Fourier transforms, convolution, and Fraunhofer diffraction

In summary, the conversation discusses the use of Fourier transforms in calculating diffraction in optics. This involves finding the Fourier transform of the convolution of simple shapes, such as circles and rectangles. The result is a pattern with a missing light in the middle and thinner bands on both sides. There is a close connection between optics and Fourier transforms, and the process involves doing a Fourier transform in each direction. The conversation also references relevant equations and explains how to calculate the Fourier transform by rows and columns.
  • #1
marnobingo665
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0
I've been exposed to this notion in multiple classes (namely math and physics) but can't find any details about how one would actually calculate something using this principle: Diffraction in optics is closely related to Fourier transforms and finding the Fraunhofer diffraction of an aperture composed of simple shapes is equivalent to finding the Fourier transform of the convolution of those shapes.

For example a professor told us that the FT of a circle is a 1st order Bessel function and that the FT of a rectangle is a sinc function. She demonstrated finding the Fraunhofer diffraction of the following aperture: (please see first reply)

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where white represents an opening and black represents and opaque region. She said it is equivalent to multiplying a circular aperture (f1) with a single opaque slit (f2) and that it would come out to be a thick band with a black dot (missing light) in the middle and thinner bands on both sides. But she never worked through how she used FTs to achieve that result: Where in this problem do you apply the concept of convolution, and what does it even mean to "take the FT of a rectangle"? How would I do this numerically instead of just sketching out intuitively what it looks like?
 
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  • #2
Image attached, couldn't figure out how to upload in original post!
 

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  • #3
Yes, this is true. Some relevant keywords: Fourier optics, spatial filtering, spatial Fourier Transform. There is a close connection between optics and Fourier transforms.
The 2-D Fourier Transform just involves doing a FT in each direction. For a digital FT, I think all you're doing is a FT along each row, and then a FT along each column.

Here's a page with the relevant equations (look for the formulas with the double summations)
http://fourier.eng.hmc.edu/e101/lectures/Image_Processing/node6.html

Note that ##\sum_{n=0}^{N-1}\sum_{m=0}^{M-1} f_{mn} e^{-j 2\pi(\frac {mk} {M} + \frac {nl} {N})} ## = ##\sum_{n=0}^{N-1} e^{-j 2\pi \frac {nl} {N}} \sum_{m=0}^{M-1} f_{mn} e^{-j 2\pi \frac {mk} {M}} ## which I believe allows you to calculate it by rows and columns in the way I described.
 

FAQ: Fourier transforms, convolution, and Fraunhofer diffraction

What is a Fourier transform?

A Fourier transform is a mathematical tool used to decompose a function into its constituent frequencies. It takes a function in the time or spatial domain and expresses it as a sum of sinusoidal waves in the frequency domain. This allows for a better understanding of the underlying frequencies present in a signal or image.

How is convolution used in signal processing?

Convolution is a mathematical operation that combines two functions to produce a third function. In signal processing, it is used to modify or filter a signal by convolving it with a kernel function. This allows for the extraction of specific information from a signal or the removal of unwanted noise.

What is Fraunhofer diffraction and how is it related to Fourier transforms?

Fraunhofer diffraction is a phenomenon that occurs when a wave passes through an aperture or around an obstacle and produces a diffraction pattern. The shape of the pattern is related to the Fourier transform of the aperture or obstacle. This means that by analyzing the diffraction pattern, we can obtain information about the Fourier transform of the object.

How are Fourier transforms and convolution related?

Fourier transforms and convolution are closely related as convolution in the time or spatial domain is equivalent to multiplication in the frequency domain. This means that by convolving two functions, we are essentially multiplying their Fourier transforms. This relationship is used in many applications, such as filtering and deconvolution.

Can Fourier transforms be applied to non-periodic signals?

Yes, Fourier transforms can be applied to both periodic and non-periodic signals. However, for non-periodic signals, the transform will result in a continuous spectrum rather than a discrete one. This means that the signal contains a continuous range of frequencies rather than specific frequencies as seen in periodic signals.

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