> Edson Brusque wrote:
>
> > > As to RMS vs averaging, that's what the original question was
> > > about.  I
> > just
> > > didn't know if doing an RMS calculation had any benifits over just
> > averaging the
> > > input signal (better noise immunity, more accurate, more linear,
> > > etc....).
> >
> >     RMS would give you a *more number.
> >
> >     Let's say you have a dimming application, where your circuit is
> > connected to mains power @ 220V and you're firing the triac at 90:.
> > You're measuring the voltage going to the load. If you average your
> > measuring, it will say 110V. RMS will say about 155V. Now, if you
> > rectify and "regulate" the energy going to the load (with full-cycle
> > rectifying, 4 diodes and a capacitor) and measure it with a
> > voltmeter, it will give you a value very close to that 155V.
>
>
> This is wrong, with 90' switching there is exactly
> half the voltage average and half the rms voltage.
> (and how do you make that "degrees" symbol??) :o)
>
> So now i'm confused! Isn't RMS just calculating the
> AC average?? I was taught that the only point to
> squaring this and then doing a square root is that
> this is only math way of making negative values
> positive for averaging.
>
> As programmers we have more power than mathematicians,
> we are not limited to:
> 1. square it
> 2. average it
> 3. unsquare it
>
> We can just do the following:
> 1. convert - values to +
> 2. average it
>
> Or have I missed something?? Does anyone think we
> MUST do square and square-root calcs in our PICs
> and why??
> -Roman
>
> --

This was also my assumption in an earlier mail on this thread. This
simplification only works with a DC value, which is of little use
when calculating RMS.

A simple example will show the difference:

Given 5 samples of a rectified voltage: 2, 5, 5, 2, 1.

Which gives the average value of (2+5+5+2+1)/5=3.

But the RMS value of this is: sqr((2^2+5^2+5^2+2^2+1^2)/5)=3.435.

This has to do with the fact that the RMS value for a voltage is the
same value that a DC voltage would have to be to produce the same
power over the same load. Power is the key word here, since
increasing the voltage gives a squared increase in power over the
same load.
==============================
Ruben Jvnsson
AB Liros Elektronik
Box 9124, 200 39 Malmv, Sweden
TEL INT +46 40142078
FAX INT +46 40947388
ruben@pp.sbbs.se
==============================

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