[sdiy] A little OT, Begging a hand

David G. Dixon dixon at interchange.ubc.ca
Wed Feb 9 04:25:47 CET 2011


> >>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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