Root Mean Square

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This page aims to describe what a root mean square value is and how different equipment measures and displays root mean square values.

Background

There are a variety of ways to quantify time-varying signals. You could look at the maximum and minimum values, the average values, and the frequency content of the signal. Another way to quantify a signal is to look at what is called the root mean square or RMS value. This value ends up being related to the power of a signal and is calculated as:

$$ \begin{align*} \E{V}{RMS}&=\sqrt{\frac{1}{T}\int_T\left(v(\tau)\right)^2\,d\tau} \end{align*}$$

where $$T$$ represents the period of a periodic signal or, for measurement devices, the sampling window it uses to determine signal qualities. Note that there is another version of RMS, which we will call the AC version of RMS, that looks at the square of the difference between the signal and its own average versus the signal value itself. If we define the average value of some signal as

$$ \begin{align*} \bar{v}&=\frac{1}{T}\int_Tv(\tau)\,d\tau \end{align*}$$

then this AC RMS version would give:

$$ \begin{align*} \E{V}{RMS, AC}&=\sqrt{\frac{1}{T}\int_T\left(v(\tau)-\bar{v}\right)^2\,d\tau} \end{align*}$$

Also,

$$ \begin{align*} \E{V}{RMS}^2&=\E{V}{RMS, AC}^2+\bar{v}^2 \end{align*}$$

Among other things, this means that $$\E{V}{RMS}$$ and $$\E{V}{RMS, AC}$$ are the same for signals with an average value of 0.

It is important to note which version of the above various measurement devices calculate! For time-varying signals,

• Keysight 34460A Multimeter:
  • Set to DCV mode, it reports $$\bar{v}$$
  • Set to ACV mode, it reports $$\E{V}{RMS, AC}$$.
  • From those two, you can determine $$\E{V}{RMS}$$ if you need it.
  • The multimeters can also report period and frequency. The more advanced model of this meter can also report maximum, minimum, and peak-to-peak values.
• Keysight DSOX1202A Oscilloscope: There are several different measurements available from the Meas button in the Measure group:
  • AC RMS: $$\E{V}{RMS, AC}$$ - use the "N Cycles" version
  • DC RMS: $$\E{V}{RMS}$$ - use the "N Cycles" version
  • The oscilloscope can also report the period, frequency, maximum, minimum, average, peak-to-peak (difference between the overall maximum and the overall minimum), amplitude (difference between the "top" and "base" values, which are more like averages of values near the maximum and near the minimum, respectively).

Examples

Single Frequency Sinusoid

A signal $$v=A\,\cos(\omega t+\phi)$$ is a periodic signal with an amplitude $$A$$, angular frequency $$\omega$$, and phase $$\phi$$. In addition to reporting the magnitude, you can find the "peak to peak" value of the signal by taking the difference between the maximum and minimum values; in this case, that is $$2A$$. The average value of this signal would be $$\bar{v}=0$$. The RMS value (which is the same as the AC RMS since the average is 0) would be:

$$\begin{align*} \E{V}{RMS}&=\sqrt{\frac{1}{T}\int_T\left(A\, \cos(\omega \tau+\phi)\right)^2\,d\tau}\\ ~&=\sqrt{\frac{1}{T}\int_T\left(A^2\, \cos^2(\omega \tau+\phi)\right)\,d\tau}\\ ~&=\sqrt{\frac{1}{T}\int_T\left(A^2\, \frac{1+\cos(2(\omega \tau+\phi))}{2}\right)\,d\tau}\end{align*}$$

Since the cosine term is now being integrated over two complete periods, that integral will be 0. The integral of a constant over a period, divided by that period, is simply that constant so an $$A^2/2$$ pops out and then you take the square root. This yields:

$$\begin{align*}\E{V}{RMS}&=\frac{A}{\sqrt{2}}=\frac{A\sqrt{2}}{2}=\E{V}{RMS, AC} \end{align*}$$

For a single frequency sinusoid, the oscilloscope's DC RMS measurement would give $$A/\sqrt{2}$$. The oscilloscope's AC RMS and the multimeter's ACV measurement would give $$A/\sqrt{2}$$. The oscilloscope's average value and the multimeter's DCV measurement would give $$0$$.

Shifted Single Frequency Sinusoid

A signal $$v=A\,\cos(\omega t+\phi) + B$$ is a periodic signal with an amplitude about its average of $$A$$, angular frequency $$\omega$$, and phase $$\phi$$. The "peak to peak" value is still the difference between the maximum and minimum values, $$2A$$. The average value of this signal would be $$\bar{v}=B$$. The RMS value would be:

$$\begin{align*} \E{V}{RMS}&=\sqrt{\frac{1}{T}\int_T\left(A\, \cos(\omega \tau+\phi)+ B\right)^2\,d\tau}\\ ~&=\sqrt{\frac{1}{T}\int_T\left(A^2\, \cos^2(\omega \tau+\phi) + 2AB\cos(\omega \tau+\phi) + B^2\right)\,d\tau}\\ ~&=\sqrt{\frac{1}{T}\int_T\left(A^2\, \frac{1+\cos(2(\omega \tau+\phi))}{2}+ 2AB\cos(\omega \tau+\phi) + B^2\right)\,d\tau} \end{align*}$$

The cosine terms are still being integrated over an integer number of complete periods, so those parts integrate to 0. This yields:

$$\begin{align*}\E{V}{RMS}&=\sqrt{\frac{1}{T}\int_T\left(\frac{A^2}{2} + B^2\right)\,d\tau}=\sqrt{\frac{A^2}{2}+B^2}\\ \E{V}{RMS}^2&=\frac{A^2}{2}+B^2=\E{V}{RMS, AC}^2+\bar{v}^2 \end{align*}$$

For a shifted single frequency sinusoid, the oscilloscope's DC RMS measurement would give $$\sqrt{\frac{A^2}{2}+B^2}$$. The oscilloscope's AC RMS and the multimeter's ACV measurement would give $$A/\sqrt{2}$$. The oscilloscope's average value and the multimeter's DCV measurement would give $$B$$.

0-to-high PWM Signal

Assume you have a pulse-width-modulated signal that is "on" (high, $$H$$ V) for some duration $$W$$ each period $$T$$ and "off" (0 V) for the rest of each period. The minimum value would be 0 and the maximum would be $$H$$. The average value of this signal would be:

$$\begin{align*} \bar{v}&=\frac{1}{T}\int_0^Tv(\tau)\,d\tau\\ ~&=\frac{1}{T}\int_0^WH\,d\tau=H\frac{W}{T} \end{align*}$$

The RMS value would be:

$$\begin{align*} \E{V}{RMS}&=\sqrt{\frac{1}{T}\int_0^T\left(v(\tau)\right)^2\,d\tau}\\ ~&=\sqrt{\frac{1}{T}\int_0^WH^2\,d\tau}\\ ~&=H\sqrt{\frac{W}{T}} \end{align*}$$

The RMS-AC value is most easily calculated using the relationship

$$ \begin{align*} \E{V}{RMS}^2&=\E{V}{RMS, AC}^2+\bar{v}^2 \end{align*}$$

or

$$ \begin{align*} \E{V}{RMS, AC}^2&=\E{V}{RMS}^2-\bar{v}^2 \end{align*}$$

so

$$ \begin{align*} \E{V}{RMS, AC}&=\sqrt{\E{V}{RMS}^2-\bar{v}^2}\\ ~&=\sqrt{H^2\frac{W}{T}-H^2\frac{W^2}{T^2}}\\ ~&=\frac{H}{T}\sqrt{WT-W^2} \end{align*}$$

If you define the duty cycle $$D$$ as the ratio of $$W$$ to $$T$$, you get:

$$\begin{align*} D&=\frac{W}{T} & \bar{v}&=H\,D & \E{V}{RMS}&=H\,\sqrt{D} & \E{V}{RMS,AC}&=H\,\sqrt{D-D^2} \end{align*}$$

Shifted PWM Signal

If a PWM signal has a low value other than 0 V, all of the above quantities would need to be adjusted. This is currently outside the scope of ECE 110. However, if you define $$H$$ as the high value, $$L$$ as the low value, $$W$$ as the "high" time, $$T$$ as the period, and $$D=W/T$$ as the duty cycle, the "low" time would be $$T-W$$ and:

$$ \begin{align*} \bar{v}&=\frac{HW+L(T-W)}{T}=HD+L(1-D)=L+(H-L)D\\ \E{V}{RMS}&=\sqrt{\frac{H^2W+L^2(T-W)}{T}}=\sqrt{H^2D+L^2(1-D)}=\sqrt{L^2+(H^2-L^2)D}\\ \E{V}{RMS,AC}&=\sqrt{\frac{W(H-L)^2(T-W)}{T^2}}=(H-L)\sqrt{D-D^2} \end{align*}$$

You can check this with the zero-to-high signal if you set $$L=0$$; you will get the expressions from the previous section.