# Dirac delta and fourier transform

1. Mar 1, 2012

### aaaa202

In my book the dirac delta is described by the equation on the attached picture. This realtion is derived from the fourier transform, but I'm not sure that I understand what it says. If u=t it is clear that one gets f(u) in the fourier inversion theorem. But why wouldn't u=t? In the derivation of the fourier transform from the discrete fourier series t was just changed to u in the expression of the coefficients to avoid confusion.
Can anyone try to picture what this expression fundamentally says? I should suspect that it is like the analogue of the ortogonality relation of the discrete fourier series, but I can't quite understand it.
And what would the situation u≠t represent?

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2. Mar 1, 2012

### jbunniii

What it is saying is that

$$\delta(t - u)$$
and
$$\exp(-iu\omega)$$

are a Fourier transform pair. A complex exponential with "frequency" equal to $u$ has a Fourier transform with all of its energy concentrated at $u$.

Last edited: Mar 1, 2012
3. Mar 1, 2012

### jbunniii

P.S. The equation

$$\delta(t - u) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \exp(i \omega(t-u) d\omega$$

is mathematically nonrigorous. The integral on the right hand side does not actually exist for any values of $u$ and $t$. What is true is that if $\mathcal{F}$ denotes the Fourier transform operator, then

$$\mathcal{F}(\delta(t-u)) = \exp(-i\omega u)$$
and
$$\delta(t-u) = \mathcal{F}^{-1}(\exp(-i\omega u))$$

in the sense of distributions. See here for more details:

http://en.wikipedia.org/wiki/Distribution_(mathematics)

Last edited: Mar 1, 2012