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Author Topic: Peak, Average and RMS  (Read 3930 times)
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k4kyv
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Don
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« on: July 30, 2009, 01:59:51 AM »

Maybe the following explanation will clear up some of the confusion over the terms peak, average and RMS voltage and current, peak versus average power, why RMS voltage and current are not the same as average voltage and current, and why average power is not calculated by multiplying average voltage times average current.

The root-mean-square (RMS) value of an alternating voltage or current is defined as the equivalent DC voltage or current that would deliver the same amount of energy to a resistor as the AC does over a complete cycle.

The formulae given in the textbooks for alternating currents assume a sine wave.  The sine waveform as seen on an oscilloscope can be thought of as the projection of a circular rotation, viewed edge-on as it moves steadily across the field of vision, to a two-dimensional plane.  Mathematically this is called a “rotating vector”.  This is where all the trigonometric calculations involving angles, degrees and π (pi) come from. 

Another term for R.M.S. is the quadratic mean. That is, the mean, or average, of a varying quantity taken to the second power.  Why the second power (square)?  Simple Ohm's law.  The power dissipated in a resistor is equal to the square of the voltage across the resistor, divided by the resistance (V˛/R), or the square of the current through the resistor multiplied by the resistance (I˛R).

For to-day’s digitally-minded in particular, this might be easier to visualise if we consider a pulsating DC voltage varying between some fixed value and zero, instead of a continually varying sinusoidal AC voltage for which we have to maintain the mental image of the rotating vector simultaneously to pondering all the peaks, averages, squares and square roots. 

Imagine that we have a battery that puts out exactly 2 volts, a 1Ω resistor and a SPST switch, as shown in the diagram. Imagine someone with superhuman reflexes opening and closing the switch precisely every millisecond, so that we have a series of square pulses, each exactly one millisecond in length, separated by a space of one millisecond.  Or think of the switch as an electronic keyer with the dit paddle held down and the speed set to 500 dits per second. See the attached drawing.

When the switch is closed, the voltage across the resistor is 2 volts. That would be the peak voltage.  But since the switch is open 50% of the time, the average voltage across the resistor is only 1 volt.  Connect an electromechanical analogue DC voltmeter across the resistor and it will read exactly 1 volt. 

But how hot does the resistor get?  Look at the power dissipated in the resistor.  While the switch is closed, we have 2 volts across 1 ohm.  As stated above, power (P) = V˛/R.  So we are dissipating 4 watts in the 1Ω resistor while the switch is closed.  This is the peak power. But since the switch is open 50% of the time, the average power dissipated in the resistor is 2 watts.

So, what is the RMS voltage across the resistor?   The equivalent steady DC voltage that would deliver the same quantity of energy per second (average power) to the resistor as does our pulsating DC, can be calculated, since we know the average power dissipated in the resistor.  What steady DC voltage would get our resistor just exactly as hot as our pulsating 2 volts?  Rearranging the equation P = V˛/R gives us V˛ = PR. We have shown that the average power is 2 watts and our resistor is 1ohm.  So V˛ = 2 watts X 1Ω, or 2 volts˛.  The RMS voltage, that is, equivalent steady DC voltage that would have the same heating effect on our resistor is the square root of V˛:  Vrms = √2, or 1.414 volts. 

So we see that our peak voltage is 2 volts, our average voltage is 1 volt and our RMS voltage is 1.414 volts.

Our peak power is 4 watts, and our average power is 2 watts.  Note that the term "RMS power" is not used.

Where does the term “RMS” (root means square) come from?  In the circuit described above we tried to clarify matters by using a 1Ω resistor, so that power (P) is simply equal to the voltage squared. RMS is the square root of the mean (average value) of the square of the varying quantity, as it is taken through one complete cycle


* RMS vs AVG.jpg (44.08 KB, 1275x1754 - viewed 475 times.)
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Don, K4KYV                                       AMI#5
Licensed since 1959 and not happy to be back on AM...    Never got off AM in the first place.

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W2DU
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Walt, at 90, Now 92 and licensed 78 years


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« Reply #1 on: July 30, 2009, 02:39:19 PM »

Best explanation of the subject I've seen, Don. Better than any text book I've seen.

Your school-teaching experience is showing!

Walt
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W2DU, ex W8KHK, W4GWZ, W8VJR, W2FCY, PJ7DU. Son Rick now W8KHK.
WA1GFZ
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« Reply #2 on: July 30, 2009, 05:17:14 PM »

Now Don can do the PDM 101 course.
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w3jn
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« Reply #3 on: July 30, 2009, 06:41:04 PM »

Quote
Imagine someone with superhuman reflexes opening and closing the switch precisely every millisecond,

Like Pete posting to defend QST?  Grin
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