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Two-port network z-parameters

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data

    Z_PARAMETERS.JPG

    2. Relevant equations

    As above.

    3. The attempt at a solution

    (z11) With the output open circuit (and therefore I2=0) z11 = V1 / I1 = (R8 + R6)I1 / I1 = 14ohms;

    (z12) With the input open circuit (and therefore I1=0) z12 = V1 / I2 = R6 = 6ohms;

    (z21) With the output open circuit (and therefore I2=0) z21 = V2 / I1 = V1 x (R6 / (R8+R6) ) x ( (R8+R6) / V1) = R6 = 6ohms;

    (z22) With the input open circuit (and therefore I1=0) z22 = V2 / I2 = R6xI2 / I2 = R6 = 6ohms

    Is the above correct? It seems so easy that I'm worried I've overlooked something and made a really basic mistake!
     

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    Last edited by a moderator: Jan 26, 2017
  2. jcsd
  3. Feb 10, 2015 #2

    BvU

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    Hello acx, welcome to PF :)

    Some things are easy :) enjoy while it lasts...
     
  4. Feb 11, 2015 #3
    Thanks BvU, I shall do just that!
     
  5. Jan 25, 2017 #4

    David J

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    Where does the value for ##I1## come from? In the first equation for ##z11## where ##I2=0## there appears to be a value of 1 for ##I1##. Then in the next equation for ##z12## and ##I1 =0## there appears to be a value of 1 for ##I2##. How are these values derived please as I cant seem to work out and my notes tell me virtually nothing. Appreciated thanks
     
  6. Jan 25, 2017 #5

    NascentOxygen

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    Can you enlarge your screen or change the default font to see print more clearly?

    z11 = V1 / I1 = (R8 + R6) I1 / I1 = 14ohms

    The formulae come from applying the z-parameter equations characterising a 2 port network.
     
  7. Jan 25, 2017 #6
    For z11 assume that the input is not open circuit. Therefore I1 is not equal to zero. I1 then cancels out from the equation leaving z11 = R8+R6. You don't need a value for I1, just that it doesn't equal zero.

    For z12 again it doesn't matter what value you have for I2. You show that z12 = R6 = 6ohms. The value of I2 becomes irrelevant.
     
  8. Jan 26, 2017 #7

    David J

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    OK thanks for the info.

    So as I understand this ##Z11## is the impedance seen looking into port 1 when port 2 (output) is open
    .

    Output is open circuit so ##I2=0##

    ##Z11=\frac{V1}{I1}=\frac{(R8+R6)(I1)}{I1}## The ##I1## cancels leaving ##R8+R6=14\Omega##

    So for ##Z22## is the impedance looking into port 2 when port 1 (input) is open

    Input is open circuit so ##I1=0##

    ##Z22=\frac{V2}{I2}=\frac{(R6)(I2)}{I2}## The ##I2## cancels leaving ##R6=6\Omega##

    The reason ##Z22## only has ##R6## to deal with is because looking into the network from port 2 we can only see the ##6\Omega## resistance

    So for ##Z12## i understand this is the ratio of the voltage at port 1 to the current at port 2 when port 1 is open. If port 1 is open the ##I1=0##

    You have used the equation ##Z12 =\frac{V1}{I2} = R6 = 6\Omega##

    My question is why does ##V1=6\Omega## this time whereas when for ##Z11,V1=R8 + R6##

    Thanks
     
  9. Jan 26, 2017 #8

    NascentOxygen

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    Inject current at top terminal 2 and it exits at lower terminal 2. With port 1 open, determine V1.

    Z_PARAMS.JPG

    Does this answer your question?
     
  10. Jan 26, 2017 #9

    David J

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    So current will flow through ##R6## and back out via the lower terminal 2 because port 1 is open. So R8 does not come into the equation so to speak??

    I have a better understanding of this now, yes. Thanks for the help
     
  11. Jan 28, 2017 #10

    rude man

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    Sneak extra info: for a passive network, x12 = x21. x can be either z or y parameters. (The rules for h, a, s parameters differ slightly).
     
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