Stability of a system from Polezero 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 splane while a pole is indicated by a X
The splane is shown in figure below
Case1
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 
Case2:
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 lefthand splane 
A system that has poles on the right hand of the splane 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 splane.
 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 nm>1, mean a pole at s= ∞ would impair the stability of the system.
Ex1: check for the stability using the polezero concept