# 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 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?

• 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." Oct 4, 2017 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, 2017 at 23:22
• yes. at more frequencies the PSD at r=1.3 is higher that the PSD is at `r=1'. Oct 5, 2017 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:

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

• No, $|x(t)|^2$ is instantaneous energy? $d/dt |x(t)|^2$ is power Feb 24, 2023 at 16:23
• Sorry @OverLordGoldDragon , you missed the mark again. Just curious, are you an Electrical Engineer? Feb 24, 2023 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? Feb 24, 2023 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: $$\frac{v^2(t)}{R} = v(t) \times \frac{v(t)}{R} = v(t) \times i(t)$$ ? Feb 24, 2023 at 22:13
• While I'm sorting this out, what would be the instantaneous energy? Feb 25, 2023 at 6:05