🚀 go-pugleaf

In article <Pine.OSF.3.91.950809073824.7679A-100000@malt.hitl.washington.edu>,
Steve White  <stevew@hitl.washington.edu> wrote:
>On 9 Aug 1995, Walter Gray wrote:
>
>> In article <408rk1$6gh@sol.ccs.deakin.edu.au>, andrew@deakin.edu.au (Andrew Cerasuolo) writes:
>> >Hi to all,
>> >Can anyone help me with calculating the average or true rms (Root Mean Square)
>> > value of a pwm signal.
>> >i.e.
>> >    	_____         _____   ___
>> >	        |   |         |   |    ^
>> >	        |   |         |   |    C	
>> >    ________|   |_________|   |________
>> >	
>> >    	|<A>|<-- B--->|<A>|     
>> >
>> >where A & B are times, f = 1/(A+B)  .
>> >C is a voltage from 0 to whatever volts.
>> >I know that as the frequency of the signal increases the average/rms is 
>> >increases but is there a mathematical formula for this relationship ?.
>> >Any answers would be appreciated, Ciao Andrew.
>> >--
>> >Andrew Cerasuolo                 The above comments are mine solely and
>> >Biol. & Chem. Sciences Workshop, not that of Deakin University.
>> >Deakin University, Geelong,     
>> >Vic 3217, Australia              andrew@deakin.edu.au
>> 
>> 
>> mean = C*A/(A+B)
>> rms = C*sqr(A*B)/(A+B)
>> 
>> To those who got the wrong answer, remember that rms is the root
>> mean of squares ABOUT THE MEAN!
>
>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).
>............................................................
>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)
>............................................................
---------------------------------------------
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) }

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
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