Re: Average, RMS and Power Factor made easy!

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "35045 by way of Terry Fritz <twftesla-at-uswest-dot-net>" <free0076-at-flinders.edu.au>



On Tue, 23 Jan 2001, Tesla list wrote:

> Original poster: "Ed Phillips by way of Terry Fritz
<twftesla-at-uswest-dot-net>" <evp-at-pacbell-dot-net>
> 
> Tesla list wrote:
> > 
> > Original poster: "35045 by way of Terry Fritz <twftesla-at-uswest-dot-net>"
> <free0076-at-flinders.edu.au>
> > 
> > On Sat, 20 Jan 2001, Tesla list wrote:
> > 
> > >
> > >       For a square wave the average equals the peak equals the RMS.
> > >
> > > Ed
> > >
> > 
> > That's just plain wrong! If the average equals the peak you MUST have DC.
> > In that case the average=peak=RMS... You are thinking of DC not the square
> > wave we are discussing.
> > 
> > Have fun,
> > Darren Freeman
> 
> 	Correct!  Delete the reference to "average" and I stand by the rest. 
> Next time I'll try to remember not to post something while my wife is
> trying to talk to me!
> 
> Ed
> 

Even if peak=RMS you must have a square wave that swings from +V to -V
with any duty cycle. The one we are talking about swings from 0 V to 1 V.
In this case peak=/=RMS...

In the square wave being discussed peak=sqrt(2)*RMS.

A simple way to add the DC offset:

Take a square wave swinging from -0.5 V to +0.5 V at 50% duty cycle. Then
RMS=peak=0.5 V  and DC component = 0 V.

Take a DC signal of +0.5 V, then RMS = 0.5 V

We would like to add them to get our square wave from 0 V to 1 V at 50 %
duty cycle.

The two signals are orthogonal under the Fourier series, in other
words they both have different sets of frequencies (square wave has
harmonics of its fundamental frequency and no DC component, while DC has
no AC components). In this case (and ONLY this case):

   RMS = sqrt(RMS1^2 + RMS2^2)

       = sqrt(0.5^2 + 0.5^2)

       = sqrt(0.5)

       = 1/sqrt(2)

       = 0.707

So I have shown for the umpteenth time using yet another method that the
RMS voltage of the square wave *that*we*are*discussing* is 0.707 V. Please
don't anyone else try to tell me that this is wrong because it only
applies to sinewaves. There are a few signals that this does work for and
our square wave is one of them.

Have fun,
Darren Freeman