# Fraunhofer diffraction and convolution of two aperture functions

1. Apr 21, 2013

### Blairo

Hello,

I'm having some trouble understanding the concept of two function convolution in Fraunhofer diffraction.

Let's say I have an aperture function in the shape of some cosine function (which is always above zero), and I want to calculate the transmission function if I only illuminate 3 such "slits" (so I capture 3 peaks of cosine aperture function). In order to do that, I was told to take the convolution of two Fourier transforms: the transmission function of cosine aperture illuminated over infinite slits and the transmission function of the Rect. function (some kind of box, which captures the 3 peaks of cosine function), which is a Sinc. function.

Is this correct? I don't quite get the physical meaning of convolution in this particular case.

Thanks for any explanations.

2. Apr 22, 2013

### Andy Resnick

If I understand you correctly, the aperture function could be written as something like cos^2(x)*rect(x/3). Assuming I kept track of the scale factors correctly, this means that 3 cycles of cos^2 (which is always positive) are illuminated.

Then, the far-field diffraction pattern is the FT of the aperture function, which would be FT(cos^2) convolved with FT(rect(x/3)), or:

(1/2 Sqrt[\[Pi]/2] DiracDelta[-2 + u] + Sqrt[\[Pi]/2] DiracDelta +
1/2 Sqrt[\[Pi]/2] DiracDelta[2 + u])**Sinc[3u],

where DiracDelta[a+u] is a delta function located at u = -a.