🚀 go-pugleaf

In article <40id4v$bq@news.scruz.net>,
Richard Steven Walz <rstevew@armory.com> wrote:
>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) }
--------------------
Posting to myself, here, I note that this, of course, simplifies to
the equation: Vrms = C * sqrt{ A/(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
----------------------------------
And indeed I noted that many others had posted this answer as well,
including someone who copied it out of the Wiley handbook. So I guess
my brain still works after all these years! ;->  (45).
-Steve Walz   rstevew@armory.com