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me375_fa10_hwk11_soln

me375_fa10_hwk11_soln - ccs Gucfi kUA 1 4 a g{‘1 u:7...

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Unformatted text preview: - ccs Gucfi kUA 1 4- a, g, {‘1 u :7 g (97.1“?) : 0 Va 1 + . h snhtu-s’) 1- . . (an. (us ‘ - , ‘ - ., r -‘5 II 1 - s u. ' - ‘ y ’ 27 t 1 - 5'0 ' = ' O ‘5 .3 'C _— 4| " A; .I i' ‘I. 1+ _' J~ . ‘7 / NM ‘A V. S 4 J! - O [a ’ ‘. I An 'Ad ‘. I .fl'l _ __-__ ‘0 7, .7 an " I. u —I& .‘. _ > - t , ‘ ’ . O '- oo . I...“ .5 =' o > ’”‘_/W 0' ’ . .II” 00 0 g f‘ .6 = a. Ft: I" ,.3' K-+K1 -. " +L + ‘ht ‘ v; z k +V51: 1 L +- A A L "' .. , G.Ls\ + .. . FLS SCSS?% ! .0... C .- 0 .0 s , v A .- f‘ ‘- u—n— 1 L * h‘ - :. - An, = L: a. = -0... 0‘ I: o + ' u i ‘ 7- ‘ , I I nc . in - l I ‘ ... o 30' l- 10.19 The steady-state ofl'set error is KP ofl'seterror=w,-wu=l— 3+KP =0.2 if K}: = 12. The time constant for this value of K p is T ___ I ___ 2 = 3 “ c + KP 3 + K p 15 The steady-state response due to the disturbance is disturbance response = c :10, = 1—; = -0.067 If both inputs are applied, the actual steady-state speed is w” = command response + disturbance response = (1 — 0.2) + (—0.067) = 0.733 Note that we can make the offset error. the disturbance response, and the time constant smaller only by making K ,2 larger than 12, but this increases the maximum required torque and probably the cost of the system. ©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 10.32 Because C = cos B, the value C = 0.707 corresponds to £3 = 45°, which is a 45° line on the complex plane. Thus C = 0.707 corresponds to a pair of roots whose real and imaginary parts have the same magnitude; that is, s = —a :l: ja. The time constant of these roots is r = l/a. Thus the roots must be 5 = -l:tj if-r = 1 and C = 0.707. These roots correspond to the polynomial equation (3+1—j)(s+1+j)=(s+1)2+1=s’+2s+2=o or 10::2 +203+ 20: 0 Compare this with the system’s characteristic equation obtained from the denominator of the transfer fimction: 1032 + (3 + Kn): + Kp = 0 Thus Kp=20and 3+KD=20, or KD=17. ©2010 McGrew-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 10.40 The system's closed-loop transfer function can be found from the block diagram. It ‘8 C(s) _ Kp + Km R(s) ~ 232 + (2+ 1(0):! + KP Thus the characteristic equation is 232 + (2 + K D)s + K p = 0. Because C is specified to be less than 1. we can write the following formula for the time constant. 2(2) _ 1 T=2+KD— This gives K D = 2. The damping ratio is given by _2+KD_ 2+2 2 — 2J2Kp — 2J2Kp = JZKP =03 C This gives K p = 2.469. If the command is a step input with a magnitude m, then R(s) = m/s, and the Final Value Theorem gives _=m cu=ali_r.r(i,sC(s)=lims( s Kp-l-KDS )m 3-00 282+(2+KD)8+KP Thus the steady state error is zero. so all three specifications are satisfied. ©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 10.45 a) The closed-loop transfer ftmction is C(s) _ 3K}: + 3Kps R(s) - s2 + 3Kps + 3Kp — 4 The damping ratio is 3K _ D = C — 2\/3KP-4 0.707 The time constant is 2 = = .1 1' 3KD 0 These two conditions give K p = 68 and K D = 20/3. b) The closed-loop transfer function with the negative rate feedka gain K 1 is 0(8) _ 3Kp R(s) — 82+3K18+3Kp-4 This denominator has the same form as the transfer function in part (a) with K D replaced by K1. Thus Kp = 68 and K1= 20/3. c) For part (a), C(s) _ 208 + 204 R(s) - 82 + 20.9 + 200 This has numerator dynamics and will overshoot more than the design in part (b), for which C(s) _ 204 R(s) — 32 + 208 + 200 ©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. ...
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