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Sci.Electronics.Basics -> Peak to Peak
There are 55 messages in this thread.
You are currently looking at messages 40 to 55.
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Author: Phil AllisonDate: 00:22 28-08-06
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"BobG" = Groper from Hell
( snip nauseating SHIT from this anencephalic )
** FUCK the HELL OFF !!
you ASD fucked TROLL !!!!
....... Phil
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Author: Tim AutonDate: 10:17 28-08-06
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BobG <bobgardner@aol.com> wrote:
> Phil Allison wrote:
> > "BobG" = Groper from Hell
> > ( snip nauseating SHIT from this anencephalic)
> > ** FUCK OFF !!
> > you ASD fucked TROLL !!!!
> ===========================================
> 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
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Author: BobGDate: 12:34 28-08-06
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Tim Auton wrote:
> 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.
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Author: Tim AutonDate: 14:43 28-08-06
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BobG <bobgardner@aol.com> wrote:
> Tim Auton wrote:
> > 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.
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
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Author: Bob MyersDate: 15:05 28-08-06
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"John Fields" <jfields@austininstruments.com> wrote in message
news:h294f2tgcpbp8fsdu2skpbaqli02ovttvc@4ax.com...
>>The RMS value of a *single* voltage measurement...
>
> ---
> That depends on knowing the shape of the waveform before the
> measurement is made.
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.
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Author: John LarkinDate: 16:12 28-08-06
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On Sun, 27 Aug 2006 22:52:43 +0100, Eeyore
<rabbitsfriendsandrelations@REMOVETHIS.hotmail.com> wrote:
>
>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
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Author: Bob MyersDate: 16:43 28-08-06
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"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in
message
news:0ej6f21cuv3mffbtjv208qjd41vaonmhta@4ax.com...
> On Sun, 27 Aug 2006 22:52:43 +0100, Eeyore
> <rabbitsfriendsandrelations@REMOVETHIS.hotmail.com> wrote:
>>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.
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.
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Author: John LarkinDate: 18:50 28-08-06
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On Mon, 28 Aug 2006 20:43:27 GMT, "Bob Myers"
<nospamplease@address.invalid> wrote:
>
>"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in
message
>news:0ej6f21cuv3mffbtjv208qjd41vaonmhta@4ax.com...
>> On Sun, 27 Aug 2006 22:52:43 +0100, Eeyore
>> <rabbitsfriendsandrelations@REMOVETHIS.hotmail.com> wrote:
>
>>>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.
>
>
>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
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Author: EeyoreDate: 18:53 28-08-06
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Bob Myers wrote:
> "John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in
message
> news:0ej6f21cuv3mffbtjv208qjd41vaonmhta@4ax.com...
> > On Sun, 27 Aug 2006 22:52:43 +0100, Eeyore
> > <rabbitsfriendsandrelations@REMOVETHIS.hotmail.com> wrote:
>
> >>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.
>
> 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
>
>
> Bob M.
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Author: EeyoreDate: 18:55 28-08-06
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BobG wrote:
> Tim Auton wrote:
> > 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.
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
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Author: EeyoreDate: 18:58 28-08-06
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BobG wrote:
> 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
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Author: EeyoreDate: 18:59 28-08-06
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John Larkin wrote:
> On Mon, 28 Aug 2006 20:43:27 GMT, "Bob Myers"
> <nospamplease@address.invalid> wrote:
>
> >"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com>
wrote in message
> >news:0ej6f21cuv3mffbtjv208qjd41vaonmhta@4ax.com...
> >> On Sun, 27 Aug 2006 22:52:43 +0100, Eeyore
> >> <rabbitsfriendsandrelations@REMOVETHIS.hotmail.com> wrote:
> >
> >>>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.
> >
> >
> >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
Good man, I knew you'd see sense !
Graham
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Author: Chris FosterDate: 10:37 21-09-06
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John Larkin <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in
news:0ej6f21cuv3mffbtjv208qjd41vaonmhta@4ax.com:
> On Sun, 27 Aug 2006 22:52:43 +0100, Eeyore
> <rabbitsfriendsandrelations@REMOVETHIS.hotmail.com> wrote:
>
>
>>
>>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
>
>
It will, It will not, It will, It will not, It will, It will not,
It will, It will not, It will, It will not, It will, It will not,
It will, It will not, It will, It will not, It will, It will not,
It will, It will not, It will, It will not, It will, It will not,
It will, It will not, It will, It will not, It will, It will not,
It will, It will not, It will, It will not, It will, It will not,
It will, It will not, It will, It will not, It will, It will not,
It will, It will not, It will, It will not, It will, It will not,
It will, It will not, It will, It will not, It will, It will not,
Nanner nanner boo bo.
--
Posted via a free Usenet account from http://www.teranews.com
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Author: Chris FosterDate: 10:40 21-09-06
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"WAZ" <wazdesign@hotpop.com> wrote in news:1156269157.387710.12830
@p79g2000cwp.googlegroups.com:
>
> When we say there is 110V AC, is that peak to peak?
>
No, that is RMS or Root Mean Square.
To calculate peak, mult RMS by 2 x (Square root of 2) (approx 1.414)
Then to get paek-to-peak, multiply peak by 2
--
Posted via a free Usenet account from http://www.teranews.com
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Author: John FieldsDate: 14:57 21-09-06
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On 21 Sep 2006 14:40:18 GMT, Chris Foster <edolan@iw.net> wrote:
>"WAZ" <wazdesign@hotpop.com> wrote in news:1156269157.387710.12830
>@p79g2000cwp.googlegroups.com:
>
>>
>> When we say there is 110V AC, is that peak to peak?
>>
>
>No, that is RMS or Root Mean Square.
>
>
>
>To calculate peak, mult RMS by 2 x (Square root of 2) (approx 1.414)
>
>Then to get paek-to-peak, multiply peak by 2
---
No.
Peak = RMS * sqrt(2)
--
John Fields
Professional Circuit Designer
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