<div>If you normalize the output of the filter you describe to the same peak amplitude as the original, it's RMS value will certainly increase. Having the peaks at a common reference point is critical.</div><div><br></div>
<div><div><div><div><br><div>-Theron</div><div>^</div><div><br><div class="gmail_quote">On Tue, Jan 11, 2011 at 10:33 AM, Mathieu Bouchard <span dir="ltr"><<a href="mailto:matju@artengine.ca">matju@artengine.ca</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><div class="im">On Mon, 10 Jan 2011, Roman Haefeli wrote:<br>
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<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
Assuming that the more compression is applied, the more the RMS amplitude [1] approaches the Peak amplitude [2] of an audio signal,<br>
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Why do you assume that ? Let's say I take a signal and divide it by its recent peak volume. The output of [osc~] will stay unchanged. A signal made of plenty of sharp spikes will have a much lower RMS/peak ratio and still be unchanged.<br>
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Are you confusing this with waveshaping such as [expr tanh($v1)] ? It may be a special case of compression, but is not what is usually meant by that.<br>
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<a href="http://en.wikipedia.org/wiki/Dynamic_range_compression" target="_blank">http://en.wikipedia.org/wiki/Dynamic_range_compression</a><div><div></div><div class="h5"><br>
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