# Relationship between Fourier and Lpalace transforms

by cocopops12
Tags: fourier, lpalace, relationship, transforms
 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?
 P: 2,237 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, $h(t)$ 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 $h(t)$ is $H(s)$. if you drive the input of that LTI with $$x(t) = e^{j \omega t}$$ then the output of the LTI system is $$y(t) = H(j \omega) e^{j \omega t}$$ same $H(s)$, just substitute $s = j \omega$. it's easy to prove, if you can do integrals.
 P: 2,237 oh, and what's easier to prove is that the Fourier transform is the same as the double-sided Laplace transform with the substitution $s = j \omega$. that's just using the definition.
P: 30

## 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.
 P: 39 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 $\sigma = 0$.

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