satbility theory

1. Ayatur rahman soomro control systems date:21-11-2014
Stability of the system
Definitions
The stability of a system relates to its response to inputs or disturbances. A system which
remains in a constant state unless affected by an external action and which returns to a
constant state when the external action is removed can be considered to be stable.
A systems stability can be defined in terms of its response to external impulse inputs..
Definition .a :
A system is stable if its impulse response approacheszero as time approaches infinity..
The system stability can also be defined in terms of bounded (limited) inputs..
Definition .b:
A system is stable if every bounded input produces a bounded output.
Routh -Hurwitz stability conditions
The Routh stability criterion provides a convenient method of determining a control systems stability.
It determines the number of characteristic roots within the unstable right half of the s-plane, and the
number of characteristic roots in the stable left half, and the number of roots on the imaginary axis. It does
not locate the roots. The method may also be used to establish limiting values for a variable factor beyond
which the the system would become unstable...
The characteristic equation being tested for stability is generally of the form
a ns n + a n-1s n-1 +....a 1s + a 0 = 0...
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2. Ayatur rahman soomro control systems date:21-11-2014
Col 1 Col 2 Col 3
Row 0 s n a n a n-2 a n-4
Row 1 s n-1
a n-1 a n-3 a n-5
Row 2 s n-2
b 1 b 2 b 3
Row 3 s n-3
c 1 ... ...
. ... ... ... ...
. ... ... ... ...
Row n-
1
s 1
y 1 y 2
Row n s 0
z 1
The elements in rows 3 downwards are calulated as shown below..
The numerator in each case is formed from elements in the two rows above the element being calculated.
Using the value in column 1(pivot column) and the value in the column to the right of the element being
calculated. The denominator is the element in the pivot column in the row above the element being
calculated The calculated element is made 0 if the row is too short to complete the calculation.
b 1= ( an-1.an-2 - an. a n-3 )/ an-1
b 2= ( an-1 .an-4 - an. a n-5)/ a n-1 ..etc..
c 1= ( b1 .an-3 - an-1. b 2 ) / b1
c 2= ( b1 .an-5 - an-1. b 3 ) / b1 .....etc.
The last row will have just one element.
Routh Stability Criteria
For system stability the primary requirement is that all of the roots of the characteristic equation have
negative real parts....
All of the roots of the characteristic equation have negative real parts only if the elements in column 1 of the
routh array are the same sign,
The number of sign changes in column 1 is equal to the number of roots of the characteristic equation with
positive real parts..
Example ..
To test the stability of a system having a characteristic equation
F(s) = s 3 + 6 s 2 + 12 s + 8 = 0
The Routh Array is constructed as follows..
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Col 1 Col 2 Col 3
Row 0 s 3
1 12 0
Row 1 s 2
6 8 0
Row 2 s 1
64/6 0
Row 3 s 0
8
Column 1 (pivot column) includes no changesof sign and therefore the roots of the characteristic equation
have only real parts and the system is stable.

3. Ayatur rahman soomro control systems date:21-11-2014
Special cases for resolving Routh Stability Array
1)....
If a zero appears in the first column 1 of any row marginal stability or instability is indicated. The normal
method of constructing the array cannot be continued because the divisor would be zero. A convenient
method or resolving this method is to simply replace the zero by a small number δ and continue as
normal. The limit as δ -> 0 is then determined and the first column is checked forsign changes..
Example.
To test the stability of a system having a characteristic equation
F(s) = s 5 + 2 s 4 + 2 s 3 + 4 s 2 + s + 1
The Routh Array is constructed as follows
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Col 1 Col 2 Col 3
Row 0 s 5
1 2 1
Row 1 s 4
2 4 1
Row 2 s 3 δ 0,5
Row 3 s 2
-1/δ 1
Row 4 s 1
0.5 0
Row 5 s 0
1
There are two sign changesin column 1 and there are therefore 2 positive roots and the system is
unstable..
2)....If a all of the elements in a row is zero (two rows are proportional)
This indicates that the characteristic polynomial is divided exactly by the polynomial one row above the all
zero row (always even-ordered polynomial). Call this polynomial N(s).
This also indicates the presence of a divisor polynomial N(s) whose roots are all symmetrically arranged
about the origin i.e. they are of the form
s = α ..or s =  j ω ..or s = - α  j ω and s = + α  j ω
An all zero row will always be associated with and odd power of s

4. Ayatur rahman soomro control systems date:21-11-2014
In order to complete the array the previous row is differentiated with respect to s and the array is completed
in the normal way..
When assessing this modified array the number of sign changesin the first column (before the all zero row)
indicates the number of roots of the remainder polynomial with positive real parts..... From the all zero row
down, each change of sign in column 1 indicates the number of roots in the divisor polynomial with positive
real roots and as the roots are symmetrical this would indicate the number of roots in the right half s plane
and the number of roots in the left- hand s plane. Root not accounted for in this wayi.e no sign change,
must lie of the imaginary axis
Example .. Consider a closed loop control system with negative feedback which hasan open loop transfer
function.
KGH(s) = K / (s (s+1). ( s 2 + s + 1) ..
The closed loop characteristic equation =
F(s) = s 4 + 2s 3 + 2s 2 + s + K = 0
The Routh Array is constructed as follows=
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Col 1 Col 2 Col 3
Row 0 s 4
1 2 K
Row 1 s 3
2 1
Row 2 s 2
3/2 K
Row 3 s 1 (3/2 -2
K)/(3/2)
In this array row 3 becomes an all zero row if K = 3/ 4 and the divisor polynomial of row 2 = (3/ 2)s 2 + 3/ 4 =0
...= 2 s 2 + 1
By dividing F(s) by (2 s2 + 1) the equation is obtained....... F(s) = ( 1/ 2 s 2 + s + 3/ 4)
N(s)= (2 s 2 + 1)
The array is completed when K = 3/ 4 bydifferentiating N(s) with respect to s.
The coefficients of N'(s) are used to replace the zero coefficients in row 3
An all zero row
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Col 1 Col 2 Col 3
Row 0 s 4
1 2 K
Row 1 s 3 2 1
Row 2 s 2
3/2 3/4
Row 3 s 1
4
Row 4 s 0
3/4
There are no sign changesup to/ including row 2 indicating the roots of the remainder polynomial are in the
right hand s plane.
As there are no changes of sign from row 2 down the roots of the divisor polynomial must lie on the
imaginary axis.
To locate these roots set s in N(s) to jω
i.e 2 ( j ω ) 2 + 1 = 0 therefore .... ω = 1/ √2 rads/ unit time

5. Ayatur rahman soomro control systems date:21-11-2014
It is sometimes required to find a range of values of a parameter for which the system is stable. This can be
achieved by use of the Routh Criteria using the method illustrated by the following example...
The system characteristic equation =
F(s) = s 3 + 3 s 2 + 3s + K = 0
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Col 1 Col 2 Col 3
Row 0 s 3
1 3 0
Row 1 s 2
1 + K 0
Row 2 s 1
(8-K)/3 0
Row 3 s 0 1 +K
In order for the system to be stable there should be no sign change in column 1. To achieve this K must be
greater than -1 and K must be less than 8. Therefore for system stability .... -1 < K < 8..
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