Power Spectrum vs. Power Spectral Density Spectrum

[paul_harris77]

Does anyone know the difference between the power spectrum of a signal and the power spectral density (PSD) spectrum of a signal?

I've read on the net lots of things ranging from:
i) They are identical
ii) Power spectrum is units of Watts, power spectral density spectrum units of Watts/Hz, so not identical (conflicts with i))
iii) PSD spectrum is the magnitude spectrum of a signal squared
iv) Power spectrum is magnitude spectrum of a signal squared (conflicts with iii))
v) Power spectrum is for periodic time signals (which are discrete in frequency domain) and PSD spectrum is for non-periodic time signals (which are continuous in frequency domain)

Which of these is correct, and what are the fundatmental differences between the power spectrum and the PSD spectrum?

Thanks!

Paul

 

[rude man]

For signals of continuous frequency content like white noise, it's Watts/Hz. But that term is also used in other disciplines. E.g. in vibration analysis the units are g²/Hz.

For signals with spot frequency content only, the units are simply Watts.

 

[paul_harris77]

Thanks for the reply. Ok, that's fine, but now I have a question about obtaining the PSD of a signal that is continuous in the frequency domain. I have read it is simply the square of the magnitude of the spectrum.

If we assume that the spectrum was obtained with the Fourier Transform, surely the units of the spectrum are Volt seconds (Vs) since we integrate the voltage signal across time. Now if this is squared we get V^2s^2. If we assume a 1ohm load, then we can take out V^2 as the power into that load (J/s). Hence we get the units J/s * s^2 = Js. Joule seconds are equivalent to Joules per Hz, so it looks like the square of the magnitude of the spectrum is in fact an energy density spectrum (ESD), not a power density spectrum (PSD)? Surely they are not equivalent, in which case why do lots of websites say the magnitude squared is the PSD instead of the ESD?

Thanks!

Paul

 

[rude man]

OK this is how the game is played.

Take a time-continuous signal y(t). It has no Fourier transform since it doesn't meet the Dirichlet conditions (finite energy, for one). So you cut off all time except -T < t < T. Call this new function x(t). It does have a Fourier transform = X(f). The energy in this signal is ∫x(t)²dt = ∫X|(f)|²df, both integrated over -∞ to ∞ (Parseval theorem).

So the power P = averaged energy = lim T→∞ of (1/2T)∫|X(f)|²df integrated from -∞ to ∞.

Define the power spectrum as G(f) = lim T→∞ of (1/2T)|X(f)|². The power between two frequencies is now given as P = 2∫G(f)df integrated from f₁ to f₂. (The reason for the "2" is based on neglecting negative frequencies in the above derivation. I hate negative frequencies! )

In sum, G(f) is the power spectrum and the power spectral density. Don't be waylaid by the terminology, concentrate on the math.



In your case, x(t) = V(t) and as you state a 1 ohm resistor is assumed.