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

What is stability and instability anyway?

Time Domain Response of Such Systems

Results of Instability

Characteristic Equation of a System

G(s)H(s)1sG

sRsY

)()()(

01 G(s)H(s) Characteristic Equation

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

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

Relation between Roots of Characteristic Equation and Stability

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

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.

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.

Example: Test the stability of the closed-loop system

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.

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

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

3563210)( 2345

sssss

sGc

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.

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 - - -

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)(

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

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

Steady-state error analysis

Consider Unity Feedback System

)()()( sCsRsE (1)

)()(1

)()( sRsGsGsC

(2)

Substitute (2) into (1)

)()(1

1)()(1

)()()( sRsG

sRsGsGsRsE

(3)

)(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

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)