The Routh Stability Criterion

When consider the design and analysis of feedback

control systems, stability is of the utmost importance.

From a practical point of view, a closed-loop feedback

system that is unstable is of little value.

Using feedback, we can stabilize unstable plants and

then with a judicious selection of controller parameters,

we can adjust the transient performance.

The Concept of Stability

A system that works at its original equilibrium states is effected by a disturbance (noise) . When the disturbance go off , the system can come back to the original equilibrium states the system is stable. Or else the system is unstable.

Definition of stability

The Concept of Stability

A stable system is a dynamic system with a

bounded response to a bounded input.

or

The unity-step response analysis of the high-order control systems

Assume : nm

the zeros of the transfer function

the poles of the transfer function

We can get the unity-step response :

)()(1

10

1

10 sRasasa

bsbsbsC

n

nn

m

mm

sssssss

zszsk

n

m 1

)())((

)()(

21

1

ssR

ttr

1)(

)(1)(

;,,

;,,

21

21

n

m

sss

zzz

n

i

ts

i

n

m ieAa

btc

1

)(

(1) If all poles are in the left of the s-plane,the system is stable .

(2) The transient portions of the response are exponentially decayed.corresponding to the poles are at the real of the left s-plane .

The transient portions of the response are the decayed oscillatory motions. corresponding to the complex poles in the left of the s-plane.

(3) The more far the poles in the left s-plane go from the imaginary axis.

The more quick the transient portions of the response are decayed.

The unity-step response analysis of the high-order control systems

(4) The more far the more poles in the left s-plane go from

the real axis. the more large the overshoot of the

response is.

(5)The zeros of the system only affect the coefficient of the

transient portions.

(6) The poles,which are the most close to the imaginary axis

of the s-plane,are of the most affecting to the transient

portions of the response.

dominant poles.

The unity-step response analysis of the high-order control systems

Suppose a transfer function

The Concept of Stability

n

nn

m

mm

asasa

bsbsb

sR

sC

1

10

1

10

)(

)(

We can always factor as

n

i

i

m

j

j

ps

zsK

sR

sC

1

1

)(

)(

)(

)(

The closed-loop system is stable if i , 0)( ipR

Note: It might turn out that there are pole-zero

cancellations, that is, z j = p i for some i. j.

If this happens, system is still unstable if 0)( ipR

The sufficient and necessary conditions of

the stability for a linear system

Im

Re

S-plane

The relationship between the systems stability and the position of poles in S-plane.

The Concept of Stability

Graphic representation

The Routh criterion represents a method of determining

the location of roots of a polynomial with constant real

coefficients with respect to the left half and right half of

the s-plane, without solving for the roots.

The Routh Criterion

The characteristic equation is:

0)( 11

10

nn

nn asasasasD

The roots are, of course nppp , , 21

Important question:

Can we tell if the system is stable, without actually

solving for the roots?

To ensure the equation does not have roots with positive real

parts, it is necessary (but not sufficient) that the following

conditions hold:

(1). All the coefficients of the equation have the same sign.

(2). None of the coefficients vanish.

These conditions are based on the laws of algebra, which

relate the coefficients of the equation .

0)( 11

10

nn

nn asasasasD all the coefficients are real.

The Routh Criterion

063.3sin101137.2237.245 ssess

The characteristic equation is short of one or more than one items The system must be unstable .

Example :

unstable, be short of the item

0663.3sin101137.2237.245 sssess

The coefficient of the characteristic equation are

different in sign The system must be unstable.

Example :

unstable, The coefficient are different in sign.

4a s

The Routh Criterion

The two necessary conditions for Equation to have no roots in the right-half s-plane

can easily be checked by inspection of the equation. However, these conditions are

not sufficient, for it is quite possible that an equation with all its coefficients nonzero

and of the same sign still may not have all the roots in the left half of the s-plane.

Routh Array

0)( 11

10

nn

nn asasasasD

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

The characteristic equation ( ) 00 a

The first two rows come directly from the polynomial D(s). Each

subsequent row is formed by operations on the two rows above:

The Routh Criterion

Routh Array

0)( 11

10

nn

nn asasasasD

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

1

302113

a

aaaac

The Routh Criterion

00 a

Routh Array

0)( 11

10

nn

nn asasasasD

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

1

504123

a

aaaac

The Routh Criterion

00 a

Routh Array

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

0)( 11

10

nn

nn asasasasD

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

1

706133

a

aaaac

The Routh Criterion

00 a

Routh Array

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

0)( 11

10

nn

nn asasasasD

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

13

23131314

c

caacc

The Routh Criterion

00 a

Routh Array

0)( 11

10

nn

nn asasasasD

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

13

33151324

c

caacc

The Routh Criterion

00 a

Routh Array

0)( 11

10

nn

nn asasasasD

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

13

43171334

c

caacc

The Routh Criterion

00 a

Routh Array

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

the number of unstable roots of D(s) is the number of

sign changes in the first column of the Routh array.

0)( 11

10

nn

nn asasasasD

nn

n

n

n

n

acs

ccccs

ccccs

aaaas

aaaas

1,1

0

44342414

3

43332313

2

7531

1

6420

0)( 11

10

nn

nn asasasasD

The Routh Criterion

00 a

Conclusion(Routh Criterion):

Necessary, Sufficient condition

All elements of the first column of the Routh-

table(array) are positive .

The system is and must is stable.

The Routh Criterion

the number of unstable roots of D(s) is the number of

sign changes in the first column of the Routh array.

The characteristic equation of a second-order system is

The Routh array is written as

Therefore, the requirement for a stable second-order

system is simply that all the coefficients be positive.

0)( 212

0 asasasD

0

0

2

0

1

1

20

2

as

as

aas

Examples

0105323 sss

0205323 sss

Stable

10

03

5

103

51

0

1

2

3

s

s

s

s

20

03

5

203

51

0

1

2

3

s

s

s

s

Example 1

Examples

Example 2

First column has two sign changes!

Unstable . 2 roots with positive real parts

0832 23 sss

8

01

082

031

0

1

2

3

s

s

s

s

Example 3

Examples

First column has two sign changes!

There are two unstable poles. In fact, the roots are:

-2.2483

0.1241 + 1.8822i

0.1241 - 1.8822i

Examples Stability vs. Parameter Range

Its much easier to use a calculator or Matlab to find roots.

So why use Routh? Routh allows us to determine

symbolically what values of a parameter will lead to

stability/instability.

Examples Stability vs. Parameter Range

For what values of k is the following system stable?

Solution: The Closed Loop transfer function is:

Ks

K

s

K

s

K

sKG

sKGsT

3

3

3

)1(

)1(1

)1(

)(1

)()(

0133)( 23 KssssD

Example 4

For stability, need first column to be positive,

so that K -1.

If K < -1, first column is + + + - , so there is 1 unstable pole.

If K > 8, first column is + + - + , so there are 2 unstable poles.

Routh Array 013323 ksss

ks

ks

ks

s

1

03

8013

031

0

1

2

3

Examples Stability vs. Parameter Range

Example 4

Possible problems:

If the first element of a row is zero, process fails.

Solution: Replace 0 by , a small positive number. (If a whole row is zero, must replace row as explained in the book. This happens whenever there are complex conjugate pairs of roots on the imaginary axis. )

The Routh Criterion

0122234 ssss

02

222

s4 1 1 1

s3 2 2 0

s2 0 1

s1 0

s0 1

Unstable. 2 roots with positive real parts.

Note:

Can use a infinitesimal >0 substituting the zero element in

the first column.

>0

Examples

Example 5

If a whole row is zero, must replace the row as explained in the book.

See 2-13 Routh_Rurwitz Criterion example 2.13.3

This happens whenever there are complex conjugate pairs of roots on

the imaginary axis.

)1025.0)(11.0()(

sss

KsG

G(s)

Determine the range of K for a stable system.

The characteristic equation

035.0025.0 23 Ksss

The range of K for a stable system is

( )(0.1 1)(0.025 1)

KG s

s s s

035.0025.0 23 Ksss

0K

14 0025.035.0 KK

140 K

Ks

Ks

Ks

s

0

1

2

3

035.0

025.035.0035.0

01025.0

Examples Axis shift

Example 6

The characteristic equation

The range of K for a stable system is

Now, it is desired that the real part of all roots should

be less than -1, what is the range of K ?

Let , then 11 ss

035.0025.0 23 Ksss

11 ss

0)2740(1511 12

1

3

1 Ksss

0.675K

8.4 , 0)2740(1511 KK

8.4675.0 K

2740

011

)2740(1511

0274011

0151

0

1

1

1

2

1

3

1

Ks

Ks

Ks

s

Example 6

Examples Axis shift