What is the Solution for a Lossless Transmission Line with Given Parameters?

AI Thread Summary
The discussion focuses on calculating the load voltage for a lossless transmission line with specific parameters, including a characteristic impedance of 50Ω and a load impedance of 80Ω. The user struggles with evaluating the input impedance and voltage due to the imaginary components in the equations. A suggestion is made to consider the source impedance as part of a voltage divider with the input impedance. The conversation also hints at using ABCD matrices for a more straightforward solution, although the user may not be familiar with this method. Overall, the user seeks clarity on handling complex calculations involving hyperbolic functions in transmission line theory.
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Homework Statement



A lossless transmission line with

characteristic impedance Z0 = 50Ω , β = 5∏x10-3,
and length l =80 meters, is terminated into a load ZL = 80Ω. The transmission line is powered by a source with 120 V and ZG = 12 Ω. Calculate the load voltage.



Homework Equations



V(z) = V+(e-jβz + \Gammaejβz

\Gamma = (ZL - Z0)/(ZL + Z0)

Zinput = Z0(ZL+Z0tanh(jβL))/(Z0+ZLtanh(jβL))

V+ = VGZinput/((eL+\Gammae-jβL)(Zinput+ZG))

The Attempt at a Solution



\Gamma = (80-50)/(80+50) = 3/13

Zinput = 50(80+50tanh(j*.4*pi)/(50+80tanh(j*.4*pi)

I can't find the input independence because my calculator can't evaluate the imaginary argument in the tanh function. I also have to same problem with solving for V+ and the imaginary exponents in the denominator. Does anyone know a way around this, or am I completely doing the wrong thing here?
 
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I'd like to add that I'm very, very confused by the concepts, and I'm basically just juggling the equations derived in class.
 
There is a simple equation for E2 for a lossless transmission line, involving only β, l, Z0 and load impedance ZLand E1.

What is tanh(jx) ) in terms of a non-hyperbolic trig function?

Unfortunartely for you, E1 is the voltage at the driven end of the line so you have to take care of the source impedance ZG. Hint: ZG forms a voltage divider with Zinput.

I can't post the equation for E2. Would violate the terms of use of this forum. It can be found in many places.

BTW the nice way to handle this problem is with ABCD matrices but I take it you haven't had those yet.
 
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