Control Engineering
Lecture# 10 & 11
30th April’2008
Stability Analysis
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BIBO: Bounded Input Bounded Output systems.
For LTI systems this requires that all poles of the closed-loop
transfer function lie in the left half of the complex plane.
Determine if the transfer function has any poles either on the
imaginary axis or in the right half of the s-plane.
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Routh-Hurwitz criterion: determine if any roots of a
polynomial lie outside the lelf half of the complex plane. It
does not find the exact locations of the roots.
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Other methods find the exact locations of the roots. For first
and second order systems, analytical method can be used.
For higher order systems, computer programs or simulation
are required.
Some basic results:
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Second order system:
P2 ( s )  s 2  a1s  a0  ( s  p1 )( s  p2 )
 s 2  ( p1  p2 ) s  p1 p2
For third order system :
P2 ( s )  s 3  a2 s 2  a1s  a0  ( s  p1 )( s  p2 )( s  p3 )
 s 3  ( p1  p2  p) s 2  ( p1 p2  p1 p3  p2 p3 ) s  p1 p2 p3
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We see that the coefficients of the polynomial are given by:
an 1  negative of the sum of all roots.
an  2  sum of the products of all possible combinatio ns
of roots taken 2 at a time.
an 3  negative of the sum of the products of all
possible combinatio ns of roots taken 3 at a time.
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Suppose that all the roots are real and on the left half plane,
then all coefficients of the polynomial are positive.
If all the roots are real and in the left half plane then no
coefficient can be zero.
The only case for which a coefficient can be negative is
when there is at least one root in the right half plane.
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The above is also true for complex roots.
1) If any coefficient is equal to zero, then not all roots are in
the left half plane.
2) If any coefficient is negative, then at least one root is in
the right half plane.
3) The converse of rule 2) is not always true.
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Example:
P( s)  s 3  s 2  2s  8  ( s  2)( s 2  s  4)
all coefficien ts are positive. But two roots
(complex) are in the right half plane.
Routh-Hurwitz Stability Criterion
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All the coefficients must be positive if all the roots are in the
left half plane. Also it is necessary that all the coefficients for
a stable system be nonzero.
These requirements are necessary but not sufficient. That is
we know the system is unstable if they are not satisfied; yet if
they are satisfied, we must proceed further to ascertain the
stability of the system.
For example,
q( s)  s  s  2s  8  ( s  2)( s  s  4)
the system is unstable yet all coefficien ts are positive.
3
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2
2
The Routh-Hurwitz is a necessary and sufficient criterion for
the stability of linear systems.
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The Routh-Hurwitz criterion applies to a polynomial of the
form:
P( s)  an s n  an 1s n 1  .......  a1s  a0
assume a0  0
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The Routh-Hurwitz array:
sn
an
an  2
an  4
an 6 ....
s n 1
an 1
an  3
an  5
an 7 ....
s n2
b1
b2
b3
b4
....
s n 3
c1
c2
c3
c4
....
.
.
.
.
.
.
s2
k1
k2
s1
l1
s0
m1
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Columns of s are only for accounting.
The b row is calculated from the two rows above it.
The c row is calculated from the two rows directly above it.
Etc…
The equations for the coefficients of the array are:
1 an an  2
b1  
an 1 an 1 an 3
1 an 1 an 3
c1  
b2
b1 b1
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1 an an  4
b2  
, .......
an 1 an 1 an 5
1 an 1 an 5
c2  
, ......
b3
b1 b1
Note: the determinant in the expression for the ith coefficient
in a row is formed from the first column and the (i+1)th
column of the two preceding rows.
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The number of polynomial roots in the right half plane is equal to the
number of sign changes in the first column of the array.
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Example:
P( s )  s 3  s 2  2 s  8  ( s  2)( s 2  s  4)
The Routh array is :
s3
1
2
s2
1
8
s1
-6
s0
8
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Since there are two sign changes on the first column, there are two roots
of the polynomial in the right half plane: system is unstable.
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Note: The Routh-Hurwitz criterion shows only the stability of the
system, it does not give the locations of the roots, therefore no
information about the transient response of a stable system is derived
from the R-H criterion. Also it gives no information about the steady
state response. Obviously other analysis techniques in addition to the RH criterion are needed.
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From the equations, the array cannot be completed if the first
element in a row is zero. Because the calculations require
divisions by zero. We have 3 cases:
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Case 1: none of the elements in the first column of the array
is zero. This is the simplest case. Follow the algorithm as
shown in the previous slides.
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Case 2: The first element in a row is zero, with at least one
nonzero element in the same row. In this case, replace the
first element which is zero by a small number  . All the
elements that follow will be functions of  . After all the
elements are calculated, the signs of the elements in the first
column are determined by letting  approach zero.
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Example:
P(s)  s 5  2s 4  2s 3  4s 2  11s  10
s5
1
2
11
s4
2
4
10
s3

6
2
s s1
12

10
6
s 0 10
When we calculate the elements :
b1  0, b 2  6, therefor e we put b1  
and calculate the other coefficien ts. You should
verify the results.
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There are 2 sign changes regardless of  is positive or
negative. Therefore the system is unstable.
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Case 3: All elements in a row are zero.
Example:
P( s)  s 2  1
s2 1 1
s1 0
s0
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Here the array cannot be completed because of the zero
element in the first column.
Another example: P ( s )  s 3  s 2  2 s  2
The array is :
s3
1 2
s2
1 2
s1
0
s0
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Case 3 polynomial contains an even polynomial as a factor.
It is called the auxiliary polynomial. In the first example,
the auxiliary polynomial is
s2 1
and in the second example, auxiliary polynomial is s 2  2
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Case 3 polynomial may be analyzed as follows:
Suppose that the row of zeros is the s i row, then the
auxiliary polynomial is differentiated with respect to s,
and the coefficients of the resulting polynomial used to
i
replace the zeros in the s row. The calculation of the
array then continues as in the case 1.
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Example:
P( s )  s 4  s 3  3s 2  2 s  2
The Routh array is :
s4
1 3 2
s3
1 2
s2
1 2
s1
0
s0
Since the s1 row contains zeros, the auxiliary
polynomial is obtained from the s 2 row :
Paux ( s )  s 2  2
The derivative is 2 s, therefore 2 replaces 0 in the
s1 row, and the Routh array is then completed.
P( s )  s 4  s 3  3s 2  2 s  2
The Routh array now becomes :
s4
1 3 2
s3
1 2
s2
1 2
s1
2
s0 2
Hence there are no roots in the right half plane. However, see
Note below.
Note : When the re is a row of zeros in the Routh array, the
system is nonstable. That is it will have roots either on the
imaginary axis (as in this example), or it has roots on the
right half plane.