Understanding Impedance: How to Find Poles and Zeros of a Network

In summary, the conversation discussed the input impedance of a network and how it can be expressed as two different equations. The first equation was derived by considering the network as a series combination of R and a parallel combination of L and C. The second equation was obtained by cross-multiplying and factoring out R, and then dividing the numerator and denominator by L. The conversation also mentioned that the initial R+ in the second equation was a typo and the correct expression should be without the R+ term.
  • #1
roam
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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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  • #2
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.
 
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  • #3
roam said:
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)}##
 
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  • #4
Your second equation has a typo (the first R). The thumbnail is correct.
 
  • #5
Thank you very much, I got the right answer now! :)
 

1. What is circuit input impedance?

Circuit input impedance refers to the measure of how much an electrical circuit resists the flow of current when a voltage is applied to it. It is represented by the symbol Z and is measured in ohms.

2. Why is circuit input impedance important?

Circuit input impedance is important because it helps determine the performance and behavior of electronic circuits. It affects the voltage and current levels in the circuit, which can impact the overall functionality and efficiency of the circuit.

3. How is circuit input impedance calculated?

Circuit input impedance can be calculated by dividing the voltage by the current at the input of the circuit. This can also be represented as the ratio of the input voltage to the input current.

4. What factors affect circuit input impedance?

There are several factors that can affect circuit input impedance, such as the type and length of the connecting wires, the type of components used in the circuit, and the frequency of the input signal. In general, the higher the frequency, the lower the input impedance.

5. How does circuit input impedance impact signal quality?

Circuit input impedance can have a significant impact on signal quality. If the input impedance is too low, it can cause reflections and distortions in the signal. On the other hand, if the input impedance is too high, it can result in a weaker signal and reduce the signal-to-noise ratio.

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