Posted by
Gluender-2 on
Dec 26, 2008; 5:30pm
URL: http://imagej.273.s1.nabble.com/FFT-filter-conversion-to-real-space-tp3694163p3694170.html
>Thank you very much for your well documented response.
>So in summary, squaring the canvas (e.g. black fill) and using FD Math
>correlate on the original images should return relatively reliable data.
That's what I would expect.
The color of the "fill" depends on the kind of your images.
>I know the polar coordinates are just a representation, but how does one
>convert an image (for instance after multiplying these two FFT
>images=correlation) back to such representation in order to be able to
>perform the inverse Fourier transform?
OK, first of all here is what concerns the spectral "multiplication":
3) What does remark 2) mean for spectral functions?
b(r) conv a(r)
| |
FT FT
| |
B(f_r) * A(f_r)
[ b(r) corr a(r) ] = [ b(r) conv aÝ(-r) ]
| | | |
FT FT FT FT
| | | |
[ B(f_r) * AÝ(f_r) ] = [ B(f_r) * AÝ(f_r) ]
* stands for multiplication and
Ý stands for "complex conjugate"
Because the Fourier-transforms of your images are
most likely complex-valued, you must take care of
the the "complex conjugate" AÝ(f_r)!
A(f_r) = Re{ A(f_r) } + i * Im{ A(f_r) };
AÝ(f_r) = Re{ A(f_r) } - i * Im{ A(f_r) };
4) How to multiply complex numbers cz = cx * cy ?
cx = rx + i ix;
cy = ry + i iy;
cz = cx * cy = (rx * ry - ix * iy) + i(rx * iy + ry * ix) = rz + i iz;
With this in mind, you have do perform the
Fourier-transforms of your images that result in
2 spectral components each (real and imaginary
parts). Then you do the complex multiplication
taking into account the "complex conjugate" of
the spectrum of one of your images. Finally you
do a Fourier-retransform of the resulting complex
"product"-spectrum which should result in a
real-valued image (but it wont!). That's all.
I just checked this procedure with FFTJ and it works perfectly!
>I checked out FFTJ as well, but it indeed is far slower, just what I wanted
>to circumvent.
>Best regards,
>Winnok
HTH further
Herbie
------------------------
<
http://www.gluender.de>
> >Dear All,
>>
>>I would like to take the opportunity to ask
>>another question regarding Fourier operations. I
>>would like to perform spatiotemporal image
>>correlation in the Fourier domain because of
>>speed and memory considerations.
>>If I understand correctly, convolution of two
>>images (i.e. correlation) in the spatial domain,
>>equals mere multiplication of the Fourier
>>transformed images in the frequency domain.
>
>Well, not perfectly but nearly...
>
>Convolution and correlation is not the same but they are quite similar.
>You must take care of the differences:
>
>1) Convolution is a commutative operation, correlation is not.
> k(rho) = [b(r) corr a(r)] != [a(r) corr b(r)] = kÝ(-rho)
>
> Ý stands for "complex conjugate" and can be ignored
> when dealing with images (that are real-valued signals).
>
>2) The relations between convolution and corelation are
> k(rho) = b(r) corr a(r) = b(r) conv aÝ(-r)
> f(rho) = b(r) conv a(r) = b(r) corr aÝ(-r)
>
>>However, when multiplying two FFT images, the
>>resulting images is no longer expressed in polar
>>coordinates and therefore cannot be transformed
>>back by inverse FFT.
>
>If you Fourier-transform an image b(r) = b(x,y),
>then it results in a (generally complex-valued)
>Fourier-spectrum in Cartesian coordinates B(f_r)
>= B(f_x, f_y).
>
>Of course, you may represent whatever
>2-dimensional function in polar coordinates but
>this has nothing to do with the
>Fourier-transformation.
>
>As far as I remember and for whatever reasons --
>though displaying Fourier-spectra in Cartesian
>coordinates -- IJ _numerically_ indicates the
>spectral coordinates as polar coordinates. Please
>correct me if I'm wrong or if this behavior has
>been changed.
>(Personally and for several reasons, I don't use
>IJ's Fourier-transformation. Instead I use the
>slower IJ-PlugIn FFTJ.)
>
>>I tried FD Math correlate, but this operation
> >requires two square images of fixed dimensions
>>and I'm not sure if enlarging the canvas size
>>goes without consequences for the correlation
>>outcome.
>
>You may add appropriate surrounding areas of
>preferably constant gray-value to your images.
>(Avoid any gray-level steps when doing so.)
>
>FD-Math in fact is based on the Fourier-approach
>and therefore requires square sized image with
>side-length being a power of 2. This is due to
>the implemented algorithm. Digital
>Fourier-transformation of arbitrarily sized
>images is much slower (the IJ-PlugIn FFTJ is able
>to do just that).
>
>>I have noticed that the latter operation appears
>>to give a similar resulting image whether one
>>uses the original images as input or the Fourier
>>transformed.
>>Could anybody please share on how to do this the right way?
>>Many thanks in advance and best wishes,
>>
>>Winnok
>>
>>ps: the archives were unaccesible, so apologies if this is a duplicate
>query.
>
>HTH
>--
>
> Herbie
>
> ------------------------
> <
http://www.gluender.de>
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