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Differences betwen Laplace and Fourier Transform

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Dear Friends,

There is a long discussion for this two kind of transforms, and i'd like to know from you what´s the concepts, spec, etc for them.

Tks
 

It is preety hard to concise the diference of the two transform, there are books dedicated to these subjects, however in a brief explanation is possible to say that the fourier transform is a particular case of Laplace transform.
In the Laplace transform ∫f(t) * exp(-st) ds where S=σ +jω when σ=o you can realize thet the Laplace transform become the fourier transform
 

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.
 

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.
Click to expand...
For 1: Laplace can also takes the whole axis (-infinity, infinity), refer to bilateral Laplace transform.
 

People who invented "bilateral Laplace transform" should be shot :) --- never mind, just a silly joke.

Seriously, have you seen any application of it? To me, it is just a name, which doesn't really make any sense. If you really understand (uni) laplace transform, you'll see that the purpose of the laplace transform is to introduce a damping factor exp(-real(s)t) for t>0, which is something the fourier transform cannot do. Now, think about it. What if we allow t<0? That damping factor is gonna explode ...
 

Dear Steve,

Who mentioned Mathematics language ? The Bilateral Laplace is well suited to problems involving noncausal signals and systems. Since you can mention that noncausal systems is not practical, It does make sense in Matemathical point of view.

As a matter in fact the damping factor is not gonna explod in t<0 if the ROC ( Region Of Convergence) is respected. The study of bilateral laplace transform is crucial in analysis of stability and causability of system and consequently in determine inverse function and minimal phase systems. So I dont think that who invented "bilateral Laplace Transform is the one who shoud be shot :)
 

Oh, well, dear claudiocamera, your appeal is accepted and those poor inventors are exempted. We will have mercy on them, but find someone else to shoot. :)

Quite honestly, I knew bilateral laplace transform long time ago, but I failed to find a right place to apply it. Or maybe more accurately, I could use either (unilateral) laplace transform or fourier transform to the examples people showed me. Noncasual system? Today's output depends on tomorrow's input? This kind of logic is distasteful to me, or perhaps I am too ignorant of it.

I will be grateful if you can show me an example to use the "bilateral" laplace transform where both the unilateral laplace and fourier transforms cannot be applied.
 

Oh, Dear Steve10, It is not my concern find someone to shot, it is not part of my culture neither from people of my country :)
The original question was already answered, and if something sound distateful for you, it cannot sound distateful for others. The concept of noncausal systems is useful to understand causal systems, as the concept of memoryless is usefull to understand memory system, what do you suggest ? Ripping out all the mathematics concepts that you dont find a practical application or either some tool that replaces it? fortunatelly, people have different views and that is what makes evolution in societies. Fortunately, most people respect different oppinions and dont thinky about shooting everybody who dont share their oppinions. Respecting different views and oppinions is a kind of behaviour that will keep me out of answering again this topic, despite of your latter replies.

Best regards...
 

Oh, buddy, take it easy. It's your freedom NOT to answer anything. But whenever you want to, you are welcome to make any comments on anything I said or I will say. We are not publishing anything here, right? We are enjoying our freedom to comment on anything we like to, right? I am not making promise whatever I said is 100 percent right. Care to show an example of noncasual system in order to show the exclusive power of bilateral laplace transform?
 

1. For some functions \[f(t)\], the integration \[\int_{-\infty}^\infty f(t) e^{-j\omega t}dt\] will not converge, so its Fourier transform does not exist.

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 \[s=(\sigma+j\omega)\].

3. It's obviously that, if the Fourier Transform \[F(\omega)\] of a function \[f(t)\] exists, then \[F(\omega)=F(s)|_{s=j\omega}\]. That means if the function is proper, the attenuation factor equals 1.
 
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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?
Click to expand...
 
Last edited by a moderator:

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?
Click to expand...

Well, it's my fault. Actually, what I meant is the situation \[\int_0^\infty f(t)e^{-st}ds\]
Click to expand...
 
Last edited by a moderator:

Steve10 is really a polite guy: first, he sugest shoot people who invented bilateral laplace transform. After, he answers a pal with a ironic "really ?" .
He struggles to say that a concept that is given in most signals and systems books is useless ( Simon Haykin, Oppenheim, Lathi etc).

As he is so smart , he should answer a very simple question, it is on page 514 Haykin Van veen book, very elementary question answered by Bilateral Laplace. But steve despize Bilateral Laplace Transform so he will not use it.

Regardless the signals are non causal, it deserves an answer.

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 ?
 

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 ?
Click to expand...


I am so sorry that you don't take jokes, or perhaps more accurately, you don't think some of what I said are jokes even though I so labeled them. Well, that's fine with me. There is one thing that I can tell you that I have not got personal with anyone on this board. I respect others, but it doesn't mean I have to agree whatever they said. Something written in the book doesn't mean it's out of challenge. Of course, I might be very well ignorant of it.

Please forgive me if you don't like what I said above.
Well, excuse me, am I supposed to answer the questions you posted above? Umm...I thought someone else would help me with those questions. By the way, are you really saying

X1(t) = exp(-2t)u(t) +exp(t)u(-t)

x2(t) = exp(-t)u(t) + exp(2t)u(-t) ?

Please notice the sign of the exponentials of the second terms on the right hands.
 

No Steve10, the signs of exponential in the right side are exactly as I posted, I mean negative.

I agree that you can challenge something written in books, but I am a little modest, I believe that a subject that is into so many books has reasons to be there.

The challenge in this case is to solve the question without using the tool that is necessary to solve it, I mean, not using the bilateral laplace transform , which acording to you is useless.

Good luck .
 

Apparently, things are getting tricky. Before moving on, I'd like to review what we are trying to accomplish.

A few posts back, I made a request that you provide an example to which the bilateral laplace transform can be applied while both the unilateral laplace and the fourier transforms CANNOT be applied. If your example is a respose to my request, then I think the following is how we should proceed:

1. You provide an example (done) and then APPLY the bilateral laplace transform to it (pending);
2. You then claim both the unilateral laplace and the fourier transforms CANNOT be applied;
3. I'll try to show you how I can apply either the unilateral laplace or the fourier transforms or both of them.

Isn't it fair?
 

Honnestly, I think other people in the forum are getting sick and tired of this dicussion.

All the point is that I've made one remark in the answer you gave and you tried to disqualify it, maybe because you just cant tolerate that someone remark your answers ( maybe)

It is simple, Laplace transform can be aplly to (-infinity, infinity) , it is the bilateral transform, it does exist, despite you like or not.

Regarding your last mail, I provided an example giving the source, Signals and Systems, Simon Hayking and Barry Van Veen 2 Edittion page 514. There you can find all you are questioning, I will not copy the stuff here, since it does not make any sense. Besides, it does not make any sense getting on this discussion, the bilateral transform is a concept and the only reason that it was questioned was a vanish crisis due to a simple remark that has no intention to hurt anybody, but coundnt imagine that would hurt a pride person.

Go to google and search for bilateral laplace transform. How many pages ? I will not loose my time with this anylonger.

Have a nice day.
 

X1(t) = exp(-2t)u(t) +exp(-t)u(-t)

x2(t) = exp(-t)u(t) + exp(-2t)u(-t)

I am sorry that the discussion has to be aborted. I really wish that we could have concentrated more on the technical issues, instead of what someone else thinks, instead of whom I am really gonna shoot, .... I believe that we are very close to the answer, which most probably favors you, but not through this example you provided.

If you apply the bilateral laplace transform, the ROC for the first equation is -1>s>-2, while the ROC for the second equation is empty (-2>s>-1). That's why I questioned you if you got wrong sign. Most importantly, regardless of the signs, I am not sure what conclusion you can draw from here. I thought you would provide an equation and solve it with the help of the bilateral laplace transform.

Anyway, you have a nice day, too.
 

After G.B. Dantzig presented his Simplex method for solving linear program:

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.
Click to expand...
 

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