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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