Power spectrum question -- seek verification

Nope. The vertical axis is in volts squared, because the frequency bin width is 1/T, where T is the total measurement time. So the vertical scaling in volts is the power spectral density (PSD) in V**2/Hz times

The horizontal axis is also 1/T per bin.

Cheers

Phil Hobbs

Considering the math, can you explain how this comes out of analysis of the units?

In a DFT is the sinewave the inputs are multiplied by unitless or the same units as the input? If unitless, the resultant values are just volts or vo lt seconds depending on how you regard the input units. The magnitudes wou ld then be volts^2 or volts^2*seconds^2 which can be related to power throu gh a constant impedance so watts or joule seconds.

I remember in college being able to come up with the units for virtually an y calculation. Sometimes it took some odd paths to simplify the units, but it always worked.

Parseval and Nyquist. The total power is the same in both domains, and the bin width in each domain is the reciprocal of the total measurement time in the other domain. (You have to leave the power spectrum two-sided for this to work.)

Cheers

Phil Hobbs

How does that relate to the units of the bins?

Although there are some versions of "power spectrum" display that actually display the amplitude of the signal at each frequency bin.

That is amplitude A determined from A.exp(i.theta) = x + iy

Rather than the true power spectrum A^2 = x^2 + y^2

Always worth checking which display your spectrum analyser is using.

Displaying A, theta plots is quite common in VLBI interferometry.

The amplitude is always going to be in volts. It's just the bandwidth/time duration that you have to worry about.

The energy is the same in the time and frequency domains. Let's say we take N samples, a[0]...a[N-1], in a time T. Parseval says

sum |a[i]|**2 = sum |A[i]|**2. i=0 to N-1 i = 1 to N-1

where {A} is the transform of {a}. An A/D converter implicitly divides its input by the reference voltage, so to convert to V**2, we need to multiply both sides by Vref**2.

It's natural to want the RHS to be normalized in terms of frequency. The frequency bin width is 1/T Hz, and the two-sided frequency range is N/T. Thus each frequency bin has the total energy in a band of 1/T Hz.

To convert the raw FFT power spectrum to PSD, besides multiplying by Vref**2, you have to multiply the frequency-domain samples by T, i.e. divide by the bin width in frequency.

PSD_2sided[i] = Vref**2 * T * |A[i]|**2, i = 0...N-1

Since the negative frequencies have the same power spectrum as the positive ones, we usually switch to the analytic signal convention, where

PSD_1sided[i] = { Vref**2 * T * |A[0]|**2 , i = 0 } { 2 Vref**2 * T * |A[i]|**2 , 0 < i < (N+1)//2 }

where // is integer division. What to do about the peculiar sample at N/2 for even N is left as an exercise for the reader. ;) (It shouldn't have anything much in it anyway, provided that you've done your antialiasing filter correctly.)

Scope and spectrum analyzer FFT power spectra are always quoted in single-sided terms.

Cheers

Phil Hobbs

Thanks for the detailed explanation. I was only referring to the raw DFT power spectrum. My simple C program does a somewhat brute force implementation of DFT, and the raw power spectrum, when plotted with gnuplot shows the peaks at the correct, expected frequencies(i.e.,fundamental, first harmonic etc.,) I will add the PSD feature soon.