Circuit Theory/Active Filters/Example90



Start with writing a node equation:
 * $$i_1 -i_2 - i_3 -i_0 = 0$$
 * $$\frac{V_{in}- V_{node}}{R_1} - V_{node}sC_1 - \frac{V_{node}-V_o}{R_3} - \frac{V_{node}}{R_2} = 0$$

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:
 * $$\frac{V_{node}}{R_2} - (0-V_osC_2) = 0$$

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:
 * $$-\frac{R_3}{R_1 + (C_2R_1R_2 + C_2R_1R_3 + C_2R_2R_3)s + C_1C_2R_1R_2R_3s^2}$$

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