🚀 go-pugleaf

In article <Pine.OSF.3.91.950814063353.9002A-100000@malt.hitl.washington.edu>,
Steve White  <stevew@hitl.washington.edu> wrote:
>On 12 Aug 1995, Richard Steven Walz wrote:
>
>> In article <Pine.OSF.3.91.950809073824.7679A-100000@malt.hitl.washington.edu>,
>> Steve White  <stevew@hitl.washington.edu> wrote:
>
>...[Clip other stuff]...
>> >Huh? I agree with the mean value you get, but how does that rms relate
>> >to the mean? I squared the function, integrated from 0 to A, multiplied
>> >by 1/(A+B), and took the square root of the whole thing to get:
>> >
>> >	rms = c*sqr(A/(A+B))
>> >
>> >Maybe I've hosed the arithmetic or something, but I can't see my error.
>> >Or maybe I just don't know what you mean by "ABOUT THE MEAN". (The
>> >method I'm using here produces the correct result for sinusoids).
>> ---------------------------------------------
>> People! My, my, my!! RMS is "the root of the mean of the square of the
>> function", the mean divisor is the abscissal duration of a cycle, and
>> the "square of the function" is what it says, f^2(x) . So simply do
>> the function piece-wise, first the zero period, B, which has value 0,
>> and then the A period, which has value C. The function then, is just
>> like doing the piecewise of a single diode rectifier, where you only
>> evaluate the integral of sin^2(x) through half a cycle, but you still
>> have a mean divisor equal to the whole cycle, thus we have to square
>> the functions for both "pieces", and that is 0^2 + C^2, which is the
>> legitimate piecewise value of the "square of the function" but we must
>> weight it when we do the mean, so (A*C^2 + B*0^2)/(A + B) is the mean,
>> which is the def. integral of the function squared divided by the cycle,
>> and so we have that the Vrms = sqrt { (A * C * C)/(A + B) }
>
>How is you're Vrms different from the rms I provided above? Since you
>included my post, I assume you were correcting my result as well, but
>the expressions we gave are equivalent.
>
>> It IS late for me, so would someone check this who has actually had
>> electronics and enough physics, as I have, to know this? I was VERY
>
>OK. Consider it checked (although I don't see where the amount of physics
>or electronics training ensures ones' answer will be correct). As Walter
>was pointing out in the part I clipped, there seems to be a more rigorous
>mathematical definition which us electronics types typically don't use.
>Whether or not Walter is correct, I have no clue.... he didn't post any
>references.
------------------------------------------------
Okay, a small bit of rigor.
The Vrms = SQRT{ Definite Integral(V^2(t))/Definite Interval }
In other words, the ROOT of the MEAN of the FUNCTION SQUARED of
Voltage, where the mean is the integrated mean. Now you were right,
though sqr is often considered the square, instead, but it seemed like
you were still puzzling about it. I was mostly responding to walter,
and I had seen four different "guesses" in a row and was tired of it!
You were right, and you SHOULD have KNOWN you were right, that's all!
;-) (And Why!)  I was just totally bleary-eyed!
-Steve Walz

>> dismayed by the disparate guesses and mathematical misconceptions I
>> saw here. I pray I have not blundered and propagated another one, but
>> at least I know what one is SUPPOSED to be, even if I erred in my
>> math. 
>> -Steve Walz   rstevew@armory.com
>............................................................
>Steve White, Lab Technician @ Human Interface Technology Lab
>e-mail: stevew@hitl.washington.edu       voice: 206.616.1411
>http://www.hitl.washington.edu/  (My page under consruction)
>............................................................
-----------------------------------------
So relax, Steve, you were right!
-Steve Walz
rstevew@armory.com