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I have read few links about difference between Fourier transform and Laplace transform but still not satisfied

Please correct me if i am wrong Simply put, the main difference between Fourier transform and Laplace transform is that real part is set to zero in Fourier transform while real part is non zero in laplace transform?

DSP_CS
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Fourier transform is an intuitive tool that's a bridge between domain of physics and mathematics, as it quantitatively describes the periodic content of the signals and also frequency response characterisation of systems that occur in physical (and engineering) applications. The use of frequencies is quite intuitive and consistent at least for stable systems...

However for unstable systems (and signals) Fourier transforms becomes mostly awkward (if not useless) to deal with. However control engineers unavoidably must make frequent use of unstable systems in their works. And for this purpuse, Fourier transform is either insufficient or awkward, hence a generalisation of the existing Fourier transform is made into the Laplace transform which conveniently yields mathematical (complex algebric) descriptions of stable as well as unstable systems which was not possible with the Fourier. The Laplace transform, therefore, includes a region of convergence parameter into it.

Another difference between the two transforms is in the time-domain transient analysis of output of LTI systems driven under nonzero initial conditions which is successfully captured in the Laplace transform only. In the sense that LCCDE with initial conditions are straightforwardly solvable by (unilateral) Laplace transforms whereas the standard FT can only solve LCCDE with zero initial conditions (initial rest)...

For one sided and two sided differences, I think Stanley has things to say...

Fat32
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  • @StanleyPawlukiewicz (see the bottom pls) – Fat32 May 18 '19 at 22:37
  • "time-domain transient analysis which is successfully captured in the Laplace transform only..." I'm not sure what that means. I've heard it a lot, and it is usually misunderstood. I'm still not sure if there's actually any truth in it. Could you explain what you mean by saying that? – Matt L. May 19 '19 at 08:55
  • @MattL. may be it's another statement for solving LCCDE with nonzero initial conditions ? (and using one sided Laplace transform). The zero input response usually decays with time and therefore called as a transient... – Fat32 May 19 '19 at 12:41
  • Yes, but I think that in its current form the statement is misleading, in the sense that it seems to suggest that the Fourier transform cannot handle transients, just steady-state behavior. I've seen this many times, especially on SE.EE, and that's plain wrong. What is true is that it's more straightforward to take initial conditions into account with the unilateral Laplace transform than with the Fourier transform. – Matt L. May 19 '19 at 13:29
  • @MattL. I think transients as initial conditions clarified it enough. I should add it to the answer too. It seems you have another definition of transients in your mind which does not fit here. And yes afaik $H(\omega)$ is the steady state behaviour of the LTI system, in the sense of page 46 of the standard text DTSP opp 2e ? And honestly I have never seen initial conditions handled by FT ? is there a unilateral FT? may be its modified similar to unilateral Laplace, but I've never seen it in practice (nor in theory). – Fat32 May 19 '19 at 14:16
  • $H(\omega)$ is a complete description of an LTI system, as simple as that. It doesn't solely describe "steady-state behavior", it describes everything the system is capable of doing. That's exactly the misunderstanding that people believe that the Fourier transform can only handle steady-state ... – Matt L. May 19 '19 at 14:27
  • @MattL. Of course $H(\omega)$ is a complete description of an LTI system as it's the Fourier transform of the impulse response $h[n]$ that should exist in the first place for an LTI system. But what I tried to say is not about the system but its output that's produced when the initial conditions are nonzero. These are different things. And (unilatrl) Laplace transform can simultaneously describe the system in terms of transfer function $H(s)$ and further provides the output component due to initials (transients) that Fourier transform cannot provide? So what's not clear here ? – Fat32 May 19 '19 at 15:15
  • It's no big deal, but you (and others) conflate initial conditions and transients, but they're not the same thing. And the misunderstanding that results (and I've seen it many times) is that people think that the Fourier transform is only useful for describing steady-state sinusoidal responses, which is - as you know - non-sense. – Matt L. May 19 '19 at 16:40
  • @MattL. yes no big deal but still important imho. A mixture of electrical circuits, signal processing and control yields an understanding that initial conditions produce transient responses, so the conflation is quite standard I believe. Can you site a reference where transients are not related to initial conditions? And probably gives a broader definition of transients then? And yes it's true that Fourier transform, almost always, is as strong as the Laplace ransform. If not stronger, as unlike Laplace it can define ideal brickwall filters, and transforms of periodic signals etc. – Fat32 May 19 '19 at 17:54
  • @MattL. for example, spikes or chirps can also produce transient responses but their analysis is quite different than what I wanted to outline here... – Fat32 May 19 '19 at 17:58
  • A transient is anything that disturbs the steady-state condition. It usually isn't, or, at least, doesn't need to be caused by non-zero initial conditions. See this Stanford page for a common-sense definition. – Matt L. May 19 '19 at 17:59
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    But I see that you've improved (in my opinion) that passage, such that it becomes clear that it is about non-zero initial conditions, not about transients in general. +1 – Matt L. May 19 '19 at 18:02
  • @MattL. yes I changed the answer to reflect those after your comments. And as my last comment on the spikes and chirps also indicate (which is also confirmed by the link you provided) those broader definition (as happens in audio signal processing) is not what I wanted to indicate. So I think it's ok now. Thanks for +1 ;-) – Fat32 May 19 '19 at 18:08
  • @Fat32 you said at end of your second paragraph "The Laplace transform, therefore, includes a region of convergence parameter into it." Does it means that there is no ROC for fourier transform? – DSP_CS Dec 25 '19 at 17:12
  • @engr Yes Fourier transforms do not define a ROC. In the continuous-time case, $H( j\omega)$ is defined as $H(s)$ on the imaginary axis $s = j\omega$ ... – Fat32 Dec 25 '19 at 17:54