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On 22.10.2010 06:05, Mathieu Bouchard wrote:
<blockquote
cite="mid:alpine.DEB.2.00.1010212253510.16612@paik.artengine.ca"
type="cite">On Fri, 22 Oct 2010, Martin Schied wrote:
<br>
<br>
<blockquote type="cite">I wanted to use a fairly high a < 1
because then the phase for all frequencies is approximately 90°
off like for the ideal a=1. Using slightly smaller factors and
comparing input / output didn't satisfy my expectations. maybe
that would't matter at all - let's experiment...
<br>
</blockquote>
<br>
How is the phase preservation an important thing for you ? I ask
because I didn't really think of it as important... (and I still
somewhat don't know when it's really important).
<br>
</blockquote>
I think it's not important for a speaker simulation either, but my
assumption was that a correct behavior in phase would be a "sign" of
a working integrator.<br>
<blockquote
cite="mid:alpine.DEB.2.00.1010212253510.16612@paik.artengine.ca"
type="cite">
<br>
<blockquote type="cite">that's indeed interesting. So the gain is
defined for a constant signal having the same input and output
samples (or in other words DC having no amplification) if I
understood it correctly.
<br>
</blockquote>
<br>
It's defined for any signal. There are different equivalent ways
to define the gain of a linear filter. In my head, I was thinking
of an input signal containing a single 1 in a sea of zeroes... but
it might be a bit easier to understand it using a constant input
signal. Then for [sig~ 1], [rpole~ 1] will diverge (as much as the
float32 number format can allow it to...) because 1/(1-1) is
undefined (it's a division by zéro). But for [sig~ 1] again,
[rpole~ 0.999] will output a constant 1000.
<br>
</blockquote>
that's the way I understood it. fine!<br>
<br>
Martin<br>
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