# Inverse Fourier Transform of a function

1. Aug 12, 2011

### CantorSet

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

This paper says if $f(t)$ is the inverse Fourier transform of a function

$f(\lambda) = e^{-2i\pi\lambda d}$

then we can "easily see" that $f(t)$ will have a peak $d$.

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 $f(x)$ be a signal function supported on $[-a,a]$. Then,

$f(\lambda) = \int_{-a}^{a}f(x)e^{-2i\pi x\lambda}dx$

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

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

$f(\lambda) = e^{-2i\pi\lambda d}$, then we have

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

Am I missing something?

2. Aug 13, 2011

### HallsofIvy

Staff Emeritus
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. Aug 13, 2011

### jackmell

First evaluate the integral:

$$\int_{-a}^a e^{2\pi i \lambda(t-d)}d\lambda$$

and then take the limit as t goes to d.

4. Aug 13, 2011

### CantorSet

Oh, I see...

So we have

$$\int_{-a}^a e^{2\pi i \lambda(t-d)}d\lambda = \frac{Sin(2a\pi (d-t)}{\pi(d-t)}$$

So the function

$$\frac{Sin(2a\pi (d-t)}{\pi(d-t)}$$

achieves its max of $$\frac{2a}{\pi}$$ when $$d = t$$.

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

$$f(\lambda) = e^{-2 i \pi \lambda d}$$

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