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Stability Analysis for Control Systems Compiled By: Waqar Ahmed Assistant Professor Mechanical UET, Taxila Sources for this Presentation 1. Control Systems, Western Engineering, University of Western Ontario, Control, Instrumentation & electrical Systems 2. ROWAN UNIVERSITY, College of Engineering, Prof. John Colton

Stability Analysis for Control Systems

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Stability Analysis for Control Systems. Compiled By: Waqar Ahmed Assistant Professor Mechanical UET, Taxila. Sources for this Presentation 1. Control Systems, Western Engineering, University of Western Ontario, Control, Instrumentation & electrical Systems - PowerPoint PPT Presentation

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Page 1: Stability Analysis for Control Systems

Stability Analysis for Control Systems

Compiled By:Waqar Ahmed

Assistant Professor MechanicalUET, Taxila

Sources for this Presentation1. Control Systems, Western Engineering, Universityof Western Ontario, Control, Instrumentation &electrical Systems2. ROWAN UNIVERSITY, College of Engineering, Prof. John Colton

Page 2: Stability Analysis for Control Systems

What is stability and instability anyway?

Page 3: Stability Analysis for Control Systems

Time Domain Response of Such Systems

Page 4: Stability Analysis for Control Systems

Results of Instability

Page 5: Stability Analysis for Control Systems

Characteristic Equation of a System

G(s)H(s)1sG

sRsY

)()()(

01 G(s)H(s) Characteristic Equation

Page 6: Stability Analysis for Control Systems

Roots of Characteristics Equations• Its very simple to find out roots of a characteristic equation• Replace ‘s’ with a factor jω

01 G(s)H(s) Characteristic Equation• Substituting the s symbol we get

0j1 ))H(jG( Characteristic Equation• Now solving this equation we get the roots• The roots will have 2 parts, one imaginary and one real• Real part is denoted with σ-axis and imaginary part with jw-

axis, as shown on next slide

Page 7: Stability Analysis for Control Systems

Real and Imaginary Axis for Roots of a Characteristic Equation

Now lets discuss why and how can we say, right hand side of this graph is unstable region and left hand side is stable?

What is inverse Laplace Transform of 1/(s-1)?What is inverse Laplace Transform of 1/(s+1)?Answer to above 2 questions will answer the query

Page 8: Stability Analysis for Control Systems
Page 9: Stability Analysis for Control Systems

Relation between Roots of Characteristic Equation and Stability

Page 10: Stability Analysis for Control Systems

A Question Here

• Is it really possible and feasible for us to calculate roots of every kind of higher order characteristic equation?

• Definitely no!• What is the solution then?• Routh-Herwitz Criterion

Page 11: Stability Analysis for Control Systems

Routh-Hurwitz Criterion

• The Hurwitz criterion can be used to indicate that a

characteristic polynomial with negative or missing

coefficients is unstable.

• The Routh-Hurwitz Criterion is called a necessary

and sufficient test of stability because a polynomial

that satisfies the criterion is guaranteed to be stable.

The criterion can also tell us how many poles are in

the right-half plane or on the imaginary axis.

Page 12: Stability Analysis for Control Systems
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Page 24: Stability Analysis for Control Systems

Routh-Hurwitz Criterion

• The Routh-Hurwitz Criterion: The number of roots of the characteristic polynomial that are in the right-half plane is equal to the number of sign changes in the first column of the Routh Array.

• If there are no sign changes, the system is stable.

Page 25: Stability Analysis for Control Systems

Example: Test the stability of the closed-loop system

Page 26: Stability Analysis for Control Systems

Solution: Since all the coefficients of the closed-loop characteristic equation s3 + 10s2 + 31s + 1030 are positive and exist, the system passes the Hurwitz test. So we must construct the Routh array in order to test the stability further.

Page 27: Stability Analysis for Control Systems

Solution (Contd)

• It is clear that column 1 of the Routh array is

• it has two sign changes (from 1 to -72 and from -72 to 103). Hence the system is unstable with two poles in the right-half plane.

103

7211

Page 28: Stability Analysis for Control Systems

Example of Epsilon Technique: Consider the control system with closed-loop transfer function:

3563210)( 2345

sssss

sGc

Page 29: Stability Analysis for Control Systems

Considering just the sign changes in column 1:

• If is chosen positive there are two sign changes. If is chosen negative there are also two sign changes. Hence the system has two poles in the right-half plane and it doesn't matter whether we chose to approach zero from the positive or the negative side.

Page 30: Stability Analysis for Control Systems

Example of Row of Zeroes

3410)( 234

ssss

sGc

s4 1 -4 3

s3 1 -1 0

s2 -3 3 0

s1 0 0 0

s0 - - -

Page 31: Stability Analysis for Control Systems

s4 1 -4 3

s3 1 -1 0

s2 -3 3 0

s1 -6 0 0

s0 3 - -

Develop an auxillary equation, A(s)

033)( 2 ssA sdssdA 6)(

Page 32: Stability Analysis for Control Systems

Steady State Error: Test Waveform for evaluating steady-state error

Page 33: Stability Analysis for Control Systems

Steady-state error analysis

G(s)

H(s)

R(s) +

-

C(s)

G(s)

R(s) +

-

C(s) Unity feedbackH(s)=1

Non-unity feedbackH(s)≠1

E(s)

E(s)

)()()( sCsRsE System error

)()()()( sCsHsRsE

Actuating error

Page 34: Stability Analysis for Control Systems

Steady-state error analysis

Consider Unity Feedback System

)()()( sCsRsE (1)

)()(1

)()( sRsGsGsC

(2)

Substitute (2) into (1)

)()(1

1)()(1

)()()( sRsG

sRsGsGsRsE

(3)

Page 35: Stability Analysis for Control Systems

)(1)()(lim)(lim)(

0 sGssRssEtee

stss

1. Step-input: R(s) = 1/s

2. Ramp-input: R(s) = 1/s2

3. Parabolic-input: R(s) = 1/s3

ps

stepss KsGee

1

1)(lim1

1)()(0

vs

rampss KssGee 1

)(lim1)()(

0

as

parabolicss KsGsee 1

)(lim1)()( 2

0

Steady-state errorFinal value theorem

Page 36: Stability Analysis for Control Systems

ExerciseDetermine the steady-state error for the following inputs for the system shown below:

a)Step-input r(t) = u(t)b)Ramp-input r(t) = tu(t)c) Parabolic-input r(t) = t2u(t)