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

  • Thread starter David J
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  • #1
David J
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Homework Statement


[/B]
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##


Homework Equations


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

The Attempt at a Solution


[/B]
##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
 

Answers and Replies

  • #2
gneill
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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!
 
  • #3
David J
Gold Member
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Appreciated. thanks again
 
  • #4
57
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Sorry to resurrect this post, just a quick simple one, Would Zin not have ohms as a unit?
 
  • #5
166
45
Yes, Zin should be in ohms.
 

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