Chapter 6, Problem 6.1P

To determine

The distributed parameters of RG-223/U coaxial cable at the given frequency.

Answer to Problem 6.1P

The distributed resistance is $3.32\Omega /\text{m}$ , distributed inductance is $223\text{nH}/\text{m}$ , distributed conductance is $560\times {10}^{-18}\text{S}/\text{m}$ and distributed capacitance is $112\text{pF}/\text{m}$ .

Explanation of Solution

Given:

The radius $a$ of inner conductor is $0.47\text{ mm}$ and the outer conductor has inner radius $b$ of $1.435\text{ mm}$ .The dielectric material is polyethylene and the conductor is copper. The given frequency is $800\text{ MHz}$ .

Concept used:

The series distributed resistance parameter $R^{\prime }$ is shown below.

  $R^{\prime }=\frac{1}{2\text{π}}\left(\frac{1}{a}+\frac{1}{b}\right)\sqrt{\frac{\text{π}f\text{μ}}{{\text{σ}}_{\text{c}}}}$   ........ (1)

Here, ${\text{σ}}_{\text{c}}$ is the conductor’s conductance.

The series distributed inductance parameter $L^{\prime }$ is shown below.

  $L^{\prime }=\frac{\text{μ}}{2\text{π}}\ln \left(\frac{b}{a}\right)$   ........ (2)

The shunt distributed conductance parameter $G^{\prime }$ is shown below.

  $G^{\prime }=\frac{2{\text{πσ}}_{\text{d}}}{\ln \left(\frac{b}{a}\right)}$   ........ (3)

Here, ${\text{σ}}_{\text{d}}$ is the dielectric’s conductance.

The shunt distributed capacitance parameter $C^{\prime }$ is shown below.

  $C^{\prime }=\frac{2\text{πε}}{\ln \left(\frac{b}{a}\right)}$

Calculation:

The conductance ${\text{σ}}_{\text{c}}$ of copper is $5.8\times {10}^7\text{S}/\text{m}$ , conductance ${\text{σ}}_{\text{d}}$ of polyethylene is ${10}^{-16}\text{S}/\text{m}$ and relative permittivity ${\text{ε}}_{\text{r}}$ of polyethylene is $2.26$ .

Substitute $0.47\times {10}^{-3}$ for $a$ , $1.435\times {10}^{-3}$ for $b$ , $8\times {10}^8$ for $f$ , $4\text{π}\times {10}^{-7}$ for ${\text{μ}}_0$ and $5.8\times {10}^7$ for ${\text{σ}}_{\text{c}}$ in equation (1).

  $\begin{array}{c}{R}^{\prime }=\frac{1}{2\text{π}}\left(\frac{1}{0.47\times {10}^{-3}}+\frac{1}{1.435\times {10}^{-3}}\right)\sqrt{\frac{\text{π}\left(8\times {10}^8\right)\left(4\text{π}\times {10}^{-7}\right)}{5.8\times {10}^7}}\\ \approx 3.32\Omega /\text{m}\end{array}$

Therefore, the distributed resistance parameter $R^{\prime }$ is $3.32\Omega /\text{m}$ .

Substitute $4\text{π}\times {10}^{-7}$ for ${\text{μ}}_0$ , $0.47\times {10}^{-3}$ for $a$ and $1.435\times {10}^{-3}$ for $b$ in equation (2).

  $\begin{array}{c}{L}^{\prime }=\frac{4\text{π}\times {10}^{-7}}{2\text{π}}\ln \left(\frac{1.435\times {10}^{-3}}{0.47\times {10}^{-3}}\right)\\ =2.23\times {10}^{-7}\\ \approx 223\times {10}^{-9}\text{H}/\text{m}\end{array}$

Therefore, the distributed inductance parameter $L^{\prime }$ is $223\text{nH}/\text{m}$ .

Substitute ${10}^{-16}$ for ${\text{σ}}_{\text{d}}$ , $0.47\times {10}^{-3}$ for $a$ and $1.435\times {10}^{-3}$ for $b$ in equation (3).

  $\begin{array}{c}{G}^{\prime }=\frac{2\text{π}\times {\text{10}}^{-16}}{\ln \left(\frac{1.435\times {10}^{-3}}{0.47\times {10}^{-3}}\right)}\\ =5.62\times {10}^{-16}\\ \approx 560\times {10}^{-18}\text{S}/\text{m}\end{array}$

Therefore, the distributed conductance parameter $G^{\prime }$ is $560\times {10}^{-18}\text{S}/\text{m}$ .

The distributed conductance in dielectric can be written as,

  $C^{\prime }=\frac{2{\text{πε}}_{\text{r}}{\text{ε}}_0}{\ln \left(\frac{b}{a}\right)}$

Substitute $2.26$ for ${\text{ε}}_{\text{r}}$ , $8.854\times {10}^{-12}$ for ${\text{ε}}_0$ , $0.47\times {10}^{-3}$ for $a$ and $1.435\times {10}^{-3}$ for $b$ in above equation.

  $\begin{array}{c}{C}^{\prime }=\frac{2\text{π}\left(2.26\right)\left(8.854\times {10}^{-12}\right)}{\ln \left(\frac{1.435\times {10}^{-3}}{0.47\times {10}^{-3}}\right)}\\ \approx 112\times {10}^{-12}\text{F}/\text{m}\end{array}$

Therefore, the distributed capacitance parameter $C^{\prime }$ is $112\text{pF}/\text{m}$ .

Conclusion:

Thus, the distributed resistance is $3.32\Omega /\text{m}$ , distributed inductance is $223\text{nH}/\text{m}$ , distributed conductance is $560\times {10}^{-18}\text{S}/\text{m}$ and distributed capacitance is $112\text{pF}/\text{m}$ .