z = 0.6 w g = z n 2 This gives w n = 1.4 Therefore, M p = e pz z /1 2 = 9.48%
SOLUTION MANUAL 91 t s = 4 zw n = 4.76 sec. 9.9 The low-frequency asymptote has a slope of – 20 dB/decade and intersects the 0 dB-axis at w = 8; 8/ s is a factor of the transfer function. The forward- path transfer function can easily be identified as G ( s ) = 81 1 2 1 1 4 1 1 8 1 1 24 1 1 36 + F H I K + F H I K + F H I K + F H I K + F H I K ss s s The phase curve can now be constructed. From the phase curve and the asymptotic magnitude curve, we obtain F M = 50º ; GM = 24 dB Phase margin of 50º gives z = 0.48 which corresponds to M p » 18%. 9.10 (a) From the Bode plot, we get F M = 12º (b) For a phase margin of 50º, we require that G ( j w ) H ( j w ) = 1 – 130º for some value of w . From the phase curve of G ( j w ) H ( j w ), we find that G ( j w ) H ( j w ) ~ - –130º at w = 0.5. The magnitude of G ( j w ) H ( j w ) at this frequency is approximately 3.5. The gain must be reduced by a factor of 3.5 to achieve a phase margin of 50º. (c) F M of 50º gives z = 0.48 which corresponds to M p ~ - 18%. 9.11 G ( j w ) = Kj j () w w + 2 2 G ( j w ) = tan –1 w 2 – 180º = – 130º This equation gives w = 2.3835 The magnitude of G ( j w ) must be unity at w = 2.3835. j . w w w + = 2 2 2 3835 = 1 This equation gives K = 1.826. Since the phase curve never reaches – 180º line, the gain margin = ¥ . 9.12 (a) From the open-loop frequency response table, we find that w f = 10 rad/sec and GM = 20 log 1 064 . = 3.88 dB; w g = 8 rad/sec and F M = 10º.

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92 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (b) For the desired gain margin of 20 dB, we must decrease the gain by (20 – 3.88) = 16.12 dB. It means that magnitude curve must be lowered by 16.12 dB, i.e., the gain must be changed by a factor of b , given by 20 log b = – 16.12. This gives b = 0.156.
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## This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.

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Chapter9 - CHAPTER 9 9.1 9.2 9.3 9.4 FEEDBACK SYSTEM...

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