new to this forum. I have a question about a certain topic I'm struggling with. I have this circuit, C1 and C2 are equal and therefore just described as C:

enter image description here

I want to deduce the transfer function Vout/Vin with the help of node voltage analysis. This is what I have as of now:

Equation 1: Vb = 0

Equation 2: (Va-Vin)/R1+Va/R2+(Va-Vb)/(1/(I*2*pifC))+(Va-Vout)/(1/(2*Pi*I)fC) = 0

I can't seem to get it to work. I define the equations in maple and solve them like a system, but I believe I must be writing the equations wrong. Can anybody give me a helping hand?

Thanks in advance!

  • \$\begingroup\$ You need the node B equation: \$\frac{0-V_a}{j \: 2\pi fC}+\frac{0-V_{out}}{R_3}=0\$ \$\endgroup\$ – Chu May 27 '16 at 21:53

You did not include R3 into your equations. Below you can find a solution using Mathematica. enter image description here

  • \$\begingroup\$ The capacitor C2 is equal to C1. e3 := (vb-va)*(I*2)*PifC+(vb-vout)/R3 = 0 e2 := (va-vin)/r1+va/r2+(va-vb)*(2*PifCI)+(-vout+va)*(2*PifCI) = 0 e1 := (Vout-Vb)/R3-(Va-Vb)*(I*2)*PifC = 0 e4 := vb = 0 solve({e1, e2, e3, e4}, [va, vout]) holy shit formatting \$\endgroup\$ – dnopas May 27 '16 at 21:53
  • \$\begingroup\$ The thing I don't understand in your picture is that you have two equations that are identical, eq1 and 3 \$\endgroup\$ – dnopas May 27 '16 at 21:56
  • \$\begingroup\$ I made an update. \$\endgroup\$ – Mario May 27 '16 at 22:01
  • \$\begingroup\$ Thanks dude! Not sure why, but it worked replacing 2PifI*C with s. Somehow maple didnt want to solve it in frequency. \$\endgroup\$ – dnopas May 27 '16 at 22:10

To perform this analysis using impedances in the frequency world, you will need to use complex algebra and make the capacitive impedances to be imaginary; that is the terms in you equations must have a "j" to indicate they are imaginary. I don't know what maple is, but the voltages and currents will have a real and imaginary component.

  • \$\begingroup\$ The impedances of the capacitors are written as 1/(2PiIf*C) where I is the imaginary unit, so basically the same. The impedance of the resistors is as they are just their value in ohms. Maple is a CAS program. By writing equations I should get x amount of equations with x amounts of unknowns and solve these. I just can't get the equations right. \$\endgroup\$ – dnopas May 27 '16 at 20:53
  • \$\begingroup\$ Sorry, I didn't realize this. If your software handles complex numbers you are all set. Just add (Vout-Vb)/R3 - (Va - Vb)/ (1/(I*2*pifC)) = 0. \$\endgroup\$ – John Birckhead May 27 '16 at 21:08
  • \$\begingroup\$ As an electrical engineer who uses maple, I recommend solving this by hand. It will likely be faster and will benefit your understanding. \$\endgroup\$ – Andrew W. May 27 '16 at 21:16
  • \$\begingroup\$ I'm just a student, but we have been teached to solve this in maple. I understand the technicalities but we just don't solve it by hand. Since you use maple, I was wondering if you could help me solve it in maple, cause it doesn't work atm. Guess I can't upload a picture of the maplesheet in the comments though. \$\endgroup\$ – dnopas May 27 '16 at 21:32

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