# Units for a PSD from square of FT?

If I am looking to do a PSD plot on a signal $$x(t)$$ with units $$m$$ then the units I am expecting in the PSD would be $$\frac{m^2}{Hz}$$ but I can't seem to get this result from the following reasoning:

If we consider the fourier transform $$F(k) = \int^{\infty}_{-\infty} x(t) e^{-2\pi i k t} dt$$ I think this should have units $$ms$$ then the PSD should be $$|F(k)|^2.$$ Which should have units $$m^2s^2$$ or $$m^2Hz^{-2}$$ which is not $$\frac{m^2}{Hz}$$. There is probably an obvious mistake here but I can't see it, can anyone help?

## 1 Answer

Your reasoning sounds correct (with unit considerations), you can see a discussion on both continuous and discrete aspects of the Fourier transform in What are the units of my data after an FFT?

What is unclear to me is why you expect the PSD to be in unit $$\frac{m^2}{Hz}$$? Powers are powers: if a quantity has unit $$\textrm{u}$$, its square has unit $$\textrm{u}^2$$. However, when dealing with physics, the quantities you mention are often somehow periodic, and evaluated via averaging over one period, or asymptotically. So basically you had a normalization factor, proportional to the inverse of a time duration, or $$\textrm{s}^{-1}$$. Hence, possibly, your $$\textrm{Hz}$$ factor.

• I am trying to recreate someone elses results using the same method and these are the units they use. However I also see these units used often in other places, here for example – Stephen Jackson Apr 29 '19 at 7:35
• That's clearer to me now. I have added an explanation – Laurent Duval May 2 '19 at 21:33