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Y and z 2-port parameters

  1. Nov 4, 2011 #1
    I am solving for the Y and Z parameters for the circuit below:
    2vaauqt.png

    For Y, I used the following equations by applying current sources at the terminals
    va is the node at the 2v1 dependent source
    (va-v1)/0.5 + (va-v2)/1 = -2v1
    I1 = v1/1 + (v1-va)/0.5
    I2 = v2/0.5 + (v2-va)/1
    I got Y = [3 -2/3; 0 8/3]

    For Z, I used KVL by applying voltage sources at the terminals
    v1 = Ia - Ib
    0 = Ib - Ia + 0.5Ib
    0 = Ic + 0.5(Ic + Id)
    v2 = 0.5(Ic + Id)
    where Ia = I1, Ib = 2v1, Ic = -2v1, and Id = I2
    I got Z = [1/3 1/6; 0 1/3]

    I am not sure what I am doing incorrectly here. Y should be equal to Z-1, but I am not getting that. I'm not sure if my Y parameters are wrong, Z parameters, or both. :\
     
    Last edited: Nov 4, 2011
  2. jcsd
  3. Nov 4, 2011 #2

    The Electrician

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    I think the major problem is where you have:

    Ib = 2v1, Ic = -2v1

    Replace those two items with Ib - Ic = 2v1

    and see if you don't get better results.
     
  4. Nov 4, 2011 #3
    when using Ib - Ic = 2v1

    I got
    v1 = Ia - Ib
    = Ia- Ic - 2v1
    = Ia + (1/3)Id - 2v1
    ====> v1/I1 = 1/3 and v1/I2 = 1/9
    plugging in values to find the remaining parameters i got v2/I2 = 1/2 and v2/I1 = 0
    so for Z parameters i got Z = [1/3 1/9; 0 1/2]

    The inverse of this is [3 -2/3; 0 2] so there's still something off :\
     
  5. Nov 5, 2011 #4

    The Electrician

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    You can't assume the voltage across the dependent source is zero; you have:

    0 = Ib - Ia + 0.5Ib
    0 = Ic + 0.5(Ic + Id)

    You need to make those:

    0 = -va + Ib - Ia + 0.5Ib
    0 = va + Ic + 0.5(Ic + Id)

    I'm not working out the complete solution using your method, so I may not see all the errors right away. I'm just pointing out what I see with a quick glance; you need to try to find errors also.

    You might need to treat the middle two meshes together as a supermesh.
     
  6. Nov 5, 2011 #5
    are my Y parameters correct? just wondering because i don't want to be trying to use them to check if my Z parameters are correct if they aren't right in the first place..
     
  7. Nov 5, 2011 #6

    The Electrician

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    Yes, they are correct. :smile:
     
  8. Nov 5, 2011 #7

    The Electrician

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    Have you had any luck deriving the Z parameters?
     
  9. Nov 5, 2011 #8
    yep, thanks. i just had a few algebraic errors that threw me off.
     
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