Timeline for What is the difference between $X(j\omega)$ and $X(\omega)$ notation?
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| May 7, 2015 at 16:45 | history | edited | Matt L. | CC BY-SA 3.0 |
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| May 7, 2015 at 14:56 | history | edited | Matt L. | CC BY-SA 3.0 |
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| May 7, 2015 at 14:50 | comment | added | Matt L. | @Jazzmaniac: OK, but the idea of the example was to show a simple case where with $X(s)$ the Laplace transform, $X(j\omega)$ does not (for all $\omega$) equal the Fourier transform. And still, both transforms exist for that example. Obviously, the ROC of $X(s)$ is $Re\{s\}>0$. The step function has a Laplace transform and a Fourier transform, that's all I mean to say when I say they both exist. I'll add two more examples later on, where one of the two does not exist, to make my point a bit clearer. | |
| May 7, 2015 at 10:46 | comment | added | Jazzmaniac | That's why the disagreement between the Laplace transform and the Fourier transform at $s=0$ or $\omega=0$ is the least you can expect. It's not much more than a fortunate coincidence following from some nice properties of the analytical continuation that you get the correct result at $s=\omega*j$ for $\omega \neq 0$. So I agree with you example, but not with the premise "...even though both transforms exist...". | |
| May 7, 2015 at 10:43 | comment | added | Jazzmaniac | I'm saying that the Laplace transform does not exist at s=0, and strictly speaking it does not even exist at real(s)=0. The improper integral does not converge there. You can fix that by either applying a Cauchy principal value assignment to the integral or by analytically continuing the Laplace transform at real(s)>0. But whatever you do, you can't fix it for s=0. So the Laplace transform does not exist everywhere on the imaginary axis, and if you're really strict, it exists nowhere on the imaginary axis in the proper sense. | |
| May 7, 2015 at 10:02 | comment | added | Matt L. | @Jazzmaniac: The Laplace transform of the step function exists. It has a pole at $s=0$. This is different from cases where the Laplace transform does not exist (such as for $x(t)=\sin(\omega t)$). If you're not saying that the Fourier transform of $x(t)=u(t)$ can be obtained from $X(s)$ by setting $s=j\omega$ (which is wrong), I'm not sure what it is that you're saying. | |
| May 7, 2015 at 9:24 | comment | added | Jazzmaniac | I disagree with your example. Clearly, the Laplace transform of the Heaviside function does not exist at s=0 (and it exists on the imaginary axis also only by analytical continuation), so your own requirements fail. Also note that the Fourier transform and the Laplace transform coincide where both are defined. | |
| May 7, 2015 at 8:00 | vote | accept | verdery | ||
| May 7, 2015 at 7:56 | history | answered | Matt L. | CC BY-SA 3.0 |