HW11_solutions - EECS 460 Winter 2008: Homework #11...

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EECS 460 Winter 2008: Homework #11 Solutions 9.28 (a) Gain crossover frequency = 2.09 rad/sec PM = 115.85 deg Phase crossover frequency = 20.31 rad/sec GM = 21.13 dB (b) Gain crossover frequency = 6.63 rad/sec PM = 72.08 deg Phase crossover frequency = 20.31 rad/sec GM = 15.11 dB (c) Gain crossover frequency = 19.1 rad/sec PM = 4.07 deg Phase crossover frequency = 20.31 rad/sec GM = 1.13 dB (d) For GM = 40 dB, reduce gain by (40 21.13) dB = 18.7 dB, or gain = 0.116 × nominal value. (e) For PM = 45 deg, the magnitude curve reads 10 dB. This means that the loop gain can be increased by 10 dB from the nominal value. Or gain = 3.16 × nominal value. (f) The system is type 1, since the slope of Gj () ω is 20 dB/decade as 0. Therefore e ss = 0. (g) Since a phase delay only affects the phase bode plot, simply subtract an angle of 0.1 ω from the phase plot (over all ω ) and recomputed the gain and phase margins as GM = 12.7 dB. PM = 109.85 deg. (h) e -T s contributes a angle of -T s ω . For instability, we want -180 o of phase at 0dB. At 0dB, the phase is currently -64.15 o ( ω = 2.09 rad/s). Therefore, we need to add 115.85 o phase lag. This corresponds π TT dd o o == = 209 11585 180 2 022 . . . rad Thus, the maximum time delay is T d = 0 9674 . sec.
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9-30 (a) Bode Plot: For stability: 166 (44.4 dB) < K < 7079 (77 dB) Phase crossover frequencies: 7 rad/sec and 85 rad/sec Nyquist Plot:
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9-30 (b) Root Loci. Note that at -180 o , ω = 7.16 rad/s and ω = 85.3 rad/s which match the results calculated in part (a). 9-39 (a) Forward-path Transfer Function: Gs Ys Es e ss s () () ( ) ( ) == ++ 2 11 0 12 5 From the Bode diagram, phase crossover frequency = 0.21 rad/sec GM = 21.55 dB gain crossover frequency = 0 rad/sec PM = infinite (b) Approximate the time delay using a second order approximation to obtain: 2 2 2 1 1 2 PM = GM=25dB 1 1 2 s s e s ⎛⎞ ⎜⎟ ⎝⎠ ≈⇒ +
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(c) Approximate the time delay using a first order approximation to obtain the following results: 2 1 PM = GM=25dB (around 0.3 rad/s) 1 s s e s ≈⇒ + -150 -100 -50 0 Magnitude (dB) 10 -3 10 -2 10 -1 10 0 10 1 10 2 -180 0 180 360 Phase (deg) Bode Diagram Frequency (rad/sec) Plant 1st order Delay 2nd order Delay
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Problem 10-20 () 2 100 10 100 Gs ss
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HW11_solutions - EECS 460 Winter 2008: Homework #11...

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