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Homework Help: Finding the Real Points of a Complex Function

  1. May 2, 2010 #1
    Hi all, I'm trying to solve the following expression for w when w is entirely real (no imaginary component). It is part of a circuits problem where I have to design a band pass filter. Note that I'm using j instead of i to denote the imaginary component.


    To be clear, I'm only looking for a relationship between R, L, and C which would cause the above expression to be entirely real. This is basically the solution for the resonant frequency of my circuit. So w at resonance equals some relationship of R, L, and C. I really appreciate any help with this.

    For anyone who is interested, this is my result for the transfer function (Vout/Vin) of an RLC combination I'm trying. The RLC configuration is L in series with R and C in parallel. In my case, the pure series RLC bandpass filter will not work as it yields an output voltage below what is required and I cannot use active components.

    Last edited: May 2, 2010
  2. jcsd
  3. May 2, 2010 #2


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    try multiplying the numerator & denominator by the complex conjugate of the denominator
  4. May 2, 2010 #3
    Tried that and didn't get too far. Actually I just managed to solve this by realizing that the phase angle of the numerator must equal the phase angle of the denominator. Thanks for the reply though! Hopefully this will help someone in the future.

    Remember, the phase angle of the numerator equals the phase angle of the denominator! By setting the two equal you can find the relation that yields a real result.
  5. May 2, 2010 #4


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    good you got an answer using phase angles

    I usullay find it easiest just to multiply as below, which makes the denominator will be real. The imaginary part is then written explicitly
    [tex]\frac{jwLR+\frac{L}{C}+\frac{R}{jwC}}{R+\frac{1}{j wC}}

    = \frac{jwLR+\frac{L}{C}+\frac{R}{jwC}}{R+\frac{1}{j wC}}

    \frac{R-\frac{1}{j wC}}{R-\frac{1}{j wC}}

  6. May 2, 2010 #5
    Yeah, when I tried that I forgot that I could bring j out of a denominator and put it in the numerator if I multiplied by -1. If you don't remember that you can't factor the j's out and set that part equal to zero.
  7. May 3, 2010 #6


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    there shouldn't be any j's in the denominator, multipying by the complex conjugate is finding the magintude square which is always a positive real number

    [tex] (R+\frac{1}{j wC})(R-\frac{1}{j wC})
    = R^2 +R \frac{1}{j wC}-R\frac{1}{j wC} -(\frac{1}{j wC})^2
    = R^2 +(\frac{1}{ wC})^2 [/tex]
  8. May 3, 2010 #7
    I meant the fractions which are left in the numerator after multiplying by the complex conjugate.
  9. May 3, 2010 #8
    Remember, 1/j = -j.
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