MOTM

> From: "Dave Bradley" <daveb@...>
> 
> You are confused, my son. Unfortunately, I must explain using a few
> engineering terms.
> 
> MOTM-110 input is linear, not exp. Its transfer function (graph of output
> over input) is a straight line. Therefore, if you input a linear voltage,
> you get a linear change out. If you input an exponential voltage, you get
an
> exponential change out. Your ear wants an exponential response, which the
> linear VCA input passes unchanged from the 800s exponential output.

Actually, this is the part I understand clearly as Paul has spelled it out
that the CV is expo and VCA is linear in the MOTM docs.  My mistake was
calling the MOTM-110 input expo since it is an expo voltage applied at that
point for proper operation.  I now see the error of my ways and that I
should call that input linear.

> Neinsky, comrade. Linear input responding to expo voltage is already the
> status quo when you hook an 800 to a 110. An input doesn't "expect"
> anything, it just responds to what is fed into it, as determined by the
> transfer function mentioned above.

By expects, I meant for the "status quo" or normal operation.  Who is
Neinsky anyhow?  Didn't he used to play for the GB Packers.  :)


> The EG has its own transfer function
> also - if you graph the voltage change over time from point A to point B,
> you get an exponential transfer function. If the VCA has an input that
also
> has an exponential transfer function, these two functions are multiplied
to
> achieve the overall transfer function that describes the total output
over
> input, which is this case is output gain over time.

Very well explained Moe.  So my original question should have been "why
would one want an "exponential" input on a VCA.  And, the correct answer is
quadratic transfer function for "whompass" attack.
:)
Larry (extra stoogy today) Hendry