analogimpedancekirchhoffs-lawstransfer function
How do I calculate the transfer function of this circuit? \$H(s) = \dfrac{V_o}{V_i}\$
I apply the Kirchhoff's law at the node \$V_x\$ but I still have \$V_x\$ as unknown :
\$\dfrac{V_x – V_i}{R1}+\dfrac{V_x – V_o}{R2}+(V_x – 0)sC = 0\$
simulate this circuit – Schematic created using CircuitLab
As jippie says, there is no single answer to this. But, think about the transfer function and how cascaded stages combine in (Laplace style) transfer functions. In this case, the transfer function is the product of 3 elements, gain, p1, and p2. Response of cascaded OpAmp stages will combine as the product of the stages. So, you could start by designing a 3 stage circuit with each stage carrying out one of the transfer function elements (gain, p1, and p2).
Look at how poles are formed by R and C combinations (Series or parallel) on the input of the OpAmp, and how they are formed by R and C combinations in the feedback path (between the OpAmp output and the negative input). Then you could look at ways to combine stages to end up with fewer stages (probably just one).
If you wish to redraw the circuit without the op-amp, try something like this:
simulate this circuit – Schematic created using CircuitLab
See the difference?
Best Answer
Since the current through R2 is zero, Vo = Vx. You can calculate Vo (and Vx) just using the known expression of a voltage divider:
$$ V_o = V_x = \frac{\frac{1}{sC}}{\frac{1}{sC} + R_1} V_i= \frac{1}{R_1Cs + 1} V_i. $$