Ch-6 RH Stability Webpage

7/31/2019 Ch-6 RH Stability Webpage

1/27

Chapter 6

Algebraic Criterionfor stability


2/27

Stability of a system:

bounded input producesbounded output

In the absence of input

output tends to zero


3/27

C

R

G

GH==== ++++1

1+GH = 0 is called thecharacteristic equation of the

system. Its roots are the closedloop poles.

If the real part of closed loop

poles are in L H P, stable response


4/27

unstablestable


5/27

Second order : as2 + bs + c = 0

product of roots =

c

a

sum of roots = ba

Both the roots will have negative real part

only when a, b, c are of the same sign.

first order :

s + = 0

s = - ,If is positive,stable condition


6/27


7/27

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 a

b

1 3 1

1

0 .

=

=b1


8/27

After completing the table,look for sign change in

FIRST COLUMN.

No sign change - stablesign change - unstable


9/27

s3

s2

s1

s0

1 11

6 11 1 6

6

. .

0

6

6 6

Routh Table

= 10

Ex: s3 + 6s2 + 11s + 6 = 0

System is STABLE


10/27

Example :

Necessary condition- fails

- unstable

s4 +10s3 + 4s + 8 = 0


11/27

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

s4s3

s2

s0

1 22

s1

14

-1


12/27

Sign change in the

first column ----

unstable


13/27

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

s4s3

s2

s0

1 22

s1

14

-1

8

2

22sign changesunstable

2 roots in RHP


14/27

s5

s4

s3

1 2

0

1 2

3

3

0

Ex:


15/27

full row becomes zero

cannot proceed furtherindicates symmetry of roots.

Form auxiliary equation using the

coefficients of the row previous to

the all-zero row.


16/27

s5

s4

s3

s1

1 2

1 2

s2

3

3

6/4

4/6

2

s0 2

No sign change

limitedly stable

A(s) = s4

+ 3s2

+ 2dA/ds = 4s3 + 6s

4 6


17/27

s5

+ s4

+ 3s3

+ 3s2

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

= ( s2 + 1) (s2 + 2) (s + 1)

Roots are :

j 1 j 2 -1, ,


18/27

GK

s s s s====

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

Ch eq:s4 + 4s3 + 13s2 + 36s + K = 0

R(s) +

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


19/27

s4

s3

s2

s0

1 K

4 0

s1

13

36

436-K

K

K

For stability 0 < K < 36


20/27

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

= (s2

+9) ( s2

+ 4 s + 4)

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

For K = 36


21/27

s

4

s3

s2

s0

1 364 0

s1

1336

4

836

36

0

Routh Table

apart from roots on imaginary axes, all

other roots have negative real parts.


22/27

s5

s4

s3

1 2

1 1

3

3

0 1

Example


23/27

first column elementbecomes zero

- method breaks down

- replace 0 by

> 0


24/27

s5

s4

s3

s0

1 21 1

s2

33

1

1

s1

3 1

1 3 1

2

1


25/27

3 1

= 3 1

As 0, 3 -

1

is negative

- unstable


26/27

Alternate method

s5

+ s4

+ 3s3

+ 3s2

+2s +1 = 0

Let s =

1

xThen

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


27/27

x5

x4

x3

x0

1 12 1

x2

33

1/2

1

x1

1

3/2

7/3

-1/7unstable

Documents

Ch-6 RH Stability Webpage