2nd order low pass filter

Discussion in 'Homework Help' started by The_Cook, Jun 1, 2014.

  1. The_Cook

    Thread Starter Member

    May 29, 2014
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    Could anyone help me find the transfer function of the 2nd order low pass filter and prove the variables are true as shown in the question?

    Any help is appreciated
     
  2. AnalogKid

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    Aug 1, 2013
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  3. The_Cook

    Thread Starter Member

    May 29, 2014
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    Would you mind explaining what a unity gain is, and if that applies to the circuit I posted.
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Unity Gain is Gain=1=Av.
    Since your Av is Av=Rb/Ra, make Rb=Ra, then Av=1 and you meet the unity gain definition of Gain=1=Av.
     
  5. The_Cook

    Thread Starter Member

    May 29, 2014
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    What does s stand for in the equation?
     
  6. shteii01

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    Feb 19, 2010
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    s=jw

    From wiki:
    "The transfer function can also be shown using the Fourier transform which is only a special case of the bilateral Laplace transform for the case where s = j\omega."
    http://en.wikipedia.org/wiki/Transfer_function
     
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  7. AnalogKid

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    If you don't know what s and unity gain are, how is it that you have a problem regarding transfer functions? The transfer function of a multi-pole filter is pretty advanced math, requiring an understanding of several layers of underlying concepts including complex algebra (imaginary numbers). I'm not criticizing you, just observing that you might be out of your depth.

    ak
     
  8. The_Cook

    Thread Starter Member

    May 29, 2014
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    Do you just write the transfer function in terms of the variables since they don't have any values given in the question or am I missing something?
     
  9. shteii01

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    Feb 19, 2010
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    No, you are not missing anything. For Part a you just write transfer function in terms of R1, R2, C1, C2, Ra, Rb. That is exactly what Part a asks you to do.
     
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes, for example
    (Vs - Va)/R1 = (Va - Vp)/R2 + (Va - Vout)*s*C1
    (Va - Vp)/R2 = Va*s*C2
    Vout = Vp * Av
    Where Av = 1 + Rb/Ra

    And you solve this for Vout/Vs
     
  11. The_Cook

    Thread Starter Member

    May 29, 2014
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    The design given on the wikipedia article does not have Ra and Rb, would you happen to have an example similar to my design circuit


    Found this example:

    http://www.daycounter.com/Filters/SallenKeyLP/Sallen-Key-LP-Filter-Design-Equations.phtml

    How would you factor in the Rb and Ra into that equation, which isn't in the example circuit but in my question
     
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