# Circuit Theory/Active Filters/Example90

Start with writing a node equation:

But the last expression for i_{o} could have also been written using C_{2}. And we need to eliminate V_{node}. So our second equation is associated with the feedback resistor:

Now these two equations can be solved for V_{o}/V_{in}:

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 V_{out}/V_{in} 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).