http://imagej.273.s1.nabble.com/Fourier-transform-and-filtering-tp3694627p3694630.html
150*160 pixels -> max=160; 160*1.5=240. Will be expanded to 256*256
180*160 pixels -> max=180; 180*1.5=270. Will be expanded to 512*512
> Michael Schmid-3 wrote:
>>
>> First, it is very important to note that the values (in pixels) in
>> "Bandpass filter" are *half* of the spatial frequencies that you can
>> read from the status line when moving the cursor over the FFT image.
>> That's because "structure size" means size of, e.g., a bright
>> particle; and the lowest strong Fourier component is that where the
>> upper half-cycle of the sinewave corresponds to the particle, the
>> negative "surrounding" correspond to the background.
>>
>> Just in case this is not clear: The values in pixels in "bandpass
>> filters" are in real space, i.e., pixels of the image. The scale of
>> the custom filter is in frequency space. 1 pixel from the origin
>> (center) corresponds to one period per image size, 2 pixels from the
>> origin corresponds to two cycles per image size, etc. For images that
>> are not 2^n*2^n, "image size" here means the lowest value of 2^n that
>> 'is not smaller than width and height.
>>
>
> Thanks for replying.
>
> Concerning the right frequencies to select, I dealt directly with the
> problem doing the following: before taking the pictures of
> interest, I shot
> one of an object of known size, so I know the actual ratio pixel/
> cm. Then,
> the smallest wave vector allowed in my system being q_min =
> 2*3,14/(image_size_cm), each pixel of the transformed image (or of the
> filter) will correspond to a wave vector of q = q_min*
> (pixel_from_center)*R.
> R has to be equal to (image_size)/(transformed_size), so on the
> borders of
> the transformed image I can retrieve half the maximum wave vector
> allowed,
> since to span the whole spectrum both -q and +q are needed.
> With this, I can arrange a custom filter at the right frequency, or
> check
> the wave vectors selected by a given bandpass filter. So, if I'm
> not doing a
> mistake, this way of using a filter should yield the right frequency.
> I tested this calibration by transforming via FFT a dummy circle, then
> obtaining an Airy frequency pattern and eventually checking the
> position of
> the first dark ring (it occurs at pixels which 'carry' a wave
> vector value
> in accord with theory).
>
>
>
>> If you run "FFT" or "Custom Filter" you will get edge artifacts,
>> usually visible in the power spectrum as horizontal and vertical
>> lines through the origin (center). The bandpass filter in ImageJ
>> continues the image outside the image area with its mirrored
>> counterpart to avoid any jumps at the edges, greatly reducing such
>> artifacts.
>> When selecting "Display Filter" of the bandpass filter, with image
>> sizes of 2^n or somewhat smaller, you will also see that the filter
>> size (equal to the size of the input image) is 2*2^n, due to the
>> space needed for the mirrored-image continuation. In this case, you
>> have to downscale the filter created by "Display Filter" by a factor
>> of 2 before using it as a custom filter.
>>
>
> I started using bandpass filtering, then I switched to custom
> filtering for
> at least two reason: one of my purposes is to effectively remove the
> 0-frequency component, and I found the filters provided rather too
> spread
> (the cutoff is too soft, as you said).
> I was unaware of the bandpass 'size duplication', but it is still a
> little
> unclear to me.
> Do you mean that the bandpass filtering actually doubles the allowed
> frequency range, so to reduce the edge noise, or that the real
> image is
> padded so to place the 'true' pixels in a range of spatial
> structures that
> are better reconstructed?
> Still, if it's a matter of avoiding artifacts, or rescaling by 2 the
> bandpass filter, I'm not so sure it can fully account for the
> inconsistency
> I see (as I simply tried to make manually the same frequency
> selection that
> 'custom filter' should be performing *as far as I know*); anyway your
> explanation has been useful, I'm going to do a bit of work on bandpass
> filtering as well.
>
> Antonio
>
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