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Having an exam next week!
I've searched a lot, couldn't find anything I could understand.

When is the Laplace variable $s$ equal to $j\omega$? Because I know that, by definition, $s = \sigma +j\omega$

Thank you!

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up vote 2 down vote accepted

$s=\sigma+j\omega $ means that $s$ is a complex variable with real part $\sigma$ and imaginary part $\omega$. When the real part is equal to zero, we have $s=j\omega$.

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Ok, I understand this, but the teacher asked someone "In which cases s=jω?", that means, in which case σ is equal to 0? Trying to search somewhere else it turns out s=jω is in Fourier transform and s=σ+jω in Laplace transform, but I didn't get that. – topoftheforts Jun 29 '12 at 18:59
@topoftheforts Ok, maybe this will help: with no constraint on $s$ you have the Fourier-Laplace transform. It becomes the Fourier transform when $\sigma=0$. It becomes the Laplace transform when $\omega=0$. – user31373 Jun 29 '12 at 19:08
Great! Thank you! – topoftheforts Jun 29 '12 at 19:42

this page compares Laplace and Fourier Transforms

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If the course is about electronics, $s=jw$ when components are assumed to be ideal meaning they have no loss factor, which makes their real part zero.

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