ROUTH STABILITY CRITERION

The routh stability criterion provides a convenient method of determining the stability of
a linear time invariant control system. This method has the following characteristics:
(i) With this method absolute stability of the system can be determined directly.
(ii) This method determines the number of characteristic roots within the stable
left half, on the imaginary axis and in the unstable right half of the s plane.
(iii) This method cannot be used for the systems containing dead time elements.
(iv) This method does not locate the roots but only tells about which part of the s
plane they lie.
(v) This stability criterion is applicable to the polynomials with only a finite
number of terms.
(vi) This method is particularly useful for higher-order systems because it does not
require the polynomial expression in the transfer function to be factored.

PROCEDURE TO CONSTRUCT ROUTH ARRAY

(i) Write down the characteristic equation of the system in s domain.
(ii) Arrange the characteristic equation in the decreasing value of power of s as
given below:
(s ) = a0 sn + a1 sn-1+ a2 sn-2 +…………………….+ an s0 = 0
(iii) Check if a0 > 0 ? If No, then multiply the characteristic equation with –1.
(iv) If all the coefficients of s in the characteristic equation are positive then
arrange them in two rows one starting with coefficients of sn while other with
coefficients of sn-1 , followed by even and odd numbered coefficients as
follows:
sn a0 a2 a4 a6 …………..
sn-1 a1 a3 a5 a7 …………..
sn-2 b1 b2 b3 b4 …………..
sn-3 c1 c2 c3 c4 …………..
. …………..
.
. …………..

1
 . …………..
s0 h1

The values of bi of Sn-2 row are calculated from the previous two rows i.e. S n and
Sn-1 . So, the values of b1, b2, b3 and so on are calculated as
a1a2  a0 a3
b1 =
a1

a1a4  a0 a5
b2 =
a1
a1a6  a0 a7
b3 =
a1
This pattern will continue until the rest of bi’s are all equal to zero. The values of ci of
Sn-2 row are calculated from the previous two rows i.e. S n-1 and Sn-2 .The values of c1,
c2, and so on are calculated as
b1a3  a1b2
c1 =
b1
b1a5  a1b3
c2 =
b1
This process continues until no more ci elements are present. Rest all rows are formed
down to s0 row.

IMPORTANT:
a) If any of the coefficient of s in the characteristic equation is negative or zero then
the system is unstable. The necessary condition for the system to be
stable is that all the coefficient of s must be present and must be
positive.
b) All the element of any row can be multiplied or divided by any number to simplify
the computational work

STABILITY ANALYSSIS:
Three cases are formed in routh array , these are:

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Case 1: When No zero in the first column
If there is No zero in the Ist column then further there are two cases
a) If all the elements of the first column are positive then all the roots of the
characteristic equation are in the left half of the s plane and the system is stable.
EXAMPLE 1
Given a system with characteristic equation
(s ) = 3 s4 + 3 s3 + 7 s2 +2 s +2 = 0
Determine the system stability using routh criterion.
Since all the coefficient of s are present and are positive so the necessary
condition for the system to be stable is fulfilled. Now construct Routh
array

s4 3 7 2
s3 3 2 0 (for missing term)

s2 (3  7)  (3  2) (3  2)  (3  0) 0
=5 =2
3 3
s1 (5  2)  (3  2) 4 (5  0)  (3  0)
= =0
5 5 5
s0 2

Since there is No sign change in the first column so the system is stable.
b) If all the elements of the first column are not positive then number of roots of
the characteristic equation in the right half of the s plane are equal to the number
of sign changes in first column of the routh array and the system is unstable.

EXAMPLE 2

Given a system with characteristic equation

(s ) = s3 + s2 +2 s +24 = 0
Determine the system stability using routh criterion.
Since all the coefficient of s are present and are positive so the necessary
condition for the system to be stable is fulfilled. Now construct Routh
array

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 s3 1 2
s2 1 24
s1 -22 0
s0 24

Since there is two sign changes from 1 to –22 and then from –22 to 24 in the first
column so the system is unstable having two roots of the characteristic equation in the
right half of the s plane.

Case 2: If there is a zero in the first column

If there is a zero in the first column then the terms in the next row all becomes infinite
and the array formation cannot continue. There are two cases under such conditions
a) There is a non zero element in the row containing zero in the first column
b) All the elements of the row containing zero in the first column are zero

a) There is a non zero element in the row containing zero in the first column
Under such circumstances follow the procedure given below:
(i) Replace zero with an arbitrary small number  and complete the routh
array.
(ii) Take the limit   0
(iii) Determine the number of sign changes in the first column of the array

Numbers of sign changes are equal to the number of roots in the right half of the s
plane.

IMPORTANT:
In the above case if, there is No sign change then a pair of roots lie on the
imaginary axis and the system is marginally stable.

EXAMPLE 3
Given a system with characteristic equation
(s ) = s5 + 3 s4 + 2 s3 + 6 s2 + 6 s + 9 = 0
Determine the system stability using routh criterion.
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Since all the coefficient of s are present and are positive so the necessary
condition for the system to be stable is fulfilled. Now construct Routh
array

s5 1 2 6
s4 3 6 9  Divide by 3
1 2 3
s3 0   3  0 is replaced by 

s2 2  3 0

3 2
1
s
3-
2  3
s0 3

2  3
When limit   0 is applied then the sign of term in the row of s2 becomes

3 2
negative and the term ( 3 - ) in the row of s1 becomes positive. So, there are
2  3
total two sign changes and two roots are on the right half of the s plane. Hence, the
system is unstable.

b) All the elements of the row containing zero in the first column are zero
If all the elements of a row are zero follow the procedure given below:
(i) Find the auxiliary polynomial [AP(s)]. The coefficients of the auxiliary
polynomial are the elements in the row directly above the zero row.
(ii) Divide the characteristic equation with the auxiliary polynomial.
(iii) Apply Routh criteria to the resultant equation obtained after the step
number (ii).
(iv) The numbers of roots in the right half of the s plane are equal to the sum
of the right half roots of the auxiliary polynomial and the roots obtained
after the Routh criteria applied to the resultant equation.

Note: Roots of Auxiliary Polynomial

For 2m degree auxiliary polynomial there are m pair of roots lying radially
opposite in s plane (i.e. real roots of equal magnitude and opposite sign and/or
complex conjugate pair. )
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EXAMPLE 4

Given a system with characteristic equation

(s ) = s5 + 2 s4 + 8 s3 + 11 s2 + 16 s + 12 = 0
Determine the system stability using routh criterion.
Since all the coefficient of s are present and are positive so the necessary
condition for the system to be stable is fulfilled. Now construct Routh
array

s5 1 8 16
s4 2 11 12
s3 5 10
2
s2 3 12
s1 0 0
s0

Since all the elements of row s1 are zero so find the auxiliary polynomial [AP(s)]. The
coefficients of the auxiliary polynomial are the elements in the row directly above the
zero row.
Auxiliary polynomial [AP(s)] = 3s2 + 12
Divide it by 3 The auxiliary polynomial = s2 + 4
Now divide the characteristic equation with the auxiliary polynomial to get the resultant
equation
(s) s 5  2 s 4  8 s 3  11 s 2  16 s  12
Resultant equation = = = s3 + 2 s2 + 4 s + 3
AP( s ) s 4
2

Apply routh criterion to the resultant equation

s3 1 4
s2 2 3
s1 5 0
2
s0 3

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Since there is no sign change in the first column of routh array, so the resultant
polynomial does not have any root in the right half of the s plane.
Solve the Auxiliary polynomial [AP(s)] = 3s2 + 12 = 0
s =  j2
Roots of (s ) = Roots of resultant polynomial + Roots of auxiliary polynomial
Now the resultant polynomial does not have any root in the right half of the s plane so
does the auxiliary polynomial so the characteristics equation (s ) does not have any
root in the right half of the s plane but the auxiliary polynomial have the roots on
the imaginary plane hence the system is marginally stable.

EXAMPLE 5
Given a system with characteristic equation
(s ) = s3 + 11 s2 + 20 s + K = 0
Determine the range of K for which the system is stable using routh criterion.
Routh array

s3 1 20
s2 11 K
s1 220  K 0
11
s0 K

220  K
From the routh array the sign in the first column must be positive so >0
11
and K > 0 . On solving, the system will be stable if and only if
220 >K > 0