1. Stability Analysis . A system is BIBO stable if all poles or roots of
characteristic equation (CE) lie in the LHP (left
half of complex plane).
Procedure to determine poles location
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
we can calculate the coefficient ai as follows
an – 1= –  pi
an – 2=  pi pj, ij
an – 3= –  pi pj pk, ij k .
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
If any coefficient ai=0 then not all roots are in the LHP
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.

2. . 6.1. Routh-Hurwitz Stability Criterion
Routh-Hurwitz criterion applies to polynomial
Qn(s) = an sn + an – 1 sn – 1 + … + a1 s+ a0
Step to apply Routh-Hurwitz criterion:
Step 1.
Create Routh array as follow
Where
Step 2.
Count the sign change in column 1 of the array.
The roots lie in the RHP = The count result.
There are 3 cases
sn an an – an – an – 6 …
sn – an – an – an – an – 2 …
sn – b b b b4
sn – c c c c4 .
s k k2
s l1
s m1
case 1 no zero appear
Example: Q(s) = s3 + s2 +2s+ 8 s s
s – 6 s
2 sign change  2 roots lie in RHP

3. 6.1. Routh-Hurwitz Stability Criterion
Case 2 first element of a row is zero
Q(s) = s5 + 2s4 + 2s3 + 4s2 +11s+ 10 s s s
The row above the zero row indicates the auxiliary polynomial. Thus the aux polynomial of s5 + s4 + s3 + s2 +2 s 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 s
/   is a small number
s –12/ 10 s s
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 s s s
A case 3 polynomial contains even polynomials, called auxiliary polynomial, as a factor.

4. 6.1. Routh-Hurwitz Stability Criterion
example
Q(s) = s4 +4 s s
We have zero row here. The aux-polynomial is
Qa(s) = s4 +4
Qa’(s) = 4s3
The array becomes s s
s 
s / s
2 sign change  2 roots lie in RHP 1 j

5. 6.1. Routh-Hurwitz Stability Criterion
example
We have control system as follow
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 s
s K
s (18-2K)/4  K<9
s K  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. K
plant
compensator
Suppose that, for constant input, we want
ess<2%.
The position error constant Kp is
Steady state error is ess =1/(1+K) < 2%, thus K > 49.

6. 6.1. Routh-Hurwitz Stability Criterion
example
To overcome the previous problem we have to use PI controller
s (18-2KP)/4 2KI  KP<9
s c
s0 2KI  KI > 0
where
plant
compensator
KP+ KI /s
Using this controller ess= 0. We will find the values of KP and KI to assure the stability.
The CE is found to be
s4 +4s3 + 5s2 +(2+2KP)5s+ 2KI = 0
The array becomes
s KI
s KP
If we choose KP=3<9 then
c = (24-8KI)/3)>0
Thus
0< KI <30

7. 6.1. Routh-Hurwitz Stability Criterion
For the first order the general CE is
Q(s) = a1s+ a0
The Routh array is
s a1
s a0
For the second order the general CE is
Q(s) = a2s2 +a1s+ a0
s a2 a0
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

8. 6.3. Stability by Simulation
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 is 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 accurate numerical integration