0
\$\begingroup\$

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!

\$\endgroup\$
  • \$\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
1
\$\begingroup\$

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

\$\endgroup\$
  • \$\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
0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.