What are the peak and RMS values of the voltage of a pulse-width-modulated signal?

A pulse-width modulated (PWM) signal is basically a signal that is turned "on" and "off" at a fixed rate, but is left "on" for varying amounts of time. In the scheme, a fixed maximum value for the pulse is selected, usually by circuit design. That will determine the peak value of the voltage (Vp) of the pulse. The pulse will have some kind of period associated with it. That's the pulse repetition rate (PRR), and it can tell us the frequency of the pulse. We have a time base or clock to set the PRR. The period of a wave (p) which is the time it takes for one complete cycle of the wave, is the inverse of the frequency of the wave. Recall that period equals one divided by frequency (f). We now look at the last factor: the pulse width.

In pulse width modulation, what we concentrate on is the width of the pulse we generate. (Peak pulse voltage and the PRR are almost always fixed.) All we have to do is decide how long to leave the pulse "on" before we turn it "off" at the end of one cycle. The pulse width (the amount of "on" time) could be quite narrow compared to the amount of "off" time. As we increase the amount of "on" time, the "off" time decreases. ("On" time plus "off" time equal the period of our PRR.)

RMS is root mean square, and it's a methematical way to "average" a sine wave. In a voltage sine wave, the area under the curve of a cycle (the RMS value of the voltage for a cycle) is tricky to calculate in a "regular" way. Hey, it's an irregular curve. Not so with a simple PWM signal. It's a modified square wave. Forget RMS. Forget integration. Go with simple math.

A PWM signal is modulated by changing the pulse width (naturally), so a fixed width (tw) must be chosen to find the average value of the output voltage. It's really very simple. Start with the peak voltage. It will have to be known. Then look at the amount of time the pulse is "on" compared to the period of the pulse. A short "on" time will result in a low average voltage. A little longer pulse width will result in a little higher average output voltage. If we reach a point where the pulse is "on" for half the period of the cycle, it will be "off" for the same amount of time. It is "on" for a full half a cycle and "off" for a full half a cycle. If the peak voltage is 100 volts, and the pulse is "on" for half a cycle and "off" for the other half, average voltage is 50 volts. It's a snap to pin it down. The actual average voltage can be calculated by using the following formula:

Vav = Vp x (tw / p)

Let's take a 50 volt peak, 10kHz clocked PWM device. The period of the clock is .1 milliSeconds. (p = 1/f = 1/10kHz = 0.1 mS)

If we have a pulse width modulated signal that has a pulse width (tw) of 1/50,000th of a second (0.02 milliSeconds) we have:

Vav = 50 volts x (0.02 mS / 0.1 mS) = 10 volts average

Finding the average value of a simple PWM device is easy if it's for a fixed pulse width. In a complex application of varying pulse widths, integration is needed, and the calculus is called for. But in the simple application, simple math works.

Not True!

"RMS" is not an "average" by any stretch of the imagination. If it were, a 50% square wave with peaks at 1 and -1 would be 0. (The RMS is actually 1). If it were, the AC from your house receptacles would be 0 volts! (The RMS is actually 120V).

RMS is represents a periodically varying waveform with its equivalent effective DC value. It can be a voltage (V) or a current (I).

For PWM:

VRMS = ( ( C x V12 ) + ( (1-C) x V22) ).5

Where: V1 and V2 are the two peak (top and bottom) voltages in respect to 0, C is the duty cycle in respect to V1 (ie C = .2, duty cycle is 20%). If you plug in the numbers for the 50% square wave above, the answer will agree. It's 1.