Relationship between Fourier and Lpalace transforms


by cocopops12
Tags: fourier, lpalace, relationship, transforms
cocopops12
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#1
Mar22-13, 04:23 PM
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P: 30
Can someone please explain WHY the statement below is valid:


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?
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rbj
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#2
Mar22-13, 05:00 PM
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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.
rbj
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#3
Mar22-13, 05:02 PM
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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.

cocopops12
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#4
Mar22-13, 05:59 PM
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Relationship between Fourier and Lpalace transforms


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.
jashua
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#5
Mar23-13, 04:17 PM
P: 43
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].


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