For 1: Laplace can also takes the whole axis (-infinity, infinity), refer to bilateral Laplace transform.steve10 said:If you talk about them in mathematics language, you can get a lot more details. Here are some major differences even an engineer should know if you work with them:
1. Laplace --- half of the axis [0, infinity),
Fourier --- whole axis (-infinity, infinity);
2. Laplace --- one dimensional,
Fourier --- multidimensional;
3. Usually use laplace on time variable t while use fourier transform on spatial variables x,y, ...;
4. The functions that have fourier transforms must not increase faster than a polynomial, while the functions that have laplace transform may increase exponentially at inifinity. For example, exp(x) does not have fourier transform, but exp(t) DOES have laplace transform;
5. The laplace tranform of a function lives in the whole complex plan, while the fourier transform of the function still lives on the real axis or real space;
6. The inverse transforms are completely different.
cevitamic said:2. For such functions, if we still want to have a certain transform domain representation of \[f(t)\], we need to make the function to be converged. It can be realized by multiplying an attenuation factor \[e^{-\sigma t}\], where \[\sigma\] is a real number. Then use the Fourier transform of \[f(t)e^{-\sigma t}\], or \[F(s)=\int_{-\infty}^\infty f(t) e^{-st}ds\], where \[s=(\sigma+j\omega)\].
Really? Are you aware that, if t<0, you are making things even worse?
steve10 said:cevitamic said:2. For such functions, if we still want to have a certain transform domain representation of \[f(t)\], we need to make the function to be converged. It can be realized by multiplying an attenuation factor \[e^{-\sigma t}\], where \[\sigma\] is a real number. Then use the Fourier transform of \[f(t)e^{-\sigma t}\], or \[F(s)=\int_{-\infty}^\infty f(t) e^{-st}ds\], where \[s=(\sigma+j\omega)\].
Really? Are you aware that, if t<0, you are making things even worse?
Well, it's my fault. Actually, what I meant is the situation \[\int_0^\infty f(t)e^{-st}ds\]
claudiocamera said:X1(t) = exp(-2t)u(t) +exp(-t)u(-t)
x2(t) = exp(-t)u(t) + exp(-2t)u(-t)
What is the ROC of these signals ?
How to solve it in s or jw domain ?
After my talk, the chairman called for discussion. For
a moment there was the usual dead silence; then a hand
was raised. It was Hotelling’s. I must hasten to explain
that Hotelling was fat. He used to love to swim in the
ocean and when he did, it is said that the level of the ocean
rose perceptibly. This huge whale of a man stood up in
the back of the room, his expressive fat face took on one
of those all-knowing smiles we all know so well. He said:
“But we all know the world is nonlinear.” Having uttered
this devastating criticism of my model, he majestically sat
down. And there I was, a virtual unknown, frantically trying
to compose a proper reply.
Suddenly another hand in the audience was raised. It was
von Neumann. “Mr. Chairman, Mr. Chairman,” he said,
“if the speaker doesn’t mind, I would like to reply for
him.” Naturally I agreed. Von Neumann said: “The speaker
titled his talk ‘linear programming’ and carefully stated his
axioms. If you have an application that satisfies the axioms,
well use it. If it does not, then don’t,” and he sat down.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?