Relationship between Fourier and Lpalace transforms

cocopops12

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?

rbj

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

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

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

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].