6.447 rad/sec and ln (% OS /100) tan Thus pc. the testing point is s 3.33 j5.519

Transcription

1 9. a. 3.33, n T ln(% OS /100) 2 2 ln (% OS /100) Thu n 6.7 rad/ec and the teting point i 3.33 j b. Summation of angle including the compenating zero i , The compenator pole mut contribute = c. Uing the geometry hown below p c tan Thu pc.985 d. Adding the compenator pole tet for j5.519 which give K=187. e. Searching the real axi egment for K=187 we find higher order pole at and f. The pole at i far away to the left. The pole at may not cancel the zero at -1. Quetionable econd order approximation the ytem hould be imulated. g.

2 A imulation of the ytem how a percent overhoot of 56.8% and a ettling time of 2.31 ec. Thi the pecification were not met becaue pole-zero cancellation wa not achieved. A redeign i required. 10. %OS -ln ( ) a. n = T = 2.; = 100 = 0.5. Thu, n =.799 rad/ and the operating %OS π 2 ln 2 ( ) 100 point i -2. ± j.16. b. Summation of angle including the compenating zero i o. Therefore, the compenator pole mut contribute 180 o o = -8.6 o. Uing the geometry hown below, tan 8.6 o. Thu, p c = pc - 2. = c. Adding the compenator pole and uing j.16 a the tet point, K = d. Searching the real axi egment for K = , we find a higher-order pole at e. Pole at i near the zero at -1. Simulate to enure accuracy of reult. f. K a = g. =.8 From the plot, T = 1. econd; T p = 0.68 econd; %OS = 35%.

3 11. a. b. and c. Searching along the = 0.8 line (13.13 o ), find the operating point at j2.38 with K =5.9. d. Since n = T, the real part of the compenated dominant pole i -. The imaginary part i tan (180 o o ) = 3. Uing the uncompenated ytem' pole and zero along with the compenator zero at -.5, the ummation of angle to the deign point, - + j3 i o. Thu, the contribution of the compenator pole mut be o = Uing the following geometry, 3 = tan 2.5 0, or pc = p c

4 Adding the compenator pole and uing + j3 a the tet point, K = e. Compenated: Searching the real axi egment for K = 105.6, we find higher-order pole at 12.32, and approximately at.71 ± Since there i no pole/zero cancellation with the zero at -6 and.5, the ytem hould be imulated to check the ettling time. f. The graph how about 2% overhoot and a 1.1 econd ettling time compared to a deired 1.52% overhoot and a ettling time of 1 econd.

5 1. Uncompenated: Search along the 0.6 line and find the operating point j1.23with K A damping factor of 0.6 correpond to a %OS=9.8%. T.36 ec K There i a higher order pole at p 0.7 Compenated: The PI controller i choen a Gc () Search along the 0.6 line and find the operating point j0.81 with K T 6.27 ec. K p The two higher order pole are at -0.6 and The ytem mut be imulated to verify performance:

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7 a. Inert a cacade compenator, uch a Gc( ). b. Program: K=1 G1=zpk([],[0,-3,-6],K) %G1=1/(+3)(+6) Gc=zpk([-0.01],[0],1) %Gc=(+0.01)/ G=G1*Gc rlocu(g) T=feedback(G,1) T1=tf(1,[1,0]) %Form 1/ to integrate tep input T2=T*T1 t=0:0.1:200; tep(t1,t2,t) %Show input ramp and ramp repone Computer repone: K = (+3) (+6) (+0.01) (+0.01) ^2 (+3) (+6) (+0.01) (+6.05) (+2.889) (+0.038) ( ) Tranfer function: 1 -

8 (+0.01) (+6.05) (+2.889) (+0.038) ( )