Chapter 10, Problem 10.1P

The parameters of a certain transmission line operating at $\omega =6\times {10}^8$ rad/s are $L=0.350\mu \text{H/m}$, C= 40 pF/m, G = 0, R = 15.0 $\Omega \text{/m}$, Find $\alpha \text{,}\beta \text{,}\lambda$, and Z_0.

To determine

The values of $\alpha ,\beta ,\lambda$ and $Z_0$.

Answer to Problem 10.1P

$\begin{array}{l}\alpha =0.0935\text{Np/m}\\ \beta =2.2447\text{rad/m}\\ \lambda =\frac{2\pi }{\beta }=2.79\text{m}\\ Z_O=93.6-j3.61\Omega \end{array}$

Explanation of Solution

Given:

Transmission line is operating atgiven frequency

   $\omega =6\times {10}^8\text{ rad/s }$

And the values of inductor, capacitor, conductance and resistance are given below-

   $\begin{array}{l}L=0.35\mu \text{H/m}\\ C=40\text{ pF/m}\\ G=\text{ 75}\mu \text{S/m}\\ R=17\Omega /\text{m}\text{.}\end{array}$

Concept used:

Calculation of unknown quantities isdone by the following formula-

   $\gamma \text{=}\sqrt{ZY}=\sqrt{(R+j\omega L)(G+j\omega C)}$

Calculation:

By using the formula,

   $\begin{array}{l}\gamma \text{=}\sqrt{ZY}=\sqrt{(R+j\omega L)(G+j\omega C)}\\ \gamma =\sqrt{(17+j0.35\times {10}^{-6}\times 6\times {10}^8)(75\times {10}^{-6}+j6\times {10}^8\times 40\times {10}^{-12})}\\ \gamma =0.0935+j2.2447{\text{m}}^{-1}\\ \text{but,}\\ \gamma =\alpha +j\beta \end{array}$

$\begin{array}{l}\alpha =0.0935\text{Np/m}\\ \beta =2.2447\text{rad/m}\\ \lambda =\frac{2\pi }{\beta }=\frac{2\times 3.14}{2.2447}=2.79\text{m}\end{array}$

The value of Zo is calculated by the following formula

   $\begin{array}{l}{Z}_0=\sqrt{\frac{Z}{Y}}=\sqrt{\frac{R\text{ + }j\omega L}{G\text{ + }j\omega C}}\\ Z_0=\sqrt{\frac{\text{17 + }j2.1\times {10}^2}{\text{75}\times {\text{10}}^{-6}\text{ + }j2.4\times {10}^{-2}}}\\ Z_o=93.6-j3.61\Omega \end{array}$