# Fourier Transform of an 2D image and associated units

I have the following rather simple problem and unfortunately I am not getting forward.

Imagine a simple 2D image with pixels and a unique value for each pixel of the image. For example, let the image be 512x512 pixels. The size of the image is given by 10 nm x 10 nm. Since I want to view the image in frequency space, I calculate the Fourier transform of the image.

Of course, the image still has 512x512 pixels, but what about the units? I would say the physical units are now 1/m, but what happens to the 10 nm?

My understanding is that 1 pixel in local space would be 10/512 nm for this example and thus 1 pixel would be 1/(10nm*512) in frequency space. Is this correct?

You are close. In the same way that a time domain signal with units seconds becomes Hertz (i.e. 1/seconds) with the Fourier transform, a spatial domain signal with units meters becomes wave number (i.e. 1/meters). The concept of frequency does not change. In both units, you can think of it as cycles per unit length of time/space. Recall that the resolution in frequency space is dependent on the length of your signal in time/space (i.e. 10 nm) and not the number of points. So for the first non-DC DFT bin, you have units of 1 cycle / 10 nm (a wave number of 0.1 nm$$^{-1}$$). For the second bin, it would be 2 cycles / 10 nm (a wave number of 0.2 nm$$^{-1}$$). Because you have a resolution of 512 pixels along each dimension, you can measure up to the Nyquist frequency, or one half the sampling rate or 256 cycles / 10 nm (25.6 nm$$^{-1}$$).
• @user61315, If your DFT has a circle with radius of 20 pixels, then you have a frequency of 20 cycles per 10 nm, or a wave number of 2 nm$^{-1}$. For the independence of number of points, consider a DFT of a time signal with sample rate $fs$ and duration $T$. The frequency resolution will be $\Delta f=1/T$ and the Nyquist frequency is $fs/2$. Note that if you increased the sample rate but kept the same duration, the number of points in your measurement, $N=T*fs$, increases but $\Delta f$ does not change.