# Why do poles in the left half of the S plane make a system stable?

A point on the S-plane (where $s=\sigma+j\omega$) represents a signal with a given frequency (given by the imaginary component) and which either decays, increases or stays stable (depending on the value on the real component).

Doing the maths, by converting $e^{-st}$ to a cosine and sine pair, I can see that points the left hand side of the plane describe signals that increase infinitely and points in the right hand side of the plane describe points that decrease to infinity.

Given this, why is it true that having poles of a transfer function (the frequency values which make the gain of the system infinite), lying in the left hand side of the S-plane (the side which makes signals increase infinity), makes a system stable ?

• H(s) verses H(-s)
– user28715
Oct 26 '17 at 15:29

An $n$th order linear system is asymptotically stable only if all of the components in the homogeneous response from a finite set of initial conditions decay to zero as time increases, or mathematically written: $\displaystyle{\lim_{t \to \infty}}\sum_{i=1}^{n}C_ie^{p_it}=0$ ($p_i$ are the system poles). As time increases (in a stable system) all components of the homogeneous response must decay to zero.If any pole has a positive real part there is a component in the output that increases without bound, causing the system to be unstable. So, in order for a linear system to be stable, all of its poles must have negative real parts (they must all lie within the left-half of the s-plane). An "unstable" pole, lying in the right half of the s-plane, generates a component in the system homogeneous response that increases without bound from any finite initial conditions.