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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!

4 Answers 4

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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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The answer will depend upon the context of the question (mathematics, engineering, electronics etc.?). For the Fourier Transform s = jω and σ = 0.

This is called the Steady State case dealing with unbounded sinusoids and no transients. The counter part is for non-zero σ with ω = 0; this covers the purely Exponential Case. Allowing both σ & ω to be non-zero produces the general case involving fixed sinusoids, and transient solutions (transient sinusoids and pure exponentials).

For example, if you wish to establish the resonant frequency of a 2nd order differential system, such as a simplified model of a spring damper for one wheel of a car, then you only require the steady state case and can set σ = 0.

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$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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    @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 $\omega=0$.2012-06-29
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this page compares Laplace and Fourier Transforms http://www.cambridge.org/us/features/chau/webnotes/chap2laplace.pdf