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I have the following circuit and I need to find the transfer function:

schematic

simulate this circuit – Schematic created using CircuitLab

Considering the position of \$V_{\text{out}}\$, it looks like I can combine the \$R\$ and \$C\$ in parallel both above and below \$V_{\text{out}}\$ and then use voltage division to get the voltage across the bottom \$R\$ and \$C\$ in parallel: $$V_{\text{out}} = V_{\text{in}}\frac{R||C}{(R||C)+(R||C)}$$ And this would leave me with a transfer function of: $$\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{1}{2}$$ However, I'm pretty sure this isn't correct. Am I looking at the circuit incorrectly when assuming I can combine the \$R\$ and \$C\$ in parallel?

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In order for the \$R\$ and \$C\$ to be in parallel, you would need \$V_{\text{out}} = 0\$ due to a short circuit. But that's not the case.

First calculate \$V_{\text{out+}}\$, the voltage at the + terminal of \$V_{\text{out}}\$. This is just a voltage divider:

$$V_{\text{out+}} = \frac{1/sC}{1/sC + R}V_{\text{in}} = \frac{1}{1 + sRC}V_{\text{in}}$$

Now calculate \$V_{\text{out-}}\$, the voltage at the - terminal of \$V_{\text{out}}\$. This is also just a voltage divider:

$$V_{\text{out-}} = \frac{R}{1/sC + R}V_{\text{in}} = \frac{sRC}{1 + sRC}V_{\text{in}}$$

Now you can calculate \$V_{\text{out}}\$:

$$V_{\text{out}} = V_{\text{out+}} - V_{\text{out-}}$$

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The issue is you're not treating the capacitance correctly as a frequency-dependent impedance. Your equation is only valid at a single frequency.

The way to do this is to write the transfer function for the low-pass section (Vout+), then write down the transfer function for the high-pass section (Vout-) then take the difference (vout = vout+ - vout-).

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  • \$\begingroup\$ Your method is correct, but the issue is not what you've stated. Assuming the Vout/Vin relationship is as given in the question, the frequency dependence cancels when simplified. The issue is that the resistor and capacitor are not in parallel, so the Vout/Vin relationship given in the question is not correct. \$\endgroup\$ – Null Apr 13 '15 at 18:30
  • \$\begingroup\$ If Ryan (the OP) had calculated each TF independently it would have been obvious they weren't in parallel, so I still think the core issue was Ryan didn't understand what a differential output was. I suspect he just wanted you to do the math for him and I'm sure he is glad you were so accommodating. \$\endgroup\$ – crgrace Apr 13 '15 at 18:48

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