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

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. enter image description here In the Picture there is the 1/Zin function.

Thanks to all for your help in solving the question!

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  • \$\begingroup\$ You seem to have a capacitor in your path which tend to conduct better at higher frequencies \$\endgroup\$ – PlasmaHH Nov 1 '16 at 16:05
  • \$\begingroup\$ You're right. I've written the exact contrary of what I meant! \$\endgroup\$ – FataMadrina Nov 1 '16 at 16:06
  • \$\begingroup\$ Vx is not −Vo(sC2R2) \$\endgroup\$ – Sunnyskyguy EE75 Nov 1 '16 at 17:39
  • \$\begingroup\$ As an alternative, you could inject a current of 1A and measure the input voltage. \$\endgroup\$ – LvW Nov 1 '16 at 20:10
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From my visual analysis,

Zin(dc)=R1+R2//R3

for f>>f-3dB

@ f = infinity Zin = R1

schematic

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

enter image description here

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  • \$\begingroup\$ I totally agree. But what I want is a symbolic expression, in Laplace domain, of the input impedance. My approach is input voltage / input current. But I'm missing something. \$\endgroup\$ – FataMadrina Nov 1 '16 at 16:16
  • \$\begingroup\$ KCL and KVL will find the correct result \$\endgroup\$ – Sunnyskyguy EE75 Nov 1 '16 at 16:19
  • \$\begingroup\$ Sure. But where? \$\endgroup\$ – FataMadrina Nov 1 '16 at 16:27
  • \$\begingroup\$ KCL on all input nodes and use Vin+-Vin-=0 with zero differential impedance ( virtual "floating" gnd) \$\endgroup\$ – Sunnyskyguy EE75 Nov 1 '16 at 16:32
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    \$\begingroup\$ SAPWIN is a symbolic analyzer which can calculate input and output impedances as well as the transfer function - all in symbolic form and (if you wish) as a function of frequency. \$\endgroup\$ – LvW Nov 1 '16 at 17:57
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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: -

enter image description here

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.

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