MOTM

A while back, when Paul was talking about the future MOTM-VCA. I remember
some discussion about having both exponential and linear inputs.  Can some
of you modular savvy guys tell me why I would want this feature in a VCA. 
Speaking from MOTM land, since that is all the modular I own, the EGs have
exponential outputs so the VCA has a exponential input and linear response.
 What would I use a linear input on a VCA for unless I was using EGs with
linear CV outs?

Larry (still trying to learn & understand this modular stuff) Hendry

The ear response to sound pressure levels (amplitude) in a log (anti-log?)
fashion is way similar to the way pitch works.

You know how, in a VCO, you have the choice of designing it to either
respond exponentially to linear CVs (the Moog/MOTM way) or to respond
linearly to exponential CVs (the Korg MS-series way). 

Same goes for a VCA. It's usually cheaper and more direct to design an
amplifier that responds linearly to an exponential control contour,
especially since ADSR envelopes are usually created from charging capacitors
which tend to exhibit the exp. curve anyway. But it doesn't have to be that
way.

However, let's say you want to use the VCA to control something other than
an audio signal. What if you want to use a CV to control the gain of another
CV, in other words, use the VCA as a voltage-controlled resistor? Then you
most likely will want to use a linear CV in and a linear response to it.

If I'm wrong, somebody hit me with a wet noodle. It's late on a Friday and
I'm not thinking properly. Something in what I just wrote doesn't ring right
but I'm too brain dead to troubleshoot it. 

I know what it is. It's ANTI-log, right? So you need the exp. response to a
linear CV if you aren't using a CV input from an AL source such as an EG.

I give up, what's the answer?



		-----Original Message-----
		From:	J. Larry Hendry [mailto:jlarryh@...]

		 What would I use a linear input on a VCA for unless I was
using EGs with
		linear CV outs?

Basically, its good to have as many buttons, knobs, switches, jacks, and
blinking lights as you can possibly cram onto a black faceplate, am I right?
Or am I right?


		-----Original Message-----
		From:	Dave Bradley [mailto:daveb@...]
		So it's good to have exp envelopes plus both exp and linear
control voltage inputs on VCAs.

1. Re: ring mod as VCA - you will quite a bit more bleed through from the X
and Y channels than you will on the "real" VCA. The spec says only 40 db
separation, which is clearly audible. I use my RM as a VCA in a pinch, but
the real VCA performance is mucho superior because of the much higher
rejection ratio.

2. To paraphrase some long ago AH sarcasm:
EG linear outputs -> VCA linear inputs makes me SAD (very unnatural sounding
decay on audio signals). EG exp outputs -> VCA linear inputs makes me GLAD
(normal state of affairs for general purpose synthesis). But EG exp
outputs -> VCA exp inputs = WHOMP-ASS punch, for really dynamic short
percussive sounds. So it's good to have exp envelopes (800s are exp, BTW)
plus both exp and linear control voltage inputs on VCAs.

Dave

> From: "J. Larry Hendry" <jlarryh@...>
>
> A while back, when Paul was talking about the future MOTM-VCA. I remember
> some discussion about having both exponential and linear inputs.  Can some
> of you modular savvy guys tell me why I would want this feature in a VCA.
> Speaking from MOTM land, since that is all the modular I own, the EGs have
> exponential outputs so the VCA has a exponential input and linear
> response.
>  What would I use a linear input on a VCA for unless I was using EGs with
> linear CV outs?
>
> Larry (still trying to learn & understand this modular stuff) Hendry
>

1) It's easier to design a linear VCA. *Much* easier!
2) For volume level-type apps, you want an exponential response. But if the
CV is expo.
(ie an ADSR) then the VCA can be linear. (log x linear = log)
3) Log on a VCA is good for percussion (log x log = quadratic response). You
get
"Super Attack Transients" this way. Also, if you CV source in linear (say, a
foot pedal)
then for volume apps you reverse the symmetry (audio goes into Log in, CV
goes to linear in).

Paul S.

-----Original Message-----
From: Tkacs, Ken <Ken.Tkacs@...>
To: 'motm@onelist.com' <motm@onelist.com>
Date: Friday, October 29, 1999 2:32 PM
Subject: RE: [motm] More VCA blabbering


>From: "Tkacs, Ken" <Ken.Tkacs@...>
>
>
>The ear response to sound pressure levels (amplitude) in a log (anti-log?)
>fashion is way similar to the way pitch works.
>
>You know how, in a VCO, you have the choice of designing it to either
>respond exponentially to linear CVs (the Moog/MOTM way) or to respond
>linearly to exponential CVs (the Korg MS-series way).
>
>Same goes for a VCA. It's usually cheaper and more direct to design an
>amplifier that responds linearly to an exponential control contour,
>especially since ADSR envelopes are usually created from charging
capacitors
>which tend to exhibit the exp. curve anyway. But it doesn't have to be that
>way.
>
>However, let's say you want to use the VCA to control something other than
>an audio signal. What if you want to use a CV to control the gain of
another
>CV, in other words, use the VCA as a voltage-controlled resistor? Then you
>most likely will want to use a linear CV in and a linear response to it.
>
>If I'm wrong, somebody hit me with a wet noodle. It's late on a Friday and
>I'm not thinking properly. Something in what I just wrote doesn't ring
right

>but I'm too brain dead to troubleshoot it.
>
>I know what it is. It's ANTI-log, right? So you need the exp. response to a
>linear CV if you aren't using a CV input from an AL source such as an EG.
>
>I give up, what's the answer?
>
>
>
> -----Original Message-----
> From: J. Larry Hendry [mailto:jlarryh@...]
>
> What would I use a linear input on a VCA for unless I was
>using EGs with
> linear CV outs?
>
>
>>

Yes, yes: we never want to use that word in connection with MOTM. "More
efficient" and "cost effective" would be acceptable. I stand corrected.

Yeah, when you talk about "expo" & "linear," it can get confusing. I always
like to talk about synthesizer functions as being "exponential _response_"
or "linear _repsonse_," because that makes clearer what aspect of the
connection is being discussed.

So an MOTM VCA is a linear response module, responding linearly to
exponential CV, whereas the VCO is exponential response to a linear CV.

		-----Original Message-----
		From:	J. Larry Hendry [mailto:jlarryh@...]
		Right, the MOTM way.  But instead of cheaper, I'll
substitute "more
		efficient."  :)

		...  I am calling the MOTM-110 CV input expo since that is
the voltage it expects even though the VCA response is linear.  Should I be
saying that different?

Oh yeah, that's good. Paul always gets right to the heart of it. That makes
sense.

		-----Original Message-----
		From:	Paul Schreiber [mailto:synth1@...]


		The VCA is linear, the CV input is linear. VCAs are not
adders, they are
		*multipliers*.

		linear x linear = linear
		log x linear = log
		log x log = quadratic polynomial

> From: "Paul Schreiber" <synth1@...>
> 
> 1) It's easier to design a linear VCA. *Much* easier!
> 2) For volume level-type apps, you want an exponential
>    response. But if the CV is expo. (ie an ADSR) then 
>    the VCA can be linear. (log x linear = log)

OK, I got that.  That's the MOTM way CV is expo and VCA is linear (ie
expects a expo CV to have a linear response).  Right?  So, I would call the
input on the MOTM-110 expo.

> 3) Log on a VCA is good for percussion (log x log =
> quadratic response). You get "Super Attack Transients"
> this way.

OK, so if your VCA was designed to have a expo curve (ie expecting a linear
CV) and you fed it a expo CV from your EG you would get that quadratic
response.  That makes sense.  So this is one good reason to have a "linear"
input on a VCA. You could put expo voltage to it for this effect.

>  Also, if you CV source in linear (say, a foot pedal)
> then for volume apps you reverse the symmetry 
> (audio goes into Log in, CV goes to linear in).

OK.  Got that, anytime you have a linear CV source.....

>From: "Tkacs, Ken" <Ken.Tkacs@...>
>
> Same goes for a VCA. It's usually cheaper and more
> direct to design an amplifier that responds linearly to
> an exponential control contour, especially since ADSR
> envelopes are usually created from charging capacitors
> which tend to exhibit the exp. curve anyway.

Right, the MOTM way.  But instead of cheaper, I'll substitute "more
efficient."  :)

> However, let's say you want to use the VCA to control
> something other than an audio signal. What if you want
> to use a CV to control the gain of another CV, in other
> words, use the VCA as a voltage-controlled resistor? 
> Then you most likely will want to use a linear CV in
> and a linear response to it. 

Ok, I understand the concept of linear + linear = linear.  But,  The
MOTM-110 does that with a "expo" input since the 110 is linear in response.
 Right?  So if you put a linear CV into the MOTM-110 you would get a linear
output.  If you put expo in you get an expo output.  Maybe terms is where I
am getting messed up.  I am calling the MOTM-110 CV input expo since that
is the voltage it expects even though the VCA response is linear.  Should I
be saying that different?

Larry (brain exponentially declining with age) Hendry

> From: "Dave Bradley" <daveb@...>

> 2. To paraphrase some long ago AH sarcasm:
> EG linear outputs -> VCA linear inputs makes me SAD (very unnatural
sounding
> decay on audio signals). EG exp outputs -> VCA linear inputs makes me
GLAD
> (normal state of affairs for general purpose synthesis). But EG exp
> outputs -> VCA exp inputs = WHOMP-ASS punch, for really dynamic short
> percussive sounds. So it's good to have exp envelopes (800s are exp, BTW)
> plus both exp and linear control voltage inputs on VCAs.

And, we all need more WHOMP ASS.  :)  Thanks Dave.  I think Paul called
this quadratic response.  HAHA..  You still reading AH ?  Too much traffic
for me.  
LH

You are confused, my son. Unfortunately, I must explain using a few
engineering terms.

> OK, I got that.  That's the MOTM way CV is expo and VCA is linear (ie
> expects a expo CV to have a linear response).  Right?  So, I
> would call the
> input on the MOTM-110 expo.
>

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.

> OK, so if your VCA was designed to have a expo curve (ie
> expecting a linear
> CV) and you fed it a expo CV from your EG you would get that quadratic
> response.  That makes sense.  So this is one good reason to have
> a "linear"
> input on a VCA. You could put expo voltage to it for this effect.
>

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. 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.

Moe

> From: "Paul Schreiber" <synth1@...>
> 
> Larry is "saying it wrong".

Thanks Paul.  I knew was messin' something up.  So the input on the
MOTM-110 would be called "linear" even though it is expecting a expo
voltage.  So, the CV input is identified by the response that the VCA
exhibits not the control voltage it is expecting.  If the MOTM-110 had a
switch on it, it would be stuck in the "linear" position then?

Larry (getting closer to "saying it right") Hendry

BTW, I think Dave B said it right when he pulled out that "Whompass" quote
from AH.  Wow!

Larry is "saying it wrong".

The VCA is linear, the CV input is linear. VCAs are not adders, they are
*multipliers*.

linear x linear = linear
log x linear = log
log x log = quadratic polynomial

Paul S.

>From: "Tkacs, Ken" <Ken.Tkacs@...>
>Basically, its good to have as many buttons, knobs, switches, jacks, and
>blinking lights as you can possibly cram onto a black faceplate, am I
right?
>Or am I right?


the absolute truth (so long as the knobs are shiny and have a good knob to
remaining panel surface area ratio:-)
cheers paulb

> 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

> From: "J. Larry Hendry" <jlarryh@...>

> 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

YES! YES! 

Moe

> From: "Tkacs, Ken" <Ken.Tkacs@...>
>
>
> Basically, its good to have as many buttons, knobs, switches, jacks, and
> blinking lights as you can possibly cram onto a black faceplate,
> am I right?
> Or am I right?

Nope, in my opinion. Crowding a panel hurts the useability a lot. One of the
bad things about Modcan and ESPECIALLY Doepfer is how crowded the panels
are, and how small the knob diameter is. Makes it real hard to set precise
values on skinny knobs with fat fingers. It's a tradeoff between compactness
of the system, features and useability.

My particular bias is towards features and useability at the expense of
compactness. It's a modular, dag nabbit! If you are going to put in
features, allow enough panel space to contain them. Also, IMHO it's better
to have a general purpose standalone module (e.g. a lag processor) than to
build a similar dedicated feature into one or more modules (e.g. glide on a
sequencer module).

Everyone has different opinions, but MOTM is a large format design, so it's
going to be hard to get a 5 or 10 space portable rack with a zillion
features.

Dave

Well, sure; I was being facetious. I'm an Industrial Designer by training,
so, of course 'human factors' is what I used to eat, sleep, and breathe. I
figured we were in a 'whomp-ass' mode of speech.

		-----Original Message-----
		From:	Dave Bradley [mailto:daveb@...]
		Sent:	Friday, October 29, 1999 4:57 PM
		To:	motm@onelist.com
		Subject:	RE: [motm] More VCA blabbering

		From: "Dave Bradley" <daveb@...>



		Nope, in my opinion. Crowding a panel hurts the useability a
lot. One of the
		bad things about Modcan and ESPECIALLY Doepfer is how
crowded the panels
		are, and how small the knob diameter is. Makes it real hard
to set precise
		values on skinny knobs with fat fingers. It's a tradeoff
between compactness
		of the system, features and useability.

>> From: "Dave Bradley" <daveb@...>

>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.  :)
>
wasn't Neinsky the one who came up with that sample rate theorem..?
I'm getting confused
I'm sure I got into music to get away from all this mathematics
and anyway, when's Paul going to give us a "whompass" filter..?
cheerspaulb (desperately trying to tackle an income spreadsheet)

> From: "Tkacs, Ken" <Ken.Tkacs@...>
>
>
> Well, sure; I was being facetious. I'm an Industrial Designer by training,
> so, of course 'human factors' is what I used to eat, sleep, and breathe. I
> figured we were in a 'whomp-ass' mode of speech.
>

Ahh so, Grasshopper.

Unable to detect irony when my hot button is pressed,

Moe

On Fri, 29 Oct 1999, Paul R Bower wrote:
> wasn't Neinsky the one who came up with that sample rate theorem..?
> I'm getting confused

Nope, Neinsky was a 17th century philosopher who claimed that God
was NOT dead, and if you stare too long into the void, you might
get bored.

> and anyway, when's Paul going to give us a "whompass" filter..?

Heck, I'd settle for a "whup-ass" filter, if there's not enough
"whompass" to go around. :-)

Dan Higdon (hdan@...)

On Fri, 29 Oct 1999, Dan Higdon wrote:

> > and anyway, when's Paul going to give us a "whompass" filter..?
> 
> Heck, I'd settle for a "whup-ass" filter, if there's not enough
> "whompass" to go around. :-)

  Didn't Paul used to hunt whomprats in his T-16? ;)

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