1
$\begingroup$

I need a little bit more explanation on technical details regarding Fourier Transform in Image Processing, mainly on the frequency aspect.

I see that we can use Discrete Fourier Transform to change an image into a signal. From spatial domain to frequency. But from this definition, suppose I want to manipulate, let's say, for example, overall intensity of an image, once my image is in frequency domain, how can I know which frequency is what in the spatial domain? How could I know which frequency belongs to what part of an image?

If there is anything seems incorrect, please correct me.

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
6
$\begingroup$

An answer to another question here provides a great explanation (with pictures) of what a single frequency looks like in the spatial domain.

Note that a single sample in the frequency domain affects all pixels in the spatial domain. That frequency produces a sinusoid with one frequency in the x dimension and another in the y dimension.

If you want to change the overall intensity of the image, you can do so by manipulating the DC value in the frequency domain. This is the value at zero frequency in x and y (upper left sample in the frequency domain). The DC value represents a wave with zero frequency in the spatial domain, so it is just a constant offset over the entire spatial image, which is basically the overall intensity.

If a single frequency affects the entire spatial image, how do you manipulate the frequency samples to do meaningful things in the spatial domain? Well, one way to think about it is that hard edges and quick intensity changes in the spatial domain correspond with higher frequencies while smooth changes in intensity in the spatial domain correspond with low frequencies.

So, if you want to emphasize edges in the image (sharpen), increase the magnitudes of higher frequencies. Similarly, if you want to smooth the edges (blur), decrease the magnitudes of higher frequencies.

Improve this answer
$\endgroup$
1
$\begingroup$

Spatial information is lost. You'd have to do the inverse transform to get it back.

You can increase the amplitude for a given frequency, but that effect would be spread out when transforming back into the spatial domain.

Improve this answer
$\endgroup$

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.