From this Wikipedia link, the fastest fequency for digital frequency is 0.5cycle/sample. It can be rewritten as 1cycle/2sample.

  • What does it mean?
  • Does it mean that at least two pixles are required to form one cycle?
  • Is there any requirement on the values of two samples?
  • One must be 0 and the other must be 255? If so, why?

Edit: The reason I have such question is that different cutoff frequencies result in different blurring degree for filtering in frequency domain.

From the link:http://www.imagemagick.org/Usage/fourier/#fft_partial enter image description here The figure shows the comparison. I hope to estimate blurring degree from cutoff frequency in digital frequency. Maybe I can estimate corresponding standard deviation of Gaussian filter from cutoff frequency.

  • $\begingroup$ no, there's no such requirement $\endgroup$ Nov 15, 2016 at 6:41
  • $\begingroup$ Does the values for the two pixels need be different? I hope to relate digital frequency to image content. If the cutoff frequency of one filter is 0.2, 5 pixels will need to form one cycle. Can I say that the pixel is affected by its neighbors within 4 pixel distance? $\endgroup$ Nov 15, 2016 at 8:11
  • $\begingroup$ By digital frequency, they are referring to spatial frequency. Since it's used for digital sensors instead of film they mention as digital frequency.. $\endgroup$ Nov 15, 2016 at 8:22

1 Answer 1


Highest digital spatial frequency is 0.5 cycles/sample. It's related to Nyquist criterion.

Does it mean that at least two pixels are required to form one cycle?

Yes. Minimum two pixels are required to form the cycle.Try to fit a one cycle of sine wave in minimum number of pixels. You will find that two minimum pixels are needed to fit the sine wave without the loss. Anything less than two pixel you can see the loss.Finally if you fit a full sine wave in single pixel,the pixel is average of whole cycle resulting in total loss.

Is there any requirement on the values of two samples?

No, there are no requirements like that.

For the edit

The blurred image after Gaussian smoothing is given by $g(x,y) = f(x,y)*h(x,y)$ ,where f(x,y) is the input image, h(x,y) is the gaussian smoothing.

In frequency domain G(u,v) = F(u,v)H(u,v)

H(u,v) = G(u,v)/F(u,v)

The Gaussian blurring function h(x,y) = ifft(H(u,v)) .

Using only the cutoff frequency from the FFT of the blurred image g(x,y) you will get the good approximate of your gaussian filter but not accurate one. The reason behind this is the input itself might not contain high frequencies but from the FFT you might think of it caused by the Gaussian smoothing. But for most natural scenarios you can get good approximate of your Gaussian smoothing from cut-off frequency itself.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.