Lossless transmission line question

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Discussion Overview

The discussion revolves around a homework problem involving a 50 Ω lossless transmission line of length 0.4λ terminated with a load of (40 + j30) Ω. Participants are exploring the calculation of the input impedance using a specific equation and discussing the implications of their findings.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents their calculations for the input impedance and questions why they arrived back at the load impedance.
  • Another participant confirms that the angle of 144° is derived from the length of 0.4λ.
  • A different participant prompts a consideration of the relationship between the magnitudes of the load and line, suggesting that differing values could lead to additional complexities in the line.
  • One participant expresses confusion about the implications of their calculations and seeks further guidance on the next steps.
  • Another participant suggests that the original poster has already reached the intended conclusion, assuming their calculations are correct.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the implications of the calculations, as there are differing interpretations of the results and the next steps to take. Some participants agree on the mathematical derivation but express uncertainty about its physical significance.

Contextual Notes

There are unresolved questions regarding the implications of the input impedance being equal to the load impedance and the potential effects of differing magnitudes of load and line on the transmission line behavior.

James123
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Homework Statement



A 50 Ω lossless transmission line of length 0.4λ is terminated in a load of (40 + j30) Ω. Determine, using the equation given below, the input impedance to the line.

Homework Equations



media-2f3c4-2f3c4045b1-a321-492c-99e4-2448c3fc32ce-2fimage-jpg.jpg


The Attempt at a Solution



Zo= 50
Zl= 50∠36.87 or 40+j30
βι= 144° (this is known due to a previous question)

Zin= 50 x ((50∠36.87)*cos144°) + (j50*sin144°)/(50*cos144°) + ((j50∠36.87)*sin144°

=50 x (-40.5∠36.87 + j29.4)/(-40.5 + j29.4∠36.87)

Here I convert to rectangular and get:

=50 x ((-32.4 - j24.3) + j29.4))/(40.5 + (j23.52 + 17.64)

=50 x (-32.4 + j5.1)/(-22.86 + 23.52)

=50 x (0.8+j0.6) = 40 +j30 Ω

This question has been done on other threads before but I couldn't make sense of them.

Can anyone tell me why I've arrived back at the Load impedance? and where am I supposed to go from here?

Any help is appreciated,

Thanks!
 

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The 144° is from 0.4λ, correct?
 
Also, notice how the magnitude of the load and line compare. What would happen if they were different?
 
Hi, yes that's where I got the 144° from.

My heads honestly scrambled on this, if they were different there would have been something between the two on the line? i.e resistance?
 
James123 said:
where am I supposed to go from here?
? You already went where you were supposed to go, assuming your math is correct.
 

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