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I would like to calculate the gain of this simple band-pass filter.

schematic

simulate this circuit – Schematic created using CircuitLab

What I'm interested in knowing is, in order to find this gain, would I have to set up a set of differential equations using Kirchoff's rules, or could I avoid this through some other trick? I'm particularly interested in the instance when the input angular frequency is equal to the characteristic frequency, $$\omega=\frac{1}{RC}$$

NOTE: I'm not asking for the answer, but rather for a hint a possible route to getting there (a trick). Also, if this question is not fit for EESE, then I apologize.

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    \$\begingroup\$ There is no "trick" necessary. The circuit is nothing else than a primary voltage divider, where one part (the most right part) is a "secondary" voltage divider. Use the capacitive impedance Z=1/(jωC) \$\endgroup\$ – LvW Feb 26 '15 at 11:20
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You need the complex frequency response, which you can get by using the voltage divider rule with complex impedances. You probably know that for a capacitor you have \$Z=1/(j\omega C)\$. This way you can express the ratio \$V_{out}/V_{in}\$ as a complex function of the independent variable \$\omega\$ (this is the complex frequency response). The gain for a given frequency is then the magnitude of the frequency response evaluated at that frequency.

If you do the math you should get for \$\omega=1/RC\$ a gain

$$G=\frac{A}{1+2A}$$

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  • \$\begingroup\$ Thank you. I can't believe I forgot that I could do that. \$\endgroup\$ – Arturo don Juan Feb 26 '15 at 20:38

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