# Relation between Power Spectral Density and RMS of the signal

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 siganl. 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? • 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 4 '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 4 '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 5 '17 at 0:27

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:

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