Positive Real Function: A function is said to be positive Real if it has all the poles and zeros on the left half of the S-plane and if the function has poles on the imaginary axis then it should be simple.
Real Part of Function :
Re[ H(jw) ] >=0 for all ‘w’
If a function F(S) is positive real then 1 / F(S) is also positive real. The highest degree and lowest degree of numerator and denominator of function may differ at the most by one. Sum of two positive real functions is also positive real.
Properties of the Reactance Function (LC) :
(i) All poles and zeros are simple and lies on imaginary axis only.
(ii) It has a pole or zero at infinity and origin.
(iii) It is the quotient of even to odd or odd to even polynomials.
(iv) The residues at all poles are real and positive.
(v) The slope dF(s) / dw is always positive.
Properties of RC Impedance [ RL Admittance ] :
(i) The poles and zeros are non-respective and lines on negative real axis.
(ii) The poles and zeros are alternate on real axis.
(iii) The lowest critical frequency is a pole.
(iv) The highest critical frequency is a zero.
(v) The residue at all its poles are real and positive
(vi) Z(0)>Z(∞)
w =0 w=∞
Properties of RL Impedance ( RC Admittance )
(i) The poles and zeros are simple lies on negative real axis.
(ii) The poles and zeros are alternate on real axis.
(iii) Lowest critical frequency is zero.
(iv) Highest critical frequency is pole.
(v) Z(∞) >Z(0)
Let us understand the above concept with the help of few examples.
Q – For the given polynomials find whether they are Hurwitz polynomials or not.
a). F1(s) = s+ 3
b). F2(s) = s2 + 5
c). F3(s) = s2 + 5s + 6
d). F4(s) = s3 + 4s2 + 8s + 4
Sol :
a). F1(s) = s + 3
s = -3
Its Hurwitz polynomial
b). F2(s) = s2 + 5
s = Ij√5
Its Hurwitz polynomial
c). F3(s) = s2 + 5s + 6
Given
s2 + 5s + 6
We can also write
s2 + 2s + 3s + 2 x 3
(s + 3)(s + 2)
F3(s) =>s = -3, -2
It is Hurwitz polynomial.
d). F4(s) = s3 + 4s2 + 8s + 4
E(s) = 4s2 + 4 O(s) = s3 + 8s
4(s2 + 1 ) s(s2 + 8)
s = 1j s = 1j2√2
Its Hurwitz polynomial
Exception Case
Q .8s6 + 4s5 + 34s4 + 16s3 + 32s2 + 12s + 6. find whether this is Hurwitz or not?
This is the case of premature termination.
F(s) = F1(s).F2(s)
From above F1(s) = 2s4 + 8s2+ 6
F2(s) = 8s6 + 4s5 + 34s4 + 16s3 + 32s2 + 12s + 6
——————————————————–
2s4 + 8s2 + 6
2s4 + 8s2 + 6 ) 8s6 + 4s5 + 34s4 + 16s3 + 32s2 + 12s + 6 ( 4s2 + 2s + 1
8s6 + 32s4 + 24s2
———————————————————
4s5 + 16s3 + 8s2 + 12s + 6 + 2s4
4s5 + 16s3 + 12s + 2s4
————————————————-
2s4 + 8s2 + 6
2s4 + 8s2 + 6
———————-
* *
F1(s) = 2s4 + 8s2+ 6 – It is Hurwitz
F2(s) = 4s2 + 2s+ 1 – Has one root or RHS
F(s) is not Hurwitz as F2(s) is non Hurwitz.
Q .Find whether the given function F(s) = s + 4 / s2 + 2s + sis positive real or not?
Sol: Now by inspection check the function
(ii) Numerator and Denominator should both be Hurwitz
(iii) Real part of function must be positive for all values of W.
N(s) = s+ 4 D(s) = s2 + 2s + s
It is Hurwitz It is Hurwitz
Finding real part of function, s= jw.
Real part
For, w1 = 1, 2, …
Then for w > 3 function becomes negative. So not positive real.