[sdiy] VCA response - continously variable between linear andexponential
cheater00 at gmail.com
Tue Mar 5 13:54:45 CET 2013
Hi Tom and David,
in case of a linear VCA where you distribute gains before the expo
converter (envelope goes to voltage divider pot, which then goes into
amp env, and into exponentiator which goes into amp env):
the output of an amplifier of gain A is:
output = A*(u*env + exp(v*env))*input
which is equal to:
A*u*env*input + A*exp(v*env)*input
in case you mix after the exponentiator (envelope goes to 50% voltage
divider resistor network, one side of that goes through expo, then
this gets mixed with the other side of the voltage divider according
to a voltage divider pot):
the output is:
output = A*(u*env + v*exp(env))*input
which is equal to:
A*u*env*input + A*v*exp(env)*input
therefore, in the simplest model, it's exactly the same.
In case of an exponential VCA:
output = exp(A+(u*log(env) + (v*env))*input
which is equivalent to:
output = exp(A) * exp(u*log(env) + v*env)*input
output = exp(A) * env^u * exp(env)^v * input
which means it's like chaining a linear and then exponential VCA and
changing their respective depth of effect with the voltage divider
The equation you are looking for is:
output = (b^(A*env)-1)/(b-1) * input
where b is variable by a pot.
see graphs here: http://wolfr.am/XntYOd
They agree exactly with what Hoshuyama posted - thanks for the info, Hoshu!
Note that you can't achieve his result by just biasing the exponential
VCA. It ends up being the same thing for every bias, but with more and
more distortion as you go in one direction, and more noise if you go
in the other:
On Mon, Mar 4, 2013 at 1:54 AM, Tom Wiltshire <tom at electricdruid.net> wrote:
> Hi Damian,
> On 2 Mar 2013, at 07:28, cheater cheater <cheater00 at gmail.com> wrote:
>> David, that may look like a good idea, but it's different - it's
>> mixing lin and exp. It's not going to give you the same range of
>> sounds you could get from a circuit which can vary the exponential
> Why not?
> I'm pretty sure David is right, since he's not mixing between a lin and a log/expo *output* from a VCA, but rather producing a CV which is a blend of lin and log going in.
> But then I thought about it some more and wondered if crossfading between the output of a lin VCA and a expo VCA isn't the same thing anyway. I haven't done the maths to prove it either way.
> Interesting question though…maybe I *will* have to do the maths.
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