Solutions_F11_ECH157_Quiz2

Solutions_F11_ECH157_Quiz2 - 11.15 Quiz II Solution Key a)...

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Unformatted text preview: 11.15 Quiz II Solution Key a) Kc K Kc K K K /(1 + K c K ) Y ( s) = 1 − τs = =c K K 1 − τs + K c K τ Ysp ( s ) − s +1 1+ c 1 + Kc K 1 − τs For stability − τ >0 1+ KcK Since τ is positive, the denominator must be negative, i.e., 1+ Kc K < 0 K c K < −1 K c < −1 / K (for K>0) Kc K K CL = 1+ Kc K Note that b) K = 10 and τ = 20 and we want or − τ = 10 1+ KcK − 20 = 10 + (10)(10) K c − 30 = 100 K c K c = −0.3 Offset: K CL = (−0.3)(10) −3 = = 1.5 1 + (−0.3)(10) − 2 ∴ Offset = +1 − 1.5 = − 50% (Note this result implies overshoot) 11-19 c) KcK (1 − τs )(τ m s + 1) K c K (1+tau_m*s) Y ( s) = = Kc K Ysp ( s ) (1 − τs )(τ m s + 1) + K c K 1+ (1 − τs )(τ m s + 1) = = K c K (1+tau_m*s) − ττ m s + (τ m − τ) s + 1 + K c K 2 K c K /(1 + K c K ) (1+tau_m*s) (standard form) ττ m τm − τ 2 − s+ s +1 1 + Kc K 1 + Kc K For stability, (1) − ττ m >0 1+ Kc K From (1) Since (2) τm − τ >0 1+ Kc K 1+ Kc K < 0 K c K < −1 1 Kc < − K Since 1 + K c K < 0 τm − τ < 0 − τ < −τ m τ > τm K = 10 , τ = 20 , Kc = –0.3 , τm = 5 From (2) For Y ( s) = Ysp ( s ) 1.5 1.5 = 2 (20)(5) 2 (5 − 20) 50s + 7.5s + 1 2.5s − s+ s +1 1− 3 (1 − 3) Underdamped but stable. 11-20 c ...
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This note was uploaded on 01/28/2012 for the course ECH 157 taught by Professor Palagozu during the Fall '08 term at UC Davis.

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