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Relationship between Fourier and Lpalace transforms

  1. Mar 22, 2013 #1
    Can someone please explain WHY the statement below is valid:
    [​IMG]

    s = σ + jω ; left hand side σ < 0
    So it basically says if all the poles have negative real parts then we can directly substitute s = jω to get the Fourier transform.

    This doesn't make sense to me, does it make sense to you? :rolleyes:
     
  2. jcsd
  3. Mar 22, 2013 #2

    rbj

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    that statement doesn't state it well. but the end result makes sense to me.

    here is what you do:

    suppose you have a Linear Time-Invariant system (LTI). then the impulse response, [itex]h(t)[/itex] fully defines the input/output characteristic of the LTI. if you know the impulse response, you know how the LTI will respond to any input.

    anyway, the double-sided Laplace transform of [itex]h(t)[/itex] is [itex]H(s)[/itex]. if you drive the input of that LTI with

    [tex] x(t) = e^{j \omega t} [/tex]

    then the output of the LTI system is

    [tex] y(t) = H(j \omega) e^{j \omega t} [/tex]

    same [itex]H(s)[/itex], just substitute [itex]s = j \omega[/itex].

    it's easy to prove, if you can do integrals.
     
  4. Mar 22, 2013 #3

    rbj

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    oh, and what's easier to prove is that the Fourier transform is the same as the double-sided Laplace transform with the substitution [itex]s = j \omega[/itex]. that's just using the definition.
     
  5. Mar 22, 2013 #4
    Thanks my friend.

    I understand that the Fourier transform is equivalent to the double-sided Laplace transform, but that doesn't explain anything clearly to me regarding the poles that have to be located on the left hand side of the s-plane in order for the substitution s = jω to be valid.
     
  6. Mar 23, 2013 #5
    A signal has its Fourier transform if and only if its ROC of Laplace transform contains the imaginary axis s=jw.

    The statement that you give is valid only for the right-hand sided signals for which the ROC is the right hand side of the poles.

    Fourier transform and Laplace transfrom (whether one-sided or two-sided) are not equivalent. Fourier transform can be considered as a special case of Laplace transform, that is, just set [itex]\sigma = 0[/itex].
     
    Last edited: Mar 23, 2013
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