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Homework Help: 2 port network Y parameters

  1. Oct 27, 2011 #1
    for the following circuit, i am trying to find the y parameters
    a4a1og.jpg

    i have found y11, y12, and y22 to match the answer that my text provides, but i am having trouble with y21

    the solution given is y11 = 1.5, y12 = -0.5, y21 = 4, y22 = -0.5

    in trying to solve for y21, i get 4.5
    this is what i attempted:
    short the voltage source at port 2
    I1 = (v1 - va)/2 - v1
    = -0.5(v1/2 - 3I1 - v1) - v1
    = -0.5(-0.5v1 - 3I1) - v1
    = 0.25v1 + 1.5I1 - v1
    -0.5I1 = -0.75v1
    I2 = 3I1
    I1 = (1/3)I2
    -0.5((1/3)I2) = -0.75v1
    y21 = I2/v1 = 18/4 = 4.5
     
  2. jcsd
  3. Oct 28, 2011 #2

    rude man

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    1. What is Va? Looks like you meant Va = output = 0, in which case your 1st equation is incorrect.
    2. I suggest writing an expression for the sum of currents at the output = 0 to start.
     
  4. Oct 28, 2011 #3
    Va is the node voltage between the two 2-Ohm resistors. why is the first equation incorrect?
     
  5. Oct 28, 2011 #4

    rude man

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    What happened to your schematic? It's been removed!

    So I don't know for sure which node you're talking about. But since there are,as I recall, only two- is Va the output node? If it is, why call it Va when it's in fact V2?
     
    Last edited: Oct 28, 2011
  6. Oct 28, 2011 #5
    2hz2160.jpg
    i obtained the correct answer by transforming the dependent current source into a voltage source in series with the 2 ohm resistor. i was doing something wrong in the node method. i'm still not sure what it is.
     
  7. Oct 28, 2011 #6

    rude man

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    That's a perfectly good way to do it.

    Labeling the resistors R1, R2, R3 from left to right and summing currents at the output:

    V1/R2 + I2 = 3I1
    V1/R2 + I2 = 3V1/R where R = R1||R2
    V1/2 + I2 = 3V1/(2/3) = 9V1/2
    I2 = 9V1/2 - V1/2 = 8V1/2 = 4V1
    I2/V1 = 4
     
  8. Oct 28, 2011 #7
    thanks for that. i just realized now you had said Va = 0 according to the way i had defined it. that cleared things up.
     
  9. Oct 29, 2011 #8

    The Electrician

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    Your assumption that I2 = 3I1 was wrong.

    The Y parameters for a two port like you show are defined like this:

    y11 V1 + y12 V2 = I1
    y21 V1 + y22 V2 = I2

    If we delete the dependent source, it's easy to derive the Y parameters for the resistors only:

    1.5 V1 - .5 V2 = I1
    -.5 V1 + 1 V2 = I2

    In other words:

    y11 = 1.5
    y12 = -.5
    y21 = -.5
    y22 = 1

    If we go back to the general Y parameter equations:

    y11 V1 + y12 V2 = I1
    y21 V1 + y22 V2 = I2

    We can simply add the dependent source to the right side of the second equation since it's a current source applied to the V2 node:

    y11 V1 + y12 V2 = I1
    y21 V1 + y22 V2 = I2-3*I1

    We see that the first equation tells us that I1 = y11 V1 + y12 V2, so our equations become:

    y11 V1 + y12 V2 = I1
    y21 V1 + y22 V2 = I2-3*(y11 V1 + y12 V2)

    which becomes:

    y11 V1 + y12 V2 = I1
    (3 y11+y21) V1 + (3 y12+y22) V2 = I2

    or we could write them as:

    y11 V1 + y12 V2 = I1
    y'21 V1 + y'22 V2 = I2

    with y'21 representing the y21 parameter after the dependent source is added.

    Substituting numerical values:

    1.5 V1 - .5 V2 = I1
    (3*1.5-.5) V1 + (3*(-.5)+1) V2 = I2

    and finally:

    1.5 V1 - .5 V2 = I1
    4 V1 -.5 V2 = I2
     
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