Transfer function of circuit composed of resistors and capacitors

It has been said that $$V_{\text{out}} = \frac{V_{\text{a}}}{2}$$ I'm not sure why this would be. Could someone please tell me why?

• Because the impedances form a voltage divider. Commented Jun 9, 2016 at 4:48
• And just what exactly would have you believe that half of the impedances are present at node Va? Commented Jun 9, 2016 at 5:35
• What exactly does that mean "half of the impedances are present at node Va?" Assuming no load at Vout, it is very obvious that Vout = Va / 2 because the two capacitors of the same value (C) form a voltage divider that cuts the input (Va) in half. Commented Jun 9, 2016 at 6:54
• It would be preferable if you had the signal flowing from left to right, as is convention. Commented Mar 12, 2021 at 18:21

At any frequency, the impedance offered by those two capacitors will be same (because both have same capacitance). Since same current flows through them, the voltage drop across them will also be the same = Va/2.

• The only two components with the same current flowing through them is the capacitor and resistor in series near Vin Commented Jun 9, 2016 at 9:24
• these two rail systems confuse me a bit but after converting it into a single line i understand now. i.e. vin-------C1------VOUT--------C2--ground Commented Jun 9, 2016 at 9:30
• Do you have any tips on how to not be confused by things like this? Commented Jun 9, 2016 at 9:34
• by vin above i actually mean Va by the way Commented Jun 9, 2016 at 9:36
• @gorge same current flows through $R_1$ and $C_1$. At node Va, this splits into two. $V_a/R_2$ flows to ground through the resistor and $V_a/(\frac{1}{jwC} + \frac{1}{jwC})$ flows through the capacitors $C_2$ and $C_3$ (left most). Commented Jun 9, 2016 at 9:46