Hello
The image that you show could be that of a phase, but it is hard to judge from what you are showing. In any case, here is how this works. Take an image for instance the moon that Matlab has pre-installed.
image1=imread('moon.tif');
imagesc(image1);colormap(gray)
Now, calculate the Fourier transform, notice that we will need to shift it to have the low frequencies in the centre:
image1_fft = fftshift(fft2(image1));
The Fourier transform returns a series of complex numbers (a + b*i):
image1_fft(1,1)
With the complex numbers you can have the real part (0.5811) the imaginary part (1.7218), absolute value, or the phase, i.e. the angle. So, let's display them. First real and imaginary:
imagesc(real(image1_fft))
imagesc(imag(image1_fft))
Now the absolute value (i.e. the magnitude)
imagesc((abs(image1_fft)))
It would seem that the image is empty, but the issue is that the centre is very very large as compared with the rest, to compensate let's use the log
imagesc(log(1+abs(image1_fft)))
Finally, let's look at the phase, or the angle
imagesc((angle(image1_fft)))
And there you go, that is the phase of the moon image.
Hope this helps. If this answers your question, please accept the answer. If not, do let me know.
