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\$C_1\$, and \$Z_2\$ to the impedance of \$C_2\$, we get a simple voltage divider with the following transfer function: $$H(s) =\frac{V_o}{V_i}= \frac{Z_2}{Z_1 + Z_2}$$ The impedances are: $$Z_1 = R … }}$$ and, depending on the ROC of the transfer function: $$\boxed{H(j\omega) = \frac{RC_1j\omega + 1}{R(C_1 + C_2)j\omega + 1}}$$ Edit: All this is valid if the circuit is initially relaxed. In this case if the capacitors are initially discharged. See s-domain equivalent circuits and impedances. …
answered May 12 '16 by Paulie