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# Lode's Computer Graphics Tutorial Fourier Transform

Lode's Computer Graphics Tutorial Fourier Transform

## An image can be blurred

An image can be blurred by, for each pixel, calculating the weighed average of that pixel and it'sneighbours, using a convolution matrix: that's the way Photoshop and Paint Shop Pro do it, forexample, you can create such a convolution matrix with the "User Defined" filter in these programs.But, since the convolution in the spatial domain is the same as a multiplication in the frequencydomain, we can also blur an image, by multiplying the spectrum with the transfer function of a lowpass filter. The transfer function of a low pass filter makes all high frequency components zero or atleast weakens them, while keeping or amplifying the low frequency ones.For very blurry blurs, a very large convolution matrix is required, and this would require a lot ofcalculations. Taking the Fast Fourier Transform, then multiplying the spectrum, and then taking theInverse FFT, goes much faster for heavy blurs. You can also create much nicers blurs easier and moreintuitively by using the spectrum.The more high frequencies you remove of the spectrum, the blurrier the image, and the morecalculations would be needed in the spatial domain:

In the above images, a circular section of the spectrum is left. You can also, for example, keep a squareone, but round ones should give nicer blurs:High Pass FiltersA high pass filter on an image keeps only the high frequency components. Since, by applying a highpass filter, the DC component of the spectrum is removed too, the resulting image can have negativepixel values that can't get drawn. To overcome this, the high pass filter of Photoshop adds 128 (half ofthe maximum value 255) to each color component the resulting pixels, so grey represents zero. Here,we'll look at the positive part of the image, the amplitude (revealing the negative values), and the imagewith the DC component added again instead.A high pass filter is exactly the opposite of the low pass filters, so now a circular section in the centerof the spectrum is removed. On the left is the resulting positive real part of the image, in the center thespectrum with the circle removed, and on the right the amplitude of the image:To see much better what's going on, add the DC component to it again:

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