Explanation of the discrete fourier transform

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Hi all,

I'm a complete novice when it comes to describing images in frequency space and i understand that it is a way of representing images as being composed of a series of sinusoids. So a horizontal striped pattern with a single spatial frequency would have a magnitude image in frequency space with 3 non zero points, the origin the two mirrored points on either side at a distance from the centre depending on the spatial frequency. However in terms of what a Fourier transform actually does to each pixel in the image can anyone explain that. So you run each pixel through a mathematical formula can anyone explain the fast and discrete Fourier transform equations in non-mathematical terms? I haven't really been able to find this online. If you were trying to explain a Fourier transform to someone who knew nothing about imaging or optics even to say the image is decomposed into a series of sinusoids could be a bit baffling..

Thanks for your help,

Matt
 
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Hi Andy,

Yes thanks for those all contain very good non-mathematical descriptions. I was trying to interpret the function though so that i could definitively say what calculation is performed on each pixel in the spatial image to yield the Fourier pixel.

F(u,v) = SUM{ f(x,y)*exp(-j*2*pi*(u*x+v*y)/N) }

One of the sites says the function can be interpreted as "the value of each point F(k,l) or pixel in the Fourier image is obtained by multiplying the spatial image with the corresponding base function and summing the result"

But what is the corresponding base function exactly?

Thanks
 
I was just thinking if I was going to take an image and use the function to calculate the Fourier component at each pixel location what numbers would i be plugging into the function. I'm sure that's something you wouldn't do as you can use software to calculate it but its just for my own understanding of what each term in the function means. I'm happy with the qualitative explanations and i can't imagine people will question me about the function itself.

Thanks for the response.