Circuit Input Impedance

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  • Thread starter roam
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  • #1
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Homework Statement



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?

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.
 

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Answers and Replies

  • #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.
 
  • #3
The Electrician
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We can further write this as:

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

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)}##
 
  • #4
rude man
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Your second equation has a typo (the first R). The thumbnail is correct.
 
  • #5
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Thank you very much, I got the right answer now! :)
 

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