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I've started working on PSD for seismic signals. In theory, PSD signal can be expressed in 2 ways. One in $(PSD=g^2/Hz)$ and other in $PSD=((meter/second^2)^2/Hz)$ and also ASD=(√PSD). Here $g$ is the acceleration.

More details have found in the link https://ucum.org/ucum.html#section-Special-Units-on-non-ratio-Scales

I want to find the amplitude of the spectrum in $(meter.second)^2/Hz$ from the expression of PSD. I don't clearly find a way to get the amplitude at a specific frequency in the unit of $(meter.second)^2/Hz$. How can I get the amplitude in the unit of $(meter.second)^2/Hz$? Probably the unit of amplitude may be derived from the unit of PSD, but I've no idea how to get the amplitude in the above unit?

OR

Without correlating the above PSD/ASD, what is the process to express amplitude (e.g. seismic signals) in terms of $(meter.second)^2/Hz$?

Answer (Confused)

So far I understand, one can find the amplitude from the $PSD$ in sunch a way that amplitude, $A = √PSD * scaling_factor$. $scaling_factor= 2/(fs*S)$ depends on the size of window and the sampling frequency of the signal (https://dsp.stackexchange.com/questions/32187/what-should-be-the-correct-scaling-for-psd-calculation-using-tt-fft/32205#32205). Am I correct?

$S$ - sum of squared samples of window function $f_s$ - sampling frequency Scaling coefficient $2$ takes into account removal of energy at negative frequencies (we drop this side of PSD).

Can anyone have any explanations?

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  • $\begingroup$ Minor comment: $PSD=ASD^2$, not $\sqrt{ASD}$. $\endgroup$
    – Andrew
    Commented Jul 4, 2022 at 14:33
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Jul 4, 2022 at 14:49
  • $\begingroup$ @Andrew, thanks for correcting me. $\endgroup$
    – Alan22
    Commented Jul 4, 2022 at 14:56

2 Answers 2

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So far I understand, one can find the amplitude from the $PSD$ in sunch a way that amplitude, $A = √PSD * scaling factor$. $scaling factor= 2/(fs*S)$ depends on the size of window and the sampling frequency of the signal (https://dsp.stackexchange.com/questions/32187/what-should-be-the-correct-scaling-for-psd-calculation-using-tt-fft/32205#32205). Am I correct? If I go wrong, I appreciate if anyone find me wrong.

$S$ - sum of squared samples of window function,

$f_s$ - sampling frequency

Scaling coefficient $2$ takes into account removal of energy at negative frequencies (we drop this side of PSD).

Now the unit of amplitude from PSD:

Unit of $PSD= ((meter/second^2)^2/Hz)$ or $ 1.m^2.s^{-3}$, therefore, the unit of amplitude could be $ASD= ((meter/second^2)/√Hz)$ or $ 1. m.s^{-3/2}$ which contradicts with the unit of amplitude $ A =(meter.second)^2/Hz$ provided in the question. In this case, this is amplitude spectral density $ASD$.

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  • $\begingroup$ The factor of $2$ depends on whether you are using a one sided or two sided power spectrum (ie, whether or not you include negative frequencies). Otherwise the analysis on the DSP stack exchange you link to is correct. $\endgroup$
    – Andrew
    Commented Jul 4, 2022 at 16:49
  • $\begingroup$ @Andrew, Thanks for your comment. Could you please clarify the unit of the $ASD= ((meter/second^2)/√Hz)$? $\endgroup$
    – Alan22
    Commented Jul 4, 2022 at 17:51
  • $\begingroup$ Yes. If the time series has units of $m/s^2$, then the ASD has units of $(m/s^2)/\sqrt{Hz}$, the PSD has units of $(m/s^2)^2/Hz$, and the Fourier transform has units of $(m/s^2)/Hz$. $\endgroup$
    – Andrew
    Commented Jul 4, 2022 at 18:46
  • $\begingroup$ Thank you very much. I've got the unit of "Amplitude " of Figure 3 in the link iopscience.iop.org/article/10.1088/1361-6382/ac348a/pdf is given $ Amplitude =((meter.second)^2/Hz) $. Once again, could you please clarify how do they find that unit or is there any typos? I cann't find the unit in that way. $\endgroup$
    – Alan22
    Commented Jul 5, 2022 at 7:55
  • $\begingroup$ Based on the units, that is a PSD. I think the authors are just being imprecise by using the word "amplitude" in the plot. $\endgroup$
    – Andrew
    Commented Jul 5, 2022 at 12:17
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This is what I figured out :

t : time (s)

a : acceleration (g)

L : number of time/acceleration samples

F : sampling frequency

  1. Amplitude (g) = 2 * abs(FFT(a)) / L

  2. PSD (g^2/Hz) = 2 * abs(FFT(a)) ** 2 / (F*L)

  3. ASD (g/Sqrt(Hz)) = sqrt(PSD)

  4. PSD or ASD in dB = 10*log(PSD or ASD)

Plot up to F/2.

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  • $\begingroup$ So the unit of amplitude is not g/Hz but g. Moreover, the zero and Nyquist frequency should not be doubled. $\endgroup$
    – Vassilis Papanikolaou
    Commented Jan 11, 2023 at 7:01

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