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I have a pressure signal at two locations (r=1 and r=1.3). I found the power spectral density (PSD) and the root mean-square (RMS) of the signal. My question is why at r=1 the PSD is higher than the PSD at r=1.3 while the RMS at r=1 is less than RMS at r=1.3.

I mean is there any relation between PSD and RMS of the signal?

enter image description here

MBaz
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Math
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  • the premise to your question is not completely correct. " at r=1 the PSD is higher than the PSD at r=1.3 [at some frequencies] while the RMS at r=1 is less than RMS at r=1.3." – robert bristow-johnson Oct 04 '17 at 20:45
  • Thank you. That is correct. Just to be clear, the second plot represents the Root Mean Square of the pressure signal Pressure_rms. You mean the RMS at r=1 is less than at r=1.3 because the PSD at r=1 is higher that PSD atr=1.3 at some points. Is that correct? – Math Oct 04 '17 at 23:22
  • yes. at more frequencies the PSD at r=1.3 is higher that the PSD is at `r=1'. – robert bristow-johnson Oct 05 '17 at 00:27

1 Answers1

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instantaneous power is:

$$ p_x(t) = |x(t)|^2 $$

mean power is:

$$\begin{align} P_x &= \lim_{T \to \infty} \quad \frac{1}{T} \int\limits_{-\frac{T}{2}}^{+\frac{T}{2}} p_x(t) \, dt \\ \\ &= \lim_{T \to \infty} \quad \frac{1}{T} \int\limits_{-\frac{T}{2}}^{+\frac{T}{2}} |x(t)|^2 \, dt \\ \end{align}$$

the relation between Power Spectral Density to mean power is:

$$ P_x = \int\limits_{-\infty}^{\infty} S_x(f) \, df $$

the relation between Root Mean Square to mean power is:

$$ P_x = \lVert x(t) \rVert^2 $$

robert bristow-johnson
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  • No, $|x(t)|^2$ is instantaneous energy? $d/dt |x(t)|^2$ is power – OverLordGoldDragon Feb 24 '23 at 16:23
  • Sorry @OverLordGoldDragon , you missed the mark again. Just curious, are you an Electrical Engineer? – robert bristow-johnson Feb 24 '23 at 17:21
  • Graduated as one, never considered myself one. I do know a thing about circuits that'll blow your mind though. At any rate, what'd I miss? – OverLordGoldDragon Feb 24 '23 at 17:54
  • So, you have a real-valued signal represented as a voltage, $v(t)$, and you apply that voltage to a resistor $R$. What is this quantity: $$ \frac{v^2(t)}{R} = v(t) \times \frac{v(t)}{R} = v(t) \times i(t) $$ ? – robert bristow-johnson Feb 24 '23 at 22:13
  • While I'm sorting this out, what would be the instantaneous energy? – OverLordGoldDragon Feb 25 '23 at 06:05
  • hol'up, $\sum |x|^2$ is integrating in continuous-time... I might be a stupid. So then do we have a discrete definition of instantaneous energy, besides like $(\int |x|^2)[n]$ (that math's probably off but you get it)? also that example's good food for thought but not justification, e.g. $\text{signal} = v \cdot i$ – OverLordGoldDragon Feb 25 '23 at 06:42
  • I spent several hours to conclude it's indeed power. It bugged me that the implication is that energy is necessarily an aggregate quantity, yet we can obviously measure e.g. instantaneous kinetic energy. Wrestling with calculus and foundations of waves from restoring force, one finds that the energy of a differential element is indeed given by $|x(t)|^2$, but any finite realization requires aggregation, hence "instantaneous energy" is zero. And, $E(t)$, as derived from $\int P(t) dt$, is implicitly aggregating from some reference. – OverLordGoldDragon Feb 25 '23 at 15:42
  • and that, a signal in abstract and general definition obeys the same logic per standard continuum, which subsumes uniformity and ordinality, plus, built on Fourier which is sines, put that all together and fix my lazy English... it's power. -- now, to tackle the discrete case, if we insist on sample-by-sample, one could argue each sample is indeed energy, in that we aggregate over the sampling period with uniformity assumption. – OverLordGoldDragon Feb 25 '23 at 15:45
  • @OverLordGoldDragon, you might wanna check out this and related stuff. Note that, in the power spectrum, we calculate the aggregate of power over all frequencies and it's still power. That's because power spectrum would be "volts-squared-per-hertz". – robert bristow-johnson Feb 25 '23 at 17:00