🚀 go-pugleaf

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.

> 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)
............................................................