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note6 - Example Prob 6.50 on page 456 K s1 s=51 s=20...

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Unformatted text preview: Example Prob. 6.50 on page 456: K : s1 + s=51 + s=20 Design a compensator such that KGs = ess  0:01 for unit ramp input. P.M.= 45o  3o. ess  1=250 for sinusoidal input with ! 0:2. Noise be attenuated by a factor of 100 for !  100. Analysis: ess to  1=250 for sinusoidal input with ! 0:2 is equivalent j1 + DGj!j  250 = jDGj!j  251 = 48dB; for all ! 0:2. Noise be attenuated by a factor of 100 for !  100 is equivalent to that jDGj!j  0:01 j1 + DGj!j that is guaranteed by jDGj!j  0:01 = ,40dB: 1 Design: ,50dB We choose K = 20 such that KGs attenuated to for !  100. It gives the Bode plot of Gain dB 50 0 −50 −100 −1 10 0 1 10 10 2 10 Frequency (rad/sec) Phase deg 0 −90 −180 −270 −1 10 0 1 10 10 2 10 Frequency (rad/sec) K = 20 implies that Kv = 20. Hence we need boost the low frequency gain by a factor of 5. Moreover it will also provide the necessary gain to achieve speci cation for ! 0:2. 2 We thus take = 5 for lag compensator. Since !c  8:6, zlag = 1; plag = 0:2 are chosen that gives s+1 : s + 0:2 We make Bode plot for KDlagG as follows: Dlags = 100 Gain dB 50 0 −50 −100 −1 10 0 1 10 10 2 10 Phase deg Frequency (rad/sec) −180 −270 −1 10 0 1 10 10 Frequency (rad/sec) It is seen that P.M. = 0. We need design lead. 3 2 10 By P:M: = 45o, we take 1 , sin 45o = 0:1716: = 1 + sin 45o However ,20 log10  = 15:3dB. Thus high frequency gain will add 15:3dB that will violate the ,40dB requirement. We thus need iterate our design. Iterative Design I: We choose a smaller K : K = 10 that gives Bode plot of KGs: Gain dB 50 0 −50 −100 −1 10 0 1 10 10 2 10 Frequency (rad/sec) Phase deg 0 −90 −180 −270 −1 10 0 1 10 10 Frequency (rad/sec) 4 2 10 We need take = 10 for lag compensator in order to get Kv = 100 as required. Since !c  6, we choose zlag = 1 and plag = 0:1. Hence we obtain s+1 Dlags = : s + 0:1 We now make Bode plots for KDlagsGs as follows: 100 Gain dB 50 0 −50 −100 −1 10 0 1 10 10 2 10 Phase deg Frequency (rad/sec) −180 −270 −1 10 0 1 10 10 Frequency (rad/sec) We note that at ! = 0:2, jKDlagsGsjs=0:2j = 227 250: 5 2 10 To get gain 250 at ! = 0:2, we set lag compensator as s + 1:1 Dlags = s + 0:1 that gives Kv = 110 100, and jKDlagsGsjs=0:2j  250: 100 Gain dB 50 0 −50 −100 −1 10 0 1 10 10 2 10 Phase deg Frequency (rad/sec) −180 −270 −1 10 0 1 10 10 2 10 Frequency (rad/sec) It is seen that the high frequency gain is much smaller than ,40dB. 6 We now design lead compensator. We choose !c  2!c = 14. Then at s = 14j , 0 6 KDlagG = ,196o: Hence P.M. = ,16o. The requirement of P.M. = 45o gives m = 45 + 16 = 61o: The is now given by We now have that 1 , sin 61o = 0:067: = 1 + sin 61o p = 54; zlead = !c= 0 plag = p !c = 3:623: 0 Hence, the lead compensator is 1 s + 3:623 Dleads = : 0:067 s + 54 7 We nally have Ds = KDleadsDlagsGs; K = 10; s + 1:1 1 s + 3:623 ; Dlags = : 0:067 s + 54 s + 0:1 The Bode plots of DsGs are as follows: Dleads = Gain dB 100 50 0 −50 −1 10 0 1 10 10 2 10 Frequency (rad/sec) Phase deg −120 −150 −180 −210 −240 −1 10 0 1 10 10 Frequency (rad/sec) 8 2 10 Noted that low frequency gain is satis ed: it is more than 48dB. For P.M. and high frequency gain, we magnify the Bode plot as follows: 40 Gain dB 20 0 −20 −40 0 10 1 10 Frequency (rad/sec) 2 10 Phase deg −120 −150 −180 −210 −240 0 10 1 10 Frequency (rad/sec) 2 10 P.M. 45o. But the gain at ! = 100 is 1:29=100 = ,37:8dB. We thus reduce value for lead compensator: = 0:067  1:29 = 0:0864 = p p zlead = 14  0:0864 = 4:116; plead = 14= 0:0864 = 47:6: 9 We now replace the original lead by 1 s + 4:116 : Dleads = 0:0864 s + 47:6 The Bode plots are given as follows. It can be seen that the P.M. and ,40dB for ! 100 are satis ed. 40 Gain dB 20 0 −20 −40 0 10 1 10 Frequency (rad/sec) 2 10 Phase deg 120 −150 −180 −210 −240 0 10 1 10 Frequency (rad/sec) 10 2 10 ...
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This note was uploaded on 01/23/2012 for the course EE 4580 taught by Professor Gu during the Fall '10 term at LSU.

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