# 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?

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.