What is the transfer function of this LC low-pass filter?

Discussion in 'General Electronics Chat' started by zhaojia, Nov 23, 2007.

  1. zhaojia

    Thread Starter New Member

    Nov 23, 2007
    7
    0
    I want to get a transfer function of the following LC low-pass filter.
    And, I want to know how it's cut-off frequency is derived.
    Thanks for your help.:)

    An LC low-pass filter:

    http://www.williamson-labs.com/images/filters-lc.gif
    (The first one, the Ui(jw) is at left for input and Uo(jw)) is the output at right.)
    [​IMG]
    The first circuit.

    What is the expression for G(jw) = Uo(jw) / Ui(jw)?
    And What is the cut-off frequency wc?
    {the value wc makes G(jwc) = 1/sqrt(2) * G(j0) }
     
  2. zhaojia

    Thread Starter New Member

    Nov 23, 2007
    7
    0
  3. zhaojia

    Thread Starter New Member

    Nov 23, 2007
    7
    0
    Sorry.
    In the thread, I said:
    And What is the cut-off frequency wc?
    {the value wc makes G(jwc) = 1/sqrt(2) * G(j0) }
    This is NOT the correct concept of cut-off frequency.

    The cut-off frequency, in fact, is a corner frequency f0 at which the attenuation characteristics emerge.
    It can be derived form the mathematically reform.
     
  4. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    fc = 1/(2*\Pi*R*C)
     
  5. zhaojia

    Thread Starter New Member

    Nov 23, 2007
    7
    0
    Thanks. But this fc is the cut-off frequency of a RC (NOT LC) low-pass filter.
     
  6. GS3

    Senior Member

    Sep 21, 2007
    408
    35
    http://en.wikipedia.org/wiki/Cutoff_frequency

     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Imagine that the inductor were replaced by a resistor, R1, and the capacitor were replaced by another resistor, R2. Then you would have a voltage divider, and Vo/Vi would be given by R2/(R1+R2). If you replace the resistance of R1 by the impedance of the inductor, j w L, and the resistance of R2 by the impedance of the capacitor, 1/(j w C), the formula for the voltage divider will become the transfer function you want, namely ZC/(ZL+ZC). You will have to do some algebraic simplification to get it to look like what you would find in a book.
     
  8. zhaojia

    Thread Starter New Member

    Nov 23, 2007
    7
    0
    Thank GS3 and The Electrician for the aproperate and nice reply.
     
Loading...