You would need to write out the equations and derive it. However, you need to decide what OPAMP model you will use. If you use the simplest OPAMP model assuming infinite gain and bandwidth, you will get the following:hey guys. wondering if anyone can help me determine the transfer function v2/v1. im a bit buggered on this, the placement of resistor one has stumped me.
much appreciated~
The major mistake that I see is that the equations for I1 and I3 are not right. For current through an impedance device, you need to take the current equal to the voltage difference, divided by the impedance. Instead, you took the voltage at one end only. You should consider the node voltage (Vn) at the connection of C1, C2 and R1.thx for such a quick response guys. sorry i forgot to mention i assumed ideal op-amp. i have attached my working. i know it is plain wrong but hopefully someone can tell me where i went wrong.
Simplifying equalities. Near the end of the analysis I replace the Xc terms with their s-domain equivalents.
\(X_{C1} = \frac{1}{sC_1}\)
\(X_{C2} = \frac{1}{sC_2}\)
Look Ma, no loops.\(V_x={\frac{{\frac{V_2}{R_1}}+{\frac{V_1}{X_{C1}}}}{\frac{1}{R_1}+{\frac{1}{X_{C1}}}+{\frac{1}{X_{C2}+R_2}}}}\)
\(V_+=\frac{{\frac{V_x}{X_{C2}}}} {{\frac{1}{X_{C2}}}+{\frac{1}{R_2}}}\)
The above equation can be solved for Vx:
\(V_+=\frac{{V_x}{R_2}} {{R_2}+{X_{C2}}\)
\(V_x=\frac{V_+(R_2+X_{C2})}{R_2}\)
Simplify the first equation at the top of this box and then set the two different expressions for Vx equal to each other and solve for V2/V1.
Oh yeah, I almost forgot. V+ = V_ = V2 due to ideal opamp operation.
by Gary Elinoff
by Gary Elinoff
by Cabe Atwell