transmission line characteristics

[fran1942]

Hello, I am trying to work out how to solve this problem.
I do not know what formula I can use to solve this problem.
If anyone can give me some help it would be much appreciated.

" If a transmission line operating at 250 MHz has a characteristic impedance Z0 of
50 Ohm and propagation coefficient γ = 0.4 + j7.5 per meter, calculate its primary
coefficients R, L, C and G. :"

 

[thylacine1975]

There are a handful of equations that encompass all of the data you have (and the unknowns) - that could be solved simultanously - listed in section 2.7 of David Pozar's "Microwave Engineering" (a worthwhile book to hunt down

Apologies for the terrible quality of the scan - but here's the relevant piece:


Cheers!

 

[fran1942]

Hello, thanks for posting that info which is very interesting, however I am still having problems.
I have since found out that the procedure required here is to use the formulas for:
1. propagation coefficient:
y=sqrt((R+jwL)(G+jwC))
and
2. formula for characteristic impedance:
Zo = sqrt(R+jwL/G+jwC)

I need to express 'R+jwL' in terms of 'G+jwC' to go from a 4 variable equation down to a 2 variable solveable equation.
Can anybody show me the process of doing that in order to solve for RLCG ?

Thanks for any assistance.

 

[albbg]

I think

1. your Zo=50 ohm is a purely real impedance, then also sqrt[(R+jwL)/(G+jwC)] must be real, that is [(R+jwL)/(G+jwC)] must be real thus

[(R+jwL)/(G+jwC)] = [(R+jwL)*(G-jwC)]/(G²+w²C²) = (RG+w²LC)/(G²+w²C²) + j(wLG-wRC)/(G²+w²C²)
(wLG-wRC) = 0 ==> L/C=R/G

2. From the previous equation Zo² = (RG+w²LC)/(G²+w²C²)

3. γ = sqrt((R+jwL)(G+jwC)) = a + jb where a=0.4 and b=7.5

γ² = (R+jwL)(G+jwC) = (a + jb)² = (a²-b²) + 2jab

but (R+jwL)(G+jwC) = RG - w²LC +jw(RC+LG) then:

RG - w²LC = a²-b²
w(RC+LG) = 2ab

then we have 4 equation, 4 unknown:

L/C=R/G
Zo² = (RG+w²LC)/(G²+w²C²)
RG - w²LC = a²-b²
w(RC+LG) = 2ab

solve them.

Please check the calculations, hope there are no errors.

 

[godfreyl]

In the more general case where Zo may not be purely resistive, it would be useful to split the 2 given equations into 4 separate equations for the real and imaginary parts of Zo and y. That will also give four equations with four unknowns.