Radio World - 27 November 1996
by Harold Hallikainen
San Luis Obispo, Calif. In the last article in this series, we discovered that the average voltage of a full cycle of a sine waveform is 0 volts (since the positive side "cancels out" the negative side), while the average of a half cycle of a sine waveform is the peak voltage times 2/pi. This was determined first by taking the average of several instantaneous voltages through the half-cycle, then by increasing the number of samples towards infinity by applying a little calculus. This time, we'll look at the RMS voltage of a sine wave.
We've heard that RMS stands for "Root Mean Square", but that expression may not be the model of clarity. If we recall that the arithmetic mean of two numbers is just the average of those two numbers, the expression starts to make a little sense. What we are doing is taking the square root of the average of the squares of the instantaneous voltages.
As a side note, I found this summer while tutoring algebra that the mean is a number that is part of a sequence of numbers that is between two other numbers. If the sequence of numbers is an arithemtic sequence (where each successive number is the previous number plus some constant), each of the elements of the sequence between two other elements of the sequence are means of the other two. For example, in the below sequence of numbers, each successive number is determined by adding three to the previous number.
1, 4, 7, 10, 13, 16
In this sequence, the means of 4 and 13 are 7 and 10. How about if there is only a single mean? That single mean is then the arithmetic mean. If we try a couple examples, we find that 7 is the arithmetic mean of 4 and 10. Further, 7 is the average of 4 and 10 (since (4+10)/2=7). So, to find the arithmetic mean of a couple numbers, just take the average.
Another kind of sequence is a geometric sequence. In a geometric sequence, each successive number is determined by multiplying the previous number by a constant (as opposed to adding a constant, which we did before). The below geometric sequence is formed using a constant of 3.
1, 3, 9, 27, 81, 243
9 and 27 are geometric means of 3 and 81 in this sequence. 9 is the geometric mean of 3 and 27 in this sequence. We normally find the geometric mean by taking the squre root of the product of the two numbers. Here, 3*27=81, and the square root of 81 is indeed 9. How about 27 and 243? 27*43=6,561 and the square root of 6,561 is 81.
Back on RMS, we'll take the square root of the arithmetic mean of the squares of the instantaneous voltages. Why? It has something to do with power. Let's see what the average power delivered by a 1 volt peak sine wave is into a 1 ohm resistor. Starting with the formula for power,
P=IV
and substituting Ohm's Law's I=V/R for I, we get
P=V2/R
Further, the instantaneous voltage of the sine wave is
V(a)=VP*sin(a)
where VP is the peak voltage and a is how far we are in to the waveform in degrees or radians (depending upon which sine function we're using). Table 1 shows the voltage at various points through a single cycle of a 1 volt peak sine wave.
| Radians | Degrees | Volts | Power (W) |
| 0 | 0 | 0 | 0 |
| 0.392699 | 22.5 | 0.382683 | 0.1464463 |
| 0.785398 | 45 | 0.707107 | 0.5 |
| 1.178097 | 67.5 | 0.92388 | 0.8535543 |
| 1.570796 | 90 | 1 | 1 |
| 1.963495 | 112.5 | 0.92388 | 0.8535543 |
| 2.356194 | 135 | 0.707107 | 0.5 |
| 2.748894 | 157.5 | 0.382683 | 0.1464463 |
| 3.141593 | 180 | 0 | 0 |
| 3.534292 | 202.5 | -0.382683 | 0.1464463 |
| 3.926991 | 225 | -0.707107 | 0.5 |
| 4.31969 | 247.5 | -0.92388 | 0.8535543 |
| 4.712389 | 270 | -1 | 1 |
| 5.105088 | 292.5 | -0.92388 | 0.8535543 |
| 5.497787 | 315 | -0.707107 | 0.5 |
| 5.890486 | 337.5 | -0.382683 | 0.1464463 |
To find the average of these instantaneous powers, we can merely add them up and divide by the number of samples. My calculator shows the sum of the powers to be 8.0000. Dividing by the number of samples (16), we find the average power is 1/2 watt (or 500 mW). This continuously varying voltage dissipates 500 mW in a 1 ohm resistor. What DC voltage would dissipate that same power?
Above, we found that P=V2/R. Solving for V, we get V=sqrt(P*R). Since, in this case, R=1 ohm, we can find the voltage by taking the square root of the power. So... A DC voltage source of sqrt(500mW) would deliver the same power to a 1 ohm load as a 1 volt peak sine wave does. Doing the square root (more popular than the Macarena?), we find the equivalent DC voltage is 707 mV. What did we do? We took the square root of the mean of the squares of the instantaneous voltages, hence RMS.
If we have a higher peak voltage, all the voltages in the volt column of table 1 would be multiplied by a constant (the peak voltage). All the powers in the power column would be multiplied by the peak voltage squared, which would result in an average of VP2/2. Taking the square root of the average, we get VP/sqrt(2) as the RMS value of any sine wave. Messing around with the equation, we find that VP=VRMS*sqrt(2). Hence, a 117 VAC power line has a peak voltage of 165.463 volts. The instantaneous voltage varies between 165.463 and -165.463 volts.
We were lucky that our average turned out as well as it did. Choosing 16 samples in a cycle gave us the exact relationship between RMS and peak voltage. Recall from our discussion of average voltage last month that we can keep increasing the number of samples towards infinity and find the true average of a continuoously varying waveform. Figure 1 shows how calculus can demonstrate the relationship between peak voltage and RMS voltage.
Harold Hallikainen designs transmitter control and lighting equipment for Dove Systems, a manufacturer serving the broadcast and entertainment industries. He also teaches electronics at Cuesta College and is an avid contra dancer. He can be reached at +1 805 541 0200 (voice), +1 805 541 0201 (fax), harold@hallikainen.com (email), and http://hallikainen.com (World Wide Web, where an archive of these articles is maintained).