Matlab Stabiliy Analysis

7/30/2019 Matlab Stabiliy Analysis

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MATLAB and Stability Analysis

DEPARTMENT OF BIOTECHNOLOGY ENGINEERING

NASSERELDEEN AHMED KABASHI

[email protected], Tel : Ext. 4524

Sem I, 2012-2013
mailto:[email protected]:[email protected]


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The issue of ensuring the stability of a closed-loop feedback system is central to

control system design. Knowing that an unstable closed-loop system is generally

of no practical value, we seek methods to help us analyze and design stable

systems. A stable system should exhibit a bounded output if the corresponding

input is bounded. This is known as bounded-input, bounded-output stability and

is one of the main topics of this chapter.

The stability of a feedback system is directly related to the location of the rootsof the characteristic equation of the system transfer function. The Routh

Hurwitz method is introduced as a useful tool for assessing system stability. The

technique allows us to compute the number of roots of the characteristic

equation in the right half-plane without actually computing the values of the

roots. Thus we can determine stability without the added computational burden

of determining characteristic root locations. This gives us a design method for

determining values of certain system parameters that will lead to closed-loop

stability. For stable systems we will introduce the notion of relative stability,

which allows us to characterize the degree of stability.

The Stability of Linear Feedback Systems

BTE 4415Dr. Nassereldeen A. Kabashi


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The Concept of Stability

A stable system is a dynamic system with a bounded response to a

bounded input.

Absolute stability is a stable/not stable characterization for a closed-

loop feedback system. Given that a system is stable we can further

characterize the degree of stability, or the relative stability.



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The Concept of Stability

The concept of stability can be

illustrated by a cone placed on

a plane horizontal surface.

A necessary and

sufficient condition for afeedback system to be

stable is that all the

poles of the system

transfer function have

negative real parts.

A system is considered marginally stable if only certain bounded

inputs will result in a bounded output.



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Overview

Linear SISO Control Systems

Differential Equations

Pole Zero Plots

Transfer Equations Routh-Hurwitz Criterion

Root Locus

Bode Plots Nyquist Criterion and Plots


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BTE 4415 Dr. Nassereldeen A. Kabashi

Linear SISO Control Systems

Closed Loop Control System

Controll

er

+--

+setpoint

load

G1


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Dr.Nassereldeen A.

Kabashi

Differential Equations

Laplace Transforms

Time-Domain to Frequency-Domain

Use Tables or MATLAB

Sample MATLAB code:syms a s t w x

laplace(t^5) returns 120/s^6

laplace(exp(a*s)) returns 1/(t-a)

laplace(sin(w*x),t) returns w/(t^2+w^2)

laplace(cos(x*w),w,t) returns t/(t^2+x^2)

laplace(x^sym(3/2),t) returns 3/4*pi^(1/2)/t^(5/2)

laplace(diff(sym('F(t)'))) returns laplace(F(t),t,s)*s-F(0)



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Transient Response Analysis(1)

Transient response refers to the process generated

in going from the initial state to the final state

Transient responses are used to investigate the time

domain characteristics of dynamic systems

Common responses: step response, impulse response,

and ramp response



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Unit step response of the transfer function system

Consider the system: 254

252

ss

sH

%*****Numerator & Denominator of H(s)

>>num = [0 0 25];den = [1 4 25];

%*****Specify the computing time

>>t=0:0.1:7;

>>step(num,den,t)

%*****Add grid & title of plot

>>grid

>>title(Unit Step Response of H(s))


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Unit step response ofH(s)

Unit Step Response of H(s)

Time (sec)

Amplitude

0 1 2 3 4 5 6 70

0.2

0.4

0.6

0.8

1

1.2

1.4



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Alternative way to generate Unit step response of the

transfer function, H(s)

If step input is , then step response is generated

with the following command:


>>num = [0 0 25];den = [1 4 25];%*****Create Model

>>H=tf(num,den);

>>step(H)

>>step(10*H)

)(10 tu



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Impulse response of the transfer function system

Consider the system: 254

252

ss

sH


>>num = [0 0 25];den = [1 4 25];


>>t=0:0.1:7;

>>impulse(num,den,t)


>>grid

>>title(Impulse Response of H(s))


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Impulse response ofH(s)

Impulse Response of H(s)

Time (sec)

Amplitude

0 1 2 3 4 5 6 7-1

-0.5

0

0.5

1

1.5

2

2.5

3



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Ramp response of the transfer function system

Theres no ramp function in Matlab

To obtain ramp response of H(s), divideH(s)by s

and use step function

Consider the system:

For unit-ramp input, . Hence

254

252

ss

sH

2

1)( s

sU

254

251

254

251222

ssssssssY

Indicate Step response

NEW H(s)


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Example: Matlab code for Unit Ramp Response

%*****Numerator & Denominator of NEW H(s)

>>num = [0 0 0 25];den = [1 4 25 0];


>>t=0:0.1:7;

>>y=step(num,den,t);

%*****Plot input & the ramp response curve

>>plot(t,y,.,t,t,b-)


>>grid

>>title(Unit Ramp Response Curve of H(s))


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Unit Ramp response ofH(s)

0 1 2 3 4 5 6 70

1

2

3

4

5

6

7Unit Ramp Response Curve of H(s)



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BTE 4415 Dr. Nassereldeen A.Kabashi

Poles & Zeros

In the s-domain a function isplotted on the imaginary axis (y)and the real axis (x)

Roots of q(s) give poles Roots of p(s) give zeros

)()(

)5)(3(210)(

sqsp

sssssY

Sample MATLAB code:>> Y=tf([10 20],[1 8

15 0])

Transfer function:

10 s + 20------------------

s^3 + 8 s^2 + 15 s

>> pzplot(Y)

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

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Pole-Zero Map

Real Axis

ImaginaryAxis


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Transfer Functions

10

S+1

1

2s+0.5

0.1

Td

_

+_

+V2

G1 G2

1.0212 VGTG d75.025.1

52

2

ssV


19/40BTE 4415 Dr. Nassereldeen A. Kabashi

Step Response of TF

Sample MATLAB code:

>> T=tf(5,[1 1.25 0.75]);

>> g=feedback(T,0.1);

>> step(g)

>> h=feedback(T,1);

>> step(h)

0 1 2 3 4 5 6 7 8 90

0.5

1

1.5

2

2.5

3

3.5

4

4.5Step Response

Time (s ec)

Amplitude

0 1 2 3 4 5 6 7 8 90

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (s ec)

Amplitude


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[kd,poles] = rlocfind(numD,denD) and find the stepresponse (the open-loop transfer function is K(s)

H(s)):

numOL = [kd 2]; denOL = [1 1 4 0];

[numCL,denCL] = cloop(numOL,denOL);

step(numCL,denCL)


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Continue

Compute the poles of G(s)=(3s+1)/(s^2+2s+1) and the response of the system

described by G(s) to a step change of magnitude 2. Plot the response.

% Computation of the poles of G(s)=(3s+1)/(s^2+2s+1)

% Enter the denominator

>>den=[1 2 1];

% Find roots of denominator. They are the poles of G(s)

>>poles=roots(den)

% You may also enter the transfer function and compute the poles directly

>>g1=tf([3 1], [1 2 1])

>>poles2=pole(g1)% Generate the time vector

>>t=0:0.1:20;

% Method 1: Use the function "step".



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Then the commands would be:% y=step(g1,t)% xlabel('time')% ylabel('Output')% title('Plot of output as a function of time to a stepinput of magnitude 2')% plot(t,y)step(g1,t);input('After inspecting the figure, close it and pressENTER to continue');pause



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Continue

% Clear the current figureclf% Method 2: Use the general simulation function "lsim".

=2*ones(size(t));y=lsim(g1,u,t);

y=step(g1,t);plot(t,y)xlabel('time')ylabel('Output')title('Plot of output as a function of time to a step input of

magnitude 2')% If you use lsim without the left side argument, Matlabgenerates the plot automatically



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Continuous Time Transfer Function(1)

Function: Use tffunction create transfer function of

following form:

Example 23

12

)( 2

ss

s

sH

>>num = [2 1];

>>den = [1 3 2];

>>H=tf(num,den)

Transfer function:2 s + 1

-------------

s^2 + 3 s + 2

Matlab Output


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Include delay to continuous time Transfer Function

Example

23

12)(

2

2

ss

sesH

s

Transfer function:

2 s + 1

exp(-2*s) * -------------s^2 + 3 s + 2

%*****Numerator & Denominator of H(s)>>num = [2 1];>>den = [1 3 2];


>>H=tf(num,den,inputdelay,2)


Matlab Output


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Function: Use zpk function to create transfer function

of following form:

Example 21

5.02

23

12)(

2

ss

s

ss

ssH

>>num = [-0.5];

>>den = [-1 -2];>>k = 2;

>>H=zpk(num,den,k)

Zero/pole/gain:2 (s+0.5)

-----------

(s+1) (s+2)

Matlab Output


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MATLAB with PID Controller

BTE

4415

(Biop

rocessControl)

C(S)=Kp

For PI C(S)=Kp + Ki/S

PID C(S)=Kp + Ki/S + Kd S


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>>Kp=50;>>Ki=1;

>>Kd=10;

>>num=[Kd Kp Ki];

>>den=[1 10 20 0 0];

>>step(num,den)



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Continue

>>Kp=500;

>>Ki=1;

>>Kd=100;

>>num1=[Kd Kp Ki];>>den1=[1 0];

>>num2=1;

>>den2=[1 10 20 0];

>>[numc,numd]=cloop(conv(num1,num2),conv(den1,den2),-1);

>>step(numc,denc)



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Stability analysis based on the location of the

closed loop poles:

The stability of the closed-loop system can easily be

determined by analyzing the location of the poles

of the closedloop

transfer function. A system is stable if the real partsof the closed-loop poles are negative, i.e. all the

poles are in

the left-hand complex s-plane.



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Is the system stable? num=[1, 5]

den=[1, -3, 4, 10, 5, -10]

T=tf(num,den)

poles=roots(den)

poles = 2.1150 + 2.1652i

2.1150 - 2.1652i

-0.9824 + 0.7214i

-0.9824 - 0.7214i

0.7348 Note that the same result is obtained using the LTI structure

[z,p,k]=zpkdata(T,'v')

The system is unstable, since there are poles with positive real part.

Plot the pole locations in the complex s-plane:

pzmap(T)



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System Stability Using MATLAB



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(a) Stability region for a and K

(b) Matlab Script


A l th N i t t bilit


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Apply the Nyquist stability

criterion:

The stability of a feedback system can also be determined

by the behaviour of the open-loop:

a. The open-loop does not have unstable poles; all poles

are in the left-hands-plane.

The closed-loop system is stable if the Nyquist curve of the

open-loop system does not encircle the (1+0j) point.

b. The open-loop does have unstable poles:

The closed-loop system is stable if for the Nyquist curve ofthe open-loop system, the number of counterclockwise

encirclements of the (1+0j) point is equal to the number of

unstable open-loop poles.



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The loop transfer function of a system is

The system is in a negative unity feedback control loop. Determine the stability of the closed-loop based on the

Nyquist criterion.

s=tf('s')

L=10/((1+10*s)*(1+s))

[z,p,k]=zpkdata(L,'v')

Does the open-loop have unstable poles? Does the Nyquist diagram ecircle the (1+0j) point? Is the closed-loop

system stable? What happens with stability if the gain 10 in the numerator of the above transfer funcion is increased?

nyquist(L),grid

Verify the result by calculating the poles of the closed-loop system. In the feedback command 1 means unity

feedback and -1 indicates the negative feedback. (The -1 can be ignored since that is the default value)

T=feedback(L,1,-1)

or

T=L/(1+L); T=minreal(T)

The minreal command cancels the zero-pole pairs:

[z,p,k]=zpkdata(T,'v')

Plot the pole locations:

pzmap(T)

As seen the system is structurally stable. The poles are left-side poles, the Nyquist diagram does not encircle the

(1+0j) point even with increased gains.



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Phase margin, Gain margin

If the system is stable the degree of stability is an

important system property. The phase margin and gain

margin tells

how far is the system from being marginally stable. The phase margin can be calculated from the phase at

the cut-off frequency

Th t f f ti f t i


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The transfer function of a system is



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Routh Array Matlab

The characteristic polynomial is

There are two sign changes positive to negative and negative to positive

As we go from row to row 3 to row 4 to row 5

Thus there are two roots in the right half plane This is verified by the

MATLAB dialogure:

Matlab Stabiliy Analysis

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