Lets say I have a transfer function $H(s)$ of a system defined in $s$-domain as: $$H(s) = \frac{1}{s - (-1-j)}$$

So I conclude that the pole on the $s$-plane is where $s = 1+j$. So far so good.

  • Now does that mean if the Laplace transform of the input to the system is $s = 1+j$ the system goes crazy/infinity/oscillate?

  • Does that mean if I take the inverse Laplace transform of $1+j$ I can find what input in time domain makes the system unstable?


2 Answers 2


Let $H(s)$ be a transfer function of the form $$H(s) = \frac{1}{s-p}$$ where $p$, which is a pole of $H(s)$, can be written as a complex number $a+jb$. Taking the inverse Laplace transform of $H(s)$ gives the corresponding impulse response $h(t)$ (that is, the output of your system when given $\delta(t)$ as input). Noting $\mathcal{L}^{-1}$ the inverse Laplace transform, we have $$h(t) = \mathcal{L}^{-1}\{H(s)\} = e^{pt} = e^{at}e^{jbt}.$$ Now let's look at what this impulse response looks like. The term $e^{at}$ is a simple exponential which will be either decaying (if $a < 0$) or growing (if $a$ > 0) with time. The term $e^{jbt}$ will be responsible for oscillations in the output of your system (remember that $e^{jbt} = \cos(bt) + j\sin(bt)$). From this, you can infer the stability of your system and understand why we need poles in the left-hand side of the $s$-plane (i.e. we need $a < 0$) for the system to be stable.

Often, the numerator and the denominator of your transfer function have real coefficients, and in this case poles appear in complex conjugate pairs. You could for example have $$h(t) = e^{at}(e^{jbt} + e^{-jbt}) = 2e^{at}\cos{bt}.$$

I like to keep this picture in mind (taken from here) which greatly summarizes this. Pole-zero plot and link with the impulse response

For more complex transfer function, partial fraction decomposition can be used to go back to simple cases as presented here.



$H(s)= \dfrac{1}{s - (-1-j)}$

then your pole is at $p_1 = -1-j$, which is in the left-hand side of the s plane, so the system is stable.

  • $\begingroup$ I have trouble with meaning of p1=−1−j here. Does that mean p1 is an input which makes the output maximum here? And if p1 is an input what is p1 in time domain? $\endgroup$
    – user16307
    Apr 29, 2017 at 18:14
  • $\begingroup$ $p_1$ is simply a designation I gave to the location of the pole in the s-plane. It is not an input to the system. What it sounds like you're looking for is the time domain impulse response of the system with the above transfer function. If you put an infinitely strong, infinitely narrow input into the system at time $t=0$, normally designated by the input function $\delta(t)$, the response will either die out or decay or both (stable system), or oscillate forever and/or grow without bound (unstable system). $\endgroup$
    – Andy Walls
    Apr 29, 2017 at 19:18
  • 1
    $\begingroup$ I forgot to mention, the impulse response of $H(s)$ is just its inverse Laplace transform, usually designated $h(t)$. Your sample $H(s)$ here has a complex pole without its complex conjugate as a pole as well, so I think $h(t)$ will end up being a complex function of $t$ in this contrived example. $\endgroup$
    – Andy Walls
    Apr 29, 2017 at 19:32

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