laboratory in automatic control lab10

8/2/2019 Laboratory in Automatic Control Lab10

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Laboratory in Automatic Control

LAB 10

System Design Using Classical Methods


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Design a Lead Compensator (1/8)

As shown in the following figure, desire a steady-state error

of less than 10% to a ramp input, Kv=10 and the leadcompensator to meet certain performance specifications: (1)

settling time (with a 2% criterion) Ts


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The steady-state error to a unit ramp input is

where

Consider a simple gain controller

then

1ss

v

eK

0lim 5 10 50

c cvs

G s G sK s

s s s

cG s K

10 500

50

v

KK K


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Solving for and using

We thus obtain the phase margin requirement:

n

21

. . % 100exp 10 0.59

43 2.26

s n

n

P O

T

60

0.01

pm

Textbook P.521


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MATLAB code

k=500;numg=[1]; deng=[1 15 50 0];sysg=tf(numg,deng); sys=k*sysg;[Gm,Pm,Wcg,Wcp]=margin(sys); % Compute phase margin.

Phi=(60-Pm)*pi/180; % Additional phase lead.alpha=(1+sin(Phi))/(1-sin(Phi)); %Compute alpha. Textbook P.591[mag,phase,w]=bode(sys);mag_save(1,:)=mag(:,:,:);M=-10*log10(alpha)*ones(length(w),1);

semilogx(w,20*log10(mag_save),w,M,'--')xlabel('Frequency (rad/sec)'),ylabel('Magnitude (dB)')hold onsemilogx([0.9 9 90],[20 0 -20],'--'),grid on


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Uncompensated

3.

25

5c

K sG s

s

10log


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MATLAB code

numg=[1];deng=[1 15 50 0];dengc=[1 25];for K=1:50:2000

numgc=K*[1 3.5];sysg=tf(numg,deng);sysgc=tf(numgc,dengc);sys1=series(sysgc,sysg);[Gm,Pm,Wg,Wc]=margin(sys1);if(Pm


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The final lead compensator design is

Resulting in a 20% steady-state error to a ramp input

1801 3.525

csG s

s


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Design a Lag Compensator (1/6)

As shown in the following figure, desire a steady-state error

of less than 10% to a ramp input, Kv=10 and the lagcompensator to meet certain performance specifications: (1)

settling time (with a 2% criterion) Ts


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Solving for and using

We thus obtain the phase margin requirement

21

. . % 100exp 10 0.59

43 2.26

s n

n

P O

T

600.01

pm

n


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MATLAB code

numg=[1]; deng=[1 15 50 0];sysg=tf(numg,deng);clf;% Clear current figurerlocus(sysg); hold onzeta=0.5912; wn=2.2555;x=[-10:0.1:-zeta*wn];y=-(sqrt(1-zeta^2)/zeta)*x;xc=[-10:0.1:-zeta*wn];c=sqrt(wn^2-xc.^2);plot(x,y,':',x,-y,':',xc,c,':',xc,-c,':')axis([-15,1,-10,10]);rlocfind(sysg)

Plot performance regions on locus.


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Compute

Select

Uncompensated K

100.1

103

10.1

comp

uncomp

v

v

K

K

zp z

p

0.1

0.01

z

p


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MATLAB code

numg=[1]; deng=[1 15 50 0];sysg=tf(numg,deng);numgc=[1 0.1]; dengc=[1 0.01];sysgc=tf(numgc,dengc);

sys=series(sysgc,sysg);clf;rlocus(sysg); hold onzeta=0.5912; wn=2.2555;x=[-10:0.1:-zeta*wn];y=-(sqrt(1-zeta^2)/zeta)*x;

xc=[-10:0.1:-zeta*wn];c=sqrt(wn^2-xc.^2);plot(x,y,':',x,-y,':',xc,c,':',xc,-c,':')axis([-15,1,-10,10]);

Compensated root locus remainsalmost unchanged.


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MATLAB code

K=100;numg=[1]; deng=[1 15 50 0];sysg=tf(numg,deng);numgc=K*[1 0.1]; dengc=[1 0.01];

sysgc=tf(numgc,dengc);syso=series(sysgc,sysg);sys=feedback(syso,[1]);step(sys);[gm,pm,wg,wp]=margin(syso);

pm


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Lab 10 Homework

Consider the control system shown in Figure. Design a lead

compensator using bode plot methods to meet the followingspecifications: (1) percent overshoot for a step input

laboratory in automatic control lab10

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