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Cascaded low-pass passive with high-pass active

How do I get my H(s) transfer function from this circuit? If there was infinite impedance between the two filters, I'd just do H_LP (s) * H_HP (s). However, we are assuming that the input/output impedance of the high/low pass components are having a non-negligible effect, call it H_z (s). So our final equation should look like: H(s) = H_LP (s) * H_z (s) * H_HP (s). I'm trying to quantify that H_z (s) term.

Solving the circuit should help with H(s). I think there's something I can do with KCL, but I'm stuck. If you could walk me through getting Vi and Vo as a function of s, R and C, that'd be appreciated as well.

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    \$\begingroup\$ In general, cascading multiple filters (without a buffer) doesn't do exactly what you'd expect because you're loading each stage. See also: electronics.stackexchange.com/q/90277/49251 \$\endgroup\$
    – Greg d'Eon
    Commented Mar 2, 2015 at 3:38
  • \$\begingroup\$ @Kynit, that's why he's asking this question because he understands that the effect is different when there's not infinite impedance between the two filters. \$\endgroup\$
    – horta
    Commented Mar 2, 2015 at 4:16
  • \$\begingroup\$ Hmm, looks like I commented too quickly. I missed the actual question and the form that he's written down. Oopsies! \$\endgroup\$
    – Greg d'Eon
    Commented Mar 2, 2015 at 4:19
  • \$\begingroup\$ The circuit resembles a very poor bandpass (bad selectivity). Is it a - more or less - academic exercise or do you really want to use the bandpass? Because there are other - better! - bandpass topologies (same parts count). \$\endgroup\$
    – LvW
    Commented Mar 2, 2015 at 8:44

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I would treat it like any other op-amp circuit.

Start at the right and work your way back.
(V_0-0)/R3=I0

That same current must flow from the - terminal to V1 so:
I0= (0-V1)/(R2+1/(sC2))

And the current going through R1 towards V1 is:
I1 = (Vi-V1)/R1

And the current flowing down from V1 is:
I2 = V1/(1/sC1)

Lastly, you know that the currents entering and leaving the nodes must be equal so at V1 you have:
I0+I1=I2

You should now have the equations to solve for everything in reference to Vo/Vi which is H(s)

Solving it all the way through I get this: $$ H(s)=\frac{V_o}{V_i}=\frac{-R_3C_2S}{(R_1C_1S+1)(R_2C_2S+1)+R_1C_2S} $$

Hopefully I didn't muck that up in the algebra...

From the looks of it, it looks like a bandpass filter due to the single order S term in the top.

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