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Problem 4. Consider the control system shown below with plant G(s) that has time con- stants T1 = 2, T2 = 10, and gain k = 0.

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Answer #1

SA- from the given inputs 6,18)= 0.1 QUE 0.1 (25+1) (10$+1) pole zero plot. There are no reos, and have a open loop poles S,ey= esa 5 i DO Too = 1 It o.1 xkc 10+1) (0+1) los !+ Olke = 100 = 20 - Olke=19 so kc = 190 () (15)= 50 (it has) = 5a (los. I

Here we can see the pole zero cancellation;

matlab code :

clc;

clear all;

close all;

s=tf('s');% definition of transfer function

g1=0.1/(1+2*s); % plant 1

g2=0.1/(1+10*s);% plant 2

step(feedback(g1*g2,1));grid % step response without controller

g=0.1/((1+2*s)*(1+10*s)); % plant for question 2)

kc=190; % proportional gain

figure

step(feedback(g*kc,1));grid % step response with ptoportional controller

gc=5*(10*s+1)/s; % PI controller

figure

step(feedback(g*gc,1));grid % step response with PI controller

figure

pzmap(feedback(g*gc,1)); % pole zero plot of closed loop ploes

File Edit View Insert Tools Debug Desktop Window Help x = DISA .. ZO E DJ Figure 1 x | ID E F D) Step Response 0.009 0.00B 0.File Edit View Insert Tools Debug Desktop Window Help xa 08H8h a my V I E Figure 2 x OBD Step Response System: untitled1 PeakFile Edit View Insert Tools Debug Desktop Window Help xa noua ķ Benny S. & I E DI Figure 3 x 1 OBD Step Response System: untiFile Edit View Insert Tools Debug Desktop Window Help xa no 8.3Ą anys 3 DE Figure 4 X O BD Pole-Zero Map 0.4 System: untitled

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