# Spatial Frequencies of the Fourier Transform

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1. Jul 16, 2015

### ecastro

The Fourier Transform transforms a function of space into a function of frequency. Considering a function $f\left(x, y\right)$, the Fourier Transform of such a function is $\mathcal{F}\left\{f\left(x, y\right)\right\} = F\left(p, q\right)$, where $p$ and $q$ are the spatial frequencies.

In numerical simulations, the function $f$ can easily be transformed by using an algorithm (built-in in the software). However, I am concerned on acquiring the spatial frequencies $p$ and $q$. Is there a way to do it?

2. Jul 16, 2015

### blue_leaf77

If the function in space domain is sampled $N$ times with sampling interval $\Delta x$, then the frequency resolution in the frequency domain is given by $2\pi/N\Delta x$.

3. Jul 16, 2015

### ecastro

I don't know if I got this right. Is the value of $N$ the number of points, $\Delta x$ is the interval between two adjacent points, and the interval between two points in the frequency domain is $\frac{2 \pi}{N \Delta x}$?

4. Jul 16, 2015

### blue_leaf77

Yes, exactly.

5. Jul 16, 2015

### ecastro

Thank you.

I have another question, how will I know the boundaries (maximum and minimum values) of the frequency domain?

6. Jul 16, 2015

### Staff: Mentor

7. Jul 16, 2015

I'll also throw in that a "spatial frequency" is more commonly called a "wavenumber".

8. Jul 19, 2015

### ecastro

The wavenumber of what? What is $\lambda$ in this case?

9. Jul 19, 2015

### olivermsun

The wavenumber of "whatever the spatial frequency belongs to."

The usual way of defining the wavenumber is exactly analogous that for frequency:
$k = 2\pi/\lambda$ in radians per length, or $\nu = 1/\lambda$ if the units are cycles per length.