Question

Problem 4. Consider the control system shown below with plant G(s) that has time con- stants T1 = 2, T2 = 10, and gain k = 0.1. 4 673 +1679+1) (1.) Sketch the pole-zero plot for G(s). Is one of the poles more dominant? Using MATLAB, simulate the step response of the plant itself, along with G1(s) and G2(s) as defined by Gl(s) = and G2(s) = sti + 1 ST2+1 (2.) Design a proportional gain C(s) = K so that the system has a steady state error of 5% for a step input. Simulate the resulting closed loop system's step response via MATLAB or Simulink. Compare the results. (3.) Now let the controller be given by the PI compensator C(s) = 50 (1+1) C(s) = 50 (+105) Determine the steady state error to a step input, as well as the steady state error to a ramp input. Simulate the step response of the system via MATLAB or Simulink. Examine the closed loop transfer function. Sketch the pole zero plot. What do you notice? Estimate the settling time and overshoot from the closed loop transfer function.

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, -0.5 Sz: -0.1 I ſ imaginary Ancie. cons - IT Real axis The pole at .ool is more dominant Than The pole at saros since the pole is more than s time in magnitude of the pole at si-ol - The open loop Transfa function is Gs) = K x Cis) (25+1) (105+1) Since this is a type-o system the steady state end to unit stop is esa 1 It G10) Cs)= kc (25+1) Clost1]

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 The open loop Transfer function Gosss 5 (105+1)x 0. s (25+1) (1os+1) since this ha type-t system The steady state end to unit step is 29o. Stead state ena to unit Rampig essa to where Ke=hto SxG) sto 's Kios teis. $* 5 605 yi) x $ 25+1) (105+1) so es= t = 2 0.5

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 0.007 0.005 - -- - - Amplitude - -- -- - 0.00 - - - 0.003 0.001 d 30 40 50 Time (seconds) Property Editor - timeplot

File Edit View Insert Tools Debug Desktop Window Help xa 08H8h a my V I E Figure 2 x OBD Step Response System: untitled1 Peak amplitude: 1.3 Overshoot (%): 37.1 At time (seconds): 3.22 System: untitled1 Settling time (seconds): 11.2 Tin . . . . . . _ ._._._._ SIL n System: untitled1 Rise time (seconds): 1.32 Amplitude 10 Time (seconds) Property Editor - Waveform

File Edit View Insert Tools Debug Desktop Window Help xa noua ķ Benny S. & I E DI Figure 3 x 1 OBD Step Response System: untitled1 Peak amplitude: 1.16 | Overshoot (%): 16.3 At time (seconds): 7.18 System: untitled1 Settling time (seconds): 16.2 -.-.-.-.-.-.-.-.- System: untitled1 Rise time (seconds): 3.28 -.- .- .- Amplitude .- .- .- .- -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- .- .- .- .- .- .- .- .- .- .- 10 15 Time (seconds) Property Editor - Waveform

File 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: untitled1 Pole : -0.25 +0.4331 Damping: 0.5 Overshoot (%): 16.3 Frequency (rad/s): 0.5 System: untitled1 Zero -0.1 Damping: 1 Overshoot (%): 0 Frequency (rad/s): 0.1 Imaginary Axis (seconds) System: untitled1 Pole : -0.25 -0.4331 Damping: 0.5 Overshoot (%): 16.3 Frequency (rad/s): 0.5 -0.4 -0.5 -0.25 -0.2 -0.15 -0.05 Real Axis (seconds) Property Editor - Figure