Peak to Peak

You're using Google Groups and are thus Google Groper.

I see you use AOL too. The combination of using Google Groups and AOL will, in the absence of other information, tend to make people think you are somewhat technically inept. Not everyone who uses one or the other is inept of course, but a startlingly high proportion seem to be. By quoting, bottom-posting and being able to spell you've already put yourself above around 99.4% of your fellow users though.

That's not why Phil called you an 'ASD f***ed TROLL' though. He just thinks you're talking shit.

Tim

===================================== I can understand why he doesn't want to talk to me... I'm the only person in the world that has tried to beat him back into his cage. I just want to know what ASD means. I can take it I think.

Check the archives. You're far from unique.

Autistic Spectrum Disorder I guess. 'Autistic' is one of his favourite insults.

Tim

No, that's not what was claimed - John L. wasn't talking about figuring out the RMS value of a periodic waveform from a single measurement; he just said he could calculate the RMS value of such a measurement. And that certainly can be done, it's just not a particularly interesting thing to do.

"Root-mean-square" just means exactly what it says - the square root of the mean of the squares of any series of values. For a sinusoidal wave, doing this process over a sufficient number of samples (and assuming we are sampling per St. Nyquist) gives you a number that happens to be 0.707 of the peak value of that sinusoid. (So would doing the whole RMS thing analytically, through the approriate use of integral calculus over at least one complete cycle - but since we're talking "instanteous measurements" here, speaking of it in terms of sampling is more relevant.) But applying the same process to a SINGLE measurement just gives you back the same number. Square the number, take the mean (gee, over ONE sample, what would that be?), and then take the square root. What do you get? See what I mean? You can run through the process, it just doesn't tell you anything you didn't already know.

Bob M.

It will not.

John

Ummm...why not? That was my first reaction, too, BUT - if I COULD get dV/dt AND the instantaneous voltage accurately in an "instantaneous" measurement, AND I can make the assumption that the waveform is, in fact, sinusoidal (and I ignore the possible effects of noise/error in either measurement), what more do I need? The dV/dt value (along with the actual voltage at the point measured) would tell me in effect "where I am" within the cycle, and from that I can calculate the peak (or any other value) from the known voltage at that point. Doesn't cover any possible DC offset, of course, but I think for the purposes of this thought experiment we are considering only "pure AC."

Bob M.

Hmmm, on second thought it does work.

John

You do need to know the frequency too of course.

I was once presented with the problem of determing the output voltage ( i.e. maximum output ) of a resolver.

The resolver could be static however. By comparing the outputs it was possible to determine the angle of rotation and then back-calculate the 'peak' output voltage.

There's usually a way to do these things.

Graham

LOL. I've done it and won too !

attention span disorder ?

Graham

If you looked at the average google groupie's post to science groups you wouldn't need to ask that !

Before google groups, the award went to webTV and before that to AOL lusers. Oops !

Graham

If you looked at the average google groupie's post to science groups you wouldn't need to ask that !

Before google groups, the award went to webTV and before that to AOL lusers. Oops !

Graham

--- I don't see how, since in the first place you'll have to measure frequency, which can't be done instantaneously.

But, even given that, as a freebie, two measurements need to be made to determine the slope of the sinusoid at the "point" of interest and, therefore, to determine its amplitude.

Consider, for example: A single measurement is made and it happens to coincide with the zero-crossing of the sine wave. is there any information there which can be used to determine the amplitude of the signal?

No.

Now let's say another single, separate measurement is made which yields a reading of 1 volt. is there any information there which will allow us to determine the amplitude of the signal?

No.

Going back to the first example, If frequency is known and a measurement is made which yields zero volts there will still be nothing known about the amplitude of the signal. However, if another measurement is made, after the first, in the smallest increment of time possible, and the time between the two measurements is known, the slope of the sine wave can be determined and its amplitude calculated.

The same is true anywhere up and down the sine wave, but the fact that _two_ measurements, displaced in time, need to be made to determine the slope means that the amplitude can't be determined from a single instantaneous measurement.

-- John Fields Professional Circuit Designer

--
but not instantaneously.

OK, if we have the voltage and slope at a single point, but we don't know the frequency, we can't determine the sine amplitude.

See, I was right after all.

John

You missed the clever way to measure dV/dt.

Graahm

In the above case yes, as in within a few tens of microseconds or so.

Graham

Such a method is indeed limited to situations where the frequency is known.

Graham

"John Fields"

** The dV/dt of a sine ( or other wave) wave at any point in time can be found by first passing it through a HPF with known 6 dB /oct haracteristic - to "differentiate " the waveform.

The output voltage of the filter is proportional to input dV/dt at any time.

** Yes - the dV/dt is at its maximum at each zero crossing.

so V peak = dV/dt divided by 2.pi.f

( snip rest of Field's drivel)

....... Phil

--
Well, at least until someone convinces you that you weren\'t?

Again? ;)