[Test-Equipment] rms voltmeter?

[J Forster]

Mr. Clark is off the mark by a mile as is often the case.

For a wave of known shape (a pure sine or square) the RMS value can be calculated from peak measurements alone. If the wave shape is not known, such calculations are impossible.

For a waveform of arbitrary shape, what I suggested provides an accurate means of calculating the RMS value of the waveform by computing the instantaneous power at each sample point and adding them up over a single period.

As to the phase between the voltage and current, it's immaterial since the question related to the measurement of  HF RF power into a pure resistive dummy load.

-John

Oh, and please define:

sqaring
immitesticle


*********************



brianclarke01 at optusnet.com.au wrote:

> Forster's information is misleading to say the least.
>
> If you take a sinewave, divide each loop, positive and negative, into two halves and divide one of each of the two halves by ten, square and integrate, the difference from sqaring and integrating the original sinewave is immitesticle.
>
> In other words, unless you need to know to the 5th electronics degree, the waveform matters little.
>
> What is far more important, and is completely missing from the original proponent's and critic's messages, is the phase between the voltage and the current.
>
> 73 de Brian, VK2GCE.
>
> > J Forster <jfor at quik.com> wrote:
> >
> > Only, if you KNOW the waveform is a pure sine wave or square wave, with
> > an
> > exception if you are using a digital scope with sampling and computation
> > capabilities for doing a point by point power measurement and
> > integrating it
> > over a cycle.
> >
> > Best,
> > -John
> >
> >
> >
> > Ron Youvan wrote:
> >
> > > [snip]
> > >    Accurate power measurements can be made by putting a "T" at the
> > > input of the load and viewing * the peak to peak (sine wave) voltage
> > > across the load and mathematically determine the power level.
> > >
> > > * Using a 'scope that is flat beyond the observed frequency if
> > > you have or can borrow one
> >
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