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!
When is Laplace variable $s =j\omega$?
4 Answers
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.
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.
$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 $s=j\omega$.
-
0@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
this page compares Laplace and Fourier Transforms http://www.cambridge.org/us/features/chau/webnotes/chap2laplace.pdf