Transmission Line Coefficients & Definitions

Click For Summary
SUMMARY

This discussion focuses on the definitions and calculations related to distortionless and lossless transmission lines, specifically at a frequency of 1 GHz. The primary coefficients provided are R = 2 Ω/m, L = 8 nH/m, G = 0.5 mS/m, and C = 0.23 pF/m. The secondary coefficients calculated include the characteristic impedance (Zo) as approximately 179.427 + j26.506 Ω, the attenuation constant (α) as 51.987 mNepers/m, and the phase constant (β) as 0.26951 Rad/m. The calculations confirm that the line is distortionless and lossless under the defined conditions.

PREREQUISITES
  • Understanding of transmission line theory
  • Familiarity with complex impedance and admittance
  • Knowledge of frequency domain analysis
  • Proficiency in using mathematical formulas for electrical engineering
NEXT STEPS
  • Study the derivation of transmission line equations
  • Learn about the impact of frequency on transmission line performance
  • Explore the significance of the characteristic impedance (Zo) in circuit design
  • Investigate the effects of resistance (R) and conductance (G) on signal integrity
USEFUL FOR

Electrical engineers, telecommunications professionals, and students studying transmission line theory will benefit from this discussion, particularly those focused on high-frequency signal transmission and impedance matching.

Jason-Li
Messages
119
Reaction score
14

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.
 
Physics news on Phys.org
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).
 
  • Like
Likes   Reactions: Jason-Li
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.
 
  • Like
Likes   Reactions: berkeman
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?
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 24 ·
Replies
24
Views
8K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 28 ·
Replies
28
Views
13K
  • · Replies 2 ·
Replies
2
Views
2K