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

[David J]

Homework Statement

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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

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

The Attempt at a Solution

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

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

 

[gneill]

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!

 

[David J]

Appreciated. thanks again

 

[Jason-Li]

Sorry to resurrect this post, just a quick simple one, Would Zin not have ohms as a unit?

 

[magoo]

Yes, Zin should be in ohms.