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Re: 2D FFT of HCP lattice -- Inconsistency with line plot?

Posted by Michael Schmid on Aug 16, 2008; 10:53am
URL: http://imagej.273.s1.nabble.com/2D-FFT-of-HCP-lattice-Inconsistency-with-line-plot-tp3695379p3695383.html

Hi Mason,

this is a general property of 2D (and 3D, etc) Fourier
transforms, not one of the algorithm used or of the special
test images:
In 2D, FFT gives the frequency corresponding to the distance
between *rows* of objects (measured perpendicular to the rows),
in 3D it gives the frequency corresponding to the distance
between *planes*.
(by the way, this is also valid for the the diffraction spots
in TEM images).

See, e.g. http://en.wikipedia.org/wiki/Reciprocal_lattice

For your application, the autocorrelation might be better -
use Process > FFT > FT Math with "Correlate" (both input images
should be the same) and "do inverse transform".

This will give you a real-space image.
The ring around the center will have a radius equal to the
typical distance of the objects.


Michael
________________________________________________________________

On 15 Aug 2008, at 18:43, Mason Guffey wrote:

> Hi Michael,
>
> Thanks so much for your help! I have to admit, I'm still a little
> confused on this, can you provide me with a reference for this? Or  
> would
> any solid state physics text have an explanation?
>
> Is this a general property of a Fourier transform (i.e in a  
> mathematical
> sense) or something particular to the 2D FFT algorithm? What I'm using
> this for is to determine the average interparticle separation for a  
> TEM
> image of a nanoparticle array. The array is a 2d close-packed (HCP)
> lattice of spheres. Does what you say hold true for arrays of spheres
> rather than the triangle example?
>
> Also, what if I wanted to use the FFT to determine the average
> interparticle separation of a disordered complex of spherical  
> particles
> (in this case the FFT yields rings rather than peaks). Would the same
> constant factor apply?
>
> I realize that this is an ImageJ forum rather than an FFT forum,  
> but I'm
> having a hard time finding any discussion of this and I think this may
> be a problem that has tripped up other people working with  
> nanoparticle
> superlattices. So your comments are very helpful.
>
> thanks again,
> Mason
>
> Michael Schmid wrote:
> > Hi Mason,
> >
> > a 2D FFT of a non-rectangular lattice gives the distance between
> > parallel *lines*, which is different from the lattice constant
> > measured along a lattice direction.
> > In physics, this is known as reciprocal lattice.
> > Also note that the reciprocal lattice vectors are perpendicular to
> > the lines of the original lattice, and their length corresponds to
> > the distance measured in that direction.
> >
> > So you get the side length of the triangles multiplied by sqrt(3)/2.
> >
> > Michael
> > ____________________________________________________________________
> >
> > not that reciprocal lattic;
> >
> > On Thu, 14 Aug 2008 15:46:05 -0500 Mason Guffey <[hidden email]>
> > wrote:
> >
> >> Hi all,
> >>
> >> I've been having a problem with ImageJ related to 2-d fourier  
> transform
> >> of ordered images. Essentially, if I take an FFT of any hexagonally
> >> close-packed lattice (e.g.
> >>
> http://content.answers.com/main/content/wp/en-commons/thumb/c/ 
> c0/180px-Tile_3,6.svg.png
> >> ) and compare that with a line plot of the objects I get  
> inconsistent
> >> results.
> >>
> >> Specifically, take a line plot down any lattice direction of  
> that image.
> >> Divide the length of the line plot by the number of triangles it  
> passes
> >> through (i.e. the total number of lattice spacings). Compare  
> that with
> >> what you get from the FFT. The line plot yields a lattice  
> constant of
> >> about 14.5 pixels per cycle while the fourier transform peaks  
> are all
> >> right around 12 pixels / cycle.
> >>
> >> Does anyone know what's going on here? This same phenomena has  
> repeated
> >> for just about any image of hexagonal close-packed circles that  
> I load
> >> into it.... The line plot method reveals X for a lattice  
> constant, while
> >> the FFT peak is at Y. There's a factor of about 1.2 between them.
> >>
> >> However, if you do the same thing with an array of straight  
> lines, the
> >> FFT and the line profile match eachother perfectly. (As is shown  
> on the
> >> imageJ website for the FFT demo).
> >>
> >> Thanks in advance
> >>
> >> --
> >> Mason Guffey | Gordon Center for Integrative Science ESB09B
> >> Scherer Group | [hidden email]
> >> Department of Chemistry | (773) 834-1877
> >> The University of Chicago | http://schererlab.uchicago.edu
> >>
>