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?
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:
$$ P_x = \lVert x(t) \rVert^2 $$
r=1the PSD is higher than the PSD atr=1.3[at some frequencies] while the RMS atr=1is less than RMS atr=1.3." $\endgroup$ – robert bristow-johnson Oct 4 '17 at 20:45r=1.3is higher that the PSD is at `r=1'. $\endgroup$ – robert bristow-johnson Oct 5 '17 at 0:27