Chapter 12, Problem 33P

(a)

To determine

Find the value of the attenuation constant due to the dielectric losses for the rectangular waveguide.

Answer to Problem 33P

The value of the attenuation constant due to the dielectric losses $\left({\alpha }_d\right)$ for the rectangular waveguide is $2.165\times {10}^{-2}\frac{\text{Np}}{\text{m}}$.

Explanation of Solution

Calculation:

Write the expression to calculate the cutoff frequency for ${\text{TE}}_{10}$ mode.

$f_c=\frac{u^{\prime }}{2a}$        (1)

Here,

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

$a$ is the inner dimension 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 (1).

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

Substitute $4.8\text{cm}$ for $a$, $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$ and $2.11$ for ${\epsilon }_r$ in above Equation.

$\begin{array}{c}{f}_c=\frac{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}{2\left(4.8\text{cm}\right)\sqrt{2.11}}\\ =\frac{\left(3\times {10}^8\right)\frac{\text{m}}{\text{s}}}{2\sqrt{2.11}\left(4.8\times {10}^{-2}\right)\text{m}}\left\{\because 1\text{c}={10}^{-2}\right\}\\ =2.151\times {10}^9\frac{1}{\text{s}}\\ =2.151\text{GHz}\left\{\begin{array}{l}\because 1\text{Hz}=\frac{1}{1\text{s}},\\ 1\text{G}={10}^9\end{array}\right\}\end{array}$

Write the expression to calculate the loss tangent.

$d=\frac{\sigma }{\omega \epsilon }$

$d=\frac{\sigma }{2\pi f{\epsilon }_o{\epsilon }_r}\left\{\because \omega =2\pi f\right\}$        (2)

Here,

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

$f$ is the operating frequency,

$\sigma$ is the conductivity and

$\omega$ is the angular frequency.

Rearrange the Equation (2) to find $\sigma$.

$\sigma =2\pi df{\epsilon }_o{\epsilon }_r$

Substitute $3\times {10}^{-4}$ for $d$, $4\text{GHz}$ for $f$, $2.11$ for ${\epsilon }_r$ and $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_o$ in above Equation.

$\begin{array}{c}\sigma =2\pi \left(3\times {10}^{-4}\right)\left(4\text{GHz}\right)\left(2.11\right)\left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)\\ =2\pi \left(3\times {10}^{-4}\right)\left(4\times {10}^9\right)\left(2.11\right)\left(8.854\times {10}^{-12}\right)\text{Hz}\cdot \frac{\text{F}}{\text{m}}\left\{\because 1\text{G}={10}^9\right\}\\ =1.4086\times {10}^{-4}\frac{1}{\text{s}}\cdot \frac{\text{F}}{\text{m}}\left\{\because 1\text{Hz}=\frac{1}{1\text{s}}\right\}\end{array}$

Simplify the above Equation.

$\begin{array}{c}\sigma =1.4086\times {10}^{-4}\frac{1}{\text{s}}\cdot \frac{\left(\frac{\text{A}\cdot \text{s}}{\text{V}}\right)}{\text{m}}\left\{\because 1\text{F}=\frac{1\text{A}\cdot 1\text{s}}{1\text{V}}\right\}\\ =1.4086\times {10}^{-4}\frac{\left(\frac{\text{A}}{\text{V}}\right)}{\text{m}}\\ =1.4086\times {10}^{-4}\frac{\text{S}}{\text{m}}\left\{\because 1\text{S}=\frac{1\text{A}}{1\text{V}}\right\}\end{array}$

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

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

${\eta }^{\prime }=\sqrt{\frac{{\mu }_o{\mu }_r}{{\epsilon }_o{\epsilon }_r}}$        (3)

Here,

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

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

Substitute 1 for ${\mu }_r$, $4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}$ for ${\mu }_o$, $2.11$ for ${\epsilon }_r$ and $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_o$ in Equation (3)

$\begin{array}{c}{\eta }^{\prime }=\sqrt{\frac{\left(4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}\right)\left(1\right)}{2.11\left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)}}\\ =\sqrt{\frac{\left(4\pi \times {10}^{-7}\right)\Omega \cdot \text{s}}{\left(18.6819\times {10}^{-12}\right)\frac{\text{s}}{\Omega }}}\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s,}\\ 1\text{F}=\frac{1\text{s}}{1\Omega }\end{array}\right\}\\ =\sqrt{67264.8055{\Omega }^2}\\ =259.53\Omega \end{array}$

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

${\alpha }_d=\frac{\sigma {\eta }^{\prime }}{2\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}$        (4)

Substitute $259.53\Omega$ for ${\eta }^{\prime }$, $1.4086\times {10}^{-4}\frac{\text{S}}{\text{m}}$ for $\sigma$, $4\text{GHz}$ for $f$ and $2.151\text{GHz}$ for $f_c$ in Equation (4).

$\begin{array}{c}{\alpha }_d=\frac{\left(1.4086\times {10}^{-4}\frac{\text{S}}{\text{m}}\right)\left(259.53\Omega \right)}{2\sqrt{1-{\left(\frac{2.151\text{GHz}}{4\text{GHz}}\right)}^2}}\\ =\frac{\left(1.4086\times {10}^{-4}\right)\left(259.53\right)\Omega \cdot \frac{\text{S}}{\text{m}}}{2\sqrt{0.7108}}\\ =\frac{0.0366\Omega \cdot \frac{{\Omega }^{-1}}{\text{m}}}{1.6862}\left\{\because 1\text{S}=1{\Omega }^{-1}\right\}\\ =2.165\times {10}^{-2}\frac{\text{Np}}{\text{m}}\end{array}$

Conclusion:

Thus, the value of the attenuation constant due to the dielectric losses $\left({\alpha }_d\right)$ for the rectangular waveguide is $2.165\times {10}^{-2}\frac{\text{Np}}{\text{m}}$.

(b)

To determine

Find the value of the attenuation constant due to the conduction losses for the rectangular waveguide.

Answer to Problem 33P

The value of the attenuation constant due to the conduction losses $\left({\alpha }_c\right)$ for the rectangular waveguide is $4.818\times {10}^{-3}\frac{\text{Np}}{\text{m}}$.

Explanation of Solution

Calculation:

Write the expression to calculate the attenuation constant due to conduction losses for the ${\text{TE}}_{10}$ mode.

${\alpha }_c=\frac{2R_s}{b{\eta }^{\prime }\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}\left(0.5+\frac{b}{a}{\left(\frac{f_c}{f}\right)}^2\right)$        (5)

Here,

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

$f_c$ is the cutoff 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,

${\sigma }_c$ is the conductivity.

Substitute $4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}$ for ${\mu }_o$, $1$ for ${\mu }_r$, $4\text{GHz}$ for $f$ and $4.1\times {10}^7\frac{\text{S}}{\text{m}}$ for ${\sigma }_c$ in Equation (6).

$\begin{array}{c}{R}_s=\sqrt{\frac{\pi \left(4\text{GHz}\right)\left(4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}\right)\left(1\right)}{\left(4.1\times {10}^7\frac{\text{S}}{\text{m}}\right)}}\\ =\sqrt{\frac{\pi \left(4\times {10}^9\right)\left(4\pi \times {10}^{-7}\right)\frac{1}{\text{s}}\cdot \frac{\text{H}}{\text{m}}}{\left(4.1\times {10}^7\right)\frac{\text{S}}{\text{m}}}}\left\{\begin{array}{l}\because 1\text{G}={10}^9,\\ 1\text{Hz}=\frac{1}{1\text{s}}\end{array}\right\}\\ =\sqrt{\frac{15791.3670\left(\frac{\Omega \cdot \text{s}}{\text{s}}\right)}{\left(4.1\times {10}^7\right)\left(\frac{1}{\Omega }\right)}}\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s},\\ 1\text{S}=\frac{1}{1\Omega }\end{array}\right\}\\ =\sqrt{3.8516\times {10}^{-4}{\Omega }^2}\end{array}$

Simplify the above Equation.

$R_s=1.9625\times {10}^{-2}\Omega$

Substitute $1.9625\times {10}^{-2}\Omega$ for $R_s$, $4.8\text{cm}$ for $a$, $2.4\text{cm}$ for $b$, $259.53\Omega$ for ${\eta }^{\prime }$, $4\text{GHz}$ for $f$ and $2.151\text{GHz}$ for $f_c$ in Equation (5).

$\begin{array}{c}{\alpha }_c=\frac{2\left(1.9625\times {10}^{-2}\Omega \right)}{\left(2.4\text{cm}\right)\left(259.53\Omega \right)\sqrt{1-{\left(\frac{2.151\text{GHz}}{4\text{GHz}}\right)}^2}}\left(0.5+\left(\frac{2.4\text{cm}}{4.8\text{cm}}\right){\left(\frac{2.151\text{GHz}}{4\text{GHz}}\right)}^2\right)\\ =\frac{0.0393}{\left(2.4\times {10}^{-2}\right)\left(259.53\right)\sqrt{0.7108}\text{m}}\left(0.5+\left(0.5\right)\left(0.2892\right)\right)\left\{\because 1\text{c}={10}^{-2}\right\}\\ =4.818\times {10}^{-3}\frac{\text{Np}}{\text{m}}\end{array}$

Conclusion:

Thus, the value of the attenuation constant due to the conduction losses $\left({\alpha }_c\right)$ for the rectangular waveguide is $4.818\times {10}^{-3}\frac{\text{Np}}{\text{m}}$.