2
\$\begingroup\$

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.

\$\endgroup\$
  • 1
    \$\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 Mar 2 '15 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 Mar 2 '15 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 Mar 2 '15 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 Mar 2 '15 at 8:44
2
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.