The Routh Hurwitz’s Criterion states that the system is stable if and only if all the elements in the first column have the same algebraic sign. If all elements are not of the same sign then the number of sign changes of elements in the first column equals the number of roots of the characteristic equation in the right half of S-plane.
Consider the following characteristic equation:
a0 Sn + a1 Sn-1 ………….an = 0
where a0,a1……… all have the same sign and are non-zero.
The coefficients in rows are arranged first
Find third row from above two rows
Continue the same procedure to find new rows.
Let’s now take an example to study this for the given polynomials below determine the stability of the system S4+2S3+3S2+4S+5=0.
1) Arranging Coefficient in Rows.
For row S2 first term
The row S2 Second term
For row S1:
The row S0
As there are two sign changes in the first column, So there are two roots or right half of the S-plane making the system unstable.
Taking another example inorder to study the special case of Routh Hurwitz,s Stability criterion.
Using Routh criterion determine the stability of the system with characteristic equation S4+8S3+18S2+16S+S = 0
Arrange in rows.
For row S2 first element
Second term =
As there is no sign change for the first column so all roots are left half of S-plane and hence the system is stable.
Special Cases of Routh Hurwitz Criterion
1. When the first element of any row is zero.
In this case the zero is replaced by a very small positive number E and the rest of the array is evaluated.
Eg.(1) Consider the following equation S3+S+2 = 0
Replacing 0 by E
Now when E→ 0, values in column 1 becomes
Two sign changes hence two roots on right side of S-planes
II) When any one row is having all its terms zero.
When array one row of the Routh Hurwitz table is zero, it is shown that the X is attests one pair of roots which lies radially opposite to each other in this case the array can be completed by auxiliary polynomial. It is the polynomial row first above row zero.
Consider following example
S3 + 5S2 + 6S + 30
For forming an auxiliary equation, selecting row first above row hang all terms zero.
dA(s)/ds = 5S2 e 30
= 10s e0.
Again forming Routh array
No sign change in column one the roots of Auxiliary equation A(s)=5s2+ 30-0
5s2+30 = 0
S2 α 6= 0
S = ± j√6
Both lie on an imaginary axis so the system is marginally stable.
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