sol_mid

sol_mid - o , by = 0 . 707. Lead compensator is needed. We...

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Solution to Midterm Test 1. For the lead compensator, we have .2 2 200 .2 20 2 20 200 degree 20 dB 45 ϖ ϖ For the lag compensator, we have .2 2 200 degree 20 dB ϖ 20 .2 2 20 200 ϖ 45 0 1
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2. (a) For the gain plot, we choose ω L = 0 . 1, and ω H = 1000. Hence we have at ω = 0 . 1, G L = 20log 10 (10 / 0 . 1) = 40dB , slope = - 20dB / dec . After marking pole/zero, we can sketch the gain/phase plots. The corrections are at the corner frequencies. The complex poles have ω n = 10, ζ = 0 . 25, which indicates the correction at ω : 20log 10 (1 / 2 ζ ) = 6dB. 1000 100 10 1 .1 1 45 90 180 135 degee 40 20 40 20 dB 1000 100 10 .1 (b) For straightline approximations, P.M.= 90 o , but the actual phase margin can be smaller. (c) The closed-loop system is stable, because P.M. > 0. 3. We notice that if ω c 50, then the phase margin is at most 5 . 7 o , due to the poles at - 1 , - 5. Hence the phase margin need be compensated for is approximately 70
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Unformatted text preview: o , by = 0 . 707. Lead compensator is needed. We compute value for the lead compensator as = 1-sin(70 o ) 1 + sin(70 o ) = 0 . 0311 . The zero and pole of lead compensator are z lead = c = 8 . 8176 , p lad = c / = 283 . 524 . 2 Hence we obtain the lead compensator as D lead ( z ) = 1 + s/ 8 . 8176 1 + s/ 283 . 524 . The magnitude increase at c = 50 is G =-10log 10 ( ) =-15dB . To compute gain K , we have (recall c = 50) K | D lead ( j c ) G ( j c ) | . 02 K/ = 1 . Hence K = 50 = 8 . 8176. Now the position constant is: K p = 10 K = 88 . 176 50 . No lad compensator is necessary. 3...
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sol_mid - o , by = 0 . 707. Lead compensator is needed. We...

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