0
\$\begingroup\$

I would like to know the conductancee,\$C\$, and capacitance,\$G\$, of my lossy transmission line which is configured as an open circuit.

In such a case, I know that the expression that defines the input impedance of the line, \$ Z_{in}\$ is given by the following expression

$$\ Z_L = \infty \Rightarrow Z_{in} = Z_{o}*coth \left( \gamma*l \right) = $$ $$\ = Z_{o}*coth \left( \ \left( \alpha+j\beta \right)*l \right) \tag 1$$

where

$$\ \gamma = \sqrt{ \left( R+j\omega L \right)* \left( G+j\omega C \right) } = $$ $$\ = \sqrt{ \left( RG-\omega^2LC \right)+ j\omega* \left(RC+LG \right) } = \alpha + j\beta $$ $$\ Z_o = \sqrt{ \frac{R+j\omega L}{G+j\omega C} } = \frac{\sqrt{ \gamma}}{G+j\omega C}$$

I can use a non-linear method to solve \$(1)\$ since I know \$ Z_{in}, L, R\$ and \$ \omega\$

My question is related to the units of \$C\$ and \$G\$ obtained when solving, are these units per length, \$\left[C\right]= \frac{F}{m}\$ and \$\left[G \right]=\frac{S}{m}\$, or the total of the line \$\left[C\right]= F\$ and \$\left[G\right]= S \$?

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

In the context of your transmission line formulas, C and G are "distributed" elements, which mean they are "per unit". So you have to deal with [C]=F/m and [G]=S/m.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.