We define the transfer function of a black box model as

\displaystyle H(s)=\frac{Y(s)}{X(s)},

where X(s) is the Laplace transform of the input and Y(s) is the Laplace transform of the output. This yields:

Y(s)=H(s)\cdot X(s).

H(s) is a rational function and therefore has zeroes and roots (removable poles and roots are removed) and therefore so does Y(s):

\displaystyle Y(s)=K\cdot \frac{\prod_{j=1}^m(s-z_j)}{\prod_{i=1}^n(s-p_i)}.

These \{z_j\} are the zeroes of Y(s) and the \{p_i\} are the poles of Y(s)This means that we have (assuming m<n) a partial fraction expansion:

\displaystyle Y(s)=\sum_{i=1}^n\frac{C_i}{s-p_i}.

Applying the Inverse Laplace transform we have an input:

y(t)=\sum_{i=1}^n C_ie^{p_it}.

These p_i are complex numbers and depending on their nature we get different behaviours.

A positive real pole is a pole p\in\mathbb{R} such that p>0. This corresponds to an output e^{pt} which tends to infinity as t\rightarrow \infty. This is divergent or unstable behaviour.

A negative real pole p=-a with a>0 corresponds to an output \displaystyle e^{pt}=e^{-at}=\frac{1}{e^{at}}\rightarrow 0 as t\rightarrow \infty. This is convergent or stable behaviour.

A zero pole p=0 yields an output e^{pt}=e^{0\cdot t}=e^0=1 which is a constant output (which is considered stable).

A purely imaginary pole p=k i is, via Euler Formula, corresponds to oscillatory behaviour:


A genuinely complex pole p=a+ik (a>0) with a positive real part (in the right-half plane) corresponds to the following output


While the e^{ikt} component is oscillations, the e^{at} goes to zero so we get behaviour that looks like:


Note that this behaviour is unstable.

When we have complex pole p=-a+ik (a>0) with a strictly negative real part (a pole in the left-half plane), then we have e^{pt}=e^{-at}(\cos(kt)+i\sin(kt)). Note that we have oscillations but modulated by a decreasing e^{-at}. This is the motion of an underdamped harmonic oscillator:


This is inherently stable behaviour.