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A ##50\Omega## lossless transmission line impedance question

  1. Jan 31, 2017 #1

    David J

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    Gold Member

    1. The problem statement, all variables and given/known data

    a ##50\Omega## lossless transmission line of length ##0.4\lambda## is terminated in a load of ##(40+j30)\Omega##. Determine, using ##Zin=Z0\frac{ZL cos \beta l+j Z0 sin \beta l}{Z0 cos \beta l +j ZL sin \beta l}## the input impedance to the line.

    ##Z0=50\Omega##

    ##ZL=(40+j30)\Omega##

    ##\beta l =2.513## radians

    ##cos\beta l =-0.809## radians or ##0.999^0##

    ##sin\beta l=0.588## radians or ##0.0438^0##


    2. Relevant equations

    ##Zin=Z0\frac{ZL cos \beta l+j Z0 sin \beta l}{Z0 cos \beta l +j Zl sin \beta l}##

    3. The attempt at a solution

    ##Zin=50\Omega\frac{(40+j30)(-0.809) +j(50)(0.588)}{50(-0.809) +j(40+j30)(0.588)}##

    So ##50\Omega\frac{(-32.36-j24.27)+(j29.4)}{(-40.45+(-17.64+j23.52)}##

    So ##50\Omega\frac{(-32.36+j5.13)}{(-58.09+j23.52)}##


    ##=50\Omega(0.5093+j0.118)##

    So ##Zin=25.465+j5.90## or ##26.14\angle13^0## or ##26.14\angle0.23## radians

    I have been over this 5 times now to check and I feel I am correct but I would like a second opinion please, if possible, to identify any errors etc.

    Thanks
     
  2. jcsd
  3. Jan 31, 2017 #2

    gneill

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    Staff: Mentor

    The only problem I can see is that the sines and cosines should be unitless. The sine or cosine (or result of any trig function) is a pure number that represents a ratio. Otherwise, well done!
     
  4. Jan 31, 2017 #3

    David J

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    Gold Member

    Appreciated. thanks again
     
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