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If we have an LTI system, with an input signal $x(t)$, impulse response $h(t)$ and output $y(t)$, I was under the assumption that if the input and impulse response were continuous in time, then you would use the FT on $x(t)$ and the Laplace transform on $h(t)$, and multiply them together to find the output $Y(f)$ in the frequency domain.

Likewise, if $x[n]$ was a discrete input signal, $h[n]$ was a discrete impulse response, and $y[n]$ a discrete output, then I though we would use the DFT on $x[n]$, z transform on $h[n]$, then multiply them together to get $Y(k)$ which is a discrete output in the frequency domain.

However I've seen people saying we would use the Fourier transform on the impulse response to get the transfer functions here so it's kinda thrown me off a little bit. Could anyone clarify this?

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    $\begingroup$ Hi Andrew. Can you support your assumptions with a reference of some kind? Most of them make no sense to me. $\endgroup$ – GKH Mar 11 at 22:33
  • $\begingroup$ @GKH Yeah, I've been self-learning some signal processing so what I've tried to learn online here about the LT. From the first 3 minutes, I assumed you would use the LT for impulse responses and because I knew FT puts x(t) into freq. dom. I thought they would multiply to give the output. And because DFT is discrete FT and ZT is discrete LT, it would be the same for discrete signals, if that makes sense? $\endgroup$ – Andrew Mar 11 at 22:44
  • $\begingroup$ That's exactly what did not make sense to me. Fat32 in his reply has put things into the right perspective. Fourier transforms work together with Fourier transforms, Laplace ones with Laplace others, Z Transforms with Z Transforms, etc. You can't mix FTs with LTs when trying to find the output of an LTI system. That was my objection to your question. $\endgroup$ – GKH Mar 12 at 0:41
  • $\begingroup$ Ah I understand, thanks for the clearing it up $\endgroup$ – Andrew Mar 12 at 1:33
  • $\begingroup$ It depends on the problem. If you look at Evans, partial differential equations, it actually explains when one is more suitable than another in theory. Like for initial value problems laplace transform is better, if your domain is unbounded fourier transform is more suitable. For numerical computations Fourier transform is always used. You can also mix the two of them to solve specific problems. Everything depends on your problem. $\endgroup$ – user8469759 Apr 30 at 19:12
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It's natural consequence of applying a transform to a convolution relation. The output $y(t)$ of an (continuous-time) LTI system is described by a convolution integral :

$$y(t) = h(t)\star x(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau $$

And when you apply a Fourier transform on this relation, it turns out to be a multiplication in the transform domain such as:

$$ Y(j\omega) = \mathscr{F}\{ y(t) \} = \mathscr{F}\{ h(t) \star x(t) \} $$ $$ Y(j\omega) = \mathscr{F}\{ h(t) \} \cdot \mathscr{F}\{ x(t) \} = H(j\omega)\cdot X(j\omega) $$

Similary you can also apply a Laplace transform on it : $$ Y(s) = \mathscr{L}\{ y(t) \} = \mathscr{L}\{ h(t) \star x(t) \} $$ $$ Y(s) = \mathscr{L}\{ h(t) \} \cdot \mathscr{L}\{ x(t) \} = H(s)\cdot X(s) $$

And exactly the same happens for discrete-time LTI systems.

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