[sdiy] Re: Leapfrog

[jhaible at debitel.net]

> All right, I've taken a closer look at the integrator
> (still numerical, not analytical), and it's an integrating
> function as it should.

... and analytically, it's an integrator, too.
With all equal resistors Ro, and only one input, it would
be H(jw) = 2/jwRoCo.

With two inputs, and 2Ro input resistors, the NIC circuit
sees Ro (parallel resistors), and the signals are both
divided by 2 because of the mutual loading of the two
2Ro resistors. So the output is 
Vout = 2/jwRoCo (1/2 vin1 + 1/2 vin2)
     = (vin1 + Vin2) /jwRoCo

(ideal opamp with infinite gain case, of course.
TL071 -> good above 1Hz, see previous mail.)

JH.



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