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-   -   What "exactly" does a filter envelope do? (http://www.infekted.org/virus/showthread.php?t=31540)

Remedial 16.09.2009 06:32 PM

What "exactly" does a filter envelope do?
 
Okay, I know, in general terms, it contains the same ADSR envelope as the amp envelope and the, rather than affecting the volume, it's effect is on the cutoff point of the filter. So, tell me if I'm right here, but, theoretically, the attack phase determines when the filter takes effect (whether it be at the start of the note or later on in it's cycle). The decay phase is the amount of time it takes for the cutoff to get to it's sustain point (which I'm guessing is the cutoff point that was dialed in using the filter cutoff knob) and the release phase is how long it takes for the filter to stop having an effect on the sound (i.e., the unfiltered sound shines through). Am I getting this right? And, now, the filter envelope amount would be how much the filter cutoff is actually affected by the dialed in filter envelope settings, right? So, at it's full amount, the sound will adhere to the filter envelope settings to a tee? I've been trying to figure this out for a while, but I haven't found any guide that "truly" explains the usage or purpose of the filter envelope, even the Sound on Sound synth series. Let me get your inputs.

HiEnergy 16.09.2009 09:41 PM

This is what happens with the filter during the lifecycle of a sound:
We are assuming an ADSR envelope. http://en.wikipedia.org/wiki/ADSR_envelope

The filter initially (when the key gets depressed) is set to its cutoff frequency (let's call it fC).
It then moves (rises or falls) depending on the envelope depth (let's call it dp for now) to a "full" value of fC+dp. dp can be positive (initially rising), zero (no envelope modulation of the filter) or negative (initially falling).
I'm not sure about negative env amounts but at least these can be achieved by use of the mod matrix.
The time that is needed to reach the apex is the attack time A.
Next the filter closes (or opens in the case of negative dp) to the sustain point (this is fC+dp*S). The time to reach the sustain point is the decay time D.
After the key is released the filter returns to fC which takes the release time R.
Now short: (for positive/negative dp)
- filter starts at fC
- rises/falls to fC+dp during the time of A
- falls/rises to fC+dp*S during the time of D
- stays on that level until key is released
- falls/rises to fC during the time of R

You are mostly right except for this is not about the balance between filtered and unfiltered sound. We're talking about rising and falling cutoff frequency. Not that the Virus couldn't use its envelopes for modulating filter balance but when speaking of the filter envelope one has cutoff modulation in mind.
After the end of the filter decay phase there won't be "unfiltered" sound but sound that has its filter open to exactly the degree that you dialed in using the cutoff knob.

Maybe somebody else can explain this more accurately and less mathematical.

stay tuned
HiEnergy

Remedial 21.09.2009 04:45 PM

Quote:

Originally Posted by HiEnergy (Post 293417)
This is what happens with the filter during the lifecycle of a sound:
We are assuming an ADSR envelope. http://en.wikipedia.org/wiki/ADSR_envelope

The filter initially (when the key gets depressed) is set to its cutoff frequency (let's call it fC).
It then moves (rises or falls) depending on the envelope depth (let's call it dp for now) to a "full" value of fC+dp. dp can be positive (initially rising), zero (no envelope modulation of the filter) or negative (initially falling).
I'm not sure about negative env amounts but at least these can be achieved by use of the mod matrix.
The time that is needed to reach the apex is the attack time A.
Next the filter closes (or opens in the case of negative dp) to the sustain point (this is fC+dp*S). The time to reach the sustain point is the decay time D.
After the key is released the filter returns to fC which takes the release time R.
Now short: (for positive/negative dp)
- filter starts at fC
- rises/falls to fC+dp during the time of A
- falls/rises to fC+dp*S during the time of D
- stays on that level until key is released
- falls/rises to fC during the time of R

You are mostly right except for this is not about the balance between filtered and unfiltered sound. We're talking about rising and falling cutoff frequency. Not that the Virus couldn't use its envelopes for modulating filter balance but when speaking of the filter envelope one has cutoff modulation in mind.
After the end of the filter decay phase there won't be "unfiltered" sound but sound that has its filter open to exactly the degree that you dialed in using the cutoff knob.

Maybe somebody else can explain this more accurately and less mathematical.

stay tuned
HiEnergy

Great explanation, HiEnergy. Exactly what I was looking for. Thanks.

Remedial 22.10.2009 03:52 AM

I also found a great video for those who needed something additional to drive to point home, as myself: http://www.youtube.com/watch?v=AZiGVSJm88U

Remedial 22.10.2009 03:15 PM

Quote:

Originally Posted by HiEnergy (Post 293417)
After the end of the filter decay phase there won't be "unfiltered" sound but sound that has its filter open to exactly the degree that you dialed in using the cutoff knob.

Maybe somebody else can explain this more accurately and less mathematical.

stay tuned
HiEnergy

Just to be sure Hi, did you mean to say "After the end of the filter release phase...", because, from what you stated, the end of the decay phase would place the filter cutoff at the sustain level that was dialed in, right, which "could" theoretically be the same as the original filter cutoff (I'm guessing if you placed the sustain level at 0) but in most cases wouldn't be. Please correct me if I'm wrong. Thanks. Also, you did a much better job of explaining the function of the release phase than did the guy in the youtube vid. My final question would be: Can you dial in a filter cutoff point below the the original filter cutoff using the sustain level or does the sustain level stop at the original value of the filter cutoff. If you can do this, then will the release phase result in the filter "rising" to the original cuttoff value? I know I'm getting a little advanced with this now, but that though just came to me. Thanks again.


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