I have recorded a micrograph of a 2-D array at a magnification of 43,000x on my DE-20 digital camera, which has a 6.4 μm pixel size and a frame size of 5120 × 3840 pixels. This magnification is correct at the position of the camera. I then compute the Fourier transform of the image. What is the spacing of the finest detail (highest resolution Fourier Coefficient) that I can hope to obtain in the computed transform, with respect to the actual particle itself in the specimen plane? What is the spacing between points in the computed Fourier transform (with respect to the original object, i.e. the crystal)?

I feel confused about the Fourier transform. So any help will be welcome. Thanks in advance!


1 Answer 1


The size of the image in frequency domain $f_{max}$ is inversely proportional to the grid spacing in real space $\Delta x$. (i.e. the finer step, the hight frequency you can sample).

And grid spacing step in frequency domain $\Delta f$ is inversely proportional to size of real space image $ x_{max}$ ( i.e. the longer interval of data you have, the more precisely you can discriminate between frequency components )

So mathematically:

$\Delta f = 2\pi / x_{max}$

$f_{max} = 2\pi / \Delta x$

Few practical notices:

But it may depend on your conventions. Sometimes you work with angular velocity (or wave vector) $ \omega = 2 \pi f$ instead of frequency. Or possibly you define your frequency domain on interval $ (-f_{max}/2; +f_{max}/2 )$ instead of $ ( 0, f_{max} )$. But this is easy to convert.

Also there is often a catch if you use typical FFT algorithm/library. Most of them shift the frequency domain periodically (i.e. roll ) by $-f_{max}/2$... so that you have low-frequency components at both ends of interval and the high frequency components in centre. Like e.g. here ( the high peaks are low frequency components. )


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