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 1 2 2 4 4 2 2 4 1 2 - + - + z z 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 2 2 + p gives z = 0.535 for M p = 0.135 p w z n 1 2 - = 0.185 gives w n = 20 The corresponding frequency response performance indices are: M r = 1 2 1 2 z z - = 1.11 ; w r = w z n 1 2 2 - = 13.25
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90 CONTROL SYSTEMS: PRINCIPLES AND DESIGN w b = w z z z n 1 2 2 4 4 2 2 4 1 2 - + - + / = 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 4 2 2 4 2 K K K t m t m m t · - · - - - - f t f t 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 t t t t 1 2 1 2 + Therefore G m × K t t t t 1 2 1 2 + = 1 This gives K = 1 1 1 1 2 G m t t + 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 4 1 2 4 2 z z z + - F M = 59.2º fi z = 0.6 w g = w z z n 4 1 2 4 2 + - This gives w n = 1.4 Therefore, M p = e - - pz z / 1 2 = 9.48%
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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 ) = 8 1 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 s s s s 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 ) = K j 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. K j 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 0 64 .
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