Re: gEDA-user: RMS Waveform of a signal

[Rubén Gómez Antolí <lobo@xxxxxxxxxxxxxx>]

Hello:

El 18/06/10 12:49, Dan McMahill escribió:
> On 6/16/2010 10:43 AM, Rubén Gómez Antolí wrote:
> > [...]

> > I'm not completly sure but, these are true when the signal is regular
> > and periodic.
>
> I'm not sure what definition you have in mind for "regular".

Sorry about, I'm so bad with english (you know) and try to make it 
simply (and in many times wrong).

Regular as without strange distorsions, and expresable with a "common" 
ecuation, like a sine wave, for example.

> [...]

> > Sincerily, I thought that this are more usual, but there are very few
> > references on web and nothing in how to obtain in "free world" (at
> > least, I can't find it).
>
> RMS is usually not defined as time-varying. Usually you'd just find
>
> sqrt(average_over_time(signal^2))
>
> and for something which is not periodic (like say a digitally modulated
> carrier with random data), you just average over enough time to where if
> you did it again later you'd get about the same answer.
>
> That RMS() function in the paper just uses a shorter window. In other
> words what they do is compute:
>
> sqrt(low_pass_filter(signal(t)*signal(t))
>
> My guess is that the lowpass filter they use for the measurement is just
> a moving average meaning the impulse response is a rectangular pulse
> with a width of say 1/10 or so of the width of the waveform display.

I'm afraid that I don't have idea how they implement these RMS, and 
possibly you are true. Perhaps we need the voodoo, how says Armin.

I can tell you that the method used in my last mail almost "fits" the 
graph that I want. Possibly is wrong in mathematical terms, and is most 
correct use the aproaching [1,2]U[3,4]U...[n-1,n] intervals and "expand" 
it by spline to all the time analysis.

In other way, the solution gave by Holger in Ngspice list, fits the 
graph wanted.


> So in octave you'd just square the signal, use the filter() function,
> and then sqrt. The trick is something like octave usually doesn't work
> on non-uniformly sampled data like what you'd get out of a simulator.

Can I ask you how to use the filter() function? I see its needs a three 
vectors... and I want to test your method.

> -Dan

Best regards.

Salud y Revolución.

Lobo.
--
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para no atar mis manos con las cadenas del soft propietario.
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