Transient response stability

System is analyzed for its:

The transient response

Steady state response

Stability

Concept of stability• Stability is the most important system

specification.• Many physical systems are inherently open-loop

unstable. Feedback control is introduced by engineer to stabilize the unstable plant

‘For open-loop stable plant, we still need feedback to adjust performance to meet the design specification.

What is stability ?

Example

Tacoma Narrows Bridge (a) as oscillation begins (b) at catastrophic failure.

Example

The M2 robot is more energy-efficient but less stable than many other designs that are well-balanced but consume much more power.

Stability• Stability is the most important system

specification.

• If the system is unstable, transient response and steady state errors are moot points. An unstable system can not be designed for a specific transient response or steady state error requirement

Stable and unstable system response • Stable state: A linear , time invariant system is stable if

the natural response approaches zero as time

approaches infinity.

• Unstable: A linear , time invariant system is unstable if the natural response approaches infinity as time approaches infinity.

• Critically stable: If the natural response neither decays nor grows but remains constant or oscillates.

A system is stable if every bounded yields a bounded output

This statement the bounded input-Bounded output (BIBO) defination of stability

Test for stability

1.Preliminary test: Check the roots of the characteristic equation to lie on the left half of s-plane.

2. Routh-Hurwitz criterion

Preliminary test

By pole configuration

Stable system-In terms of pole configuration

• Stable :A linear system stable , if all poles of its transfer function lie in the left-half plane

Unstable system-In terms of pole configuration

• Instability A linear system is only unstable, if at least one pole of its transfer function lies in the right-half plane, or, if at least one multiple pole (multiplicity ) is on the imaginary axis of the plane

Marginally stable system-In terms of pole configuration

• Marginally stable A linear system is critically stable, if at least one single pole exists on the imaginary axis, no pole of the transfer function lies in the right-half plane, and in addition no multiple poles lie on the imaginary axis

•

All cases

The concept of stability

• Pole location and stability

Routh-HurwitzStability Criterion

The Routh-HurwitzStability Criterion

• The Routh-Hurwitz stability criterion provides a method to examine the system stability without the need to solve for poles of a system.

• Using Routh-Hurwitz Criterion one can find how many poles are in the left half-plane,right half-plane, and on the imaginary axis.

• However, using this method, one cannot find

the exact coordinates of the poles.

The Routh-HurwitzStability Criterion

• The method requires two steps:

• - Generate data table called as Routh table

• - Interpret the data table to find how many poles are in the right half-plane of the complex plane

The Routh-Hurwitzstability Criterion

• The closed loop gain/Transfer function

P

Z

s

sN

sHsG

sG

sR

sYsT

)(

)(

)()(1

)(

)(

)()(

• N(s) is the numerator, its roots are called as zeros• (s) (or G(s)) is the denominator or characteristic equation, its roots are called as poles.

• The characteristic function in the Laplace variable is writtenas:

013

21

1)( asasasasasD nn

nn

nn

Routh Table• Consider the characteristic equation

013

21

1)( asasasasasD nn

nn

nn

Create Routh Table• Consider closed loop system with transfer

function

01

12

23

34

4

)(

asasasasa

sN

Initial Routh Table

s4 a4 a2 a0

S3 a3 a1 0

s2 b1 b2

s1 c1 c2

s0 d1

00

To find other terms Routh Table

•

From the Routh’s criterion

• If all the elements of the first –column of the Routh’s array are of the same sign, the roots of the polynomial are all in the left half of the s-plane. The system is stable.

• If there are changes of sign in the first column, the number of sign changes indicates the number of roots with positive real part, i.e. the number of roots that lie on the right half of the s-plane. The system is unstable

Example: D(s) = s3 + 2s2 + 4s + 3

• Initial Routh table

Calculate the other coefficients

All the elements of the 1st columnAre of the same sign, Hence all roots lie on the left half of s plane.System is said to be stable.

Example: D(s) = s4 + 5s3+5s2 + 10s + 12

• The Routh’s array

The 1st column has two sign changes and so there are two roots of the characteristic Equation that lie on the right half of the s-plane. The system is said to be unstable.

Example:

Two sign change=2 poles on right half Plane, systemunstable

Exercise 1: Test the stability of the closed-loop system

•

Routh’s array

• Take the common factor

The 1st column is

The two sign changes,poles on right half of s-planeUnstable system

Special case.

•A zero element in the first column

A zero element in the first column• If 1st element of the column is zero, replce with null

element with small positive number epsilon( ) and proceed with the construction of the array

Equation

0522)( 234 sssssP

5

0/)52(

05)0(

021

521

0

1

2

3

4

s

s

s

s

s

Now assume, Epsilon = -ve ,means there are two sign changes-s3-s2 &s2-s1,so unstableepsilon = +ve=If it is 1 & 2 , there two sign changes =+ve =if it is > 2, it becomes stable

Epsilon = -ve, 2 sign changes , unstableEpsilon = -ve, check what value the column does changes sign=unstable

ExampleExample

Example• Routh’s arrayS4 1 11 K 0S3 6 6 0s2 60-6/6= 10 6k-0/6=k 0S1 60-6k/10 0S0 [(60-6k)/10]*K/((60-6k)/10)=k

To make system stable,1st column should not have any sign changeIf K=0,Last row S0=0If K=10, s1 row =0IfK>0 and K<10 , system is stableIf K=-ve, unstable.

SoFor the system to be stable, 0<K<10

From this lesson you learn

• Stability

• Stable system-In terms of pole configuration

• Stability-Routh’s Criterion

Any Question ?

Steady state error

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