Procedure for Routh test for stability and limitations(Beak Down of Routh test):( Necessary Conditions,Theorems of the Routh Test)

Procedure for Routh test for stability and limitations:( Theorems of the 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 .

Necessary Conditions

1.  All coefficients 1 2 1 0 a , a , a ........., a , a n n - n - should be present , ie none of them should be zero
2.  All the coefficients 1 2 1 0 a , a , a ........., a , a n n- n - should be of the same sign. If these necessary conditions (not sufficient) are satisfied, the sufficiency test can be conducted based on a Routh’s table constructed as follows.

 

 

ROUTH TEST:

The Routh test is a purely algebraic method for determining how many roots of the characteristic equation have positive real parts; from this it can also be determined whether the system is stable, for if there are no roots with positive real parts, the system is stable.

The procedure for examining the roots is to write the characteristic equation in the form

a_o Sⁿ + a₁ S^n-1 + a₂ S^n-2 + a₃ S^n-3 +………….. a_n-1 S +a_n = 0………………………(1)

Required observations from the polynomial for Routh test:

Where a₀ is positive. (If a₀ is originally negative, both sides are multiplied by -1.)

In this form, it is necessary that all the coefficients a_o,a₁,a_2,a₃,…………..,a_n-1,a_n be positive if all the roots are to lie in the left half plane.

If any coefficient is negative, the system is definitely unstable, and the Routh test is not needed to answer the question of stability. (However, in this case, the Routh test will tell us the number of roots in the right half plane.)

 If all the coefficients are positive, the system may be stable or unstable. It is then necessary to apply the following procedure to determine stability.

Procedure to determine stability:

Routh Array

Arrange the coefficients of Eq. into the first two rows of the Routh array, as follows:

The array has been filled in for n = 7 in order to simplify the discussion. For any other value of n, the array is prepared in the same manner. In general, there are (n + 1) rows. For n even, the first row has one more element than the second row.

The elements in the remaining rows are found from the formulas

The elements for the other rows are found from formulas that correspond to those just given. The elements in any row are always derived from the elements of the two preceding rows. During the computation of the Routh array, any row can be divided by a positive constant without changing the results of the test. (The application of this rule often simplifies the arithmetic.) Having obtained the Routh array, the following theorems are applied to determine stability.

Theorems of the Routh Test

The necessary and sufficient condition for all the roots of the characteristic equation to have negative real parts (stable system) is that all elements of the first column of the Routh array (a₀, a₁, b₁, c₁, etc.) be positive and non zero

If some of the elements in the first column are negative, the number of roots with a positive real part (in the right half plane) is equal to the number of sign changes in the first column.

If one pair of roots is on the imaginary axis, equidistant from the origin and all other roots are in the left half plane, all the elements of the nth row will vanish and none of the elements of the preceding row will vanish. The location of the pair of imaginary roots can be found by solving the equation

  

where the coefficients C and D are the elements of the array in the (n - 1)^th row as read from left to right, respectively. We shall find this last rule to be of value in the root-locus method presented in the next chapter.

The algebraic method for determining stability is limited in its usefulness in that all we can learn from it is whether a system is stable. It does not give us any idea of the degree of stability or the roots of the characteristic equation.