chapter 4...chapter 4 contents introduction system stability routh-hurwitz criterion construction of...

Linear System StabilityChapter 4

Contents

Introduction

System Stability

Routh-Hurwitz Criterion

Construction of Routh Table

Determining System Stability

Routh-Hurwitz Criterion (Special Cases)

1. Zero only in First Column

2. Zero for Entire Row

Stability via Routh Hurwitz

Introduction

Stability is the most important system specification. If a system is unstable, the transient response and steady-state errors are in a moot point.

Definition of stability, for linear, time-invariant system by using natural response:

A system is stable if the natural response approaches zero as time approaches infinity.

A system is unstable if the natural response approaches infinity as time approaches infinity

A system is marginally stable if the natural response neither decays nor grows but remains constant or oscillates.

)()()( tctctc naturalforced

Introduction

Definition of stability using the total response bounded-input,

bounded-output (BIBO):

i. A system is stable if every bounded input yields a bounded output.

ii. A system is unstable if any bounded input yields an unbounded

output.

Absolute & Relative Stability

Absolute Stability:

i. The absolute stability indicates whether the system is stable or not.

ii. This is indicated by the presence of one or more poles in RHP.

Relative Stability:

i. Relative stability refers to the degree of stability of a stable system

described by above.

ii. This depends on the transfer function of the system, which is

represented by both the numerator (that yields the zeros) and

denominator (that yields the poles).

iii. This can then be referred to in the study of system response either in

time or frequency domain.

System Stability

Stable systems have closed-loop transfer functions with poles only

in the left half-plane.

System Stability

Unstable systems have closed-loop transfer functions with at least

one pole in the right half-plane and/or poles of multiplicity

greater than 1 on the imaginary axis.

System Stability

Marginally stable systems have closed-loop transfer functions with

only imaginary axis poles of multiplicity 1 and poles on the

imaginary axis.


To determine stability of a given system, we have to consider the

manner in which the system is operating, whether open-loop or

closed-loop.

i. If the system is operating in open-loop, first find the closed loop

transfer function.

ii. Find the closed-loop poles.

iii. If the order of the system is 2 or less, factorise the denominator of

the transfer function. This will provide the roots of the polynomial, or

the closed-loop poles of the system.

iv. If the system order is higher than 2nd-order, use construct Routh

table and apply Routh-Hurwitz Criterion.

v. Any poles that exist on the RHP will indicate that the system is

unstable.

Routh–Hurwitz Criterion

Routh-Hurwitz Stability Criterion:

The number of roots of the polynomial that are in the right half-

plane is equal to the number of changes in the first column.

Systems with the transfer function having all poles in the LHP is

stable.

Hence, we can conclude that a system is stable if there is no change

of sign in the first column of its Routh table.

Routh–Hurwitz Criterion

If a polynomial is given by:

The necessary conditions for stability are:

i. All the coefficients of the polynomial are of the same sign. If not, there are poles on the right hand side of the s-plane.

ii. All the coefficient should exist accept for a0.

0.....)( 011

1

asasasasTn

nn

n

Where,

an, an-1, …, a1, a0 are constants

n = 1, 2, 3,…, ∞


For the sufficient condition, we must form a Routh-array.

1

321

1

31

2

1

n

nnnn

n

nn

nn

a

aaaa

a

aa

aa

b

1

541

1

51

4

2

n

nnnn

n

nn

nn

a

aaaa

a

aa

aa

b


For the sufficient condition, we must formed a Routh-array.

1

2131

1

21

31

1b

baab

b

bb

aa

c nn

nn

1

3151

1

31

51

2b

baab

b

bb

aa

c nn

nn

1

2121

1

21

21

1i

ihhi

i

ii

hh

j

21 ik

Construction of Routh Table

Equivalent closed-loop transfer function

Initial layout for Routh table

Completed Routh table


Example: How many roots exist on RHP?

Routh–Hurwitz Criterion (Special Cases)

However, special cases exists when:

1. There exists zero only in the first column.

2. The entire row is zero.

1.Zero only in First Column

If the first element of a row is zero, division by zero would be

required to form the next row.

To avoid this, an epsilon, , is assigned to replace the zero in the first column.

Example: Consider the following closed-loop transfer function T(s).

35632

10)(

2345

ssssssT

Zero only in First Column

To determine the system stability, sign changes were observed

after substituting with a very small positive number or

alternatively a very small negative number.

Zero Only in First Column

Exercise: For the following closed-loop transfer function T(s), determine the number of poles that exist on RHP.

123232

1)(

2345

ssssssT

2. Zero for Entire Row

An entire row of zeros will appear in the Routh table when a

purely even or purely odd polynomial is a factor of the original

polynomial.

Example: s4 + 5s2 + 7 has an even powers of s.

Even polynomials have roots that are symmetrical about the origin.

i. Roots are symmetrical & real

ii. Roots are symmetrical & imaginary

iii. Roots are quadrantal

Zero for Entire Row

Zero for Entire Row

Example:

Differentiate with respect to s:

5684267

10)(

2345

ssssssT

86)( 24 sssP

0124)( 3 ss

ds

sdP

Zero for Entire Row

Example:How many poles are on RHP, LHP and jω-axis for the closed-loop system below?

Zero for Entire Row

Exercise: For the following closed-loop transfer function T(s), determine the number of poles that exist on RHP, LHP and the jω-axis

20384859392212

20)(

2345678

sssssssssT

Stability Via Routh-Hurwitz

Example: Find the range of gain K for the system below that will

cause the system to be stable, unstable and marginally stable,

Assume K > 0.

Closed-loop transfer function:

Ksss

KsT

7718)(

23




Assume K > 0.

Forming the Routh table:




Assume K > 0.

If K < 1386:

All the terms in 1st column will be positive and since there are no

sign changes, the system will have 3 poles in the left-half plane

and are stable.

If K > 1386:

The s1 in the first column is negative. There are 2 sign changes,

indicating that the system has two right-half-plane poles and one

left-half plane pole, which make the system unstable.




Assume K > 0.

If K = 1386:

The entire row of zeros, which signify the existence of jω poles. Returning to the s2 row and replacing K with 1386, so we have:

P(s)=18s2 +1386

chapter 4...chapter 4 contents introduction system stability routh-hurwitz criterion construction of...

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