Making the units of a analytically calculated PSD consistent with the units of an FFT One thing that has always puzzled me is making the units of an analytically derived Power Spectrum Density (PSD) consistent with the units of an FFT.
Let's say we record the output of a frequency generator, which we just set to output a signal of the form
$$v(t) = V_{\rm{pk}} \cos (2 \pi f_0 t)$$
for some finite amount of time, $T$, and take the FFT of this  acquired signal. We might then expect to see something like this:

Now as I understand it the FFT is a Linear Amplitude Spectrum (LAS) which will have units of Volts, or, a Power Spectrum (PS) which will have units of Volts-squared, depending on what we choose do with the resultant FFT. But it is clear the FFT can only have units of volts because that is what the units of the time-transient signal are.
Now lets say I want to fit this FFT with some realistic function. This is straight forward as I have the transient signal and I can calculate, analytically a PSD from a time truncated Fourier transform:
$$\frac{1}{T}\left|\int_{0}^{T} v(t) e^{-2 \pi j f t} \ {\rm{d}}t \right|^{2}$$
which will be in units of Volts-squared per Hertz ($\rm{V^{2}/Hz}$).
So its clear that the units of the FFT and the units of my function I want to fit to it don't match up! Multiplying by some quantity in units of $\rm{Hz}$ fixes the problem for example either the bandwidth of the FFT, $\Delta f = 1/ T$ or the Effective Noise Bandwidth (ENBW).
How can i reconcile the units of my analytic expression and the units of my FFT?

The objective is to simply fit my FFT data, so I given that all this should only really affect the amplitude, I could of course allow all this to be absorbed by an arbitrary constant -- but it would be nicer to have a consistent result.
You do this by doing two things:

First establish what the relationship of scale exists between your A/D converter and the numerical values that go into your DFT.  Find out how many volts corresponds to a value of $1.0$ in the samples that go into the DFT.

The next thing you gotta do is express your integral above without that $\frac{1}{T}$ factor (which is an error) as a Riemann summation having equal-width little rectangles of width equal to the sampling period.  That will become identical to the summation of the DFT.


That should spell it out.