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) 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:

$e^{pt}=e^{ikt}=\cos(kt)+i\sin(kt)$.

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

$e^{pt}=e^{(a+ik)t}=e^{at+ikt}=e^{at}e^{ikt}=e^{at}(\cos(kt)+i\sin(kt))$.

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.