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# The Convolution Theorem and Diffraction

Module by: Paul Padley. E-mail the author

Summary: We show the convolution theorem and show how it can be used to solve complex diffraction problems.

To handle more complex cases of diffraction using Fourier transforms we need to know the convolution theorem. Say g ( x ) g ( x ) is the convolution of two other functions f f and h h . Then g ( x ) = f h = f ( x ) h ( x x ) x g ( x ) = f h = f ( x ) h ( x x ) x It is probably best to illustrate convolution with some examples. In each example, the blue line represents the function h ( x x ) h ( x x ) , the red line the function f ( x ) f ( x ) and the green line is the convolution. In the animation; follow the vertical green line that is the point where the convolution is being evaluated. Its value is the area under the product of the two curves at that point.

It might be easier to picture what is going on if we capture a couple of frames.

Here is a slightly more complicated example

Finally it is interesting to note what happens when we spread out a few functions, that is in this case, f f is a step function in a couple of places.

The convolution theorem states that if G ( k ) = Ϝ { g ( x ) } G ( k ) = Ϝ { g ( x ) } F ( k ) = Ϝ { f ( x ) } F ( k ) = Ϝ { f ( x ) } and H ( k ) = Ϝ { h ( x ) } H ( k ) = Ϝ { h ( x ) } and if g ( x ) = f h g ( x ) = f h then G ( k ) = F ( k ) H ( k ) . G ( k ) = F ( k ) H ( k ) . We can easily show this G ( k ) = g ( x ) e i k x x = e i k x f ( x ) h ( x x ) x x = h ( x x ) e i k x x f ( x ) x G ( k ) = g ( x ) e i k x x = e i k x f ( x ) h ( x x ) x x = h ( x x ) e i k x x f ( x ) x now set w = x x w = x x then x = w + x x = w + x G ( k ) = h ( w ) e i k w w e i k x x = h ( w ) e i k w w f ( x ) e i k x x = H ( k ) F ( k ) G ( k ) = h ( w ) e i k w w e i k x x = h ( w ) e i k w w f ( x ) e i k x x = H ( k ) F ( k )

Now say we want to consider the case of two long slits with width a a . This can be described by the convolution of one slit with two delta functions. Unfortunately it is not possible to animate this since the delta function is infinitely narrow. However an extremely narrow Gaussian is a good approximation to the Dirac delta function and I have used that for the animation below.

So two slits of a finite width can be described by the convolution of two delta functions and rectangular aperture function. Then the Fraunhofer diffraction pattern is just the product of the two Fourier transforms.

To sumarize: Fraunhofer diffraction patterns are the Fourier transform of the aperture function. The Fourier transform of the convolution of functions is the product of the Fourier transforms of the individual functions. each of our complex diffraction cases could be considered the convolution of simpler cases, hence the resulting patterns were the products of those simpler cases.

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