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 Fri, 19 Jan 2001, Tesla list wrote:

> Original poster: "Charles Hobson by way of Terry Fritz
<twftesla-at-uswest-dot-net>" <charles.a.hobson-at-btinternet-dot-com>
> 
> Hello Dave and Terry,
> 
> A neat symetrical square wave consists of its fundemental frequency and the
> odd harmonics of the fundamental. These are all sinusoidal waves obeying the
> law of superposition to form the square wave. The fourier series describes
> this and provides the relative amplitudes these frequencies. Voltage
> Amplitude is regarded as Peak Voltage of any wave form. I think the rms of a
> square wave is equal to its peak value. (The .7071 factor to get rms as you
> say is good only for  sine waves)

If you want to use Fourier analysis then check out Parseval's Power
Theorem and use it on the square wave we are discussing.. You will get
0.707...  It is far easier to calculate the RMS value by hand though as
there are only two voltage levels to do the averaging over.

> I suspect that different makes of instruments for measuring voltages will
> give different results because their  frequency responses are not the same.
> That is to say that the HP instrument probably has a wider band width than
> the Tek. I hope this is of some help.

That didn't turn out to be the answer, it was the AC coupling of the HP
and the DC coupling of the Tek.. So the Tek gave the right answer to our
question since our question had a DC component.

> 
> Regards Chuck
> 

Have fun,
Darren Freeman