Stability Analysis
characteristic equation (CE) lie in the LHP (left
we can calculate the coefficient ai as follows
an – 1= –  pi
half of complex plane).
an – 2=  pi pj, ij
A system is BIBO stable if all poles or roots of
an – 3= –  pi pj pk, ij k
Procedure to determine poles location
1.
2.
3.
Routh-Hurwitz criterion determines if any
roots of polynomial lie outside the LHP
without calculating the roots.
Calculation of the exact location of the
roots analytically.
Calculation of the exact location of the
roots using digital computer.
Let us consider a real polynomial
Qn(s)=(s–pn) (s–pn – 1) (s–pn – 2)…(s–p0)
=
sn +
an – 1
sn – 1 +
an – 2
sn – 2 …
+ a1 s+ a0
.
.
.
a0 = (– 1)n  p0 p1 p3…
It is obvious that if all pi lie in the LHP the all
coefficient must be positive. In other words
1.
If any coefficient ai=0 then not all roots
are in the LHP
2.
If any coefficient ai<0 then at least one
root is in the LHP
However, it does not mean that if all
coefficients are positive then all pi lie in the
LHP.
6.1. Routh-Hurwitz Stability Criterion
Routh-Hurwitz criterion applies to polynomial
Qn(s) = an sn + an – 1 sn – 1 + … + a1 s+ a0
Where
b1  
1 an an2
an1 an1 an3
b2  
1 an an  4

an 1 an 1 an 5
c1  
1 an 1
b1 b1
c2  
1 an 1 an 5

b3
b1 b1
Step to apply Routh-Hurwitz criterion:
Step 1.
an 3
b2
Create Routh array as follow
Step 2.
sn
an
an – 2
an – 4
an – 6 …
sn – 1
an – 1
an – 3
an – 5
an – 2 …
Count the sign change in column 1 of the array.
The roots lie in the RHP = The count result.
sn – 2
b1
b2
b3
b4
There are 3 cases
sn – 3
c1
c2
c3
c4
case 1 no zero appear
.
.
.
Example: Q(s) = s3 + s2 +2s+ 8
s2
k1
s1
l1
s0
m1
k2
s3
s2
s1
s0
1 2
1 8
–6
8
2 sign change  2 roots lie in RHP
6.1. Routh-Hurwitz Stability Criterion
Case 2 first element of a row is zero
Q(s) = s5 + 2s4 + 2s3 + 4s2 +11s+ 10
s5 1
2
11
s4 2
4
10
s3 /0 
6  is a small number
s2 –12/ 10
s1 6
s1 10
2 sign change  2 roots lie in RHP
Case 3 all elements of a row is zero
Q(s)= s5 + s4 + s3 + s2 +2 s +2
s5
s4
s3
1
1
0
1
1
0
2
2
A case 3 polynomial contains even
polynomials, called auxiliary polynomial,
as a factor.
The row above the zero row indicates the
auxiliary polynomial. Thus the aux polynomial of
s5 + s4 + s3 + s2 +2 s +2 is Qa = s4 +s2 +2.
Two method available to solve this problem.
Method 1
First factorized the polynomial and analyze each
factor individually. For the above polynomial we
have
s5 + s4 + s3 + s2 +2 s +2 = (s+1)(s4 +s2 +2)
We then create Routh array for s+1 and for
s4+s2+2
Method 2
Differentiate the aux polynomial. The coefficient
of the result replaces the zero row. Thus the
coefficient of 4s3 + 2s replaces the zero or
s3
4
2
6.1. Routh-Hurwitz Stability Criterion
example
s4 +4
Q(s) =
s4 1
0
4
s3 0
0
We have zero row here. The auxpolynomial is
Qa(s) = s4 +4
Qa’(s) = 4s3
The array becomes
s4 1
0
s3 4
0
s2 
4
s1 -16/
s0 4
4
2 sign change  2 roots lie in
RHP
j
1
6.1. Routh-Hurwitz Stability Criterion
example
We have control system as follow
compensator
K
plant
2
s 3  4 s 2  5s  2
However we have to check the stability, and
we will use Routh-Hurwitz criterion
The CE is found to be
s3 + 4s2 +5s+ 2+2K = 0
The array becomes
Suppose that, for constant input, we want
ess<2%.
The position error constant Kp is
2


K p  lim K 3
K
2
s 0
 s  4 s  5s  2 
Steady state error is ess =1/(1+K) < 2%, thus
K > 49.
s3
1
5
s2
4
2+2K
s1
(18-2K)/4  K<9
s0
2 + 2K
 K>-1
Here for stability requirement we must have
-1<K<9
We conclude that, using proportional
compensator, our requirement to have ess<2%
can not be satisfied.
6.1. Routh-Hurwitz Stability Criterion
example
To overcome the previous problem we have
to use PI controller
compensator
KP+ KI /s
plant
2
s 3  4 s 2  5s  2
stability.
The CE is found to be
(18-2KP)/4
s1
c
2KI
where

4
18  2 K p
(2  2 K p )(18  2 K p ) 

8
K

 I

4



4
(1  K P )(9  K p )  8K I
18  2 K p
If we choose KP=3<9 then
s4 +4s3 + 5s2 +(2+2KP)5s+ 2KI = 0
The array becomes
c = (24-8KI)/3)>0
Thus
s4
1
5
s3
4
2+2KP
2KI
 KP<9
 KI > 0
s0 2KI
c
Using this controller ess= 0. We will find
the values of KP and KI to assure the
s2
0< KI <30

6.1. Routh-Hurwitz Stability Criterion
For the first order the general CE is
Q(s) = a1s+ a0
The Routh array is
s1 a1
s0 a 0
For the second order the general CE is
Q(s) = a2s2 +a1s+ a0
The Routh array is
s2 a 2 a 0
s1 a 1
s0 a 0
Hence for the first and second order system, the necessary and sufficient condition for
stability is that all the coefficient of characteristic polynomial must be of the same sign
6.2 Roots of the CE
•
The RH criterion allows us to determine
stability without finding the roots of the
CE.
•
The stability can also be determined by
finding the roots of the CE
•
For first and second order finding the roots
is trivial.

•
For third order algebraic methods are
available

•
For higher order numerical methods are
required
•
One disadvantage of numerical methods is
that all parameters must be assigned
numerical value
•
Finding the range of parameter becomes
more difficult when numerical methods is
used.
•
The 1st and 2nd order system, its
stability can be determined by the
sign of the coefficient.
6.3. Stability by Simulation




An obvious method for determining
stability is by simulation.
First system is simulated then the output
observed for typical inputs.
The stability then become evident.
For nonlinear and time varying system,
simulation may be the only way.
There are no general way for analyzing
nonlinear and/or time varying system.
Care must be taken to ensure the accurat
numerical integration