http://imagej.273.s1.nabble.com/FFT-filter-conversion-to-real-space-tp3694163p3694171.html
(?). The resulting image (+ swap quadrants) corresponds with the one
generated by FD Math as well as Bob's convolve 3D plugin. The latter
actually runs very fast and works on rectagular images. Thanks for
>> 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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assistent prof. P. Van Oostveldt
Dep. Moleculaire Biotechnologie
Fac. Toegepaste Landbouwwetenschappen, Ugent
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