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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?

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  • $\begingroup$ One sided or 2 side Laplace? $\endgroup$ – Stanley Pawlukiewicz May 18 at 20:25
  • $\begingroup$ Please kindly guide about both cases if ezi for u $\endgroup$ – engr May 18 at 20:30
  • $\begingroup$ Very related: this answer over at EE. $\endgroup$ – Matt L. May 19 at 9:02
  • $\begingroup$ @MattL. thanks dear $\endgroup$ – engr May 19 at 9:06
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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...

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  • $\begingroup$ @StanleyPawlukiewicz (see the bottom pls) $\endgroup$ – Fat32 May 18 at 22:37
  • $\begingroup$ "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? $\endgroup$ – Matt L. May 19 at 8:55
  • $\begingroup$ @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... $\endgroup$ – Fat32 May 19 at 12:41
  • $\begingroup$ 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. $\endgroup$ – Matt L. May 19 at 13:29
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    $\begingroup$ 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 $\endgroup$ – Matt L. May 19 at 18:02

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