Circuit Theory/Active Filters/Example90

Active filter ... find transfer function

Start with writing a node equation:

$i_1-i_2-i_3-i_0=0$

$\frac{V_{in}-V_{node}}{R_1}-V_{node}s{C}_1-\frac{V_{node}-V_o}{R_3}-\frac{V_{node}}{R_2}=0$

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

$\frac{V_{node}}{R_2}-(0-V_{o}s{C}_2)=0$

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:

$-\frac{R_3}{R_1+(C_2{R}_1{R}_2+C_2{R}_1{R}_3+C_2{R}_2{R}_3)s+C_1{C}_2{R}_1{R}_2{R}_3{s}^2}$

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