1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two-port network models

  1. Oct 9, 2013 #1
    [/SUB]1. The problem statement, all variables and given/known data

    Design a ∏ section symmetrical attenuator to provide a voltage attenuation of 15 dB and a characteristic impedance of 600 Ω.

    2. Relevant equations

    ZS = ZL = Z

    R1 = R3 = Z (K + 1 / K -1)

    R2 = Z (K2 - 1 / 2 K)

    3. The attempt at a solution

    Z = 600Ω
    K = 15dB = 15(15/20) = 5.6234

    R1 = R3 = 600 (5.6234 + 1 / 5.6234 - 1)

    = 859.54 Ω


    R2 = 600 (5.62342 - 1 / 2 x 5.6234)

    = 600 (30.6226 / 11.2468)

    = 1633.67 Ω

    Can someone please confirm if I am on the right lines? Thanks
     

    Attached Files:

    Last edited: Oct 9, 2013
  2. jcsd
  3. Oct 10, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Not what I got. I don't know where you got your formulas. I just wrote a set of equations to force Zin = 600 ohms from both sides, and an attenuation of 15 dB = gain of 0.1778. My R1 was not too much below yours but my R2 was exactly double yours. Interesting coincidence!

    Maybe I'm misunderstanding what the problem asks for. My pi network looks like 600 ohms from both directions and the attenuation is 15 dB. I'm not assuming a source nor a load impedance.
     
  4. Oct 10, 2013 #3
    Hi, when talking about current or voltage, it is db = 20log(ratio).

    If you are taking the ratio of powers, then it is dB = 10log(ratio).

    We are using first one, which gives us 15db= 20 log (n),

    10 (15/20) = n

    10 ( 3/4) = n

    4√ 10 (3) = 5.6234 = k

    I have found similar example but still want to get confirmation is it right or not
     
  5. Oct 10, 2013 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    If it's attenuation, it's -15dB, not 15 dB, which is 10^(-15/20) = 01778 = k.
     
    Last edited: Oct 10, 2013
  6. Oct 10, 2013 #5
    I know that attenuation means the reduction of signal strength during transmission. But how you can tell its with this example?
    Your calculations would be right, but i dont know why is minus there.
     
  7. Oct 10, 2013 #6
    Thanks but it is actually the same example that I was using, but I still can't find a minus there in that example!
     
    Last edited: Oct 10, 2013
  8. Oct 10, 2013 #7

    rude man

    User Avatar
    Homework Helper
    Gold Member

    If you go by the link, attenuation is considered a positive quantity.

    Follow their example.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Two-port network models
  1. Two port network model (Replies: 9)

Loading...