Laboratory in Automatic Control

Lab 2

Simulation of System

Function (1/2)

• Syntax– w = conv(u,v)– r = roots(c)– p = poly(r)– v = polyval(p,x)

• Matlab code

3 2

( ) 1 1 1

( ) 3 3 1

( 5) 64

p s s s s

q s ss s

p

Example

p=conv([1 1],conv([1 1],[1 1]))r=roots(p)q=[1 3 3 1]roots(q)poly(r)polyval(p,-5)

Function (2/2)

Transfer function (1/11)

• Syntax– sys = tf(num,den)

• Matlab code

2

( )( )

( )

10

5

Y s numG s

U s den

s

Method 1num=[10];den=[1 0 5];G1=tf(num,den);

Method 2G2=tf([10],[1 0 5]);

Example

U(s) Y(s)System

Transfer function (2/11)

( 1)( 2)( )

( 2 )( 2 )( 3)

s sH s

s i s i s

n1=[1 1];n2=[1 2];d1=[1 2*i];d2=[1 -2*i];d3=[1 3];% d1=[1 2*j];d2=[1 -2*j];d3=[1 3];num=conv(n1,n2);den=conv(d1,conv(d2,d3));H=tf(num,den)

Example

Transfer function (3/11)

• Syntax– step(sys)– step(sys,t)

– step(sys1,sys2,...,sysN)– step(sys1,sys2,...,sysN,

t)– step(sys1,'PlotStyle1',...,

sysN,'PlotStyleN')

( ) 10( )

( ) 5

Y s kG s

U s s

Example

U(s) Y(s)System

U(s): unit step signal

K=1,2,5

Transfer function (4/11)

• Matlab code

k=1;G1=tf(10*k,[1 5])k=2;G2=tf(10*k,[1 5])k=5;G3=tf(10*k,[1 5])step(G1,'-',G2,'*',G3,'o')

Transfer function (5/11)

• Matlab code

k=1;G1=tf(10*k,[1 5])k=2;G2=tf(10*k,[1 5])k=5;G3=tf(10*k,[1 5])step(G1,'-',G2,'*',G3,'o',5)

Transfer function (6/11)

• Syntax– p = pole(sys)– z = zero(sys)

– pzmap(sys)– pzmap(sys1,sys2,...,sysN)

Transfer function (7/11)

2

1( )

3 2

sG s

ss

• Matlab code

G=tf([1 1],[1 3 2])pole(G)zero(G)pzmap(G)

Example

Transfer function (8/11)

Transfer function (9/11)

• Syntax– [z,p,k] = tf2zp(num,den)– [num,den] = zp2tf(z,p,k)

• Matlab code

Example

% Transfer function convert to pole-zero[z,p,k] = tf2zp([1 1],[1 3 2])% Pole-zero convert to transfer function[num, den] = zp2tf(z,p,k)transfer_function=tf(num, den)

Transfer function (10/11)

• Syntax– sysr = minreal(sys)– Minimal realization and pole-zero cancellation

• Matlab code

G=tf([1 6 5],[1 3 2])minreal(G)pole(G)zero(G)

Transfer function (11/11)

Series and parallel connection

• Syntax– sys = series(sys1,sys2)

– sys = parallel(sys1,sys2)

The feedback function (1/5)

• Syntax : initial sign=-1– sys = feedback(sys1,sys2)– sys =

feedback(sys1,sys2,sign)

– Note: sign=

If sign=-1 then system was negative feedback.

If sign=+1 then system was positive feedback.

1Sys1

Sys2

sign

+

The feedback function (2/5)

• Matlab codeExample

n1=[2 5 1];n2=[1 2];d1=[1 2 3 ];d2=[1 10];G=tf(n1,d1);H=tf(5*n2,d2);F1=feedback(G,H)F2=feedback(G,H,-1)F3=feedback(G,H,1)

相等2

2

5 12( )2 3

5( 2)( )

10

ssG sss

sH s

s

The feedback function (3/5)

Result:

Showing F1, F2, and F3, respectively

The feedback function (4/5)

• Unity feedback• Matlab code Example

n1=[2 5 1];n2=[1 2 3];G=tf(n1,n2);f1=feedback(G,1)f2=feedback(1,G)

2

2

5 12( )2 3

ssG sss

The feedback function (5/5)

Result:

Showing f1 and f2, respectively

Lab Assignments

• Solve Problems P1 and P2

• Lab report should at least contain:– The MATLAB code and the plot for P1– The plot generated for P2 (Please plot all 3

curves on the same plot)

Lab HomeworkP1.

Consider the feedback system depicted in the figure.

(a) Compute the closed-loop transfer function using the series and feedback functions.

(b) Obtain the closed-loop system unit step response with the step function, and verify that final value of

the output is 2/5.

1

1s

Controller

2

3

s

s

Plant

R s Y s

Lab Homework

2

P2.

A system has a transfer function

15 /

3 15

Plot the response of the system when is a unit

step for the parameter 3,6, and 12.

X s z s z

R s s s

R s

z