| Re: FFT and ROTATE [message #62326] |
Fri, 05 September 2008 14:04  |
wheinz
Messages: 2
Registered: September 2008
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Junior Member |
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Thanks for the suggestions. I am still confused by a few things, but
with Wox's code I think I can clarify my question. The problem is that
although Wox's code shows that the FFT of the rotated image and the
rotated FFT of the original image give you back images that look like
the original when inverse -transfored, the values of the coefficients
in those two FFTs are not the same. I added some print statements to
Wox's code to show what is confusing me.
pro rotFFTtest3
;;load an image
fn = filepath('md1107g8a.jpg',SUBDIRECTORY='examples/data')
image=bytarr(251,251)
image90=bytarr(251,251)
image[0,0]= read_image(fn)
image90[0,0] = rotate(read_image(fn),1)
;display the images
window,0
tvscl,image,0
tvscl,image90,1
n = size(image,/dim)
nfreq=n/2+1 ; # positive freq in each dim
nfreq_m=nfreq-1-(~(n mod 2)) ; # negative fequencies in each dim
;;take fft of image
f = fft(image)
;;shift it
f = shift(f,-nfreq[0],-nfreq[1])
;;rotate the fft 90 degrees
f90_1 = rotate(f,1)
;;shift it back
f90_1 = shift(f90_1,nfreq[0],nfreq[1])
;;take the fft of image90 -- the rotated image
f90_2 = fft(image90)
;;**********new code to print sorted values of coefficients**********
;;get the real parts of the ffts
fr =real_part(f)
f90_1r = real_part(f90_1)
f90_2r = real_part(f90_2)
;;now we look at the set of real coefficiencts of each fft
sf = fr[sort(fr)]
s1 = f90_1r[sort(f90_1r)]
s2 = f90_2r[sort(f90_2r)]
print,'The sorted list of real coefficients.'
print,' fft(image):',sf[0:4]
print,' rotated fft:',s1[0:4]
print,'fft(image90):',s2[0:4]
;;**********end new code to print sorted values of
coefficients**********
tvscl,fft(f90_1,/inverse),2
tvscl,fft(f90_2,/inverse),5
end
The program prints the following:
IDL> rotFFTtest2
The sorted list of real coefficients.
fft(image): -36.3499 -36.3499 -26.5806 -26.5806
-5.83106
rotated fft: -36.3499 -36.3499 -26.5806 -26.5806
-5.83106
fft(image90): -36.3499 -36.3499 -26.6964 -26.6964
-5.65933
So, you can see that the values differ, and (to answer ken's question)
the differences in the values are greater than the machine's
precision. Here I only printed the real parts, but the same thing
happens with the imaginary parts.
The fact that the inverse transformations return the original image
suggests that there is a phase shift introduced somewhere in the
transform. I can understand that if the array has an even number of
elements in x or y, then a rotation forces a translation of the
pixels. But an array with an odd number of elements in x and y should
not, especially if we are only looking at 90 degree rotations.
Why aren't the sorted lists of the real parts of the coefficients from
the rotated fft and the fft of the rotated image equal?
The article Vince linked to states that different implementations of
the FFT use different indexing and normalization approaches. Is it
possible that these differences could come from whatever scheme IDL
uses for its FFT?
Thanks,
Will
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