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Circuit Input Impedance

  1. Mar 12, 2014 #1
    1. The problem statement, all variables and given/known data

    I have some trouble understanding the worked examples below:

    Considering the input impedance of the network below:

    3Ecj7kA.jpg

    ##z_{in} = R+ \frac{sL/sC}{sL+(1/sC)}##

    ##z_{in} = R \left[ \frac{s^2+s/(RC)+1/(LC)}{s^2+1/(LC)} \right]##

    Where ##s=j\omega##.

    How did they get from the first expression to the second expression?

    3. The attempt at a solution

    Clearly looking into the network R is in series with the parallel combination of L and C, so we have ##R + L \parallel C## which is the first expression:

    ##z_{in} = R+ \frac{sL/sC}{sL+(1/sC)}##

    We can further write this as:

    ##z_{in} = R+ \frac{sRL + (R/sC)+(sL/sC)}{sL+(1/sC)}##

    I'm really confused. Where does the second expression given above come from? :confused:

    Any help is greatly appreciated.

    P.S. This is part of a problem about finding poles and zeros of the network. The quadratics in the numerator and denominator of the 2nd expression can be factorized to give the poles and zeros.
     

    Attached Files:

  2. jcsd
  3. Mar 13, 2014 #2

    rude man

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    Just algebra!
    1. cross-multiply
    2. factor out R, this means putting a 1/R coefficient in one term.
    3. divide numerator & denominator by L.
     
  4. Mar 13, 2014 #3

    The Electrician

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    The above expression is incorrect; the initial R+ shouldn't be there. The expression should be:
    ##z_{in} = \frac{sRL + (R/sC)+(sL/sC)}{sL+(1/sC)}##
     
  5. Mar 13, 2014 #4

    rude man

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    Your second equation has a typo (the first R). The thumbnail is correct.
     
  6. Mar 13, 2014 #5
    Thank you very much, I got the right answer now! :)
     
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