Chapter 12, Problem 31P

To determine

Find the value of the attenuation constant due to the conduction losses and the distance travelled by the waveguide which is attenuated by $30\%$.

Answer to Problem 31P

The value of the attenuation constant due to the conduction losses $\left({\alpha }_c\right)$ and the distance travelled $\left(z\right)$ by the waveguide which is attenuated by $30\%$ is $0.0314\frac{\text{Np}}{\text{m}}$ and $6.5445\text{m}$.

Explanation of Solution

Calculation:

Given that the complex permittivity is ,

${\epsilon }_c=16{\epsilon }_o\left(1-j{10}^{-4}\right)$

${\epsilon }_c=16{\epsilon }_o-j\left(16{\epsilon }_o\times {10}^{-4}\right)$        (1)

Refer to Equation (12.57) in the textbook for the complex permittivity.

${\epsilon }_c=\epsilon -j\frac{\sigma }{\omega }$        (2)

Compare the Equations (1) and (2).

$\epsilon =16{\epsilon }_o$

$\frac{\sigma }{\omega }=16{\epsilon }_o\times {10}^{-4}$

Here,

${\epsilon }_o$ is the permittivity of the free space which is $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$,

$\sigma$ is the conductivity and

$\omega$ is the angular frequency.

Write the expression to calculate the cutoff frequency.

$f_c=\frac{u^{\prime }}{2}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$        (3)

Here,

$m$ and $n$ are integers,

$u^{\prime }$ is the phase velocity of uniform plane wave in dielectric medium and

$a$ and $b$ are the inner dimensions of the waveguide.

Write the expression to calculate the phase velocity of uniform plane wave in the lossless dielectric medium.

$u^{\prime }=\frac{c}{\sqrt{{\epsilon }_r}}$

Here,

$c$ is the speed of light in vacuum which is $3\times {10}^8\frac{\text{m}}{\text{s}}$ and

${\epsilon }_r$ is the permittivity of the medium.

Substitute $\frac{c}{\sqrt{{\epsilon }_r}}$ for $u^{\prime }$ in Equation (3).

$\begin{array}{c}{f}_c=\frac{\left(\frac{c}{\sqrt{{\epsilon }_r}}\right)}{2}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}\\ =\frac{c}{2\sqrt{{\epsilon }_r}}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}\end{array}$

Substitute $4\text{cm}$ for $a$ and $b$, $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$, $2$ for $m$, $1$ for $n$ and $16$ for ${\epsilon }_r$ in above Equation.

$\begin{array}{c}{f}_c=\frac{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}{2\sqrt{16}}\sqrt{{\left(\frac{2}{4\text{cm}}\right)}^2+{\left(\frac{1}{4\text{cm}}\right)}^2}\\ =\frac{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}{8}\sqrt{{\left(\frac{2}{4\times {10}^{-2}\text{m}}\right)}^2+{\left(\frac{1}{4\times {10}^{-2}\text{m}}\right)}^2}\left\{\because 1\text{c}={10}^{-2}\right\}\\ =\left(0.375\times {10}^8\frac{\text{m}}{\text{s}}\right)\sqrt{2500{\text{m}}^{-2}+625{\text{m}}^{-2}}\\ =\left(0.375\times {10}^8\right)\left(55.9017{\text{m}}^{-1}\right)\frac{\text{m}}{\text{s}}\end{array}$

Simplify the above Equation.

$\begin{array}{c}{f}_c=2.0963\times {10}^9\frac{1}{\text{s}}\\ =2.0963\times {10}^9\text{Hz}\left\{\because 1\text{Hz}=\frac{1}{1\text{s}}\right\}\\ =2.0963\text{GHz}\left\{\because 1\text{G}={10}^9\right\}\end{array}$

Given that the operating frequency of the waveguide is $10\%$ above the cutoff frequency. That is, the operating frequency is $110\%$ of the cutoff frequency. It is expressed as follows:

$\begin{array}{c}f=\left(110\%\right)f_c\\ =\left(\frac{110}{100}\right)\left(2.0963\text{GHz}\right)\\ =1.1\left(2.0963\text{GHz}\right)\\ =2.3059\text{GHz}\end{array}$

Write the expression to calculate the intrinsic impedance of a uniform plane wave in the medium.

${\eta }^{\prime }=\sqrt{\frac{\mu }{\epsilon }}$

${\eta }^{\prime }=\sqrt{\frac{{\mu }_o}{16{\epsilon }_o}}\left\{\because \epsilon =16{\epsilon }_o\right\}$        (4)

Here,

${\mu }_o$ is the permeability of the free space which is $4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}$.

Substitute $4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}$ for ${\mu }_o$ and $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_o$ in Equation (4)

$\begin{array}{c}{\eta }^{\prime }=\sqrt{\frac{\left(4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}\right)}{16\left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)}}\\ =\sqrt{\frac{\left(4\pi \times {10}^{-7}\right)\Omega \cdot \text{s}}{\left(141.664\times {10}^{-12}\right)\frac{\text{s}}{\Omega }}}\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s,}\\ 1\text{F}=\frac{1\text{s}}{1\Omega }\end{array}\right\}\\ =\sqrt{8870.5462{\Omega }^2}\\ =94.2478\Omega \end{array}$

Write the expression to calculate the attenuation constant due to conduction losses for the ${\text{TM}}_{21}$ mode.

${\alpha }_c=\frac{2R_s}{b{\eta }^{\prime }\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}$        (5)

Here,

$R_s$ is the real part of the intrinsic impedance of the conducting wall,

$f_c$ is the cutoff frequency and

$f$ is the operating frequency.

Write the expression to calculate the real part of the intrinsic impedance of the conducting wall.

$R_s=\sqrt{\frac{\pi f\mu }{{\sigma }_c}}$

$R_s=\sqrt{\frac{\pi f{\mu }_o{\mu }_r}{{\sigma }_c}}$        (6)

Here,

${\mu }_r$ is the permeability of the medium and

${\sigma }_c$ is the conductivity of brass.

Substitute $4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}$ for ${\mu }_o$, $1$ for ${\mu }_r$, $2.3059\text{GHz}$ for $f$ and $1.5\times {10}^7\frac{\text{S}}{\text{m}}$ for ${\sigma }_c$ in Equation (6).

$\begin{array}{c}{R}_s=\sqrt{\frac{\pi \left(2.3059\text{GHz}\right)\left(4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}\right)\left(1\right)}{\left(1.5\times {10}^7\frac{\text{S}}{\text{m}}\right)}}\\ =\sqrt{\frac{\pi \left(2.3059\times {10}^9\right)\left(4\pi \times {10}^{-7}\right)\frac{1}{\text{s}}\cdot \frac{\text{H}}{\text{m}}}{\left(1.5\times {10}^7\right)\frac{\text{S}}{\text{m}}}}\left\{\begin{array}{l}\because 1\text{G}={10}^9,\\ 1\text{Hz}=\frac{1}{1\text{s}}\end{array}\right\}\\ =\sqrt{\frac{9103.3283\left(\frac{\Omega \cdot \text{s}}{\text{s}}\right)}{\left(1.5\times {10}^7\right)\left(\frac{1}{\Omega }\right)}}\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s},\\ 1\text{S}=\frac{1}{1\Omega }\end{array}\right\}\\ =\sqrt{6.0689\times {10}^{-4}{\Omega }^2}\end{array}$

Simplify the above Equation.

$R_s=0.0246\Omega$

Substitute $0.0246\Omega$ for $R_s$, $4\text{cm}$ for $b$, $94.2478\Omega$ for ${\eta }^{\prime }$, $2.3059\text{GHz}$ for $f$ and $2.0963\text{GHz}$ for $f_c$ in Equation (5).

$\begin{array}{c}{\alpha }_c=\frac{2\left(0.0246\Omega \right)}{\left(4\text{cm}\right)\left(94.2478\Omega \right)\sqrt{1-{\left(\frac{2.0963\text{GHz}}{2.3059\text{GHz}}\right)}^2}}\\ =\frac{0.0492}{\left(4\times {10}^{-2}\right)\left(94.2478\right)\sqrt{0.1735}\text{m}}\left\{\because 1\text{c}={10}^{-2}\right\}\\ =0.0314\frac{\text{Np}}{\text{m}}\end{array}$

From the comparisons of Equations (1) and (2),

$\frac{\sigma }{\omega }=16{\epsilon }_o\times {10}^{-4}$

Rearrange the above Equation.

$\begin{array}{c}\sigma =16\omega {\epsilon }_o\times {10}^{-4}\\ =16\left(2\pi f\right){\epsilon }_o\times {10}^{-4}\left\{\because \omega =2\pi f\right\}\end{array}$

Substitute $2.3059\text{GHz}$ for $f$ and $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_o$ in above Equation.

$\begin{array}{c}\sigma =16\left(2\pi \times 2.3059\text{GHz}\right)\left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)\times {10}^{-4}\\ =16\left(2\pi \times 2.3059\times {10}^9\right)\left(8.854\times {10}^{-12}\right)\times {10}^{-4}\text{Hz}\cdot \frac{\text{F}}{\text{m}}\left\{\because 1\text{G}={10}^9\right\}\\ =2.0525\times {10}^{-4}\frac{1}{\text{s}}\cdot \frac{\text{F}}{\text{m}}\left\{\because 1\text{Hz}=\frac{1}{1\text{s}}\right\}\end{array}$

Simplify the above Equation.

$\begin{array}{c}\sigma =2.0525\times {10}^{-4}\frac{1}{\text{s}}\cdot \frac{\left(\frac{\text{A}\cdot \text{s}}{\text{V}}\right)}{\text{m}}\left\{\because 1\text{F}=\frac{1\text{A}\cdot 1\text{s}}{1\text{V}}\right\}\\ =2.0525\times {10}^{-4}\frac{\left(\frac{\text{A}}{\text{V}}\right)}{\text{m}}\\ =2.0525\times {10}^{-4}\frac{\text{S}}{\text{m}}\left\{\because 1\text{S}=\frac{1\text{A}}{1\text{V}}\right\}\end{array}$

Write the expression to calculate the attenuation constant due to the dielectric losses.

${\alpha }_d=\frac{\sigma {\eta }^{\prime }}{2\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}$        (7)

Substitute $94.2478\Omega$ for ${\eta }^{\prime }$, $2.0525\times {10}^{-4}\frac{\text{S}}{\text{m}}$ for $\sigma$, $2.3059\text{GHz}$ for $f$ and $2.0963\text{GHz}$ for $f_c$ in Equation (7).

$\begin{array}{c}{\alpha }_d=\frac{\left(2.0525\times {10}^{-4}\frac{\text{S}}{\text{m}}\right)\left(94.2478\Omega \right)}{2\sqrt{1-{\left(\frac{2.0963\text{GHz}}{2.3059\text{GHz}}\right)}^2}}\\ =\frac{\left(2.0525\times {10}^{-4}\right)\left(94.2478\right)\Omega \cdot \frac{\text{S}}{\text{m}}}{2\sqrt{0.1735}}\\ =\frac{0.0193\Omega \cdot \frac{{\Omega }^{-1}}{\text{m}}}{0.8331}\left\{\because 1\text{S}=1{\Omega }^{-1}\right\}\\ =0.0231\frac{\text{Np}}{\text{m}}\end{array}$

To calculate the distance over which the wave is attenuated by 30%,

$\begin{array}{c}{E}_o{e}^{-\left({\alpha }_c+{\alpha }_d\right)z}=\left(70\%\right)E_o\\ =\left(\frac{70}{100}\right)E_o\\ =0.7E_o\end{array}$

Simplify the above Equation.

$\begin{array}{c}{e}^{-\left({\alpha }_c+{\alpha }_d\right)z}=0.7\\ \left({\alpha }_c+{\alpha }_d\right)z=\ln \left(\frac{1}{0.7}\right)\end{array}$

Rearrange the Equation.

$\begin{array}{c}z=\frac{1}{{\alpha }_c+{\alpha }_d}\ln \left(\frac{1}{0.7}\right)\\ =\left(\frac{1}{0.0314\frac{\text{Np}}{\text{m}}+0.0231\frac{\text{Np}}{\text{m}}}\right)\ln \left(\frac{1}{0.7}\right)\\ =\left(\frac{1}{0.0545\frac{\text{Np}}{\text{m}}}\right)\ln \left(\frac{1}{0.7}\right)\\ =6.5445\text{m}\end{array}$

Conclusion:

Thus, the value of the attenuation constant due to the conduction losses $\left({\alpha }_c\right)$ and the distance travelled $\left(z\right)$ by the waveguide which is attenuated by $30\%$ is $0.0314\frac{\text{Np}}{\text{m}}$ and $6.5445\text{m}$.