Prewitt filters are popular filters in image processing for edge detection http://en.wikipedia.org/wiki/Prewitt_operator

Can anyone give a proof on why Perwitt and other edge detecting filters are high pass?

  • 4
    $\begingroup$ Is your question "How do I prove that Prewitt is high-pass?" or "Why is it that all edge-detectors are high-pass filters? $\endgroup$ Dec 17, 2012 at 11:55
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    $\begingroup$ What about taking Fourier transform of the filter kernel? This will reveal which frequencies are preserved and which are attenuated. $\endgroup$
    – Libor
    Dec 18, 2012 at 1:45
  • $\begingroup$ I tried to take fft of the kernel (3x3). Here is what I get. How do I interpret that? 0 0 0 1.5000 - 0.8660i 1.5000 - 0.8660i 1.5000 - 0.8660i 1.5000 + 0.8660i 1.5000 + 0.8660i 1.5000 + 0.8660i $\endgroup$
    – mkuse
    Dec 18, 2012 at 11:20

2 Answers 2


That depends on the definition of high-pass filter. If you define a high-pass filter as a filter that has high response in the high frequencies in frequency domain, then the easiest way is to take a look at the magnitude of Fourier transform, (by definition).

Applying Fourier transform (in Matlab)

 A = fftshift(abs(fft2(padarray([-1 -1 -1; 0 0 0; 1 1 1],[10 10]))));

results in the following image:

enter image description here

Now the interpretation - The part in the middle is the low frequencies. It has low response. There are 2 high responses, both with zero X frequency and some high Y frequency. That is not surprising since we took a filter that detects edges in Y direction.

Why does it make sense to define high-pass filter in this way?

Because a convolution can be thought as multiplication in Frequency domain. That is, if you have a signal $S$ and filter kernel $f$,

$F[S**f] = F[S] * F[f] $

Or, put in another way:

$ S**f = F^{-1} [ F[S] * F[f] ] $

where convolution is denoted by $**$

There is another, more intuitive way, that does not involve Fourier transform. A response to a linear filter is strong when the underlying signal "looks" like the filter itself. Therefore, there should be a strong response to edges.


A simple explanation: How would you try to implement a differentiation in circuit design? The prewitt operator is simply a digitalization of discrete differentiation. If it was implemented in a circuital form it would attentuate low frequency signals and only allow high frequency signals to pass through.

Which is precisely the definition of a high-pass filter.

  • $\begingroup$ I think you meant "diminish", not "accentuate" low frequency signals. $\endgroup$
    – Jim Clay
    Dec 18, 2012 at 14:58
  • $\begingroup$ Thanks for checking. I meant 'attenuate' but ended up typing 'accentuate'. $\endgroup$
    – Naresh
    Dec 19, 2012 at 3:59

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