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I am assuming this configuration:
-----------|Capacitor 2|----------|--------|-------
| |
Vin [R] [C1] Vout
| |
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-------------------------------------------------------
1. Using the Laplace transformer, C1 impedance equals 1/ sC1 (similar for C2).
2. Then treat C1 as you would a resistor to reduce the network:
For two parallel resistors, the combined resistance = R1*R2 / (R1 + R2)
Zparallel R and C1: R*1/sC1 / (R + 1 / sC1) = R / (RsC1 + 1)
3. Then to determine how the input varies from the output, Vout / Vin, using voltage divider
Voltage divider for two resistors: Vout = Vin * R1 / (R1+R2)
For the above: Vout = Vin * [R/RsC1 + 1] / ([R/RsC1 + 1] + 1/sC2),
If you do a little manipulation, you will get: Vout = Vin * RsC1 / (RsC2 + RsC1 + 1)
What else can I help you with?
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