# Fourier transform of a sinc function

• Dec 4th 2011, 07:01 AM
halfnormalled
Fourier transform of a sinc function
I'm trying to show the fourier transform of a since function:

$f(x) = 2 sinc (2x)$

I can't figure out how to show this. I could just work through the integral of the fourier transform, but that seems difficult for me... I think I could show it more easily using the duality property. i.e, finding the fourier transform of a rectangle function, then applying the duality property to show that transform the other way around. But I can't figure out what the rectangle function should be!

Can anyone help out or give some hints? Am I on the right track? Thanks!
• Dec 8th 2011, 12:18 AM
halfnormalled
Re: Fourier transform of a sinc function
OK, I've been working on this... I've got a few approaches. The first is this:

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From theory, we know that the fourier transform of a rectangle function is a sinc:

$rect(t) => sinc(\frac{w}{2\pi})$

So, if the fourier transform of $s(t)$ is $S(w)$, using the symmetry property (duality):

$s(t) => S(w)$
$S(t) => 2\pi \cdot s(-w)$

We can get

$rect(t) => sinc(\frac{w}{2\pi})$
$sinc(\frac{t}{2\pi}) => 2\pi \cdot rect(-w)$

From the similarity property of the fourier transform, we get
$s(t \cdot a) => \frac{S(\frac{w}{a})}{|a|}$

so, scaling t in $sinc(\frac{t}{2\pi})$ by a gives

$sinc(\frac{t \cdot a}{2\pi}) => \frac{2\pi}{a} \cdot rect(\frac{-w}{a})$

so to find $sinc(2t)$, substitute $4\pi$ for a in the previous line

$sinc(\frac{t \cdot 4\pi}{2\pi}) => \frac{2\pi}{4\pi} \cdot rect(\frac{-w}{4\pi})$

$sinc(2 \cdot t) => \frac{1}{2} \cdot rect(\frac{-w}{4\pi})$

To find 2sinc(st) from the original question, multiply both sides by two using the linearity property.

So, if the fourier transform of $s(t)$ is $S(w)$, using the symmetry property (duality):

$s(t) => S(w)$
$S(t) => 2\pi \cdot s(-w)$

We can get

$rect(t) => sinc(\frac{w}{2\pi})$
$sinc(\frac{t}{2\pi}) => 2\pi \cdot rect(-w)$

From the similarity property of the fourier transform, we get
$s(t \cdot a) => \frac{S(\frac{w}{a})}{|a|}$

so, scaling t in $sinc(\frac{t}{2\pi})$ by a gives

$sinc(\frac{t \cdot a}{2\pi}) => \frac{2\pi}{a} \cdot rect(\frac{-w}{a})$

so to find $sinc(2t)$, substitute $4\pi$ for a in the previous line

$sinc(\frac{t \cdot 4\pi}{2\pi}) => \frac{2\pi}{4\pi} \cdot rect(\frac{-w}{4\pi})$

$2 \cdot sinc(2 \cdot t) => rect(\frac{-w}{4\pi})$

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This works for me. Do you think this would suffice to answer the question posed above?

The other approach.. I'll split into a seperate post later on, got to get to work!