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I am confused with 2d magnitude plot of frequency spectra. So we have 2 images, the first one is shown at the top, and the dilated or enlarged version of the white box is shown at the second row.

enter image description here

For the dilated version, in the spatial domain when travelling in the $x$-direction, you see a fast change in intensity, and when you travel in the $y$ direction, the change is slower, so that the $v$ direction frequency component is smaller. In the frequency domain, $x$ maps to $u$, and $y$ maps to $v$.

I do understand that when you enlarge in one domain, you must do the opposite to the other domain. The thing I don't understand is that in the dilated version, when you travel in $x$, and you still see the same rate of change as in the original image. Since $u$ maps to $x$ in the frequency domain, we expect that the frequency spectrum looks the same in the $u$ direction. However, the result is that, in the $u$ (or the horizontal) direction of frequency domain, its frequency component shrinks; the $v$ direction remains intact. But we just change its $y$ direction!


Update: To be more precise, please refer to the area of the red box. In the second figure, it gets shrink.

enter image description here

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  • $\begingroup$ well, In the y-direction, consider is as a rectangular function with scale change. $\endgroup$ – Eric Zhou Oct 15 '14 at 1:25
  • $\begingroup$ Your question seems wrong to me. The scale in the $u$ axis seems identical to me in both FFT images. $\endgroup$ – Jim Clay Oct 15 '14 at 18:04
  • $\begingroup$ @JimClay, please check my update. $\endgroup$ – kuku Oct 15 '14 at 22:17
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    $\begingroup$ @kuku Again, the $u$ axis has not changed in the red box, it is the $v$ axis that has "shrunk", which is exactly what you would expect from expanding the rectangle in the y axis. $\endgroup$ – Jim Clay Oct 16 '14 at 3:16
  • $\begingroup$ How many number of samples in x any y? $\endgroup$ – Creator Jul 13 '15 at 3:21
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The scaling property of the Fourier transform tells you that if $\mathcal{F}(x(t)) = X(f)$, then $\mathcal{F}(x(at)) = \frac{1}{|a|}X(\frac{f}{a})$, with $a\neq 0$. So when you dilate in time, you contract in frequency, and vice-versa.

Now, for continuous 1D signals, the Fourier transform of a uniform box will be a cardinal sine. If you dilate the box, the cardinal sine will be contracted. This could be coarsely observed from the location of its zeros.

Here, you deal with a 2D discrete signal. So the result should be a discrete (or periodic) version of a cardinal sine. Its spread in one direction can be visually estimated by the horizontal and vertical black lines, since your 2D discrete Fourier transform is separable.

Now look at one horizontal line in your red boxes. The blackest dots that separate the horizontal gray/white bumps have the same location on both images. Now look at several lines. The black dots align vertically. So you see that the vertical lines are spaced in the same manner on the top and the bottom, which is exactly what you expect.

The result in the vertical direction is the controverse. Horizontal black lines are closer to each other in the bottom row.

What is sometimes misleading in Fourier plots is that some features in the frequency domain look orthogonal to those in the spatial domain. You can see more of this effect in An Intuitive Explanation of Fourier Theory.

If you want to gain additional conviction, crop a line from each spatial image, and an horizontal line from the spectrum, and superimpose the plots.

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The frequency component shrinks in the $v$ direction in your image. For better understanding refer to the following MATLAB code.

x=zeros(200,200);
x(80:120,100)=1;
X=fft2(x);
y=zeros(200,200);
y(20:180,100)=1;
Y=fft2(y);
figure(1);imshow(x);
figure(2);imshow(y);
figure(3);imshow(fftshift(log(abs(X)+1)),[]);
figure(4);imshow(fftshift(log(abs(Y)+1)),[]);
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You increased the length of the white box. Now when you traverse in x direction , the rate of change remains the same. But now you have more number of pixels where intensity changes from black(0) to white (255). This has resulted in more energy along that direction. check your image. The u component is more brighter in expanded image.

This energy can be represented using few harmonic coefficients. Check your image. You would see a denser grid along u direction.

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