[sdiy] A little OT, Begging a hand

[Barry Klein]

Is this an off the shelf sensing widget or a homebrew?
Luckily I don't have the ability or education to address this "solution".
I don't even know where you guys are going if you get one (or is it
several?).  I'd just hook this up on the bench and play with it.  Probably
be done by now.

Barry


-----Original Message-----
From: synth-diy-bounces at dropmix.xs4all.nl
[mailto:synth-diy-bounces at dropmix.xs4all.nl] On Behalf Of David G. Dixon
Sent: Tuesday, February 08, 2011 7:26 PM
To: 'Ian Fritz'; 'Veronica Merryfield'; 'synth DIY'
Subject: Re: [sdiy] A little OT, Begging a hand

> >>The problem is soluble.  We know that G8 = 1/2k2, and we have six
> equations
> >>for six conductances:
> >>
> >>G1 - G6 = 0
> >>
> >>G1*(VA - V1) + G2*(VA - VB) + G4*(VA - VC) = 0
> >>
> >>G2*(VB - VA) + G3*(VB - V2) + G57*(VB - VC) = 0
> >>
> >>G6*(Vout - VC) + G8*(Vout - V3) = 0
> >>
> >>G57*(VC - VB) + G4*(VC - VA) + G6*(VC - Vout) = 0
> >>
> >>1/G1 + 1/G2 + 1/G3 = 1/K
> >>
> >>This will give us G1, G2, G3, G4, G57, G6 and G8.
> 
> You are forgetting that you have to also solve for V0, since it is a
> dependent variable (not an applied voltage).  So you have one more
> variables than equations.
> 
> You have, in fact, just proved that the system is not solvable.

It is if we know VA, VB, VC and Vout, in addition to the applied voltages
V1, V2 and V3.  Otherwise, we have 9 or 10 unknowns, and the situation is
quite hopeless.

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