# transfer function confusion

It has been said that Vout = Va/2 I'm not sure why this would be. Could someone please tell me why?

• Because the impedances form a voltage divider. – Ignacio Vazquez-Abrams Jun 9 '16 at 4:48
• And just what exactly would have you believe that half of the impedances are present at node Va? – gorge Jun 9 '16 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. – mkeith Jun 9 '16 at 6:54

• @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). – nidhin Jun 9 '16 at 9:46