Root Mean Square

🎯 Learning Objectives:

In this topic we will discuss the widely used metric of RMS (Root Mean Square) calculations and fully understand the origins of the value, why it is needed in the first place in the domains of analog signals and what it means in the context of audio signals.

In previous topics we took a peek at the peak value, the peak to Peak value and the average value of a sinusoidal signal. In the domain of electrical engineering and in audio as well averages are quite important, But the instantaneous peaks? well, not so much. Though we looked at a simple average calculation of a sinusoid in the earlier topic that is not how the effective average of a signal is calculated. The reason for this is basically dependent on what we are trying to find when calculating the average.
A very important aspect of a signal is the energy that it carries over time. That is the power of the signal within a given period of time. RMS value or the root mean square value is needed when calculating the energy associated with the given signal.

Let us back up here before we get carried away we will put a few bullet points here about the topics that we need to discuss to fully understand what we just termed as RMS value. We will get to each one of these in turn. The very first thing that we need to talk about is just a brief summary of some electrical theory to discuss:
The power drawn from a resistive load in the circuit.

Derive the equation for what we refer to as the RMS value of voltage and will visualize what it means and what it looks like
The RMS values for some common waveforms and signals including sinusoids, square waves, triangle waves and some others.
What an RMS value or average value means in the context of audio signals and the sounds that we listen to

Let us understand with example. So let us say, there is a DC voltage source of around 5 volts applied to a resistive load like a light bulb.

So why is RMS calculations needed in the first place?

The light bulb would draw some amount of current from source, which is around 2 amps. We can calculate that the bulb draws a power of 10 watts ( $P=VI$ ), so the bulb is rated to 10 watts. We can do this because this is a DC source and the voltage drop is constant and does not change and we can assume that the current drawn is constant as well. The instantaneous power drawn from the source that is the power drawn at any given time is always the same.

However, what if we were to replace the DC source with an AC source. The AC source is by definition an alternating source of current and in simple cases is a sinusoid of 50 Hz to 60 Hz, much like our test signal that we have been discussing the topic with.

What is the power drawn by the bulb? How do we calculate that?

Well if we apply the formula as ( $P=VI$ ), the instantaneous power would keep changing as the voltage and current oscillates. Since AC is typically a periodic function it makes more sense to calculate the average power drawn instead.

So the average power or the mean power is basically the mean of the instantaneous power.

P_{mean}=mean(P_{inst})

P_{mean}=mean(V_{inst} ⋅ I_{inst})

Since we are assuming that it is a purely resistive load by using Ohm's law we can substitute current here with voltage divided by the resistance.

P_{mean}=mean(V_{inst} ⋅ \frac{V_{inst}}{R_{inst}})

Since resistance is constant and the average does not really apply to the resistance we have the following formula

P_{mean}=mean(\frac{V_{inst}^2}{R})

In essence what we are theoretically trying to do here is to replace an AC source with an equivalent DC source such that the average power drawn by the bulb would be the same in both cases.

Signal domain illustration — **Figure 1.** Signal domains: physical, analog, and digital.

If we go by that principle the equivalent DC source will be equal to the square root of the mean of the squares of the instantaneous ac voltage and this is where the term RMS comes from.

V_{DC}^2=mean(V_{inst}^2)

V_{DC}=\sqrt(mean(V_{inst}^2))

V_{DC}=mean(V_{inst}^2)

V_{DC}=V_{RMS}

It is the root of the mean of the squares, so basically RMS voltage value is just the DC equivalent of an AC source.

So there is a bit of theory for you from an electrical engineering perspective on where the RMS value comes from. It might not seem very relevant to you from an audio engineering perspective but RMS measurements are dominant everywhere.
So let us look at the formula for calculating the RMS value and here it is.

RMS=\sqrt{\frac{\int_0^T \\f(t)^2 \\dt}{T}}

Examples:

Below are example of three function and their RMS values. We can notice that each of these function have different RMS values.

We can think about audio signals as electrical signals in fact they are electrical signals. Their voltage is time bearing rarely are audio signals have a pure sinusoids or even harmonics. For that matter real world audio signals are complex. So to find the power associated with these signals RMS calculations are the way to go but as we can imagine complex real-world signals like human speech do not have any discernible periodicity associated with them.

Now the question is how is this applicable to audio and what does an RMS value signify in the context of audio signals?

For these sort of signals we need to apply time weighting, but why calculate RMS at all. When we listen to sounds we do not perceive the loudness of sounds based on the Peaks that they contain.
Here are two clips as show in the video below. One is an uncompressed drum signal (labeled as signal_01) and other is compressed metal (labeled as signal_02). They are both normalized such that their peaks are equal.

As you could hear though the Peaks are normalized the perceived loudness of the second track is far greater than that of the first. The drum track has large transients, when the drums are hit but they decay fast. The amount of time that you experience the transient is really important for our apparent perception of loudness. Large but short transients do not sound loud to us whereas high levels sustained over larger lengths of time do in fact sound loud.

If You observe the meters in the second video below, here the light green indicates the RMS levels (the audio is muted, so that yo can focus on RMS green bars). As as you can see when they are both played side by side their RMS levels are way off. The RMS level, though not a perfect metric of subjective loudness, is the simplest way to get an idea of how loud a signal is. It is easy to calculate and easy to implement. It does not take into account the frequency aspect of the sound source but it does take into account the temporal aspect of the sound source.

🧪 Interactive Examples

Loading Desmos calculator…

📝 Key Takeaways

RMS (Root Mean Square) is a measure of the effective value of a varying signal — it represents the DC-equivalent voltage that would produce the same power in a resistive load.
It is calculated as the square root of the mean of the squares of instantaneous values:
$V_{RMS} = \sqrt{\frac{1}{T}\int_0^T v(t)^2\,dt}$
RMS gives a more realistic measure of signal power or loudness than peak values.
In audio, RMS corresponds to how loud a sound feels over time — not just how high its peaks go.
Typical RMS-to-peak relationships:
- Sine wave: 0.707 × Peak
- Square wave: 1.0 × Peak
- Triangle wave: 0.577 × Peak
RMS metering is used across audio engineering, power systems, and signal processing to represent consistent energy or loudness.

🧠 Quick Quiz

Test your understanding - select and submit an answer.

1) What does the RMS value of an AC signal represent?

A) The average of instantaneous voltages
B) The DC-equivalent voltage that produces the same power
C) The peak-to-peak voltage
D) The instantaneous current

2) The RMS value of a sinusoidal voltage with a peak value of 10 V is:

A) 10 V
B) 7.07 V
C) 5 V
D) 14.14 V

3) Which waveform has the same RMS and peak value?

A) Sine wave
B) Square wave
C) Triangle wave
D) Sawtooth wave

4) The RMS value of a 12 V peak-to-peak sine wave is approximately:

A) 4.24 V
B) 6 V
C) 8.48 V
D) 12 V

5) For a purely resistive circuit, instantaneous power is proportional to:

A) Current (I)
B) Voltage (V)
C) Square of voltage or current
D) Frequency of the waveform

6) What is the RMS current if a 10 Ω resistor dissipates 50 W of power?

A) 5 A
B) 3.16 A
C) 2.5 A
D) 7.07 A

7) A triangle wave has a peak voltage of 12 V. Its RMS value is approximately:

A) 8.48 V
B) 6.93 V
C) 6.78 V
D) 4.24 V

8) The RMS value of a waveform is calculated by:

A) Averaging the waveform directly
B) Squaring → Averaging → Taking the square root
C) Doubling the peak value
D) Dividing the peak by √3

9) In audio, RMS level better represents:

A) Maximum loudness peaks
B) Perceived loudness over time
C) Silence between peaks
D) The noise floor

10) If an AC source provides an RMS voltage of 230 V, what is the peak voltage?

A) 162.6 V
B) 325.3 V
C) 460 V
D) 230 V