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SIGNALS AND CONTROL SYSTEMS

Instructor : Dr. Raouf Fareh

Fall Semester 2014/2015

Week 10_2 Stability of Linear Feedback Systems

Introduction

� The stability of a feedback system is directly related to the location of the roots of 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.

� It was discovered that all coefficients of the characteristic polynomial must have the same sign and non-zero if all the roots are in the left-hand plane.

� These requirements are necessary but not sufficient. If the above requirements are not met, it is known that the system is unstable. But, if the requirements are met, we still must investigate the system further to determine the stability of the system.

The Routh-Hurwitz criterion is a necessary and sufficient criterion for the stability of

linear systems.

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The Routh-Hurwitz Stability Criterion

( )( ) ( )

31

31

11

51

4

13

31

2

11

3211

10

5313

5312

5311

42

012

21

1

1

1

1

0

−−

−−

−−

−−

−

−−

−−

−

−−

−−−−

−

−−−−

−−−−

−−−−

−−

−−

−−

−=

−=

−=−=

••••••••••••

=+++++

nn

nn

nn

nn

nn

nn

nn

nn

nn

nnnnn

n

nnnn

nnnn

nnnn

nnnn

nn

nn

nn

bb

aa

bc

aa

aa

ab

aa

aa

aa

aaaab

hs

cccs

bbbs

aaas

aaas

asasasasa LCharacteristic equation, q(s)

Routh array

The Routh-Hurwitz criterion statesthat the number of roots of q(s)with positive real partsis equal tothe number ofchanges in signofthefirst column of the Routh array.

���� =�������������

���or

���� =������ �������

���or

���� =���������������

���or

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:

b1

a1 a0⋅ 0( ) a2⋅−

a1

a0

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

0

0

10

11

022

bs

as

aas

Example: Second-Order System

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Example: Third-Order System

• The characteristic polynomial of a third-order system is:

The polynomial satisfies all the necessary conditions because all the coefficients exist and are positive.

Utilizing the Routh array, we have:

Because there are two changes in sign in the first column, we find that two roots of D(s) lie in the right- hand plane, and our prior knowledge is confirmed.

6

4 3 22 3 5 10 0s s s s+ + + + =Routh's Tabulation

2 sign change 2 RHP roots⇒

4

3

2

1

0

2 3 10

1 5

1 3 2 57 10

17 5 1 10 45

07 710

s

s

s

s

s

× − × = −

− × − × =−

Example: Fourth-Order System

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Special Case1: Zero only in the first column

Zero only in the first columnWhen forming the Routh table, replace the zero with a small number � and evaluate the first column for positive values of �.

Problem: Determine the stability of the closed loop transfer function:

0 0, for column first of Sign →ε>ε

system unstable

RHPin poles two

changessign two

⇒

⇒

3s2s1s0s

5s4s

+

+

+

+

+

−

Special Case 2: Entire row is zeroProblem: Determine the number of right hand plane poles in the closed-loop transfer function:

Now we are faced with the problem of zeros in the third row.

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1. Form a new polynomial using the entries in the row above zeros. The polynomial will start with power of s in that row, and continue by skipping every other power of s, i.e.

2. Next we differentiate the polynomial with respect to s and obtain

(2)

3. Finally the row with all zeros in the Routh table is replaced with the coefficients in Eq.(2), and continue the table.

Special Case 2: Entire row is zero

We see no sign changes hence no RPH poles.

Special Case 2: Entire row is zero

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• Some polynomial only have roots symmetrical about the origin.

• Routh table from the even polynomial ��� → ��� is a test of the even polynomial.• The rows of zeros indicates the possilbility of jw roots.

• Why does an entire row of zeros occur?When a purely odd or even polynomial is a factor of the original polynomial. (�� �6�� � 8 is an even polynomial as it only has even power of s.)

Special Case 2: Entire row is zero

Examples

Example 1

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Examples

As an example, consider

Two sign changes, hence unstable with two RHP poles

Example 2

Examples

Example 3

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Examples

Example 4

Examples

Example 5

Given the system below find the range of the gain K that will lead to stability, unstability

and marginal stability.

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Examples

Example 5

• If K is positive (assumed) all the elements in the first column are positive except that the �� entry may be positive/nagative.

• If K < 1386, the system will remain stable.

• If K > 1386, the system will remain unstable.

• If K = 1386, a row of zeros, so using

Since there are no sign change from �� to ��

the even polynomial has no unstable roots,only j! roots, therefore the system is marginally stable when K = 1386.