Transmission Line Coefficients & Definitions

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Discussion Overview

The discussion revolves around the definitions and calculations related to distortionless and lossless transmission lines, including the determination of secondary coefficients such as Zo, α, and β at a frequency of 1 GHz based on given primary coefficients.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant defines a distortionless transmission line as one that attenuates all signal frequencies equally and shifts them in time uniformly, while a lossless transmission line is characterized by R and G being equal to zero.
  • The participant provides calculations for β, α, and Zo using the given coefficients, presenting detailed mathematical steps.
  • Another participant confirms the calculations and provides an alternative expression for α, suggesting it checks with the original computation within a few percent, using approximations for R0 and G0.
  • Subsequent posts reiterate the confirmation of calculations and express gratitude for feedback, while also noting a potential error in the final computation of Zo, which is later reassured as correct.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the calculations presented, although there is some uncertainty expressed regarding the final computation of Zo, which one participant initially questions but later confirms as accurate.

Contextual Notes

The discussion includes assumptions about the approximations used for R0 and G0, and the potential for minor discrepancies in the calculations due to these approximations.

Jason-Li
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Homework Statement



(a) State what is meant by a ‘distortionless’ and a ‘lossless’ transmission line.(b) A transmission line has the primary coefficients as given below. Determine the line’s secondary coefficients Zo, α and β at a frequency of 1 GHz.
R = 2 Ω/m
L = 8 nH/m
G=0.5 mS/m
C=0.23 pF/m

The Attempt at a Solution



I've seen some old threads with this question, but it's more the workings, methodology and definitions - if someone could be so kind to have a look over.
a)
Distortionless:
The transmission line in order to be 'distortionless' must both attenuate all signal frequencies in the same proportion and shift them in time by the same amount.
Loseless:
A transmission line is known as loseless when there are no energy loses along the line due to coeffecients R & G. This occurs when both of the aforementioned coefficients equal 0.

b)
So:
R &= 2 Ω/m
L &= 8 nH/m = 8x10^{-9} H/m
G=0.5 mS/m = 0.0005 S/m
C &=0.23 pF/m = 0.23*10^{-12} F/m

ω=2πf = 1x10^{9} *2π = 6.283*10^{9)

Due to high frequency the formulas:
β = ω√(LC)
= 6.283*10^{9} * √ ( 8*10^{-9) * 0.23*10^{-12}
= 0.26951 Radm-1

α = R/2 * √(C/L)+G/*√(L/C)
= 2/2 * √(0.23*10^{-12}/8*10^{-9})+0.0005/2*√(8*10^{-9}/0.23*10^{-12})
= 1 * 0.00536+0.00025*186.501
= 0.05198714308
= 51.987 mNepers m-1

Zo = √((R+jωL)/ (G+jωC))
= √((2+j6.283*10^{9}*8*10^{-9}) / (0.0005+j6.283*10^{9}*0.23*10^{-12}))
= √ ( (2+j16π) / (0.0005+j(0.00046π) )
= √ (31491.630+j9511.802)
= 179.427+j26.506Ω

I appreciate any and all help. If anyone else is doing this question and wants advice I can try to help also.
 
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This all looks good.

An exact value of α = 0.5(R/R0 + G/G0)
where R0 = Re{Z0} and G0 = Re{Y0} and Y0 = 1/Z0
which checks with your computation (within a few %; I approximated Z0 = R0 and G0 = 1/R0).
 
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rude man said:
This all looks good.

An exact value of α = 0.5(R/R0 + G/G0)
where R0 = Re{Z0} and G0 = Re{Y0} and Y0 = 1/Z0
which checks with your computation (within a few %; I approximated Z0 = R0 and G0 = 1/R0).

Thank you very much for the feedback.
 
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rude man said:
This all looks good.

An exact value of α = 0.5(R/R0 + G/G0)
where R0 = Re{Z0} and G0 = Re{Y0} and Y0 = 1/Z0
which checks with your computation (within a few %; I approximated Z0 = R0 and G0 = 1/R0).

Hi again Rudeman,

Was looking over my old working and found that in the final line of working is incorrect, unless I'm just being an idiot!

= √ (31491.630+j9511.802)
= 179.427+j26.506Ω

Just to update this post.
 
Jason-Li said:
Hi again Rudeman,

Was looking over my old working and found that in the final line of working is incorrect, unless I'm just being an idiot!

= √ (31491.630+j9511.802)
= 179.427+j26.506Ω

Just to update this post.
Well, I checked that last computation and found it correct. So you don't meet the qualifications for idiot. Sorry! :smile:

As I said, I had previously checked your answer against my approximate one and found it correct also.
So I'm not sure what problems are still extant?
 

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