Inverse Fourier Transform of a function

[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?

 

[HallsofIvy]

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?

 

[jackmell]

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

Am I missing something?

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.

 

[CantorSet]

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.

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 $$L^2[-a,a]$$.