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# Fourier transform and Laplace transform

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#### Karkas

##### Member
I'm not very sure if this is the right forum to post this, but didn't find a better one.

I would like to know if the Fourier transform can always be achieved by finding the Laplace transform, I'm asking because I find in a Signals & Systems book, a problem that said Find the Fourier transform using the Laplace transform, well, actually it was the solution manual, and what they did was find the Laplace transform and change S for jω. Is that always possible?

Yes, from what I can remember my studies in differential equations, about 20 years ago.

well most of the time but not always because for example laplace transform of the step frunction is 1/s evaluating at s=0+j*2*pi*f but that wont give you the right trans form value. the unit step transform is 1/(j2pi*f)+(1/2)delta(f)

Also i was wondering if any one knew the transform of sinc(t)u(t)

Thanks for the answer, but the unit step transfomr is not 1/jω + πδ(ω)?

But anyway you're right, it is not always the same, is that the only case you know? because if there are more, then is not a very sure thing to do.

i thought that was the the transform for unit step . thats what my book says ..........

I don't have a book here but I looked it up in the web and I found it that way in two sites, and nonde of them is wikipedia.

well most of the time but not always because for example laplace transform of the step frunction is 1/s evaluating at s=0+j*2*pi*f but that wont give you the right trans form value. the unit step transform is 1/(j2pi*f)+(1/2)delta(f)

Also i was wondering if any one knew the transform of sinc(t)u(t)

I think its rect(w)

Fourier transform have a property called duality, which basically means that if you have
if the following is a transform pair s(t)<-->S(w) then S(t)<--> s(w).

The fourier transform of a rect pulse is a sinc function, so by duality, a sinc funtion would have a rect as its fourier transform.

well u can derive it for unit pulse

u know that transform of delta is 1 and derivative of unit pulse is delta
so by integration rule
taking integral of delta
integral of g(t) transforms to (G(f)/(j2*pi*f)) + ((G(0))/2 )*delta(f)
now for
values of G(F) and G(0) are 1 because delta transforms to 1 so thas how i got that equation

Tresca
i dont want sinc (t)
i want transform of sinc(t)u(t)

also i just wana check if i am getting duality right
g(t) <--->G(f)
and g'(t) <----> j2*pi*f*G(f)
so is t *G(t) =j g'(f)

*= multiplication not convolution ...

sorry = j*f*G(f)

The fourier and laplace transform are pretty much the same thing except σ is nonzero when talking about "laplace" transforms . Fourier transform is laplace evaluated with σ=0.

Fourier transform is the jω axis of the laplace transform. If you "draw" the laplace space. And only take a look at the slice which goes along the jω axis then that is the fourier "space".

Fourier transformation is only a "thin slice" (the jω axis) from the s plane. Stable systems contain laplace transforms that converge on the jω axis and thus have a fourier transforms assuming dirichlet conditions are fullfilled. If you have a laplace transform for a stable system. You can simply put σ=0 and volla you have your fourier transform.

Keep in mind that S = σ + jω. So puting σ=0 you have.....jω .
You can only do this for systems that have laplace transforms that converge for the jω axis.

Last edited:
That was the answer I was looking for, thank you very much steinar96, and the other posters too.

The fourier and laplace transform are pretty much the same thing except σ is nonzero when talking about "laplace" transforms . Fourier transform is laplace evaluated with σ=0.

Fourier transform is the jω axis of the laplace transform. If you "draw" the laplace space. And only take a look at the slice which goes along the jω axis then that is the fourier "space".

Fourier transformation is only a "thin slice" (the jω axis) from the s plane. Stable systems contain laplace transforms that converge on the jω axis and thus have a fourier transforms assuming dirichlet conditions are fullfilled. If you have a laplace transform for a stable system. You can simply put σ=0 and volla you have your fourier transform.

Keep in mind that S = σ + jω. So puting σ=0 you have.....jω .
You can only do this for systems that have laplace transforms that converge for the jω axis.

This is the best explanation in this topic.

Can we say that, if a linear system is stable we can directly replace "s" by "jω" to find the Fourier transform?

Hi,

The unit step can be looked at like this too...

K => 2*pi*K*d(w), where d is the curly delta, and
sgn(t) => 2/jw

and

u(t)=0.5+0.5*sgn(t)

so we would get:

u(t) => 0.5*2*pi*d(w)+0.5*2/jw

which simplified comes out to:
u(t) => pi*d(w)+1/jw

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