# Power Spectral Density computation and units

I want to make some calculs of power spectral densité of signal. For example a real voltage signal (physical unit : $V$) in time $g(t)$, its fourier transform $G(f)$ and $S_g(f)$. As far as I know, the power spectral density units for g(t) is $V^2/Hz$. However, I find in various sources the following equation to calculate the power spectral density : $S_g(f)=|G(f)|^2=G(f)G(-f)$

I do not understand this "definition" (of course the most fundamental definition of PSD is given by the Wienner-Kintchine theorem) , as it leads to a spectral density in $V^2/Hz^2$.

Regards, Mike

• Are you essentially asking how to convert a discrete time series of data to a power spectral density? Commented Jun 17, 2015 at 1:56
• I also started a very similar question, but was not aware of this one at the time. Here is the link to my question for reference: dsp.stackexchange.com/q/65963/38419 Commented Mar 31, 2020 at 18:34

I don't know if I get your question, but here is something you can look into.

Fourier transform is basically converting a time domain function to frequency domain. That being said, Fourier transform basic formula gives you the amplitudes of various components of a given signal in frequency domain. Note the word amplitude. That's what Fourier transform provides you, amplitudes. These amplitudes are functions of, of course, frequency. By exponential Fourier transform formula, we get these amplitudes as complex quantities. That might be your $g(t) \iff G(f)$, where $g(t)$ gives the amplitudes of the signal vs. time, and $G(f)$ gives the amplitudes of the different components of the signal vs. their individual frequencies.

Now you are interested in PSD. That's power. So now you don't want amplitudes vs. frequency, you want power vs. frequency, right? Power is proportional to the square of amplitude. So how do you convert amplitude to power? You square the amplitudes.

But, amplitudes occur as complex quantities in exponential formula, remember? So if you multiply one complex quantity with its conjugate, you get the square of its amplitude. That's why you might use

$$S_g(f)=\lvert G(f)\rvert^2=G(f)G(-f)$$

Here, $G(-f)$ is the conjugate of the complex quantity $G(f)$. At $f = 0$, you of course get a real quantity. In that case, you have,

$$G(f)G(-f)=G(f)G(f)=\lvert G(f)\rvert^2$$

Hope that helps.

Thanks for your answers. I perhas was not clear enough, the problem is not converting $Hz$ to $T^{-1}$ ... The physical dimension of the expression $|G(f)^2|$ is $V^2T^2$ (notice the square after $T$) because $G(f)$ is a fourier transform of a voltage time signal. However the power spectral density unit shoud be $V^2T$. For that reason, $|G(f)^2|$ cannot be a good formula for the computaion of $S_g(f)$ . However, I found various references in which $S_g(f)$ is defined by $|G(f)^2|$. So what's wrong with PSD estimation ?

• Please post comments as comments, not as answers. You can just add a comment to your original question or if you don't have enough rep to do that, just hit the "edit" button to add things to the original question itself. Commented Jun 17, 2015 at 1:55

The power spectral density (PSD) indicates the strength of the energy as a function of frequency. In other words, it shows at which frequencies energy variations are strong and at which frequencies they are weak. The unit of PSD is energy per bandwidth and you can obtain energy within a specific frequency range by integrating PSD within that frequency range. Computation of PSD is done directly by the FFT or computing autocorrelation function and then transforming it.

Here there is a detailed explanation of how it is calculated.

• "The power spectral density (PSD) indicates the strength of the energy [...]", not really: Power and energy are different. PSD is not ESD. "PSD is done directly by the FFT", DFT gives amplitude and phase, not PSD.
– mins
Commented May 5, 2023 at 14:07