RMS voltage of a pulse width mod. waveform.

How do I calculate the RMS voltage of a pulse width?
I have a Ton 5.4us, Toff 10.2us, frequency 64kHz, and the peak (measured on
a scope) is 14 volts.
A meter reads 5.6vdc.
How do I calculate the pulse to get my 5.6v?
Thanks in advance.
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Is your meter true RMS capable? Many are only accurate for sinewaves, and tend to be frequency limited to boot.
Let's say your pulse peak is Vp, the on time is Ton, the off time is Voff. The total energy delivered to a purely reistive load R during one cycle is then:
E = (Vp^2/R)*Ton
and the average power over the cycle is:
P = E/(Ton + Toff)
= (Vp^2/R)*Ton/(Ton + Toff)
For DC voltage the power is V^2/R. The RMS voltage for your pulse would give an equivalent average power, so we can equate:
Vrms = Vp*sqrt(Ton/(Ton + Toff))
In your case that would work out to about 8.2V
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Greg Neill wrote:

Would help if the OP would furnish information on the meter.
As has been established the OPs meter is a Fluke 87 and is "true RMS"
But Fluke states the bandwidth is 20kHz.
The OP states he is measuring a pulse with a frequency of 64kHz. And a pulse has frequency components far higher than 64kHz.
--
bud--

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Steve wrote:

RMS and DC voltages are not the same for time-varying signals.
RMS = Root mean square. Calculate the mean (time-average) value of the amplitude-squared waveform over one period and find the square root of the result. Here's a start:
+-----+ <-- 14*146 | | --+ +--------+ <-- 0 0 5.4 15.6
A DC voltmeter should report the average voltage measured over one period. Same wave as above, except peak amplitude is 14, so: Vavg = (14 * 5.4)/15.6 = 4.84.
If you want 5.6 "vdc", you must adjust the duty cycle, peak voltage, or both.
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For any waveform, the RMS voltage is the square root of the mean square voltage. In concept, you take the voltage as a function of time, square that, integrate it over one cycle of the waveform, divide by the period of one cycle, and then take the square root.
In the case where the waveform in question is a square wave whose voltage is 0 for part of the period and Vp for the rest of the period, evaluating the above is particularly simple. You ought to work it out for yourself to see just how simple, but the answer is
    Vp * sqrt(Ton / (Ton + Toff))

So the true RMS voltage is 8.24 V.

If it is a true RMS reading meter, then either your meter or your scope is out of calibration. On the other hand, most meters do *not* read RMS voltage, they read either peak voltage or average voltage, and then scale that by a correction factor so they read RMS voltage *for a sine wave*. Their reading isn't valid for non-sine waveforms.

Do you really want 5.6 V RMS output? If so, you need to adjust Ton and Toff, or possibly Vp.
    Dave
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snipped-for-privacy@cs.ubc.ca (Dave Martindale) wrote in writes:

Hmmm what does a Fluke 87 measure in? I must have lost certain brain cells and begining to confuse myself.
The peak of (US) AC outlets is 170v peak. The value the Fluke measures is 120 or (0.707) * 170.
So I should be measuring the RMS of the pulse.
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It's a true RMS meter, isn't it?

That's correct. Although integrating a sin^2 function is somewhat more difficult than integrating a square wave (I recommend looking it up in a table of integrals, or (these days) checking online), it turns out that the integral of sin^(t) over one cycle is t/2, or the mean square of a sine function with peak 1 is 0.5. That means the RMS voltage of a sine wave is 1/sqrt(2) times the peak.
Interestingly, that's *also* true for a square wave with a low voltage of 0 V, high voltage of 1 V and a 50% duty cycle. So RMS is 1/sqrt(2) times peak for this waveform too. But the *average* voltage is not the same - the mean absolute value for the sine wave is 108V, while the square wave's mean value will be 85. So an average-reading meter that measures 108 V and "converts" it to 120 V for display will be fooled by the 85 V mean of the square wave and display a bogus voltage, not 120 V.
    Dave
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snipped-for-privacy@cs.ubc.ca (Dave Martindale) writes:

This is getting somewhat off topic, but I saw a particularly elegant explanation on the Web about why the integral of sin^2(t) over one cycle is t/2:
* Because of Pythagoras, and the definition of sin and cos,
    sin^2(x) + cos^2(x) = 1, for any x
* So, the integral of (sin^2(t) + cos^2(t)) dt is just t
* So the definite integral of this sum over any interval 2*pi in length is just 2*pi.
* But cos(x) is really the same function as sin(x) with a phase shift. The same is true of cos^2 and sin^2. If you integrate over one period of the waveform, the integral is always the same no matter what start and end points you pick.
* So the sin^2 and cos^2 terms contribute equally to the integral (as long as you integrate over one period), and each one is half the total.
* So the integral of sin^2(t) dt over a 2*pi period is just 1*pi, and the mean value of the amplitude is 1/2.
    Dave
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Are you saying that the Fluke 87 reads 5.6 VDC on *this* waveform?

The Fluke displays the correct RMS voltage when measuring a 60 Hz sine wave with 170 V peak. But virtually *any* meter will do that, even if it's average-reading, doesn't work above 100 Hz, and is accurately calibrated on only the 200 V scale.
Now, the Fluke probably is accurately calibrated on all scales, and it is a RMS-reading meter. But as someone else pointed out, it can only read the RMS voltage of a waveform accurately under limited conditions: the peak-to-average ratio of the waveform has to be within some maximum, and all the significant harmonics of the waveform have to be within the meter's measuring bandwidth. The *fundamental* of your waveform is 64 kHz, and the meter bandwidth is only 20 kHz, so you shouldn't expect it to provide an accurate measurement of even a sine wave at that frequency.
It looks like you need something with a bandwidth of 1 MHz or so to accurately measure the RMS voltage of this waveform. For example, digital storage scopes can capture the waveform, and some of them have built-in functions that display the RMS voltage measured over one cycle of a waveform.
    Dave
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snipped-for-privacy@cs.ubc.ca (Dave Martindale) wrote in writes:

conditions:
maximum,
the
it
cycle
OK, this is more interesting. Yes, the meter reads 5.6V DC on the waveform I provided details for.
It's a pulse width modulator driving a FET.
I'm trying to learn how one would associate this with say a motor. Possibly I don't care about voltage and only the feedback of the motor's speed. I assume if I have a motor with a 5-volt max, I can have it spin at half speed at 2.5v - ignoring any thoughts about starving the motor for a moment.
So my inital curiosity is what type of pulse width I would provide to supply a DC voltage of 2.5v
This is where my questions came from - pure curiosity, however, it appears I'm getting in-depth (and partially confusing) answers.
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