# Circuit Theory/Active Filters/Example90

Active filter ... find transfer function

${\displaystyle i_{1}-i_{2}-i_{3}-i_{0}=0}$
${\displaystyle {\frac {V_{in}-V_{node}}{R_{1}}}-V_{node}sC_{1}-{\frac {V_{node}-V_{o}}{R_{3}}}-{\frac {V_{node}}{R_{2}}}=0}$

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:

${\displaystyle {\frac {V_{node}}{R_{2}}}-(0-V_{o}sC_{2})=0}$

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:

${\displaystyle -{\frac {R_{3}}{R_{1}+(C_{2}R_{1}R_{2}+C_{2}R_{1}R_{3}+C_{2}R_{2}R_{3})s+C_{1}C_{2}R_{1}R_{2}R_{3}s^{2}}}}$

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