what is the PSD unit by using FFT method

Question

I'm just doing a power spectral density analysis of a signal in time domain. I'm following the fft method described in :

http://www.mathworks.com/support/tech-notes/1700/1702.html

It gives the real physical unit for the PSD. However, the unit is "power", is that mean "V^2/Hz"?

If I take 10*log10(power) or 10*log10(V^2/Hz), do I get the unit of "dB/Hz"?

Then how can I convert it to dBm/MHz?

Apologetics

This is actually a really difficult question, especially if you include a window before doing the FFT, which you probably should. Is your signal a transient or are you averaging multiple frames of a long signal? If the former, PSD probably doesn't make sense anyway. You may want to ask this on dsp.stackexchange.com – Volnay

I’m voting to close this question because it is not about programming – Ulibarri

Answer 1

It depends on the unit of your timeseries. Often we think of this as just "amplitude", but if your timeseries is a series of voltage amplitude vs. time, then your PSD estimate will be Volts^2/Hz. This is because the PSD is the Fourier Transform of the autocorrelation of your original signal: The autocorrelation has units of Volts^2, and running it through the Fourier Transform decomposes these units over frequency, instead of time, resulting in units of Volts^2/Hz. This is commonly referred to as Watts/Hz, but the conversion from Volts^2 to Watts is not very physically meaningful, as W = V^2/R.

10*log10(power) will result in a unit of dB/Hz, but remember that decibels are always a comparison between two power levels; you are quantifying a ratio of powers. A better definition of decibels is 10*log10(P1/P0), as explained here. If you simply plug a PSD bin estimate into this equation, you are setting your PSD bin to P1 and implicitly comparing it to a P0 value of 1. This may be what you want, and it may not be. For visualization purposes, this is fairly typical, but if you have a standard reference power you should be comparing to, you should use that for P0 instead.

Assuming that you are attempting to plot a dB Power Spectral Density estimate, to convert from Hz to MHz, you simple rescale the x-axis of your frequency graph. Remember that a MHz is just 1 million Hz, so the only difference is that 240000Hz = 0.24MHz

EDIT The point brought up by mtrw is a very valid one; if you are dealing with large amounts of data and are averaging FFT vectors, I highly suggest the Multitaper method; it's a much more statistically sound method of sacrificing frequency resolution for greater confidence on your PSD estimate.

Slicer

your answer is very informative and satisfactory. one question tough, regarding the "not very physical meaningful" transition between Volts^2 to Watts can be done by declaring a 1Ohm impedance or so I have been told. Is there any troth to that? what are the implications in the sense of the 10log10(P1) discussed? – Occupation

Yes, assuming you know the load that the voltage is applied across, you can then perform the conversion from Volts^2 to Watts. The units question is separate from the issue of a reference power level when using logarithmic units. As stated previously, oftentimes the reference level is "1", (1 Watt, 1 Volt, etc...) however in some cases it's not. As an example, in mobile wireless communications, it's often expected to use 1mW as the reference power level, since that power level is much more reasonable for a small mobile device than a full watt of power, and results in smaller dB numbers. – Slicer

Answer 2

If you have a PSD in W/Hz i.e. 100 W/Hz then you have 50 dBm/Hz. dB/Hz or is often vaguely and generically used instead of dBm/Hz. Audacity uses dB as shorthand for dBFS (not dBFS/Hz, because it is computing a DFT, and discrete frequencies use a power spectrum and not a density) . A digital signal that reaches 50% of the maximum level has an amplitude of −6 dBFS, which is 6 dB below full scale – the removal of the MSB, hence the 6dB/bit figure (because 50% of maximum level is 25% of maximum power; 1/4 = - 6dB)

dBm is the logarithmic ratio of the power with respect to 1mW, you divide the power by 1mW to get a unitless ratio, and then take the logarithm to get dB units, which in this case makes more sense to be clarified as dBm.

dBc/Hz is the ratio with respect to the carrier power, which is a ratio of two dBm/Hz values, meaning you subtract them and you get dBc/Hz; you get the same result if you divide the two linear power levels in W and then convert the ratio to dB (or more appropriately dBc).

dB-Hz is a logarithmic measure of bandwidth with respect to 1Hz and

dBJ is a measure of spectral density as a logarithmic ratio to 1 joule, seeing as W/Hz is indeed J.

Power spectral density is a density function, so you need to integrate it to get the actual quantity, like a line Integral of a V/m electric field, or a probability density of probability per x. This does not make sense for discrete quantities and instead the power spectrum is used akin to a probability mass function. If you see dB (which should be used for the discrete frequency domain) instead of dBm/Hz then it's wrong, but if you see it instead of dBm then it's right, as long as it's made clear what the reference is.

Cubical