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/(s^{2}+ω^{2})

F(sin(w)) = w/(s^{2}+ω^{2})

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+σ_{0}/{(s+ σ_{0})2 +ω_{0}^{2}}

Exponentials f1(t) =Me^{-σ1t} and f2(t) = Me^{-σ1t}

F1(s) =1/(s+ σ_{1}) and F2(s) = 1/(s+ σ_{2})

Sinusoidal

F1(s) = ω_{1}/(s^{2}+ω_{1}^{2})

F1(s) = ω_{2}/(s^{2}+ω_{2}^{2})

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:

Any function F(s) cannot have poles on the right hand of the s-plane.

There should not be multiple poles on the jω axis

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