Circuit Theory/Active Filters/Example90

Start with writing a node equation:

Active filter ... find transfer function

But the last expression for io could have also been written using C2. And we need to eliminate Vnode. So our second equation is associated with the feedback resistor:

Now these two equations can be solved for Vo/Vin:

solve([(vin-vnode)/r1 - vnode*s*c1 - (vnode-vout)/r3 - vnode/r2, vnode/r2 - (0-vout*s*c2)],[vout,vin])

The transfer function is going to be the ratio of Vout/Vin so:

vout := -vnode/(c2*r2*s)
vin := (vnode*(r1 + c2*r1*r2*s + c2*r1*r3*s + c2*r2*r3*s + c1*c2*r1*r2*r3*s^2))/(c2*r2*r3*s);
vout/vin
-r3/(r1 + c2*r1*r2*s + c2*r1*r3*s + c2*r2*r3*s + c1*c2*r1*r2*r3*s^2)

Yields this:

Which looks like a low pass filter (inverted because of the op amp).