In The Name of Allah The Most Beneficent The Most Merciful

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ECE 4545:Control Systems

Lecture:Stability

Engr. Ijlal HaiderUoL, Lahore

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Definitions of stability in various domains

Ecological stability, measure of the probability of a population returning quickly to a previous state, or not going extinct

Social stability, lack of civil unrest in a society

Quotes: “Every time I try to define a perfectly stable person, I am appalled by the dullness of that person.”

− J. D. Griffin

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Stability Examples of stable and unstable systems are the spring-

mass and spring-mass-damper systems. These two systems are shown next. The spring-mass

system (Figure 1(a)) is unstable since if we pull the mass away from the equilibrium position and release, it is going to oscillate forever (assuming there is no air resistance). Therefore it will not go back to the equilibrium position as time passes.

On the other hand, the spring-mass-damper system (Figure 1(b)) is stable since for any initial position and velocity of the mass, the mass will go to the equilibrium position and stops there when it is left to move on its own.

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Stability• Conceptually, a stable system is one for

which the output is small in magnitude whenever the applied input is small in magnitude.

• In other words, a stable system will not “blow up” when bounded inputs are applied to it.

• Equivalently, a stable system’s output will always decay to zero when no input is applied to it at all.

• However, we will need to express these ideas in a more precise definition.

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Control Systems

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Characteristic root locations criterion

A system is stable if all the poles of the transfer function have negative real parts.

Stability in the s-plane.

Control Systems

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Stable systemA necessary and sufficient condition for a feedback system to be stable is that all the poles of the transfer function have negative real parts.

This means that all the poles are in the left-hand s-plane.

Control Systems

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The Routh-Hurwitz rule

Rule If any of the coefficients ai, i = 0,1,2,…,n-1 are zero or negative, the system is not stable. It can be either unstable or neutrally stable.

ansn an 1s

n 1 an 2sn 2 ...... a1s a0 0, an 0

Control Systems

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Routh-Hurwitz CriterionThe Routh-Hurwitz criterion states

that the number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array.

Control Systems

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Equivalent closed-looptransfer function

Control Systems

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Initial layout for Routh table

Control Systems

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Ordering the coefficients of the characteristic equation.

ansn an 1s

n 1 an 2sn 2 ...... a1s a0 0

5

4

3

2

11

n

n

n

n

n

nn

n

aa

aa

aa

ss

Control Systems

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The Routh-Hurwitz array

snsn 1

sn 2

sn 3

s0

anan 1

bn 1

cn 1

hn 1

an 2an 3

bn 3

cn 3

an 4an 5

bn 5

cn 1

Control Systems

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The algorithm for the entries in the array

bn 1 an 1an 2 anan 3

an 1

1an 1

an an 2

an 1 an 3

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4

13

1

nn

nn

nn aa

aaa

b

31

31

11

1

nn

nn

nn bb

aab

c

Control Systems

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Completed Routh table

Control Systems

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Four distinct cases of the first column array1. No element in the first column is zero.

2. There is a zero in the first column and in this row.

3. There is a zero in the first column and the other zero in this row.

4. As in the previous case, but with repeated roots on the j-axis.

The Routh-Hurwitz Stability CriterionCase One: No element in the first column is zero

Example 6.1 Second-order system

The Characteristic polynomial of a second-order system is:

q s( ) a2 s2 a1 s a0

The Routh array is w ritten as:

w here:

b1a1 a0 0( ) a2

a1a0

Therefore the requirement for a stable second-order system is simply that all coef f icients be positive or all the coefficients be negative.

00

10

11

022

bsas

aas

The Routh-Hurwitz Stability CriterionCase Two: Zeros in the first column while some elements of the row containing a zero in the first column are nonzero

If only one element in the array is zero, it may be replaced w ith a small positive number that is allow ed to approach zero after completing the array.

q s( ) s5 2s4 2s3 4s2 11s 10

The Routh array is then:

w here:

b12 2 1 4

20 c1

4 2 6

12

d1

6 c1 10

c16

There are two sign changes in the first column due to the large negative number calculated for c1. Thus, the system is unstable because two roots lie in the right half of the plane.

00100001006

10421121

01

11

21

3

4

5

sdscsbs

ss

The Routh-Hurwitz Stability CriterionCase Three: Zeros in the first column, and the other elements of the row containing the zero are also zero

This case occurs when the polynomial q(s) has zeros located symetrically about the origin of the s-plane, such as (s+)(s -) or (s+j)(s -j). This case is solved using the auxiliary polynomial, U(s), w hich is located in the row above the row containing the zero entry in the Routh array.

q s( ) s3 2 s2 4s K

Routh array:

For a stable system we require that 0 s 8

For the marginally stable case, K=8, the s^1 row of the Routh array contains all zeros. The auxiliary plynomial comes f rom the s^2 row.

U s( ) 2s2 Ks0 2 s2 8 2 s2 4 2 s j 2( ) s j 2( )

It can be proven that U(s) is a factor of the characteris tic polynomial:

q s( )U s( )

s 2

2 Thus, w hen K=8, the factors of the characteristic polynomial are:

q s( ) s 2( ) s j 2( ) s j 2( )

00

241

02

81

2

3

Kss

Kss

K

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The auxiliary polynomial The equation that immediately precedes the zero

entry in the Routh array.

The Routh-Hurwitz Stability CriterionCase Four: Repeated roots of the characteristic equation on the jw-axis.

With simple roots on the jw-axis, the system will have a marginally stable behavior. This is not the case if the roots are repeated. Repeated roots on the jw-axis will cause the system to be unstable. Unfortunately, the routh-array will fail to reveal this instability.

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Absolute StabilityA closed loop feedback system

which could be characterized as either stable or not stable.

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Relative StabilityFurther characterization of the

degree of stability of a stable closed loop system.

Can be measured by the relative real part of each root or pair of roots.

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Root r2 is relatively more stable than roots r1 and r1’.

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Axis shift464)( 23 ssssq

14)1(6)1(4)1( 2323 nnnnnn ssssss

sn3

sn2

sn1

sn0

1101

1100

)1)(1())((1)( 2 jsjsjsjsssU nnnn

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Stability versus Parameter RangeConsider a feedback system such as:

The stability properties of this system are a function of the proportional feedback gain K. Determine the range of K over which the system is stable.

Control Systems

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Stability versus Two Parameter Range

Consider a Proportional-Integral (PI) control such as:

Find the range of the controller gains so that the PI feedback system is stable.

)K,K( 1

Control Systems

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Example 4 (cont’d)

The characteristic equation for the system is given by:

021

11 1

)s)(s(sKK

Thank You!

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