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Characteristic Impedance for a (∏) two port network

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate the Characteristic Impedance (Z0) of the two port network?


    2. Relevant equations

    When port 2 open-circuit, I2 = 0

    So, Z11 = V1 / I1

    = 40(60) / 100 = 24 Ω

    When I2 = 0, thus V2

    V2 = 40 / 20+40 V1 = 0.666 V1

    Z21 = V2 / I1

    = 0.666 V1 / (V1 / 40-20)

    = 0.666 V1 / (V1/20) = 0.666 V1 : V1/20

    = 0.666 V1 x 20/v1

    = 0.666 V1 x 20 / V1

    = 13.32 V1 / V1

    = 13.32 Ω

    Now, when I1 = 0, so:

    Z22 = V2 / I2

    = 40(60) / 100

    = 2400 / 100

    = 24 Ω

    When port 2 open circuit, I1 - 0, so:

    V1 = 40 / (20+40) V2

    = 40 / 60 V2

    = 0.666 V2

    I2 = V2 / 24

    Z12 = V1 / I2

    = 0.666 V2 / (V2 / 24)

    = 0.666 V2 : (V2 / 24)

    = 0.666 V2 (24/ V2)

    = 15.984 Ω

    AM I CORRECT???

    Can someone please help???
     

    Attached Files:

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    Last edited: Sep 26, 2013
  2. jcsd
  3. Sep 26, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Write the equations defining Z parameters, then set i2 = 0.

    I have never heard the term 'characteristic impedance' apply to z parameter networks. It's used in transmission lines.
     
  4. Sep 26, 2013 #3
    When your network is symmetrical, Z12 has to equal Z21.

    Why are you going to all the trouble of calculating the Z parameters, and the voltage transfer ratios, when (apparently) all you want is the characteristic impedance?

    And, if you want the characteristic impedance, why did you stop your calculations before you calculated it?
     
  5. Sep 29, 2013 #4
    Z0 = √ R1 x R2² / R1 + (2 x R2)

    Z0 = √ 20 x (40)² / 20 + (2 x 40)

    Z0 = √ 20 x 1600 / 20 + 80

    Z0 = √ 32000 / 100

    Z0 = √ 320

    Z0 = 17.8885

    Z0 = 17.88 Ω

    Can someone please confirm that I am correct? Much appreciated!!!!
     
  6. Sep 30, 2013 #5
    Your final answer is correct, but your arithmetic procedures leading to it are not unambiguous.
     
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