CS Matlab Exercises

8/8/2019 CS Matlab Exercises

http://slidepdf.com/reader/full/cs-matlab-exercises 1/16

University of Newcastle upon Tyne

School of Electrical, Electronic and Computer Engineering

CONTROL SYSTEMS

Control Systems and MatlabIntroduction

Consider the OLTF with negative feedback:)4.0)(25)(15(

)20(18)(

+++

+=

s s s

s sG . To define it type

separately the numerator and the denominator as two different polynomials. The code is written inan m-file.% prg1% This m-file creates and prints the transfer function% num stands for numerators% den stands for denominatornum=18*[1 20]den=conv(conv([1 15],[1 25]), [1 0.4])printsys(num,den,'s')

» prg1num =

18 360den =

1.0000 40.4000 391.0000 150.0000num/den =

18 s + 360----------------------------s^3 + 40.4 s^2 + 391 s + 150

To find the CLTF we use the command feedback(numg,deng,numh,denh) % prg3% This m-file creates and prints the transfer function% numol stands for numerator of the open loop TF

% denol stands for denominator of the open loop TF% numcl stands for numerator of the close loop TF% dencl stands for denominator of the close loop TF numol=18*[1 20]denol=conv(conv([1 15],[1 25]), [1 0.4])[numcl,dencl]=feedback(numol,denol,1,1)printsys(numcl,dencl,'s')» prg3

numol =18 360

denol =1.0000 40.4000 391.0000 150.0000

numcl =0 0 18 360

dencl =1.0000 40.4000 409.0000 510.0000

num/den =18 s + 360

----------------------------s^3 + 40.4 s^2 + 409 s + 510

Control Systems and Matlab 1





CONTROL SYSTEMS

Step Response

To find the step response of a system use the command:

step(numerator, denominator).

1

s +4s2

Transfer FcnSu m

Step

-K-

Gain

The CLTF isK ss

K

sR

sC

GH

G

sR

sC H

ssG ++

=⇔+

==

+= 4)(

)(

1)(

)(2

1

4

12

So if K=1:% prg6% step 1 % This m-file finds the step responsenum=[1];den=[1 4 1];step(num,den);gridxlabel('time')ylabel('output')title('step response')

Time (sec.)

A m p l i t u d e

Step Response

0 5 10 15 20 250

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

step response

time

o u t p u t

The command step takes a default time for the transient response to be finished (i.e. 25s here)

If we want to define the time we can use the next m-file:

% prg7 % step 2 % This m-file finds the step response t=0:0.1:30;






CONTROL SYSTEMSnum=[1];den=[1 4 1];step(num,den,t);gridxlabel('time')ylabel('output')

title('step response')

Time (sec.)

A m p l i t u d e

Step Response

0 5 10 15 20 25 300

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

step response

time

o u t p u t

For more complex systems it is possible to combine the commands step & plot :% prg8 % step 3 % This m-file finds the step response t=0:0.1:30;

num=[1];den=[1 4 1];c=step(num,den,t);plot(t,c,'or')gridxlabel('time')ylabel('output')title('step response')

0 5 10 15 20 25 300

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time

o u t p u t

step response

More than one graph can be plotted on the same figure using a combination of plot and step:






CONTROL SYSTEMS

Note that when we found the step response we used as an argument the time. If we do not then step

will take a default number for time and so later when using the command plot, an error message

will be given because the two vectors do not have the same size.% prg9 % step 4 % This m-file finds the step response t=0:0.1:20;%k=1 num1=[1];den1=[1 4 1];%k=100 num2=[100];den2=[1 4 100];c1=step(num1,den1,t);c2=step(num2,den2,t);plot(t,c1,'or',t,c2)grid

xlabel('time')ylabel('output')title('step response')

0 5 10 15 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

time

o u t p u t

step response

To find the impulse response we use the command impulse. This is used exactly as the command

step. To find the ramp response we have to find the step response of the function s

s)(G

Frequency Response Bode Diagrams

To plot the Bode diagram we use the command bode(num,den) .Π.χ. :% prg16 %bode 1clcclear all

clfnum=10;den=[1 5 5 0];






CONTROL SYSTEMSbode(num,den);

Frequency (rad/sec)

P h a s e ( d e g ) ; M a g n i t u d e ( d B )

Bode Diagrams

-150

-100

-50

0

50

10-2

10-1

100

101

102

-300

-200

-100

0

As with the root locus plot we can chose the frequencies:%prg17 %bode 2clcclear allclfnum=10;den=[1 5 5 0];w=logspace(-1, 4,100);bode(num,den,w);

Frequency (rad/sec)


Bode Diagrams

-300

-200

-100

0

100

10-1

100

101

102

103

104

-300

-200

-100

0

To find the gain and phase margins we use the command margin(num,den) since it is very difficult

to calculated directly from the above figures with any accuracy. This command plots the Bode

diagram and calculates the margins:% prg18 %bode 3clcclear all

clfnum=10;den=[1 5 5 0];margin(num,den);






CONTROL SYSTEMS

Frequency (rad/sec)


Bode Diagrams

-150

-100

-50

0

50

Gm=8.0 dB (Wcg=2.2); Pm=25.5 deg. (Wcp=1.3)

10-2

10-1

100

101

102

-300

-200

-100

0

Nichols diagrams

To find the Nichols plot use the command nichols(num,den).% prg19 clcclear allclf num=10;den=[1 5 5 0];

nichols(num,den);ngrid

Open-Loop Phase

Open-

LoopGai

Nichols

- - - - - - -50-40

-20

0

20

40

60

80

63

1

0.50.25

0

-1

-3

-6

-12

-20

-40

Pm

Gm

With the Nichols plots we use the ngrid command.






CONTROL SYSTEMS

Nyquist diagrams

As with Nichols to create a Nyquist diagram we use the command nyquist(num,den):% prg20%nyquist 1

clcclear allclfnum=10;den=[1 5 5 0];nyquist(num,den);

Real Axis

I m a g i n a r y A x i s

Nyquist Diagrams

-2.5 -2 -1.5 -1 -0.5 0-250

-200

-150

-100

-50

0

50

100

150

200

250

The command nyquist takes both negative and positive values for the frequency. For this reason we

see a curve which is symmetrical to the x-axis. If we want only positive frequencies then:% prg22%nyquist 3 clcclear allclfnum=10;den=[1 5 5 0];w=0:0.1:100;[re,im,w]=nyquist(num,den,w);

plot(re,im);grid






CONTROL SYSTEMS

AGm

0

1=

5

A 0

0

-5

-10

-15

-2 -20

-1.8 -1.6 -1.4 -0.2 0 -1.2 -1 -0.8 -0.6 -0.4

All the above commands can be used for systems that are described in State Space form – see later

Root Locus

The design of the root locus plot of a system is very easy and fast with the use of Matlab. This is

more obvious in more complex systems than the examples used here. To find the root locus we use

the command rlocus(xxx,yyy) where xxx is the numerator and yyy is the denominator of the OLTF

e.g.:

1

s +5s2

Tran sfer FcnSum

k

Ga in

% prg12

ocus 1 % rlclc

llclear anum=1;den=[1 5 0];rlocus(num,den)






CONTROL SYSTEMS

-6 -5 -4 -3 -2 -1 0 1 2-1.5

-1

-0.5

0

0.5

1

1.5

Real Axis

I m a g A x i s

The above root locus plot is for values of gain from 0 to a specific value to allow conclusions about

the behaviour and stability of the system. If we would like to define the values of gain then:% prg13

%rlocus 2

clc

clear all

clfk=0:1:100;

num=1;

den=[1 5 5 0];

rlocus(num,den,k)

-5 -4 -3 -2 -1 0 1 2-4

-3

-2

-1

0

1

2

3

4

Real Axis

I m a

g A x i s

or % prg14

%rlocus 3

clc

clear allclf

k1=0:2:10;

k2=11:0.5:90;

k3=91:10:500;

k=[k1 k2 k3];

num=1;

den=[1 5 5 0];

rlocus(num,den,k)






CONTROL SYSTEMS

-5 -4 -3 -2 -1 0 1 2-8

-6

-4

-2

0

2

4

6

8

Real Axis

I m a g A x i s

For k=11 to 90 we have more samples of the locus.

To find the poles and a specific point of the locus we use the command rlocfind(xxx,yyy)% prg15%rlocus 4 clcclear allclfk1=0:2:10;k2=11:0.5:90;k3=91:10:500;k=[k1 k2 k3];num=1;

den=[1 5 5 0];rlocus(num,den,k);[k,poles]=rlocfind(num,den)

After the execution of the m-file the point of interest can be defined with the mouse:» prg15Select a point in the graphics window

selected_point =

-0.4330+ 1.1730i

k =

6.4034

poles =

-4.1691-0.4154+ 1.1676i-0.4154- 1.1676i






CONTROL SYSTEMS

-5 -4 -3 -2 -1 0 1 2-8

-6

-4

-2

0

2

4

6

8

Real Axis

I m a g A x i s

If that point were on the imaginary axis then this would produce the maximum stable gain.

The commands rlocus, rlocfind can be used for systems that are described in State Space form – seelater

State Space

To describe a system in State Space we must define the matrices A,B,C,D:

% prg2% This m-file creates and prints a system, which is defined in state %space a=[-40.4 -391 -150; 1 0 0; 0 1 0];b=[1 0 0]';c=[0 18 360];d=[0];printsys(a,b,c,d)

» prg2a =

x1 x2 x3

x1 -40.40000 -391.00000 -150.00000x2 1.00000 0 0x3 0 1.00000 0

b =u1

x1 1.00000x2 0x3 0

c =x1 x2 x3

y1 0 18.00000 360.00000d =

u1y1 0






CONTROL SYSTEMS

State Space to Transfer Function

From A, B, C, D to find the TF use the command ss2tf( Α ,B,C,D) % prg4% This m-file converts a second order system

% form state space to transfer function% num stands for numerator% den stands for denominatorA=[1 2; 4 5];B=[1 0]';C=[1 0];D=[0];[num,den]=ss2tf(A,B,C,D);printsys(num,den,'s')

» prg4

num/den =s - 5-------------s^2 - 6 s – 3

Transfer Function to State Space

From the TF can find the matrices A,B,C,D with the command tf2ss(num,den) % prg5% This m-file converts a second order system% form transfer function to state space% num stands for numenator% den stands for denominatornum=[1 -5];den=[1 -6 -3];[A,B,C,D]=tf2ss(num,den);printsys(A,B,C,D)

prg5a =

x1 x2x1 6.00000 3.00000x2 1.00000 0

b =u1

x1 1.00000x2 0

c =x1 x2

y1 1.00000 -5.00000

d =u1

y1 0






CONTROL SYSTEMS

Step in State Space

The command step can be used for state space defined systems: %prg 9ba = [6 3;1 0];

b = [1;0];c = [1 -5];d = [0];step(a,b,c,d)

Time (sec.)

A m p l i t u d e

Step Response

0 0.16 0.32 0.48 0.64 0.80

1

2

3

4

5

6

7

Consider a system that can be described by the following equations:

x

x

x

x

u

u

y

y

x

x

1

2

1

2

1

2

1

2

1

2

1 1

6 5 0

1 1

1 0

1 0

0 1

•

•

=− −

+

=

.

The step response is:

⇒−==⇒−=⇒

⇒

=

−=⇒

=

+=⇒

=

+=

−−

−•

B AsI C sGsU

sY sBU AsI C sY

sCX sY

sBU AsI s X

sCX sY

sBU s AX ssX

CX y

BU AX X LAPLACE

11

1

)()()(

)()()()(

)()(

)()()(

)()(

)()()(

5.65.6

)()(,

5.6)()(

5.6

5.7

)(

)(,

5.6

1

)(

)(

)(

)(

5.6

5.6

5.6

5.7

5.65.6

1

)(

)(

655.7

1

5.6

1

01

11

5.6

11

10

01

22

22

2

1

21

2

21

1

2

1

22

22

2

1

2

1

++=

++=

++

+=

++

−=

⇒

++++

+

++++

−

=

⇒

⇒

+−

++=

−

+

⇒

−

sssU sY

sss

sU sY

ss

s

sU

sY

ss

s

sU

sY

sU

sU

ssss

s

ss

s

ss

s

sY

sY

s

ss

sss

s






CONTROL SYSTEMS

So we expect four graphs:» a=[-1 -1;6.5 0];» b=[1 1;1 0];» c=[1 0;0 1];» d=[0 0;0 0];

» step(a,b,c,d)

Time (sec.)

A m p l i t u d e

Step Response

-0.4

-0.2

0

0.2

0.4

From: U1

T o : Y 1

From: U2

0 4 8 120

0.5

1

1.5

2

T o : Y 2

0 4 8 12

If the step response for only one input was wanted:

» step(a,b,c,d,1) or » step(a,b,c,d,2)

Time (sec.)

A m p l i t u d e

Step Response

-0.4

-0.2

0

0.2

0.4

T o : Y 1

0 2 4 6 8 10 120

0.5

1

1.5

2

T o : Y 2

Time (sec.)

A m p l i t u d e

Step Response

-0.2

-0.1

0

0.1

0.2

0.3

T o : Y 1

0 2 4 6 8 10 120

0.5

1

1.5

2

T o : Y 2






CONTROL SYSTEMS

Lab Exercises:

1. Consider the following system:

1.1 Find and plot the step response of the CL system.

1.2 Find and plot the step response of the system for k=30,60,90.

1.3 Describe the system in SS.

1

s +6s +11s+63 2

Tran sfer Fcn

Su m

k

Gain

2.1 Find the root locus plot and the maximum stable gain. Relate the step response to the CL

pole locations.

1

s +6s +11s+63 2

Transfer Fcn

Su m

k

Gain

2.2 Plot the root locus and find the maximum stable gain. Relate the step response to the CL

pole locations.

1

s +1.1s +10.3s +5s4 3 2

T ransfer Fcn

Sum

k

Gain






CONTROL SYSTEMS

C l S d M l b 16


pole locations:

s +5s+62

s +s2

Transfer FcnSu m

k

Gain


pole locations:

1

s +6s +25s3 2

Transfer FcnSu m

k

Gain

2.5 Find the gain and phase margin of the above OLTF

3 Produce the Bode, Nyquist and Nichols diagrams for the following systems.

)164)(2(

)1(320)(

21+++

+=

ssss

ssG

4.0,5.1,4,)2)(1(

)(2 =++

= k sss

k sG

100,10,)5)(1(

)(3 =++

= k sss

k sG

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CS Matlab Exercises