# 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?

#### BrownOut

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

#### tau17

##### New Member
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)

#### Karkas

##### Member
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.

#### tau17

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

#### Karkas

##### Member
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.

#### tresca

##### Member
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.

#### tau17

##### New Member
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)

#### tau17

##### New Member
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)

#### steinar96

##### New Member
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.

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#### Karkas

##### Member
That was the answer I was looking for, thank you very much steinar96, and the other posters too.

#### hkBattousai

##### Member
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?

#### MrAl

##### Well-Known Member
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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