1

Chapter 6

Algebraic Criterion for stability

Stability of a system means

• bounded input produces bounded output

Stability of a system means

• In the absence of inputoutput tends to zero

CR

GGH

====++++1

2

1+GH = 0 is called the characteristic equation

of the system. Its roots are the closed

loop poles.

If the real part of closed loop poles

are in L H P, stable response

G (s) = transfer function of a system

g (t) = L-1 G(s)= impulse

response

××××- a

e -a t

stable

3

××××

stable××××

e- a t sin bt

-a

-jb

jb××××

unstable××××

e+ a t sin bt

a

jb

-jb

××××

limitedly stable

××××

s in ωωωω tjω

-jω

××××

unstable

××××t sin ωωωωt

××××

××××

jω

-jω

4

limitedly stable

×××× u -1 (t)

unstable

×××× t u -1 (t)××××

unstablestable

first order :

s + αααα = 0 s = - αααα , is the pole.

5

If αααα is positive, stable

condition

Second order :

a s 2 + bs + c = 0

product of roots =ca

sum of roots =−−−− ba

Both the roots will have negative

real part only when a, b, c are of

the same sign.

6

Therefore necessary and

sufficient conditions for

negative ness of real part

is all coefficients must be

non zero and of the same

sign.

First & Second order.

Necessary & sufficient condition

- all coefficients must bepresent and of same sign

If the order of the system is more than two….The condition is necessary

but not sufficient.

Routh stability test

is for checking thenegativeness of the realpart of the roots.

7

Routh Stability Criterion:

Necessary condition satisfied.

s3 + 6s2 + 11s + 6 = 0

s3 + 6s2 + 11s + 6 = 0

s3

s2

s1

s0

1 11

6 11 1 66

. .−−−− 0

6

6 6

Routh Table

= 10

s3

s2

s1

s0

a0 a2

a a a aa

1 2 0 3

1

−−−−0

a1 a3

a0 s3 + a1 s2 + a2 s1 + a3 = 0

b a ab

1 3 1

1

0−−−− .

=

=b1

After completing the table, look for sign change in

FIRST COLUMN.

No sign change - stable

sign change - unstable

8

s3 + 6s2 + 11s + 6 = 0

s3

s2

s1

s0

1 11

6 11 1 66

. .−−−− 0

6

6 6

Routh Table

= 10

No sign change in first column

System is STABLE

Hence Routh Test is a Go/No Go test. Does not reveal any other

quantitative performance

measures.

s4 +2s3 + 8s2 + 4s + 3 = 0Ex :

s4

s3

s2

s0

1 32 0

s1

84

633

3

stable

9

Example :

Necessary condition - fails- unstable

s4 +10s3 + 4s + 8 = 0s4 +2s3 + s2 + 4s + 2 = 0

Ex :

s4

s3

s2

s0

1 22

s1

14

-1

Sign change in the first column ----

unstable

s4 +2s3 + s2 + 4s + 2 = 0Ex :

s4

s3

s2

s0

1 22

s1

14

-182

2�2sign changes �unstable �2 roots in RHP

10

s5

s4

s3

1 2

0

1 2

3

3

0

Ex:

�full row becomes zero

�cannot proceed further

�indicates symmetry of roots.

Form auxiliary equation

using the coefficients of

the row previous to the

all-zero row.

s5

s4

s3

1 2

0

1 2

3

3

0

11

Form Auxiliary equation A(s)

A(s) = s4 + 3s2 + 2dA/ds = 4s3 + 6s

s5

s4

s3

s1

1 21 2

s2

33

φ 46/44/6

2

s0 2

φ 6

No sign change limitedly stable

s5 + s4 + 3s3 + 3s2 + 2s + 2= (s4 + 3s2 +2) (s+1)= ( s2 + 1) (s2 + 2) (s + 1)

Roots are :

±±±± ±±±±j 1 j 2 - 1, ,

R(s) +

-G(s) C(s)E(s)

12

GK

s s s s====

++++ ++++ ++++4 3 24 13 36

Characteristic equation is

s4 + 4s3 + 13s2 + 36s + K = 0

s4

s3

s2

s0

1 K4 0

s1

1336

436-K

K

K

For stability

0 < K < 36

When K = 36 ,s1 row becomes zero.

∴∴∴∴ A (s) = 4s2 + K = 0 or 4s2 + 36 = 0 s2 + 9 = 0 s = ±±±± j 3

13

s4

s3

s2

s0

1 364 0

s1

1336

4φ 836

360

Routh Table Routh test indicates thatapart from roots on

imaginary axes, all otherroots have negative real

parts.

s4 + 4s3 + 13s2 + 36 s +36

= (s2 +9) ( s2 + 4 s + 4)

∴∴∴∴ Roots are ±±±± j 3 , - 2 , - 2 .System limitedly stable

For K = 36

s5

s4

s3

1 21 1

33

0 1

Example

14

first column elementbecomes zero- method breaks down- replace 0 by

∈ > 0

s5

s4

s3

s0

1 21 1

s2

33

11

s1

φφφφ ∈∈∈∈3 1∈∈∈∈−−−−

∈∈∈∈ 13 1

2

−−−−∈∈∈∈∈∈∈∈−−−−

1

3 1∈∈∈∈−−−−∈∈∈∈

= 31

−−−−∈∈∈∈

As ∈→∈→∈→∈→∈∈∈∈

0, 3 -1

is negative

- unstable

Alternate method

s5 + s4 + 3s3 + 3s2 +2s +1 = 0

Let s = 1x

Then

x5 + 2x4 + 3x3 + 3x2 + x + 1 = 0

15

s5

s4

s3

s0

1 12 1

s2

33

1/2

1s1

1

3/2

7/3-1/7

unstable