Simplest way is take a reading of the mains electricity in your house. If you're in europe it should come out at about 230-240VAC and the meter (if it's true RMS) will agree with that. I think in America and Japan the domestic mains voltages are very different but you will no doubt know what it should be in your own home country and a true RMS meter will agree with it.
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They are all off base here. Actually I am surprised Phil didn't mention it.
Compare the readings with the same peak to peak value o a sine wave and a s quare wave. If they read the same it is not true RMS. If it is true RMS on
has the resolution to discern 1 volt from 0.71 volt.
You do have a scope right ? If not you'll have to figure out a way to clip a sine wave. You don't need any power behind it, a Zener or stack of diode s will do. A scope is the best way though. If you have to wing it come back and I will try to figure out the easiest way.
That should be all he needs then. Some sort of variable AC supply and some diodes or something to match the P-P value. Ba da bing, ba da boom. (well not boom hopefully, he only needs a volt, and one volt because it is convenient)
Trying to express it seems to be the problem. Both the positive and negativ e sides of the waveform contribute, so it is not really half. But if you ha ve a 1 volt P-P square wave, it is only going to read half a volt because a one volt symmetrical waveform would have the voltage going from half a vol t negative to half a volt positive.Similarly if it is going from zero to on e volt positive, it will read 50 % of that volt. That is where the one half comes in. Perhaps it would have been easier just to call it the peak value and forget peak to peak. But then the statement might be wrong on non-symm etrical (+ to -) waveforms. They can just treat it all as positive.
Words - dammit !
to call ground? "
Not ground - COMMON. The emitter is common to both the input and the output . You put current to the B - E junction and then the transistor starts cond ucting from C - E. thereofre the E is always the common. How you ground it in the circuit will determine input and output impedances as well as voltag e and/or current gain. But TO THE TRANSISTOR, the emitter is always common.
Another thing they teach wrong is that 180 out of phase is the same as inve rting the waveform. Well they should not teach it that way because it is on ly true for waveforms that are symmetrical on both axes. For example it is absolutely not true of a sawtooth wave. It IS true of a triangle wave but t hat is different. The sawtooth is not symmetrical in both axes, only one, i f it even is actually.
Regardless, the point is that you can detect the difference in the reading to determine if the meter is true RMS. It seems like the sine and square co mparison would be the easiest way to do it. the needed waveforms are the mo st easily attainable.
Anyway, I probably should've wrote the peak rather than half the peak to pe ak. It would be a more accurate statement.