Question

what is inverse laplace or fourier transform of a shifted dirac delta function?

Answer 1

@kainui

Answer 2

What have you tried and also why are you asking just curious.

Answer 3

just curious, my friend had some question on circuits, and a unitary exponential input fourier transform looked just like a fourier transform of a constant 1, but with the frequencies shifted... and i saw the fft of 1 is this dirac delta function, so i was wondering how they got that

Answer 4

not really sure what you mean, have you seen how the dirac delta is derived to begin with, maybe that'll help if I show that?

Answer 5

yeah show me that

Answer 6

http://www.wolframalpha.com/input/?i=fourier+transform&rawformassumption=%7B%22F%22,+%22FourierTransformCalculator%22,+%22transformfunction%22%7D+-%3E%221%22&rawformassumption=%7B%22F%22,+%22FourierTransformCalculator%22,+%22variable1%22%7D+-%3E%22t%22&rawformassumption=%7B%22F%22,+%22FourierTransformCalculator%22,+%22variable2%22%7D+-%3E%22w%22&rawformassumption=%7B%22C%22,+%22fourier+transform%22%7D+-%3E+%7B%22Calculator%22%7D

Answer 7

and i mean that, why is ft of 1 a diract delta fn

Answer 8

Yeah that's cause an exponential is just a "shifted" constant. So I guess I'll show you the derivation of the dirac delta now. Basically it's the identity matrix, so it's what you get when you fourier transform and then do the inverse transform.
Like in matrices you would write it like this, acting on a test vector \(\vec v\).
\[\mathcal{F}^{-1} \mathcal{F}\vec v = I \vec v = \vec v\]

Makes sense so far what I'm talking about? The rest is downhill from here for the most part, just filling in details.

Answer 9

This is what we want out of our Fourier transform, that if you transform in one direction and transform back, you get what you started with:

\[\left(\int_{-\infty}^\infty dx e^{-ixy}\right) g(x)=G(y)\]\[\left(\int_{-\infty}^\infty dy e^{iyz}\right) G(y)=g(z)\]

You could plug one into the other to get:
\[\left(\int_{-\infty}^\infty dy e^{iyz}\right) \left(\int_{-\infty}^\infty dx e^{-ixy}\right) g(x)=g(z)\]

Answer 10

I'm trying to keep it like operators or matrices acting on stuff to the right of it, that ways it helps to make it clearer what I'm actually doing.

Answer 11

is that g(x) in the integral too

Answer 12

Everything to the right of an integral is in that integral.

Answer 13

Look at the dummy indices to know what is "in" the integral, since the dx and dy tell you specifically what is being summed over.

Answer 14

ok got it

Answer 15

Now by definition of the dirac delta we have:

\[\left(\int_{-\infty}^\infty dx \delta(x-z)\right) g(x)=g(z)\]
So now plug it in on the right of this:
\[\left(\int_{-\infty}^\infty dy e^{iyz}\right) \left(\int_{-\infty}^\infty dx e^{-ixy}\right) g(x)=g(z)\]

""plugs in sexually""
\[\left(\int_{-\infty}^\infty dy e^{iyz}\right) \left(\int_{-\infty}^\infty dx e^{-ixy}\right) g(x)=\left(\int_{-\infty}^\infty dx \delta(x-z)\right) g(x)\]
Oh hey it's like pure matrices now if we remove g(x)

\[\left(\int_{-\infty}^\infty dy e^{iyz}\right) \left(\int_{-\infty}^\infty dx e^{-ixy}\right) =\left(\int_{-\infty}^\infty dx \delta(x-z)\right) \]

But this is just an aside, we haven't actually derived anything yet.

Answer 16

Careful, you have an f(x) outside if you've integrated over a!

Answer 17

ok i see now only the g(z) will remain

Answer 18

what i put before is wrong because it simplfies to the same integral doesnt it