# Image Reconstruction:Phase vs. Magnitude

Figure 1.(c) shows the Test image reconstructed from MAGNITUDE spectrum only. We can say that the intensity values of LOW frequency pixels are comparatively more than HIGH frequency pixels.

$$f(x,y)= ∑_(u=0)^(U-1)∑(v=0)^(V-1)\ |F(u,v)| *exp^(1j*2Π*xu)/M) * e^(1j*2Π*(vy)/N) /$$

Figure 1.(d) shows the Test image reconstructed from PHASE spectrum only. We can say that intensity values of HIGH frequency (edges,lines) pixels are comparatively more than LOW frequency pixels.

Why this magical contradiction of intensity change (or exchange) is present between Test image reconstructed from MAGNITUDE spectrum only and Test image reconstructed from PHASE spectrum only, which when combined together form the original Test image?

also i want to know that, in phase only reconstruction part, why do i get only edges or lines,why not low frequency components?? because from the 2nd equation i am not getting any idea that only edge like features are emphasized $$f(x,y)= ∑_(u=0)^(U-1)∑(v=0)^(V-1)\ exp(j*angle(u,v)) *exp^(1j*2Π*xu)/M) * e^(1j*2Π*(vy)/N) /$$

• Exactly the same question as dsp.stackexchange.com/questions/16995/… – Batman Jun 27 '14 at 20:17
• Sagar, from which paper have you copied the pictures for your question? I'm trying to find the source to cite it in my work. – Kate Jul 2 '15 at 9:23
• @Kate Why do you think that I copied the above pictures from somewhere ? I have not copied pictures from any source . I have obtained the above results myself. I have done MATLAB coding and if you want, I will send you my code. – sagar Jul 3 '15 at 15:07

One way to think about it is in terms of what you are discarding. The majority of the power in the image's spectrum is in the low frequency bins, right? So when you ignore the magnitude of the frequency samples by setting them all to, say, unity, you are attenuating those samples whose magnitude was higher than unity and amplifying those whose magnitudes were lower than that.

Specifically, if a frequency sample has the value $a e^{j\alpha}$, where $a,\alpha \in \mathbb{R}$ and $j=\sqrt{-1}$, then we ignore its magnitude by dividing it by $a$. This way every frequency sample in the image's spectrum has the same magnitude (i.e., $1$), and when we apply the inverse DFT we get the so-called phase-only image. However, this means that those frequency samples with the most energy (large values for $a$) are also attenuated the most. In other words, the more energy a frequency sample has, the more energy it is that gets ignored. Thus, since for typical optical images (e.g., the one you are looking at), where most the spectral energy is low frequency, the phase-only image is devoid of low frequency content.

As for the magnitude-only image, remember that location in the image domain is tightly coupled with phase in the frequency domain. When you discard the phase of an image's spectrum, one of the things you are discarding is the location of all that energy in the image domain. Thus, when you go back to the image domain from the magnitude-only spectrumn, you essentially have uninterpretable garbage.

• could you give any mathematical justification to your sentence "remember that location in the image domain is tightly coupled with phase in the frequency domain" ? – devraj Jun 24 '15 at 16:48
• any mathematical equation or simple example solved with mathematical formulas? – devraj Jun 24 '15 at 16:58
• @devraj See here. – AnonSubmitter85 Jun 24 '15 at 20:53

This problem relates much more to the properties of the human eye than the properties of the Fourier transform. The human eye measures local contrasts - local high pass filters, a global Fourier transform doesn't. It is true that any locality information in the Fourier transform is contained in the phase, but also true that every single complex exponential spreads over the whole image.

So we would need to look at linear combinations of the basis vectors - but that is exactly what we would like to avoid by using a transformation in the first place!

• sir,I didn't get your sentence "but also true that every single complex exponential spreads over the whole image." – sagar May 14 '15 at 8:28
• I'm not a sir. The dual basis functions of the fourier transform are the complex exponentials, which don't have finite support. So there is no "locality" information in any part of individual coefficients. – mathreadler May 14 '15 at 13:21