# Laplace Transfer Function

Discussion in 'Analog & Mixed-Signal Design' started by lcesero, Oct 1, 2018.

1. ### lcesero Thread Starter New Member

Dec 4, 2017
3
0
Hi All,

I am trying to figure out on how I can eliminate E3 in both of the node equations so that I can have one equation with just E1 and E2.
I am thinking that I can solve the node E3 equation for E3 and substitute that back into my equation. This will give me an equation with just E1 and E2 so that I can compute the transfer equation. Am I on the correct path with this thinking.

Please see attachments for Schematic, Node equations, and Transfer Equation

Regards,

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2. ### drc_567 AAC Fanatic!

Dec 29, 2008
585
84
Generically speaking, the easiest approach is to draw 3 current loops, say one clockwise in each window of the circuit. Doing this will give 3 equations, with the 3 currents ... I1, I2, and I3. You follow each current loop clockwise, using the individual voltage for each component. For components that are members of two loops, you will have a current difference ... something like R(I1-I2).... You don't have to bother with node voltages ... just solve for the individual I's ... then take a look at what is required after you solve for the currents. ...There should be some example problems someplace.

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,666
473
Don't you trust your algebra? Here's what I get:

4. ### lcesero Thread Starter New Member

Dec 4, 2017
3
0
Hi DRC_567,

Thanks for the response. I will take a look at doing the mesh analysis and solve for the transfer function.

5. ### lcesero Thread Starter New Member

Dec 4, 2017
3
0
Hi Electrician,

I do trust my algebra. Lol. I can solve this equation using Mathlab or Octave but I wanted to solve this equation using paper just to brush up on my laplace transforms and algebra. Thanks for the suggestion for using the inverse laplace. I will look into that also.