RE(2): Linear vs. Exp = Pitch vs. Frequency

[jh]

Two remarks:

(1) Linear detuning is a very desirable effect indeed. (See separate reply to this
thread.) If you ever come near Nuernberg, give me a call and I'll show you.

(2) Linear detuning (i.e. constant term in beat rate) can *not* be achieved by using
the linear_FM input of an exponential VCO. The reason is that any modulation from
the linear_FM input is transposed (multiplied) by the keyboard output that controls
the expo input of the exponential converter. So removing a capacitor doesn't help
at all. What you need is injecting a tiny amount of offset *current* directly into the
CCO (current controlled oscillator) core which is part of any VCO.

JH.

-----Original Message-----
From:	Florian Anwander [SMTP:fa_diy at fa.camelot.de]
Sent:	Wednesday, April 21, 1999 11:44 PM
To:	synth-diy at mailhost.bpa.nl
Subject:	Re: Linear vs. Exp = Pitch vs. Frequency

Hello Michael,

>indicated to me that he might need a little more explanation about the
>logarithmic nature of the relationship between Pitch and Frequency.
Sorry if my english is that bad, that people misunderstand me as you and  
others did. Of cause I know the difference between Pitch and Frequency and  
between V/Hz and V/Oct (I earned fifteen years my money with explaining it  
..) and I know also you do understand it. So we have to find out where is  
the misunderstanding.

So let me try another way:

You have two VCOs, not synced, not crossmodulated, pure mixed sound. If you  
send a constant voltage into a linear FM-input of a VCO (and the FM-input is  
_not_ capacitor coupled as the most FM-inputs are...(*)) you will hear a  
detune with an defined beating rate which depends of the constant voltage.  
You can reach the same effect with an constant voltage put to the log-Input;  
the only difference is, that you will need an other amount of voltage to  
reach the same beatrate. We are speaking up to now always of constant  
voltages, no keyboard playing, no LFO...

If you give a changing voltage (lets say an LFO) to linear inputs of both  
VCOs and add the constant voltage to the linear input of one of the VCOs, the  
beating won't change with the "movement" of the LFO.
If you give a changing voltage (lets say an LFO) to LOGARITHMIC inputs of  
both VCOs and add the constant voltage to the LOGARITHMIC input of one of the  
VCOs, the beatrate won't change with the "movement" of the LFO again.

It is an offset which does not change! That is what I meant when I said:
>>Sorry, but I do not understand what you mean. A Detune is an _constant_
>>offset and there is no relevance for logarithmic or linear thinking becaus
>>you dont have differential aspects.

Something completly different is, when you say:
>[...] and they would always beat at a rate of one time per second, no
>matter what note or what octave you played.
Did you ever try this? Exactly the opposite is the fact. If you say "nomatter  
what octave you play" you mean: you put a voltage which changes in 1-Volt- 
steps (=keyboard voltage) to the log-input of both VCOs and a constant offset  
voltage to the lin-input of the second VCO. If you do this, the beating will  
change with every octave. Just try.



(*): I was "alpha tester" for Doepfer and I hated him for putting in all  
those coupling capacitors in the modulation pathes; I removed three years ago  
also the C4 in my A111-prototype to reach exactly this change of beatrate and  
you want to spoil it by telling me that effect wont work... :))))


Gruss,
 Florian Anwander
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Internet:  fa at fa.camelot.de           http://www.camelot.de/~fa/
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