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 ?

  • $\begingroup$ H(s) verses H(-s) $\endgroup$ – 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.


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.

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    $\begingroup$ Also see bounded-input bounded-output (BIBO) stability for a good definition of stability. $\endgroup$ – Olli Niemitalo Oct 26 '17 at 15:31
  • $\begingroup$ 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? $\endgroup$ – Tim Mottram Oct 26 '17 at 15:45

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