Zeros/poles at Laplace and at Fourier Transform

Question

I recently started "relearning" the Laplace transform, and I noticed something. It seems to me that the intuitive idea of poles and zeros is different between these two transforms!

For example, in case of Fourier transform, we can write something like this for the input signal of the system: let the input be a complex sinusoid $e^{j \omega_0 t}$, then $$y(t)=\int_{-\infty}^{+\infty}d\tau \, h(\tau)x(t-\tau)=e^{j \omega_0 t} \cdot \int_{-\infty}^{+\infty}d\tau \, h(\tau)e^{-j \omega_0 \tau}=e^{j \omega_0 t} \cdot H(\omega_0),$$ where $H(\omega_0)$ is the Fourier transform of the impulse response of the system $h(t)$ on a frequency $\omega_0$. The same thing can be seen in frequency domain, where Dirac distribution $2 \pi \delta(\omega-\omega_0)$ essentially "probes" the transfer function $H(\omega)$ at $\omega=\omega_0$. In other words, the transfer function is acting upon the complex sinusoid and changes it accordingly.

Zeros and poles in this case come naturally as those values of frequency where transfer function zeroes out or explodes. Hence, when we multiply $H(\omega)$ with a complex sinusoid at one of these frequencies, the result is clearly visible; the sinusoid either disappears or becomes really big (infinite).

Now, let's focus on the Laplace transform; I am not managing to gain the same insight here. As far as the transfer function $H(s)$ is concerned, zeros and poles are still the same thing: characteristic values of $s$ where the function zeroes out or explodes, but similarity ends here; if I choose a causal complex sinusoid as an input of a causal system, i.e. $e^{s_0t}u(t)$, I can write

$$y(t)=\int_{-\infty}^{+\infty}d\tau \, h(\tau)x(t-\tau)=e^{s_0t} \cdot \int_{0}^{t}d\tau \, h(\tau)e^{-s_0\tau} \neq e^{s_0t} \cdot H(s_0),$$ so, in short, I can't apply the same logic with zeros/poles as I did with Fourier transform. It's similar if I look at the situation in the $s$-domain $$Y(s)=H(s) \cdot \frac{1}{s+s_0,}$$ and it's obvious that it's not like $H(s)$ "probes" the complex sinusoid $e^{s_0t}u(t)$ and changes only its amplitude/phase, as was the case with Fourier transform.

So, my conclusion is that, if I feed the system with signal $e^{s_zt}$, where $s_z$-value is the value where, for example, $H(s)=0$, I can't just say "$H(s)$ is going to act upon it and change it accordingly - in this case, nullify it". The same conclusion goes for poles.

My question is, how should I interpret the poles and zeros of the Laplace transform?

Thanks!

Answer 1

You can think of it as though the Fourier transform is giving you the "steady state" behavior after the system has been exposed to that sinusoid for an infinite amount of time. Hence there isn't much to say except the magnitude and phase because everything other than the input frequency will be gone by then.

On the other hand, the Laplace transform expresses the causal response in detail, including all the details of the signal dying off or diverging. Those details are why you get all the extra "junk" surrounding the singularity of $1/(s-s_0)$ instead of having just a clean $\delta(s-s_0)$. Note that the difference between a $\delta$ function and a $1/s$ function is exactly the difference between the idiomatic "constant"/"DC" function for each of the transforms: $f(x) = 1$ vs. $f(x) = u(x)$.

So the answer to your question is that you can interpret the poles and zeros in basically the same way. Think of the Fourier transform as shifting the time origin of the Laplace transform infinitely far to the left.