# How does the Fourier Transform invert units?

I don't really understand how units work under operations like derivation and integration. In particular, I am interested in understanding how the Fourier transform gives inverse units (i.e. time transforms into frequency), and if there are physical interpretations of the transform that involve units other than time.

If you integrate $$\int f(x)\textrm{d}x = F$$ then $F$ has the units of $f$ times the units of $x$.

Similarly if you differentiate, $$\frac{\textrm{d}f}{\textrm{d}x}$$ has units of $f$ divided by units of $x$.

If you look at the simple example of integrating and differentiating with respect to time to go between position, velocity, and acceleration, you'll see this works. For example, differentiate position (meters) with respect to time (seconds) to get velocity $(\frac{\textrm{m}}{\textrm{s}})$.

If you have a function of time and you Fourier-transform it, and then perform the inverse $$f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(\omega)e^{-i\omega t}\textrm{d}\omega$$ you should have the same units on both sides. $\omega$ has units of $\textrm{s}^{-1}$ (and so $\textrm{d}\omega$ has the same) and $e^{-i\omega t}$ is dimensionless. That means $F$ needs to have the same units as $f$, but multiplied by $\textrm{s}$. So the Fourier transform of a time series does not have units of frequency. It actually has units of inverse-frequency added on, so that when we multiply it by a frequency, we get the same units back that we started with. For example, if $f$ is measuring volts, $F$ has units of volts-seconds. Note that we are not transforming the units of $t$ and $\omega$ at all. We are choosing to express a function either in terms of time or in terms of frequency, then figuring out what units follow along.

Yes, Fourier transforms can be applied to data aside from time-series. For example, we can Fourier-transform a spatial pattern to express it in wavenumber-space, that is, we can express any function of space as a sum of plane waves. Physically, this Fourier transform is performed (for example) by a diffraction grating, which Fourier-transforms the spatial pattern of the grating.

• For the diffraction grating example, the model of a pure Fourier transform doesn't seem quite right, because then the units indeed don't agree: we get light in and light out. Would you agree? Maybe a discrete Fourier transform would somehow fit better, given the finiteness of a grating — then it's just a weighted summation, instead of integration. I don't have enough understanding of gratings... – Evgeni Sergeev Dec 26 '17 at 3:51
• @EvgeniSergeev No, it is not light in and light out. It is intensity in intensity-meter out. – Matt Jun 19 at 8:43

I think the Fourier transform transforms units just like an integral - it adds the unit of whatever it is you are integrating to the units of the function being integrated. If we are integrating $$G \textrm{d}x$$ then the units are the units of $$G$$ times the units of $$x$$.

For a Fourier transform, we are integrating over time so our function has units of whatever it had, times time. It is a function of frequency, but that doesn't mean the transform has units of frequency. Just as if $$G$$ was a position function as a function of time, it has units of position, not time.

The inverse Fourier is an integral over frequency. So we undo that unit multiplication and we are back to our original units.