Does the RMS value of a signal yields its root-power?

I'm relatively new to the field of signal processing and might be a bit confused about terminology here. I'm trying to understand what determines whether I deal with a root-power or simple power quantity and how that relates to the RMS value of the signal.

I understand that the RMS of a signal is proportional to the signal's power. At the same time, this answer seems to heavily imply that RMS is equal to the root mean power (take sqrt on both sides of last equation). However, when looking at the wikipedia article on the calculation of decibels, they make it sound like whether I'm dealing with a power or root power is a property of the underlying signal, not the RMS value at all:

"As sound pressure is a root-power quantity, the appropriate version of the unit definition is used"

This leaves me confused. How do I determine whether I'm dealing with a power or root-power quantity, and what is their relation to RMS?

Given that RMS means Root Mean Square the answer is rather obvious: it's a root power quantity. The RMS of a signal has the same units that the signal itself.

Saying that "RMS is proportional to power" is just sloppy phrasing. Correct would be "the square of the RMS is proportional to power".

If you by "power" you mean "actual physical power in Watts" things get a lot more complicated. The RMS value of a signal may or may not be proportional or even just related to the actual physical power.

EDIT PER COMMENT

Signal power is NOT the same as physical power. Let's consider a voltage signal of $$2V_{RMS}$$. The signal power would simply be $$4V^2$$. It's already obvious from the units that this not a physical power, which would be measured in Watts.

The physical power really depends on what the voltage signal is connected to. Physical power is always based on the product of two physical root power quantities which are related by some sort of an impedance. In this case it's voltage and current.

If the signal is applied to 1 Ohm resistor than the current will be $$2A_{RMS}$$ in phase with the voltage and the power will indeed be $$4W$$. If it's connected to a 2 Ohm resistor, the current will be $$1A$$ and the power will be only $$2W$$. If the signal is connected to nothing, the current is 0 and so is the power. If the signal is connected to a 1 Henry Inductor, the current will be $$2A_{RMS}$$ but the power will still be 0 since voltage and current are 90 degrees out of phase.

• Thanks for your answer. I'm a bit confused about your use of "proportional". According to the second link I referenced, it seems like the RMS is exactly equal to the square root of power? Also, your last paragraph seems to impy that power =/= power, could you elaborate on that? Jun 17 at 14:36
• ... and if the signal is connected to back-to-back diodes or other nonlinear device(s), then things get complicated! Jun 17 at 19:41