# A Candle in the Dark

A look on science, politics, religion and events

## FT of some common functions

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In this post, let’s look at the fourier transform of some functions which are quite useful.

(i) The rectangle function
We’ll start with the rectangular function, also called the box function. It’s defined as

$f_a(x) = \begin{cases} 0, \quad |x| > a/2 \\ 1, \quad |x| \leq a/2 \end{cases}$

Now, consider the fourier pair of $f(x)$. We have,

$\phi(p) = \int\limits_{-\infty}^{\infty} f(x) e^{2 \pi i p x}dx$

$= \int\limits_{-a/2}^{a/2} e^{2 \pi i p x}dx$

$= \left(e^{2 \pi i p a/2} - e^{-2 \pi i p a/2}\right)/ \left(2 \pi i p \right)$

$= a \sin(\pi p a)/(\pi p a)$

$= a \text{sinc} (\pi p a)$

So, the fourier transform of the boxcar function is the sinc function!

I’ll revist this fourier pair again, while discussing the wave theory of light. In fact, we can use this fourier pair to show that the interference pattern we get in a double slit experiment is infact the sinc function!

(ii) The Gaussian function

Next, we’ll look at the gaussian function. The gaussian function has the interesting property that it’s fourier pair is also a gaussian function! Consider,

$f(x) = e^{-x^2/a^2}$

Let’s caculate the fourier pair, $\phi(p)$.

$\phi(p) = \int\limits_{-\infty}^{\infty} e^{\frac{-x^2}{a^2}} e^{2 \pi i p x} dx$
$= \phi(p) = \int\limits_{-\infty}^{\infty} e^{-\left(x/a - a \pi i p\right)^2} e^{-pi^2 p^2 a^2} dx$

To evaluate this integral, use the substitution,
$x/a - a \pi i p=r$

After evaluating the integral and substituting the limits, the expression for $\phi(p)$ is obtained as,

$\phi(p) = a \sqrt{\pi} e^{-\pi^2 p^2 a^2}$

which is also a gaussian.

Also, if you try plotting the fourier pairs for different values of a (and hence, different widths of the gaussian), you’ll notice that the wider the gaussian in x-space, the narrower it is in p-space (ie, the transform space), and vice versa.

(iii) The delta function

The dirac delta function (although, not strictly a function), can be represented as
$\delta(x) = \begin{cases} 0, \quad x \neq 0 \\ \infty, \quad x = 0 \end{cases}$

Now let’s apply the fourier transform to the delta function. We get,

$\phi(p) = \int\limits_{-\infty}^{\infty} \delta(x) e^{2 \pi i p x} dx$

and by the property of the delta function, this is,

$= e^{2 \pi i p 0}$

$= 1$

Therefore, we find that the fourier pair of the delta function is unity. That is $\delta(x) \rightleftharpoons 1$

Also, notice that
$\delta(x-a) \rightleftharpoons e^{2 \pi p a}$
$\delta(x+a) \rightleftharpoons e^{- 2 \pi p a}$

and hence,
$\delta(x+a) + \delta(x-a) \rightleftharpoons 2 \cos(2 \pi p a)$

(iv) The Shah function

The shah function, also known as a Dirac comb, is an infinite combination of evenly spaced dirac functions.

$f_a(x)= \sum_{n=-\infty}^{\infty} \delta(x-an)$

The fourier transform of the shah function is also another shah function, with a period of 1/a. You’ll find that the shah function is quite invaluable in convolutions, where it’s role is to create infinte “copies” of the original function, with period equal to the spacing between the teeth of the comb.

In my next post, I’ll explain more about convolution, especially the convolution theorem which is a real time saver in performing transforms, and other theorems relating to fourier transforms

Note: If you find any errors, please do inform me, and I’ll correct them. Also, click on the thumbnail images to get a detailed graph

Written by parseval

June 22, 2007 at 2:26 pm

Posted in mathematics