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I'm trying to draw an approximate gain vs frequency curve for this RC circuit:

RC Circuit

I can solve for specific points, i.e. when \$\omega=\frac{1}{RC}\$ the Gain scale will be at -3dB but I'm not sure what values to plug in for R & C since I have two resistors and two capacitors in this circuit. Am I just allowed to treat them like they're in series and use \$ C=C_1C_2\div(C_1+C_2)\$ and \$R=R_1+R_2\$

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Am I just allowed to treat them like they're in series and use \$ C=C_1C_2\div(C_1+C_2)\$ and \$R=R_1+R_2\$ ?

No, you can't do that.

To solve this, the easiest way (IMO) is to solve for the transfer function in the s-domain:

  • First, find the s-domain equivalents of all the circuit elements (hint: resistors are \$R\$, capacitors are \$\frac{1}{sC}\$).

  • Combine R1, C2; then combine that with R2. Now you have a voltage divider, you can solve for the voltage at the junction of C1, R1, C2.

  • Go back to the uncombined R1, C2. You know the voltage on the left of R1, so now R1, C2 is another voltage divider. You now have transfer function of the circuit, and can easily plot the frequency response from that.

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    \$\begingroup\$ Just my 2 cents, but I think nodal analysis is easier to do than doing multiple voltage dividers / combined impedances. It's a little more generic and formulaic. \$\endgroup\$ – Swarles Barkely Feb 22 '16 at 2:02
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Your formula assumption doesn't work, but you can solve it analytically quite easily. Capacitors are just complex resistors.

Use an impedance of R for resistors, and 1/(j w C) for capacitors, and solve the circuit as normal. You'll need to keep track of all the complex numbers. You should be able to find an expression for Vout/Vin in terms of R1, R2, C1, C2 and w quite easily. Take the absolute value if you're only interested in the magnitude of the response.

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  • \$\begingroup\$ Is the problem simplified at all if I can treat the circuit as being made up of two independent RC filters? \$\endgroup\$ – Craig Feb 22 '16 at 4:27
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    \$\begingroup\$ No - because both sections are not electrically decoupled. \$\endgroup\$ – LvW Feb 22 '16 at 10:20

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