Transmission Line Coefficients & Definitions

  • #1
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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.
 

Answers and Replies

  • #2
rude man
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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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Likes Jason-Li
  • #3
57
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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).
Thank you very much for the feedback.
 
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Likes berkeman
  • #4
57
8
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.
 
  • #5
rude man
Homework Helper
Insights Author
Gold Member
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792
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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