hwsln-421-s06-04

hwsln-421-s06-04 - ECE 421 Spring 2006 Solutions to HW...

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ECE 421, Spring 2006 Solutions to HW Assignment #4 Problem B-5-9 For the original con f guration, ω n = 10 = 3 . 162 r/s and ζ =1 / 2 ω n =0 . 158 . In the second con f guration, the equivalent open-loop transfer function is G eq ( s )= 1 s · 10 s +1+10 K h = ω 2 n s ( s +2 ζω n ) (1) The undamped natural frequency is unchanged, but now the damping ratio is changed. To solve for the gain to get ζ . 5 we equate the coe cients of like powers of s in Eqn. (1). 1+10 K h =2 n K h = 2 n 1 10 = ω n 1 10 . 2162 . (2) MATLAB can be used to generate the plots. The following code can be used. t = linspace(0,10,1001); Kh = 0.2162; [ncl1,dcl1] = feedback(10,[1 1 0],1,1,-1); [ncl2,dcl2] = feedback(10,[1 1+10*Kh 0],1,1,-1); ys1 = step(ncl1,dcl1,t); ys2 = step(ncl2,dcl2,t); yr1 = step(ncl1,[dcl1 0],t); yr2 = step(ncl2,[dcl2 0],t); er1 = t’ - yr1; er2 = t’ - yr2; figure(1),clf,subplot(211),plot(t,ys1,t,ys2),grid,xlabel(’Time (s)’),ylabel(’Amplitude’),. .. title(’Step Responses’),. .. subplot(212),plot(t,er1,t,er2),grid,xlabel(’Time (s)’),ylabel(’Amplitude’),. .. title(’Errors in Ramp Responses’) Problem B-5-10 G ( s K s ( s +2+ Kk ) = ω 2 n s ( s n ) = 16 s ( s +5 . 6) (3) K = ω 2 n K =16 (4) 2+ n k = 2 n 2 K . 225 (5) Problem B-5-11 G ( s 16 s ( s +0 . 8+16 k ) = ω 2 n s ( s n ) ω n =4 r/s (6) 1
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0 . 8+16 k =2 ζω n k = 2 n 0 . 8 16 =0 . 200 (7) T r = π cos 1 ( ζ ) ω n p 1 ζ 2 . 6046 sec ,T p = π ω n p 1 ζ 2 . 9069 sec (8) T s = 4 n =2 sec ,P O = e πζ/ 1 ζ 2 · 100% = 16 . 3% or 0 . 163 (9) Problem B-5-27 For the original system G ( s )= K/J s 2 = 4 s 2 = ω 2 n s ( s +2 n ) (10) so ζ and ω n = 4=2 r/s. For the second con
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This note was uploaded on 10/25/2010 for the course ECE 421 taught by Professor Cook,g during the Spring '08 term at George Mason.

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hwsln-421-s06-04 - ECE 421 Spring 2006 Solutions to HW...

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