Question

steps

R(s) E(s) C(s) G(s) FIGURE P9.1 FIGURE P9.2 9. Consider the unity feedback system shown in Figure P9.1 with [Section: 9.3] K G(s) (s+4)3 a. Find the location of the dominant poles to yield a 1.6 second settling time and an overshoot of 25%. b. If a compensator with a zero at -1 is used to achieve the conditions of Part a, what must the angular contribution of the compensator pole be? c. Find the location of the compensator pole d. Find the gain required to meet the requirements stated in Part a. e. Find the location of other closed-loop poles for the compensated system f. Discuss the validity of your second-order approximation

Answer 1

c(s) K EA RiA Ceat trauafer fnctian ndanped R ay wt the lamping fator aud the the dominant ples be matural npictivaly and Wn 4 16 3aY2.5 setling time, ts -n O.25 FOvershoot, ose 0.25 Ni-82 m(0.25 N-AY 0.)9472 (-ty 019472 O.16298 119472 0.4037 2.5 -6.1927. 0.4037 Aue deninant pdles Loattar - 20htjanT- 2.5j 250 I 5.6656 SAu

(b) let kst1) augle Cantributian G) Ge) at s-2.5js.cuse) The Ahould be -Igo. C154356655) AE2.51js) 2599.5 S.I 3 tau the augle Contributian of where is the Compinsator ale - 3 00-683-fp1 4p=-120.683. &O9 he Compeniontr ole should be Angular Camti butian -120. 683 us L1599 120%683 (-120.683) - 1.6853 tru LCP-2.5) 5.6656 tru p-2.5 (P-2.5)=(S.6656 I6853 p 5.8617 -S.8617 Locati n tue

d) = I. A=(-2.5+) 5.66 56) k(-25f) 56456 +1) 2.5135 CasC+ 5.8017) 168 t X16)k - L 4.41916 K= 226.287 the reguiraneut statek n Part(a) ik = 226- 287. CAu Gain Aeqirone ured tmeet np characteristics egp r- KIATI)KIAPI) 1t + 226-237 ()-0 +22G.287 (At1= O t 1264 4884 044)+ 8617570.3404 375.1488)+ 226.287(st)-0 ( 17-8617t18.3404571.69863+0 36166t of 14364,-14973, 2.5j5 6c 57 2outian the ab 2uation etseo clones Loop aes for tue Camperrsatid yatam: (Amy

ples BA11-4973 Rnot a dminant ple So etfect f 55 6657 pde ifecting tarle not 1.364 he 2.5 8--11-4973 prominent in the Respamae of Second-oror Rppox čmatian 4he Agl: poler zero map of tue clored lorp ysten appreximated second arden sten fg.2: to shown in The hene Ghe G-1927 rndfs 20.4037. RIS 8(s4250n fz Aetnd ordev approxinuatin the givew ysten Transfer fnction f riginal nysteu and Lerived below'- the approximalzd yteum the ahe riginel 4pproxincated GcSs) GA) KAF TaA)-TAT 25 226.287 &P423 S4S.8617 Shere ki=f 22 6.287 Gol 4358 I+ Gt 226. 287 = O.37C24 226.287 (6+1) (178617 +118.34048+591.64863 G01.4358) ALthed mltans pprexinntan The MAT LA8 are Sep reaponae uf buth the Tne alidity tae hown at end the auser to Lne saript and plot ane from tue respomses, though the almet same fr buth the canes,the % seenn thaE can be sett ting Reak overshoot and nse time are dterent. Ths, the toangient time the appreximate d rysten doeo not the'approinaled sten. Wence, the approximatin in ughhy valid lorely frllers that

MATLAB SCRIPT:

clear all;
clc;
num1=[226.287 226.287];
den1=[1 17.8617 118.3404 571.6486 601.4358];
sys1=tf(num1,den1);
num2=[14.4286];
den2=[1 5 38.35];
sys2=tf(num2,den2);
step(sys1);hold on;grid
step(sys2)
legend('response of the original system','response of the approximated system')

Response of the Systems:

Step Response 0.6 response of the original system response of the approximated system 0.5 0.4 0.3 0.2 0. 0.5 1.5 2.5 Time (seconds) Amplitude