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Zeros in a circuit transfer function

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Complete the circuit in the figure in order to get a transfer function as: H(s) = k*[s(s^2+10^10)]/[(s+10^3)(s+10^5)(s^2+1.41*10^5s+10^10)].

    3. The attempt at a solution
    If I split the circuit in four stages: H1(s), H2(s), H3(s) and H4(s), I obtain:

    H1(s) = -(1/(R1*C1))/(s+1/(R2*C1)). One single pole.
    H3(s) = -(1/(R3*C2))/(s+1/(R4*C2)). One single pole.
    H4(s) = (1/(LC3))/(s^2+(1/(R5*C3))s + 1/(LC3)). Complex pole.

    How can I build an H2(s) stage to include a simple zero and a double zero? How can I build a circuit like that? It cannot introduce poles!

    Thank you.

    Attached Files:

  2. jcsd
  3. Sep 14, 2011 #2

    rude man

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    Highly impractical, but you can add 1 op amp stage with a C input & R feedback (giving you the zero at the origin), and a second stage with a parallel L-C input and L feedback (giving you the complex-conjugate zero pair on the imaginary axis).

    In real life you would not have two single-pole stages like you have to deal with ....
  4. Sep 14, 2011 #3
    Thank you.

    Curiously, what happens in real life if you build a circuit like this?
  5. Sep 14, 2011 #4

    rude man

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    First, realize that your transfer function is not realizable in any practical way. That's because it calls for zero gain at w = sqrt(10^10) = 1e5 rad/s or 1e5/2pi Hz. That is impossible to get and still have finite gain at other frequencies. So this problem is a "textbook" problem & so it's OK to use impractical circuits like what I gave you.

    The s in the numerator would be made part of a Ts/(Ts + 1) circuit so you'd get you zero at the origin plus your pole at s = - 1/T. That's just a series R-C in the input and R in the feedback:
    Vout/Vin = -Zf/Zi = -Rf/(Ri + 1/sC) = -(Rf/Ri)sRiC/(sRiC + 1).

    There are fancy networks (like 3 R's and 2C's in the feedback etc.) for synthesizing complex-conjugate zeros (or poles) but never if the poles or zeros have to be right on the imaginary axis, since that implies an infinite-Q circuit.
  6. Sep 15, 2011 #5

    The Electrician

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    It's not difficult to place a complex zero right on the jw axis (or close to the jw axis on either side). Twin-T and bridged-T networks are one way to do it.
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