Inverse Fourier Transform of a function

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Hi everyone, this is not a homework question but from my reading of a signals processing paper.

This paper says if [itex]f(t)[/itex] is the inverse Fourier transform of a function

[itex]f(\lambda) = e^{-2i\pi\lambda d}[/itex]

then we can "easily see" that [itex]f(t)[/itex] will have a peak [itex]d[/itex].

Part of the issue here is my shaky of the Fourier transform, which up til this point, I understand as a frequency decomposition of a signal. That is, let [itex]f(x)[/itex] be a signal function supported on [itex][-a,a][/itex]. Then,

[itex]f(\lambda) = \int_{-a}^{a}f(x)e^{-2i\pi x\lambda}dx[/itex]

is the Fourier transform with the property that [itex]|f(\lambda)|[/itex] quantifies the "amount" of frequency [itex]\lambda[/itex] in the original signal function [itex]f(x)[/itex].

But returning to my original problem, if we take the inverse Fourier transform of

[itex]f(\lambda) = e^{-2i\pi\lambda d}[/itex], then we have

[itex]f(t) = \frac{1}{2\pi}\int e^{-2i\pi\lambda d} e^{2i\pi\lambda t} d \lambda = \frac{1}{2\pi}\int e^{2i\pi\lambda (t-d)} d \lambda [/itex]. But I can't see how [itex]|f(t)|[/itex] is maximized at [itex]t=d[/itex], as it becomes the integral of 1.

Am I missing something?
 

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  • #2
HallsofIvy
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It might make more sense if you did not use "f" both for the function and its Fourier transform. And what are the limits of integration in the final integral?
 
  • #3
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[itex]f(t) = \frac{1}{2\pi}\int e^{-2i\pi\lambda d} e^{2i\pi\lambda t} d \lambda = \frac{1}{2\pi}\int e^{2i\pi\lambda (t-d)} d \lambda [/itex]. But I can't see how [itex]|f(t)|[/itex] is maximized at [itex]t=d[/itex], as it becomes the integral of 1.

Am I missing something?
First evaluate the integral:

[tex]\int_{-a}^a e^{2\pi i \lambda(t-d)}d\lambda[/tex]

and then take the limit as t goes to d.
 
  • #4
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First evaluate the integral:

[tex]\int_{-a}^a e^{2\pi i \lambda(t-d)}d\lambda[/tex]

and then take the limit as t goes to d.
Oh, I see...

So we have

[tex] \int_{-a}^a e^{2\pi i \lambda(t-d)}d\lambda = \frac{Sin(2a\pi (d-t)}{\pi(d-t)} [/tex]

So the function

[tex] \frac{Sin(2a\pi (d-t)}{\pi(d-t)} [/tex]

achieves its max of [tex]\frac{2a}{\pi} [/tex] when [tex] d = t [/tex].

Thanks for the help. By the way, was there an easy way to see this without having to integrate? The original function

[tex] f(\lambda) = e^{-2 i \pi \lambda d} [/tex]

is a Fourier basis elements on [tex] L^2[-a,a] [/tex].
 
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