Peak to Peak

[Phil Allison]

"BobG" = Groper from Hell


( snip nauseating SHIT from this anencephalic )



** **** the HELL OFF !!


you ASD fucked TROLL !!!!





........ Phil

 

[Tim Auton]

===========================================
Any Aussies care to step up to your buddy here and explain his 'Strine
to me? Is a Groper Someone Who Gropes?

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 fucked TROLL' though. He just
thinks you're talking shit.


Tim

 

[BobG]

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

=====================================
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.

 

[Tim Auton]

=====================================
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.

Check the archives. You're far from unique.

I
just want to know what ASD means. I can take it I think.

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


Tim

 

[Bob Myers]

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.

 

[John Larkin]

An 'instantaneous' measurement of V and dV/dt will tell you all you need to know
about the magnitude of a sinusoidal waveform.

It will not.

John

 

[Bob Myers]

On Sun, 27 Aug 2006 22:52:43 +0100, Eeyore


It will not.

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.

 

[John Larkin]

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

 

[Eeyore]

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."

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

 

[Eeyore]

=====================================
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.

LOL. I've done it and won too !

I just want to know what ASD means. I can take it I think.

attention span disorder ?

Graham

 

[Eeyore]

Is a Groper Someone Who Gropes?

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

 

[Eeyore]

Hmmm, on second thought it does work.

John

Good man, I knew you'd see sense !

Graham

 

[John Fields]

Hmmm, on second thought it does work.

---
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]

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.

 

[John Larkin]

---
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.

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

 

[Eeyore]

---
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.

You missed the clever way to measure dV/dt.

Graahm

 

[Eeyore]

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

Graham

 

[Eeyore]

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.

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

Graham

 

[Phil Allison]

"John Fields"

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.

** 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.

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?

** 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

 

[John Fields]

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.