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,
     
  2. drc_567

    Senior Member

    Dec 29, 2008
    418
    50
    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
    2,621
    468
    Don't you trust your algebra? Here's what I get:

    TransfX.png
     
  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.
     
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