Chapter 12, Problem 25P

(a)

To determine

Find the value of the attenuation constant due to the dielectric losses and due to the conduction losses of the copper waveguide for ${\text{TE}}_{10}$ mode.

Answer to Problem 25P

The value of the attenuation constant due to the dielectric losses $\left({\alpha }_d\right)$ and due to the conduction losses $\left({\alpha }_c\right)$ for ${\text{TE}}_{10}$ mode is $0.0185\text{Np}/\text{m}$ and $0.033\text{Np}/\text{m}$ respectively.

Explanation of Solution

Calculation:

Given dimensions $a\times b$ for the copper waveguide is $1\text{cm}\times 2\text{cm}$.

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 $3\times {10}^8\text{m}/\text{s}$ 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 $1\text{cm}$ for $a$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2.6$ for ${\epsilon }_r$ in above Equation.

$\begin{array}{c}{f}_c=\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{2\left(1\text{cm}\right)\sqrt{2.6}}\\ =\frac{\left(3\times {10}^8\right)\text{m}/\text{s}}{2\sqrt{2.6}\left(1\times {10}^{-2}\right)\text{m}}\left\{\because 1\text{c}={10}^{-2}\right\}\\ =9.3026\times {10}^9{\text{s}}^{-1}\\ =9.3026\text{GHz}\left\{\begin{array}{l}\because 1\text{Hz}=\frac{1}{1\text{s}},\\ 1\text{G}={10}^9\end{array}\right\}\end{array}$

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

$\begin{array}{c}{\eta }^{\prime }=\sqrt{\frac{\mu }{\epsilon }}\\ =\frac{377}{\sqrt{{\epsilon }_r}}\end{array}$

Substitute $2.6$ for ${\epsilon }_r$ in above Equation.

$\begin{array}{c}{\eta }^{\prime }=\frac{377}{\sqrt{2.6}}\Omega \\ =233.81\Omega \end{array}$

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

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

Substitute $233.81\Omega$ for ${\eta }^{\prime }$, ${10}^{-4}\text{S}/\text{m}$ for $\sigma$, $12\text{GHz}$ for $f$ and $9.3026\text{GHz}$ for $f_c$ in Equation (2).

$\begin{array}{c}{\alpha }_d=\frac{\left({10}^{-4}\text{S}/\text{m}\right)\left(233.81\Omega \right)}{2\sqrt{1-{\left(\frac{9.3026\text{GHz}}{12\text{GHz}}\right)}^2}}\\ =\frac{\left({10}^{-4}\right)\left(233.81\right)\Omega \cdot \text{S}\cdot {\text{m}}^{-1}}{2\sqrt{0.39904}}\\ =0.0185\Omega \cdot {\Omega }^{-1}\cdot {\text{m}}^{-1}\left\{\because 1\text{S}=1{\Omega }^{-1}\right\}\\ =0.0185\text{Np}/\text{m}\end{array}$

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

${\alpha }_c=\frac{2R_s}{b{\eta }^{\prime }\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}\left(0.5+\frac{b}{a}{\left(\frac{f_c}{f}\right)}^2\right)$        (3)

Here,

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

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

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

$R_s=\sqrt{\frac{\pi f{\mu }_o}{{\sigma }_c}}\left\{\because \mu ={\mu }_o\right\}$

Substitute $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_o$, $12\text{GHz}$ for $f$ and $5.8\times {10}^7\text{S}/\text{m}$ for ${\sigma }_c$ in above equation.

$\begin{array}{c}{R}_s=\sqrt{\frac{\pi \left(12\text{GHz}\right)\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)}{5.8\times {10}^7\text{S}/\text{m}}}\\ =\sqrt{\frac{\pi \left(12\times {10}^9\right)\left(4\pi \times {10}^{-7}\right){\text{s}}^{-1}\cdot \text{H}\cdot {\text{m}}^{-1}}{5.8\times {10}^7\text{S}/\text{m}}}\left\{\begin{array}{l}\because 1\text{G}={10}^9,\\ 1\text{Hz}=\frac{1}{1\text{s}}\end{array}\right\}\\ =\sqrt{\frac{47374.10113\Omega \cdot \text{s}\cdot {\text{s}}^{-1}}{5.8\times {10}^7{\Omega }^{-1}}}\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s},\\ 1\text{S}=\frac{1}{1\Omega }\end{array}\right\}\\ =\sqrt{8.1679\times {10}^{-4}{\Omega }^2}\end{array}$

Simplify the above Equation.

$R_s=0.02858\Omega$

Substitute $0.02858\Omega$ for $R_s$, $1\text{cm}$ for $a$, $2\text{cm}$ for $b$, $233.81\Omega$ for ${\eta }^{\prime }$, $12\text{GHz}$ for $f$ and $9.3026\text{GHz}$ for $f_c$ in Equation (3).

$\begin{array}{c}{\alpha }_c=\frac{2\left(0.02858\Omega \right)}{\left(2\text{cm}\right)\left(233.81\Omega \right)\sqrt{1-{\left(\frac{9.3026\text{GHz}}{12\text{GHz}}\right)}^2}}\left(0.5+\left(\frac{2\text{cm}}{1\text{cm}}\right){\left(\frac{9.3026\text{GHz}}{12\text{GHz}}\right)}^2\right)\\ =\frac{0.05716}{\left(2\times {10}^{-2}\right)\left(233.81\right)\sqrt{1-\left(0.601\right)}}\left(0.5+\left(2\right)\left(0.601\right)\right)\\ =0.033\text{Np}/\text{m}\end{array}$

Conclusion:

Thus, the value of the attenuation constant due to the dielectric losses $\left({\alpha }_d\right)$ and due to the conduction losses $\left({\alpha }_c\right)$ for ${\text{TE}}_{10}$ mode is $0.0185\text{Np}/\text{m}$ and $0.033\text{Np}/\text{m}$ respectively.

(b)

To determine

Find the value of the attenuation constant due to the dielectric losses and due to the conduction losses of the copper waveguide for ${\text{TM}}_{11}$ mode.

Answer to Problem 25P

The value of the attenuation constant due to the dielectric losses $\left({\alpha }_d\right)$ and due to the conduction losses $\left({\alpha }_c\right)$ for ${\text{TM}}_{11}$ mode is $0.02344\text{Np}/\text{m}$ and $0.0441\text{Np}/\text{m}$ respectively.

Explanation of Solution

Calculation:

Write the expression to calculate the cutoff frequency for ${\text{TM}}_{11}$ mode.

$f_c=\frac{u^{\prime }}{2}{\left[\frac{1}{a^2}+\frac{1}{b^2}\right]}^{\frac{1}{2}}$

Substitute $\frac{c}{\sqrt{{\epsilon }_r}}$ for $u^{\prime }$ in above equation.

$\begin{array}{c}{f}_c=\frac{\left(\frac{c}{\sqrt{{\epsilon }_r}}\right)}{2}{\left[\frac{1}{a^2}+\frac{1}{b^2}\right]}^{\frac{1}{2}}\\ =\frac{c}{2\sqrt{{\epsilon }_r}}{\left[\frac{1}{a^2}+\frac{1}{b^2}\right]}^{\frac{1}{2}}\end{array}$

Substitute $1\text{cm}$ for $a$, $2\text{cm}$ for $b$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2.6$ for ${\epsilon }_r$ in above Equation.

$\begin{array}{c}{f}_c=\left(\frac{3\times {10}^8\text{m}/\text{s}}{2\sqrt{2.6}}\right){\left[\frac{1}{{\left(1\text{cm}\right)}^2}+\frac{1}{{\left(2\text{cm}\right)}^2}\right]}^{\frac{1}{2}}\\ =\frac{\left(3\times {10}^8\right)\text{m}/\text{s}}{2\sqrt{2.6}}\left(111.803{\text{m}}^{-1}\right)\\ =10.4\times {10}^9{\text{s}}^{-1}\\ =10.4\text{GHz}\left\{\begin{array}{l}\because 1\text{Hz}=\frac{1}{1\text{s}},\\ 1\text{G}={10}^9\end{array}\right\}\end{array}$

Substitute $233.81\Omega$ for ${\eta }^{\prime }$, ${10}^{-4}\text{S}/\text{m}$ for $\sigma$, $12\text{GHz}$ for $f$ and $10.4\text{GHz}$ for $f_c$ in Equation (2).

$\begin{array}{c}{\alpha }_d=\frac{\left({10}^{-4}\text{S}/\text{m}\right)\left(233.81\Omega \right)}{2\sqrt{1-{\left(\frac{10.4\text{GHz}}{12\text{GHz}}\right)}^2}}\\ =\frac{\left({10}^{-4}\right)\left(233.81\right)\Omega \cdot \text{S}\cdot {\text{m}}^{-1}}{2\sqrt{0.2489}}\\ =0.02344\Omega \cdot {\Omega }^{-1}\cdot {\text{m}}^{-1}\left\{\because 1\text{S}=1{\Omega }^{-1}\right\}\\ =0.02344\text{Np}/\text{m}\end{array}$

Write the expression to calculate the attenuation constant due to conduction losses for the ${\text{TM}}_{11}$ mode.

${\alpha }_c=\frac{2R_s}{b{\eta }^{\prime }\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}\left[\frac{{\left(\frac{b}{a}\right)}^3+1}{{\left(\frac{b}{a}\right)}^2+1}\right]$        (4)

From part (a), the real part of the intrinsic impedance of the conducting wall is,

$R_s=0.02858\Omega$

Substitute $0.02858\Omega$ for $R_s$, $1\text{cm}$ for $a$, $2\text{cm}$ for $b$, $233.81\Omega$ for ${\eta }^{\prime }$, $12\text{GHz}$ for $f$ and $10.4\text{GHz}$ for $f_c$ in Equation (4).

$\begin{array}{c}{\alpha }_c=\frac{2\left(0.02858\Omega \right)}{\left(2\text{cm}\right)\left(233.81\Omega \right)\sqrt{1-{\left(\frac{10.4\text{GHz}}{12\text{GHz}}\right)}^2}}\left[\frac{{\left(\frac{2\text{cm}}{1\text{cm}}\right)}^3+1}{{\left(\frac{2\text{cm}}{1\text{cm}}\right)}^2+1}\right]\\ =\frac{0.05716}{\left(2\times {10}^{-2}\right)\left(233.81\right)\sqrt{0.24889}}\left[\frac{8+1}{4+1}\right]\\ =0.0441\text{Np}/\text{m}\end{array}$

Conclusion:

Thus, the value of the attenuation constant due to the dielectric losses $\left({\alpha }_d\right)$ and due to the conduction losses $\left({\alpha }_c\right)$ for ${\text{TM}}_{11}$ mode is $0.02344\text{Np}/\text{m}$ and $0.0441\text{Np}/\text{m}$ respectively.