What are the limitations for Routh test for stability (Beak Down of Routh test)?

Limitations:

1. The test is limited to systems that have polynomial characteristic equations. This means that it cannot be used to test the stability of a control system containing a transportation lag.
2.  It gives only information about absolute stability of the system. The degree of stability of a stable system cannot be obtained
3. The Criterion can be applied only if the characteristic equation has constant coefficients and cannot be applied if they are not real or contain exponential terms as in the case of systems with dead time.
   

SPECIAL CASES AND THE SOLUTION FOR THEM

There are two special cases in which the simple Routh’s test breaks down. These and the  remedies are given below.

Case 1.

The first element in a row becomes zero, whereas some of the other elements are nonzero.

Remedy

1. Replace the zero elements by a small positive number e and continue the tabulation. While determining the sign of the elements of the first column, find the sign as the value of e tends to zero.
2. Replace s in the given equation by s = (1/z) and do the tabulation in the equation in z , rather than in s. The information obtained will be the same as for the equation in s

 Example :
Consider the equation :

s5 + s 4 + 2s 3 + 2s 2 + 3s + 5 = 0

Both the necessary conditions are satisfied. Form Routh’s table

 

 The remedy is to replace the first element of the third row by e and continue the tabulation as follows

 

The sign of the elements of the first column, we see that e > 0 ( as per assumption)
(2e + 2)/e > 0 as in the limit (2e + 2) à 2 and 2 / e à a very large +ve quantity and 

{-2. (2e + 2)/e + 2e}/{(2e + 2)/e} tends to a negative quantity as it tends to -2 as e > 0

So the elements in the first column are +, +, +, +, - and + with the result that there are two
sign changes and hence two roots of the equation on the right half of the s-plane.
The second method also can be attempted. But sometimes, the second method will not
solve the difficulty .

Case 2

When all the elements in a particular row become zero before completing the tabulation This happens when there are pairs of roots which are equal in magnitude but opposite in sign like when there are two roots on the imaginary axis or when two pairs of roots are symmetrically placed around the origin in a quadruple.

Remedy
Write down the equation using the coefficients immediately previous to the all zero row. This equation is called auxiliary equation. The roots which are of equal in magnitude and opposite direction can be obtained by solving this equation. To complete the tabulation, differentiate the auxiliary equation with respect to s once and use the coefficients in place of the all zero row.

Example.

s5 + 2s 4 + 24s 3 + 48s 2 - 25s - 50 = 0

Rouths array

 

The auxiliary equation is : 2s 4 + 48s 2 - 50 = 0
Differentiate this with respect to s giving 8s3 + 96s
In the Routh’s table, the all zero row is replaced by these coefficients

 

There is one sign change in the elements in the first column and therefore there is one root on the right half of the s-plane.

Solving the aux equation A(s) = 2s 4 + 48s 2 - 50 = 0 , which can be factorized as:
 

(s 2 + 25)(s 2 -1) = 0 so that the roots are at s = +1, -1 and s = +5j and -5j
 

Obviously both pairs are of equal magnitude, but opposite sign. The roots on the right half plane is identified as +1 .