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Transfer function and Bode plot

  1. Nov 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the transfer function for the filter on the attached picture and sketch it in Bode diagram, if you know ##Q_0 =\frac{\omega _0L}{R}>>1## and ##\omega _0^2=\frac{1}{LC}##.
    Captureg.PNG

    2. Relevant equations


    3. The attempt at a solution
    If I am not mistaken, we defined the transfer function as $$H(i\omega )=\frac{U_{out}(\omega )}{U_{in}(\omega )}$$
    where ##U_{out}(\omega )=IZ##. $$\frac 1 Z =\frac{1}{R+i\omega L}+\frac{1}{i\omega C}$$ $$Z=\frac{-\omega ^2 LC + i\omega RC}{R+i\omega (L+C)}$$ So finally $$H(i\omega )=\frac{U_{out}(\omega )}{U_{in}(\omega )}=\frac{Z}{Z+R}=\frac{-\omega ^2LC+iR\omega C}{R^2-\omega ^2LC+i\omega R(L+2C)}$$ But I highly suspect this is wrong from the very start. I have no idea how to solve this kind of problems.
     
  2. jcsd
  3. Nov 25, 2014 #2

    vela

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    The calculation is almost right. Your expression for Z can't be right because it involves summing L and C, which you can't do because they don't have the same units. In your first expression for 1/Z, you're adding ZC instead of 1/ZC.
     
  4. Nov 26, 2014 #3
    Thank you!

    Well, the end result is not that important (Except that ##1/Z_c## part, that is important). I was just not sure if this: $$H(i\omega )=\frac{U_{out}(\omega )}{U_{in}(\omega )}=\frac{Z}{Z+R}$$ is right. I did the rest of the calculations using Mathematica.
     
  5. Nov 26, 2014 #4

    vela

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    Yeah, that's fine. That's just a plain-old voltage divider. It holds for impedances, not just resistances.
     
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