CSE

1. CONTROL SYSTEMS
ENGINEERING
Course Code: MEB 4101
Course Credit 4
Stability Criteria and Analysis
Prepared by;
Masoud Kamoleka Mlela
BSc, Electromechanical Eng. (UDSM)
MSc, Renewable Energy (UDSM)
PhD, Mechanical Engineering (HEU, CN).

2. Stability Criteria and Analysis
Roots and Polynomials -Poles and Zeros
Transfer Function
Ignoring the origins for now, define a transfer function as a
function with a polynomial numerator and a polynomial
denominator.
Some examples are
Assume the denominator order is greater than or equal to the numerator
order in general for control purpose

3. Stability Criteria and Analysis
Roots and Polynomials -Poles and Zeros
 A pole is a root of the denominator
 A zero is a root of the numerator.
Find the poles and zeros of the following and marks
in the complex plane, ‘o’ for zeros and ‘x’ for poles.

4. Stability Criteria and Analysis
Roots and Polynomials -Poles and Zeros
Left half plane (LHP) and right half plane (RHP)
These definitions are simple as they seem.
 They refer to the right of the imaginary axis is the RHP
 Any thing to the left of the imaginary axis is the LHP.

5. Stability Criteria and Analysis
Roots and Polynomials -Poles and Zeros
Left half plane (LHP) and right half plane (RHP)
Problems with LHS and RHP
 How many RHP poles does the following transfer function include?
This equation does not need to consider the zeros,
The RHP is defined as the real part of the root being positive. The imaginary part is
unimportant.
The poles here are clearly : 3, -1±2j
So only the 3 is in the RHP
Answer is ONE
Im
Re

6. Stability Criteria and Analysis
Roots and Polynomials -Poles and Zeros
Left half plane (LHP) and right half plane (RHP)
Tutorial Problem
 How many RHP poles and Zeros does the following transfer function include?

7. • What is Stability?
A system is said to be stable, if its output is under control.
Otherwise, it is said to be unstable. A stable system produces a
bounded output for a given bounded input.
The following figure shows the response of a stable system.
Stability Criteria and Analysis
This is the response of first order control system for
unit step input.
 This response has the values between 0 and 1.
So, it is bounded output.
 We know that the unit step signal has the value of
one for all positive values of t including zero. So, it
is bounded input.
 Therefore, the first order control system is stable
since both the input and the output are bounded.

8. Stability Criteria and Analysis
The stability of the system can be obtained from the
coefficients s of its characteristic equation. let us
consider a system with a characteristic equation
1 2 1 0
0 1 2 1
... 0
n n n
n n
a s a s a s a s a s
 

      (1)
An assumption that has been made in equation (1).
Even when , the equation (1) can be multiplied by -1
and can made positive
a) All the coefficients of the characteristic equation should be positive
b) No term should be missing in the characteristic equation
0 0
a 
0 0
a 

9. Stability Criteria and Analysis
a) All the coefficients of the characteristic equation should be positive
b) No term should be missing in the characteristic equation
Conditions (a) and (b) are necessary for a system to be
stable. If some of the coefficients are zero or negative it can
be concluded that the system is NOT STABLE. If all the
coefficients of the characteristics equation are positive or
negative, there is a possibility of stability of the system to
exist and let us now proceed to examine the sufficient
conditions of system stability.

10. Stability Criteria and Analysis
5 4 2
4 2 0
s s s s
    
4 3 2
5 2 3 2 16 0
s s s s
    
3
12 10 0
s s
  
Example: Investigate whether the following systems
represented by the characteristic equations may be stable:
(a)
(b)
(c)
Solution:
,
a) Given characteristic equation in which the s3 term is missing
and so the system cannot be stable.
b) The characteristic equation is , in which the s3 term
alone has a negative coefficient while all the other coefficients are positive.
Therefore, the necessary condition is not satisfied and the system is
UNSTABLE.
c) The characteristic equation is , in which the s2 term is missing and
also the s term has a negative coefficient while the other terms have positive
coefficients. Therefore, both the necessary conditions for system stability are
not satisfied and so the system becomes UNSTABLE
5 4 2
4 2 0
s s s s
    
4 3 2
5 2 3 2 16 0
s s s s
    
3
12 10 0
s s
  

11. Stability Criteria and Analysis
Routh-Hurwitz Stability Criterion
1 2 1 0
0 1 2 1
... 0
n n n
n n
a s a s a s a s a s
 

     
The Routh-Hurwitz criterion was developed independently by A. Hurwitz (1895) in
Germany and E.J Routh (1892) in the United States.
Then the Routh’s array S obtained as follows
Let the characteristics polynomial (the denominator of the transfer function after
cancellation of common factors with numerator) be given by
n
s 0
a 2
a 4
a 6
a
1
n
s 
1
a 3
a 5
a 7
a
2
n
s 
1 2 3 0
1
1
a a a a
b
a

 1 4 5 0
2
1
a a a a
b
a

 1 6 7 0
3
1
a a a a
b
a


3
n
s 
1 3 2 1
1
1
b a b a
c
b

 1 5 3 1
2
1
b a b a
c
b


1
s
0
s n
a
Where the first two rows are obtained from the coefficients of the characteristic equation

12. Stability Criteria and Analysis
While preparing the Routh table for a given polynomial, some of the elements may
not exist. In calculating the entries in the line that follows, these elements are
considered to be zero.
 For a system to be stable, it is necessary and sufficient that each entry of the first
column of the Routh array of its characteristic equation be POSITIVE, if . If this
condition is not satisfied, the system is unstable
0 0
a 
Example:
Determine the stability of the system whose characteristic equation is given by
4 3 2
6 23 40 50 0
s s s s
    
6 5 4 3 2
5 8 12 20 100 150 200 0
s s s s s s
      
(a)
(b)

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