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Op-Amp Filter Circuit Transfer Function

Discussion in 'General Electronics Discussion' started by LordSputnik, May 1, 2013.

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  1. LordSputnik

    LordSputnik

    45
    0
    Aug 11, 2011
    Hi!

    I've been having some trouble deriving the transfer function of the following circuit:

    [​IMG]

    Currently, I start by saying T(s) = -Zf(s)/Zi(s), where T(s) is the transfer function in the complex frequency domain, Zf(s) is the feedback impedance and Zi(s) is the input impedance.

    Zf(s) = 1/sC + R = (1 + sCR) / sC = (1 + (2.63x10^-3) s) / (470x10^-9 s)

    Zi(s) = 1/sC + R = (1 + sCR) / sC = (1 + (22x10^-6) s) / (22x10^-9 s)

    Then:

    T(s) = (1 + (2.63x10^-3) s) / (1 + (22x10^-6) s) * (22/470)

    This seems to be alright to me, but when I try to work out the unit step response by splitting the numerator into two parts (one looks like 1/(1 + sT), the other is s/(1+sT), so the differentiation rule for Laplace transforms can be used), I get:

    v(s) = (22/470) * (1 + 118.5 e^(-t / 22 x 10^-6) )

    Which doesn't seem to match the output from my simulator exactly (about 10% difference).

    Any ideas?

    Thanks!
     
  2. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    I derived the transfer function and found the corresponding form in a table of Laplace transforms. Not sure what you mean by splitting the numerator. I just substituted 1/s for Vi as the unit step function.

    Of course you realize this is only a theoretical exercise since an op amp circuit as shown would not work properly with no input bias current.
     

    Attached Files:

  3. LordSputnik

    LordSputnik

    45
    0
    Aug 11, 2011
    Thanks for your help! It seems that substituting numbers into your solution gives the same results as mine - I expect it was just an error in the simulation.

    I only had 1/(s+a) in the very simple table I was using, so by splitting the numerator, I mean:

    1/s * (s+a)/(s+b) = 1/s * (s/(s+b) + a/(s+b))

    Which allowed me to work out the time domain response with the one equation I had.

    The circuit is part of a pre-built board, so I expect a resistor has been included on the board to bias the input, but this has been left off the diagram to reduce the complexity. I'll check for this today.
     
  4. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    Are the feedback components connected in parallel rather than in series?
     
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