How to find Fourier transform of u(t)????

[purnapragna]

fourier transform of u(t)

Hey,

I wish to find the Fourier transform of the signal \[u(t)\] which is unit step using the property of differentiation. i.e.,

\[\frac{d}{dt}u(t)=\delta(t)\]. Thus applying the differentiation property we get

\[j\omega F\left[u(t)\right]=1\] and thus,

\[F\left[u(t)\right]=\frac{1}{j\omega}\] which is obviously wrong!!!

But we already know that

\[F\left[u(t)\right]=\frac{1}{j\omega}+\pi \delta(\omega)\]

So what is the wrong with the property?? is that we should not use it as we wish???

help me out in this!!!

 

[bulx]

Nice question!

I think the porblem is that you have apllied the diffrentiation property in the wrong direction. lt gives you a way of finding out the FT of signal's derivative, if you know the FT of the signal.

Here you have come the other way. You take the FT of the derivative the signal and try to find the FT of the signal using the diff property, which is not how the property is to be used.

The correct direction will be to write the delta function and its FT which we know is 1. Then integrate it to get the step function. now get its FT by applying the integration property.

Then you get the exactly result you have written.
-b

 

[purnapragna]

That is a good explanation!!

But one more thing how about \[sgn(t)\]? I mean if I apply the differentiation property to \[sgn(t)\] then now, if I integrate I wont get the FT of it. How to resolve???

 

[bulx]

why not?
sgn(t) <-> 2/jw

d/dt(sgn(t)) = 2δ(t)

FT of differential of sgn(t) = jw * FT of sgn(t) = 2 * 2/jw = 2

2 is the FT of 2δ(t).

to go the other way:
2δ(t) <-> 2

∫2δ(t) = sgn(t) + 1 (unless you add 1 there will be an offset)

FT {sgn(t) + 1} = 2/jw + Π * G(w=0) * δ(w) = 2/jw + 2Π * δ(w)
FT {sgn(t)} + FT {1} = 2/jw + 2Π * δ(w)
FT {sgn(t)} + 2Π * δ(w) = 2/jw + 2Π * δ(w)
FT {sgn(t)} = 2/jw

-b

 

[purnapragna]

ha ha thanks anyway!!