Solving for Q factor

Discussion in 'Homework Help' started by psycoadam, Oct 22, 2011.

  1. psycoadam

    Thread Starter New Member

    Oct 18, 2011
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    The problem asks to calculate the Q factor of the filter. I have read up on solving for the Q factor using the inductance. With Zin = Vin/Iin = ZL = jωL.

    After doing some circuit evaluation I came up with my Zin being
    s^2*C2*C1*nR^2

    after solving it two times I came up with the same answer twice. Then setting s = jω and then Zin = jωL I do not get all of the "s" to cancel. The problem does not offer me a frequency only the values of R,m,n,C1,C2.

    I am unsure of what to do from this point, or if I have even came at the problem the right way from the beginning. Any help would be appreciated, I also can assume I probably did not calculate the right Zin.
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    469
    You don't want to calculate Zin. Instead you want to calculate the transfer function of output voltage over input voltage. You will find it is a second order system. You then put the transfer function into standard form to match up the Q factor, and identify it.

    EDIT: Thinking about it further, your expression for input impedance doesn't look right. I say this because I just noticed that Iin=(V2-Vo)/R, hence the expression for Vo (which relates to the transfer function T=Vo/V2) is included in the input impedance.

    Hence, the input impedance can be found by Iin=V2(1-T)/R, which leads to Zin=R/(1-T). Note that if T is a second order system, then Zin will be much more complicated than what you wrote. It should look something like the following form.

    Y_{in}={{1}\over{Z_{in}}}={{1}\over{R}}{{s^2+\omega_B s+\omega_o/(m+1)}\over{s^2+\omega_B s+\omega_o}}

    where the omegas are parameters with units of radians/s, which can be related to the circuit component values. Note that the Q can be related to the  \omega_B value.
     
    Last edited: Oct 22, 2011
  3. psycoadam

    Thread Starter New Member

    Oct 18, 2011
    19
    0
    I think I understand what you are saying. I am still really confused where you got your Zin equation. if Zin = R/(1-T) how did you come up with

    [​IMG]

    I can see the general form of a second order transfer function but I am at a loss as to how R went from being in the numerator to the denominator. Another question I have is when writing the transfer function am I solving for Vo or Iin?
     
  4. psycoadam

    Thread Starter New Member

    Oct 18, 2011
    19
    0
    The more I work on this problem, the most lost I become. I would really appreciate a more indepth description. Thanks!
     
  5. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    I apologize for causing more confusion. I made a mistake above, and you are correct to note that the impedance shouldn't have units of 1/Ohms. I accidentally wrote the formula for input admittance Yin=1/Zin. I have now corrected it above.

    So let's go through a few steps to clarify.

    First, how did I get Zin=R/(1-T)? Note that the input current will equal the input voltage minus the opamp negative input terminal voltage divided by the resistance R.

     I_{in}={{V_2-V_m}\over{R}}

    But, the Vm voltage is about equal to the Vlp voltage on the other input terminal of the opamp. Hence, we get ...

     I_{in}={{V_2-V_{lp}}\over{R}}

    Now, lets define the voltage gain transfer function to be T=Vlp/V2, and substitute this in to the above equation.

     I_{in}={{V_2-T\;V_2}\over{R}}

    Factor out the V2 to get ...

     I_{in}={{V_2\;(1-T)}\over{R}}

    Now input impedance is defined to be Zin=V2/Iin which leads to

     Z_{in}={{R}\over{1-T}}

    OK, Let's end this post and continue in the next post.
     
    Last edited: Oct 22, 2011
  6. steveb

    Senior Member

    Jul 3, 2008
    2,433
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    Next, lets consider how to determine the transfer function for voltage gain. There are a number of ways to do this, but it all amounts to writing out some linear circuit equations and then solving for Vlp/V2.

    I prefer to start at the output and work backwards, so I do the following.

    V_{lp}={{-I_2}\over{sC_2}}, where I2 is the current through C2.

     I_2={{V_{bp}}\over{nR}} since the negative terminal of A2 is near ground potential.

    V_{bp}=V_{lp}-{{I_1}\over{sC_1}}, where I1 is the current through C1.

    I_1={{V_2-V_{lp}}\over{R}}-{{V_o}\over{mR}}

    Now you have the task of solving 4 linear equations with 4 unknowns (Vlp, I2, Vbp and I1).

    When I solve this I get.

    T={{1/(nC_1C_2R^2)}\over{s^2+s/(nC_2R)+(m+1)/(mnC_1C_2R^2)}}

    Now we can use Zin=R/(1-T) to get

    Z_{in}=R\Biggl({{s^2+s/(nC_2R)+(m+1)/(mnC_1C_2R^2)}\over{s^2+s/(nC_2R)+1/(mnC_1C_2R^2)}}\Biggr)

    Double check my work and keep an eye out for more mistakes ... I'm old! :p
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
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    I derived the same transfer function as steveb shows. I simplified the math somewhat by first reducing the source and the R, mR voltage divider to its Thevenin equivalent.

    I would finally calculate the Q as being

    Q=\sqr{\frac{n(m+1)C_2}{mC_1}}

    That also needs checking as I'm also in the older age bracket.
     
    Last edited: Oct 25, 2011
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