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, $h(t)$ 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.

 

[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 $s = j \omega$. 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 $\sigma = 0$.