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Image Reconstruction:Phase vs. Magnitude 
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#1
Jun2314, 06:49 AM

P: 26

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.
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? 


#2
Jun2314, 07:34 AM

Mentor
P: 17,526

The magnitude and the phase contain different information from each other, and together they contain the same information as in the original complex image. So if the original image has information A,B,C, and D and the magnitude spectrum has A and D then B and C must be in the phase spectrum.



#3
Jun2714, 04:12 AM

P: 26

@Mentor sir but can u tell me in some detail what is happening in figure 1.(c) and 1.(d)actually?



#4
Jun2714, 09:06 AM

Sci Advisor
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PF Gold
P: 12,251

Image Reconstruction:Phase vs. Magnitude
If you want to understand what is happening in 1c, to account for the lack of any apparent image, what is happening is that it is the result of the sum of a whole set of spatial harmonics that are only 'in phase' at point (0,0) and produce a massive maximum there and nearby. (Their phases are all set to zero)
In 1d, you are starting with a whole lot of spatial harmonics with the same amplitude  producing a more or less uniform brightness over the image. But there are certain places on the original scene (edges) where the relative phases go to produce a sum with big discontinuities. Even when the amplitudes of the harmonics are kept the same, this still gives identifiable and abrupt changes in resultant amplitude in the same places as the edges in the original. It's interesting to note that the signal analysis done in the eye is very sensitive to phase distortion of a signal (the phase is what we are looking at because that tells you where the edges are) whereas the ear is much more sensitive to amplitude frequency distortion (you can hear speech, for instance when the audio signal has been subjected to all sorts of phase distortions in audio compression systems). 


#5
Jun2714, 09:07 AM

Mentor
P: 17,526

Image d can be understood as a filter. Since the phase is preserved and the magnitude is set to 1 this image is the same as the original image with a filter which is inversely proportional to the kspace magnitude. Since the kspace magnitude is high in the center and low on the edges, this amounts to a high pass filter. Visually, you see the high pass filtering also due to the preservation of edges and the loss of contrast.
I don't know a simple way to understand image c. EDIT: just noticed sophiecentaur's approach for understanding image c, which seems good to me. 


#6
Jun2714, 03:20 PM

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Whoops. Where did that post go? I was all ready to have a go at answering it.
I am not sure what was meant by low and high frequency pixels. A pixel is the same width over all the picture. I think you could use the term low and high frequency spatial variation. 


#7
Jun2814, 01:58 AM

P: 26

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) is image function and F(u,v) is its 2D Fourier transform
f(x,y)= ∑_(u=0)^(U1)∑(v=0)^(V1)\ F(u,v) exp^(1j*2Π*xu)/M) * e^(1j*2Π*(vy)/N) (1) 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. f(x,y)= ∑_(u=0)^(U1)∑(v=0)^(V1)\ exp(j*angle(u,v)) *exp^(1j*2Π*xu)/M) * e^(1j*2Π*(vy)/N)(2) i want to ask 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...... 


#8
Jun2814, 02:00 AM

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#9
Jun2814, 05:11 AM

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To account for the visibility of edges, you need to realise that, in the frequency domain, the high and low frequency components are not only there at the edges; they are everywhere (that's what Fourier is all about. The reason that you 'see' a step or impulse is that they happen to add in those places to produce a visible (but very slight) change in brightness. Over most of the picture, the frequency components add up to produce an 'average' brightness with no visible change  hence the midgrey appearance.
You need to take into account our subjective appreciation of a scene as well as the Maths involved. Imo, the reason that it works for our eyes is that our vision system is constantly searching for edges and outlines. Out in the wild, it's the best way to recognise food and threat. It's the outline of an elephant, against a grey wall that allows us to spot it and not the slight change in greyness. Likewise, we are good at spotting small amounts of rapid movement but we can ignore the variation in light levels as a cloud goes over the sun etc. Having written all this, I still have to agree with you that the results you showed are not what you'd expect intuitively. When I was shown the effect, years ago, I was just as confused as you have been! 


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