Let's say I have a transfer function which is a unit step function.

$H(s) = \int_{0}^{∞}e^{-st}dt$

But when we write,

$H(s) = 1/s$ it is only true when $Re(s) > 0$

So after we derive the Laplace Transform for the unit step, we forget all about how we derived it and how $H(s) = \infty$ for all $Re(s) \leq 0$

What makes the pole $s=0$ so special. It does not lie in the ROC like every other point in the left half plane, so why is it the only pole for our system?

Shouldn't the system have infinite poles then on the left half of the s-plane? I guess the more generalized question would be isn't every point outside the ROC of a transfer function its pole?


If you can accept that the above proposition is TRUE then consider the following contradiction.

Consider a simple RC filter whose transfer function is

$H(s) = \frac{1}{s+1}$

This system is said to have 1 pole at $s=-1$

Applying the inverse Laplace transform, we now the corresponding time domain function

$h(t) = \exp(-t)$

But we know that Laplace Transforms for exponential such as $\exp(at)$ is only defined when $s>a$

In our case the Laplace transform could only be defined for $s>-1$. Therefore, to say that there is only one pole at $S=-1$ would be incorrect as everything to the left of $s=-1$, including the line at $Re(s)=1$ would also be undefined and out of the ROC.

So the question again becomes why do we, when solving such systems only consider them to have ONLY one pole at $s=-a$?

Is that not an incomplete answer?


1 Answer 1


No. Every point that's not in ROC doesn't have to be a pole. ROC is a region where the transform attains a finite value, but this ROC also has to "produce back" the time domain signal via the inverse transform, or else it does not converge to that sequence in time domain. So you have to take both the finiteness of the transform as well as the reproducibility of the time domain signal via the inverse transform into account to define a "region that converges and is consistence for both aspects" while defining an ROC for a signal. I am sure you know that the same transfer function with different ROCs produces different time domain signals. So specifying a transfer function does not uniquely define a time domain signal. Only when an ROC is defined along with the transfer function can we associate to a unique time domain sequence. So the left over region that's not part of the ROC doesn't have a pole, since clearly if we shift the ROC to be that side now, we can define a different sequence.

  • 1
    $\begingroup$ That clears things up. Actually, I looked up the proper definition of the Inverse Laplace Transform and as you said, it does specify the ROC by setting Re(s) > Re(right most pole) for the limits of the integration. And that should neatly reverse the Laplace transform without confusion about unspecified regions. $\endgroup$
    – underdog
    Apr 17, 2020 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.