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I have been studying about Frequency Domain from many days but still i am not able to clear one of the confusion. We say that in frequency domain an image is represented as waves. What I am able to think is that if we take out an intensity profile column wise (i.e one row of intensity values), we have pixel location and at each pixel location we have some intensity value. So using these values with instensity values on y axis and location on x axis, we can draw a wave for one particular row. Now we find DFT coefficients. These DFT coefficients are multiplied with various waves of different frequencies and added. Ultimately we get the original image. Similarly we do it for 2nd row , 3rd row and so on. But when I saw the following image I got confused.

enter image description here

If we take its fourier transform and see its image in frequency domain we get 3 dots in the middle at frequency location 4. I know we get these dots at frequency location 4 because the image has frequency 4. Now the question is why do I get only one row of 3 dots? Why not each row contains 3 dots as I have discussed earlier? Please explain. Thank You.

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You only get one row of three dots because the image is constant in the vertical direction. That means that vertically speaking, the image only has 0 Hz content. Thus, you get the three dots in row 0 (the 0 Hz row), and nothing in any of the others. If you made one of the image's rows different than the others then you would see differences in the FFT's rows as well.

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  • $\begingroup$ Jim Thanks for your comments but yes along vertical direction the intensity is constant but if we consider row wise, the intensity values are changing like the first row. Then why not 3 dots in second row? Please explain. $\endgroup$ – Navdeep Oct 7 '14 at 16:58
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    $\begingroup$ The problem is that you think the FFT values in the second row correspond to the pixel values in the second row. They do not. All FFT values apply to ALL of the pixels in the image. They represent image "waves". There is no "waviness" (i.e. no change) in the vertical direction, so there is only 0 Hz content in the vertical direction. $\endgroup$ – Jim Clay Oct 7 '14 at 17:22
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    $\begingroup$ @Navdeep I highly recommend that you read and seek to understand the answer that nikie linked to above. When you understand it you will have the answer to your question. $\endgroup$ – Jim Clay Oct 7 '14 at 17:27

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