Transmission line secondary coefficients

Click For Summary
SUMMARY

The discussion focuses on calculating the secondary coefficients of a transmission line with primary coefficients R=2 ohm/m, L=8 nH/m, G=0.5 mS/m, and C=0.23 pF/m at a frequency of 1 GHz. The calculated characteristic impedance (Zo) is approximately 26.4316 + j0.0000327906 Ω, while the propagation constant (β) is 246.2946 rad m-1 and the attenuation constant (α) is 482.0029 Np m-1. The calculations confirm that the line does not meet the small-attenuation condition, as G/ωC is 0.35, indicating significant attenuation.

PREREQUISITES
  • Understanding of transmission line theory
  • Familiarity with complex impedance calculations
  • Knowledge of angular frequency calculations
  • Proficiency in using formulas for characteristic impedance and propagation constants
NEXT STEPS
  • Study the derivation of transmission line equations
  • Learn about the impact of frequency on transmission line performance
  • Research methods for minimizing attenuation in transmission lines
  • Explore advanced topics in electromagnetic wave propagation
USEFUL FOR

Electrical engineers, telecommunications professionals, and students studying transmission line theory will benefit from this discussion, particularly those involved in high-frequency circuit design and analysis.

Help_required
Messages
3
Reaction score
0

Homework Statement


A transmission line has the primary coefficients R= 2 ohm/m, L=8 nH/m, G= 0.5 mS/m and C= 0.23 pF/m. Determine the lines secondary coefficients Zo, α and β at a frequency of 1 GHz.

Homework Equations


ω= 2 π f

See uploaded formulas document

The Attempt at a Solution


I'm not sure if I've approached this right but this is what I've done:

I started by calculating the angular frequency from the given 1GHz frequency

ω= 2 π f = 2π*1exp^9 = ω= 6283185307 rad/s

Using this I went on to calculate Zo

Sqrt ((2+j6283185307*8exp^-9)/(0.5exp^-3+j6283185307*0.23exp^-12))

Zo= 26.4316 + j0.0000327906 or Zo= 26.4316 θ 0.0000711° Ω

then using the above formulas I calculated the following

β=sqrt(1/2*(6283185307^2*8exp^-9*0.23exp^-12-2*0.5exp^-3+sqrt(2^2+6283185307^2*(8exp^-9)^2)*((0.5exp^-3)^2+6283185307^2*(0.23exp^-12)^2))

=1.65952exp^5

β=246.2946 rad m^-1α=sqrt (1/2*(2*0.5exp^-3-6283185307^2*8exp^-9*0.23exp^-12+sqrt(2^2+6283185307^2*(8exp-9)^2)*((0.5exp-3)^2+6283185307^2*(0.23exp^-12)^2))

= 3.24771exp^5

α=482.0029Np m^-1

Am i along the right lines with this? (hope you can understand the above workings...It got complicated with all the brackets etc)
 

Attachments

Physics news on Phys.org
Just checking Z0 using the simplified formula: <br /> Z_{0}\approx \sqrt{\frac{L}{C}}=\sqrt{\frac{8\cdot 10^{-9}}{0.23\cdot 10^{-12}}}=\sqrt{34782.6087<br /> }=186.5.
 
You need to use symbols instead of numbers until the very end. Keeping track of all those numbers gives me and a lot of other helpers a splitting headache!

I went as far as checking your formula for Z0 (but not your math) which looks OK.
 
Svein said:
Just checking Z0 using the simplified formula: <br /> Z_{0}\approx \sqrt{\frac{L}{C}}=\sqrt{\frac{8\cdot 10^{-9}}{0.23\cdot 10^{-12}}}=\sqrt{34782.6087<br /> }=186.5.
This is not a small-attenuation line. In particular, the requirement G/ωC << 1 is not met. (= 0.35). But I agree, the OP's number looks off.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 24 ·
Replies
24
Views
8K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K