Stability of a system from Pole-zero concept

Given a polynomial N(s) = P(s) / Q(s)

By factorizing the numerator and denominator polynomials, we can easily show that the polynomial becomes zero when the ‘s’ terms in the numerator polynomial have the values s=0, -z1,-z2…….-zn, thus the roots of numerator define the zeros. Also, the roots of the denominator polynomial define the poles of the function. Zeros are marked by ‘O’ on the s-plane while a pole is indicated by a X

The s-plane is shown in figure below

Case-1

Unit step function F(s) =1/s	F(cos(w)) = s/(s2+ω2)	F(sin(w)) = w/(s2+ω2)
The function has a only one pole at s=0	The function has a zero at s=0 and two poles at +/- jω	There are two poles at    +/-jω
In all the three functions we see that the poles and zeros are on the imaginary axis only

Case-2:

Damped cosine function F(s) = s+σ0/{(s+ σ0)2 +ω02}	Exponentials f1(t) =Me-σ1t and f2(t) = Me-σ1t F1(s) =1/(s+ σ1) and F2(s) = 1/(s+ σ2)	Sinusoidal F1(s) = ω1/(s2+ω12) F1(s) = ω2/(s2+ω22)
Again we observe that the poles and zero in all these three cases are either on an imaginary axis or on the left-hand s-plane

A system that has poles on the right hand of the s-plane is basically unstable because such a system gives rise to exponentially increasing transient response.

The necessary condition for the stability of the network function therefore are:

1. Any function F(s) cannot have poles on the right hand of the s-plane.
2. There should not be multiple poles on the jω axis
3. The degree of the numerator polynomial cannot exceed the denominator polynomial by more than one. As, if n-m>1, mean a pole at s= ∞ would impair the stability of the system.

Ex-1: check for the stability using the pole-zero concept