# 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
Commented Oct 26, 2017 at 15:29
• It isn't the presence of poles on the left half-plane that makes a system stable but rather the absence of any poles in the right half-plane. The presence of just one pole in the right-half plane cannot be overcome by a gazillion poles in the left half-plane all doing their best to make the system stable. Commented Oct 9, 2022 at 2:36

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.

Let us consider a pole at left half of $$s$$-plane $$\frac{1}{s + a}$$. In time domain it is $$e^{-a t}$$ that is exponential decaying means it has decaying such that meets at a point on time axis as well as output is finite.

But for pole on right side $$\frac{1}{s - a}$$ its Laplace is $$e^{at}$$ simply like linear there is no definite input and output.

So for a stable system it has BIBO System so pole should be on left side of $$s$$-plane.

I think I have answered this: If the points on the S plane that make the system respond infinitely lie in the left hand side of the S plane then, because points in the left hand side of the plane increase in amplitude as time progresses, as time progresses for the system a larger and larger signal is required to make the system respond infinitely. And, conversely, if a pole exists in the right hand side of the plane, as time increases the system will respond infinitely to smaller and smaller amplitudes of this frequency meaning that as time proceeds and infinitely small amount of this frequency will cause the system to saturate.

• Also see bounded-input bounded-output (BIBO) stability for a good definition of stability. Commented Oct 26, 2017 at 15:31
• Ok, thanks Olli, I had had a look but there are a lot of terms in there that I don't understand. Was my answer essentially true? Commented Oct 26, 2017 at 15:45

On left hand side as t approached infinity output approaches infinity(BIBO). But on Right hand side of the s-plane apparently as t approaches 0 output approaches infinty which is not an ideal case in real life signals ultimately make the control systems unstable.