Input impedance butterworth filter with multiple feedback

Question

I have a problem finding the input impedance of a Butterworth filter. The circuit is shown in the figure below:

I have calculated the transfer function between input and output; but now I have to find the symbolic expression of the input impedance seen from the \$V_{\text{in}}\$ generator. I've tried a \$V_{\text{in}}/I_{\text{in}}\$ approach. Since

$$I_{\text{in}} = \frac{(V_{\text{in}} - V_{\text{x}})}{R_1} $$

and $$ V_{\text{x}} = -V_{\text{o}}(sC_2R_2) $$

I find that $$ Z_{\text{in}} = \frac{R_1}{1+ W(s) sC_2R_2} $$ Where \$W(s)\$ is the transfer function $$ W(s) = \frac{V_{\text{o}}}{V_{\text{in}}} $$ This sounds wrong to me, because the input impedance should decrease at high frequency, and in my case it's increasing. Where am I wrong? Thank for the precious help!

EDIT : I've checked it with SAPWIN, and it looks like the expression above is correct. In the Picture there is the 1/Zin function.

Thanks to all for your help in solving the question!

Answer 1

From my visual analysis,

Zin(dc)=R1+R2//R3

for f>>f-3dB

@ f = infinity Zin = R1

^{simulate this circuit – Schematic created using CircuitLab}

DC gain is -R3/R1 where Vx(f=0)=0

For visual aid on Vx see Attenuation, phase shift of xy for VI plot using Java with sweep. You can change any parameter

Answer 2

I would consider using superposition and yes, you have to factor in the transfer function because R3 (and the output) imposes a significant degree of complexity on the input impedance. First the easy stuff; you can forget C2, and R2 can be set in parallel with C1 because the op-amp has a virtual earth. So, it boils down to: -

Now you have two voltage sources and a few fixed value impedances so use superposition to calculate what the voltage at "X" is then you can calculate the current through R1 and then you have the input impedance.