Properties of the Routh Hurwitz Polynomial:
- A polynomial to be Hurwitz when-
- P(s) is real when s ir real
- The roots of P(s) have real parts which are zero or negative.
- Properties of Hurwitz polynomial P(s) = ansn + an-1sn-1 +…………….a1s + a0 are:
- Coefficient of s must be positive
- Both odd and even parts of Hurwitz polynomials should have roots on the imaginary axis only
- The continued fraction expansion of the ratio of even to odd (if even part is of higher degree) or odd to even(if the odd part is of the polynomial is of higher degree) of Hurwitz polynomial should be positive quotients
- If a polynomial is Hurwitz to even multiplicative factor W(s) i.e. P1(s) =W(s)P(s), if P(s) is Hurwitz then P1(s) must also be Hurwitz.
- If the ratio of a polynomial P(s) and its derivative P’(s) gives all positive quotients during partial fraction expansion, the polynomial is Hurwitz.
How to test whether the polynomial is Hurwitz or not
- By physically checking the terms of the polynomial P(s)
- All coefficients of the polynomial must be positive and real.
- There should not be any missing terms in powers of ‘s’ unless the polynomial is completely even or completely odd polynomial.
- Analytical Testing
- Find the continued fraction of the polynomial then
a) All quotients must be +ve
b) The process should not abruptly end because of some common factors. It does so, then do D(s) /D’(s) where D(s) is the common polynomial which was the last divisor that resulted in zero remainder
Routh-Hurwitz Criterion of stability of network function
- Given a network function in the form of N(s) = P(s) / Q(s)
- Separate the even and odd coefficient
- Write even coefficient in the first row
- Write the odd coefficients in the second row
- Complete the array for each power of s starting highest power in 1st row
Example Of Routh Hurwitz Polynomial
Q(s) = b0s5 + b1s4 + b2s3 + b3s2 + b4s1 + b5
Even coefficient terms b0,b2, b4s,
Odd coefficient terms b1,b3, b5,
Writing the coefficients in an array form for m=5, the array will contain m+1 =6 elements
|S5||b0 b2 b4|
|S4||b1 b3 b5|
Find other coefficients c1,c2,d1,d2,e1,f1 using
c1 = – (b0b3-b1b2)
c2 = – (b0b5 – b1b4)
d1 = – (b1c3 – c1b3)
d2 = – (b1*0 – c1b5)
e1 = -(c1d2 – d1c2)
f1 = – (d1*0 – e1d2)
Note the pattern carefully
According to the Routh Hurwitz stability criterion, the system is said to be stable, if and only if, there is no change in sign of the first column of the array.
This gives the roots with –ve real parts and hence gives the condition for stability.