Chapter 13, Problem 13.1P

The conductors of a coaxial transmission line are copper $\left({\sigma }_c=5.8\times {10}^7\text{S/m}\right)$, and the dielectric is polyethylene $\left(\epsilon \text{r'=2}\text{.26,}\sigma \text{/}\omega ϵ\text{'=0}\text{.0002}\right)$. If the inner radius of the outer conductor is 4 mm, find the radius of the inner conductor so that (a) $Z_0=50\Omega$; (b) C = 100 pF/m; (c) $L=0.2\mu \text{H/m}$. A lossless line can be assumed.

To determine

(a)

The radius of the inner conductor so that $Z_0=50\Omega .$

Answer to Problem 13.1P

The radius of the inner conductor so that $Z_0=50\Omega$ is $a=1.142\text{mm}$.

Explanation of Solution

Given:

Conductivity of copper of a coaxial transmission line ${\sigma }_c=5.8\times {10}^{-7}\text{S/m}$.

The dielectric is polyethylene:.

   $\begin{array}{l}{{\epsilon }^{\prime }}_r=2.26\\ \frac{\sigma }{\omega {\epsilon }^{\prime }}=0.0002\end{array}$.

   $b=4\text{mm}$.

Characteristics impedance is $Z_0=50\Omega$.

Calculation:

The characteristic impedance of the transmission line is given as:.

   $Z_0=\frac{1}{2\pi }\sqrt{\frac{\mu }{{\epsilon }^{\prime }}}\ln \left(\frac{b}{a}\right)$.

Where,

   $\mu$ is the permeability..

   ${\epsilon }^{\prime }$ is the dielectric permittivity..

   $b$ is the inner radius of the outer conductor..

   $a$ is the radius of the inner conductor..

Thus,

   $\begin{array}{c}{Z}_0=\frac{1}{2\pi }\sqrt{\frac{\mu }{{\epsilon }^{\prime }}}\ln \left(\frac{b}{a}\right)\\ 50=\frac{1}{2\pi }\sqrt{\frac{377}{2.26}}\ln \left(\frac{b}{a}\right)\\ \ln \left(\frac{b}{a}\right)=\frac{50\times 2\pi \sqrt{2.26}}{\sqrt{377}}\\ \ln \left(\frac{b}{a}\right)=1.25\end{array}$.

Taking antilog on both sides, we get:.

   $\begin{array}{l}\left(\frac{b}{a}\right)=e^{1.25}\\ \frac{b}{a}=3.50\\ b=4\text{mm}\\ a=\frac{4}{3.50}\\ a=1.142\text{mm}\end{array}$.

Conclusion:

Thus, the radius of the inner conductor so that $Z_0=50\Omega$ is $a=1.14\text{mm}$.

To determine

(b)

The radius of the inner conductor so that $C=100\text{pF/m}\text{.}$

Answer to Problem 13.1P

The radius of the inner conductor so that $C=100\text{pF/m}$ is $a=1.14\text{mm}$.

Explanation of Solution

Given:

Conductivity of copper of a coaxial transmission line ${\sigma }_c=5.8\times {10}^{-7}\text{S/m}$.

The dielectric is polyethylene:.

   $\begin{array}{l}{{\epsilon }^{\prime }}_r=2.26\\ \frac{\sigma }{\omega {\epsilon }^{\prime }}=0.0002\end{array}$.

   $b=4\text{mm}$.

Capacitance is $C=100\text{pF/m}$.

Calculation:

The capacitance of the transmission line is given as:.

   $C=\frac{2\pi {\epsilon }^{\prime }}{\ln \left(\frac{b}{a}\right)}$.

Where,

   ${\epsilon }^{\prime }$ is the dielectric permittivity..

   $b$ is the inner radius of the outer conductor..

   $a$ is the radius of the inner conductor..

Thus,

   $\begin{array}{l}C=\frac{2\pi {\epsilon }^{\prime }}{\ln \left(\frac{b}{a}\right)}\\ C=100\text{pF/m}\\ C={10}^{-10}\text{F/m}\\ {10}^{-10}=\frac{2\pi {\epsilon }^{\prime }}{\ln \left(\frac{b}{a}\right)}\\ \ln \left(\frac{b}{a}\right)=2\pi \times 2.26\times 8.854\times {10}^{-2}\\ \ln \left(\frac{b}{a}\right)=1.257\end{array}$.

Taking antilog on both sides, we get:.

   $\begin{array}{l}\left(\frac{b}{a}\right)=e^{1.257}\\ \frac{b}{a}=3.51\\ b=4\text{mm}\\ a=\frac{4}{3.51}\\ a=1.14\text{mm}\end{array}$.

Conclusion:

Thus, the radius of the inner conductor so that $C=100\text{pF/m}$ is $a=1.14\text{mm}$.

To determine

(c)

The radius of the inner conductor so that $L=0.2\text{μH/m}$

Answer to Problem 13.1P

The radius of the inner conductor so that $L=0.2\text{μH/m}$ is $a=1.472\text{mm}$.

Explanation of Solution

Given:

Conductivity of copper of a coaxial transmission line ${\sigma }_c=5.8\times {10}^{-7}\text{S/m}$.

The dielectric is polyethylene:.

   $\begin{array}{l}{{\epsilon }^{\prime }}_r=2.26\\ \frac{\sigma }{\omega {\epsilon }^{\prime }}=0.0002\end{array}$.

   $b=4\text{mm}$.

Inductance is $L=0.2\text{μH/m}$.

Calculation:

The inductance of the transmission line is given as:.

   $L=\frac{{\mu }_0}{2\pi }\ln \left(\frac{b}{a}\right)$.

Where,

   ${\mu }_0$ is the permeability

   ${\epsilon }^{\prime }$ is the dielectric permittivity

   $b$ is the inner radius of the outer conductor

   $a$ is the radius of the inner conductor

Thus,

   $\begin{array}{l}L=\frac{{\mu }_0}{2\pi }\ln \left(\frac{b}{a}\right)\\ L=0.2\text{μH/m}\\ L=0.2\times {10}^{-6}\text{H/m}\\ 0.2\times {10}^{-6}=\frac{4\pi \times {10}^{-7}}{2\pi }\ln \left(\frac{b}{a}\right)\\ \ln \left(\frac{b}{a}\right)=\frac{0.2\times {10}^{-6}\times 2\pi }{4\pi \times {10}^{-7}}\\ \ln \left(\frac{b}{a}\right)=1\end{array}$.

Taking antilog on both sides we get,

   $\begin{array}{l}\left(\frac{b}{a}\right)=e^1\\ \frac{b}{a}=2.718\\ b=4\text{mm}\\ a=\frac{4}{2.718}\\ a=1.472\text{mm}\end{array}$.

Conclusion:

Thus, the radius of the inner conductor so that $L=0.2\text{μH/m}$ is $a=1.472\text{mm}$.