Chapter 12, Problem 33P

To determine

Find the value of the phase constant, attenuation constant due to the dielectric losses, attenuation constant due to the conduction losses, phase velocity, group velocity and the wavelength of a rectangular waveguide.

Answer to Problem 33P

The value of the phase constant $\left(\beta \right)$, attenuation constant due to the dielectric losses $\left({\alpha }_d\right)$, attenuation constant due to the conduction losses $\left({\alpha }_c\right)$, phase velocity $\left(u_p\right)$, group velocity $\left(u_g\right)$ and the wavelength $\left({\lambda }_c\right)$ of a rectangular waveguide is $199.94\text{rad}/\text{m}$, $1.481\times {10}^{-2}\text{Np}/\text{m}$, $0.02183\text{Np}/\text{m}$, $2.357\times {10}^8\text{m}/\text{s}$, $1.689\times {10}^8\text{m}/\text{s}$ and $5\text{cm}$ respectively.

Explanation of Solution

Calculation:

Given dimensions $a\times b$ for the waveguide is $2.5\text{cm}\times 1.5\text{cm}$, since $a>b$, the dominant mode is ${\text{TE}}_{10}$ mode.

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 relative 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 $2.5\text{cm}$ for $a$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2.26$ for ${\epsilon }_r$ in above Equation.

$\begin{array}{c}{f}_c=\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{2\left(2.5\text{cm}\right)\sqrt{2.26}}\\ =\frac{\left(3\times {10}^8\right)\text{m}/\text{s}}{2\sqrt{2.26}\left(2.5\times {10}^{-2}\right)\text{m}}\left\{\because 1\text{c}={10}^{-2}\right\}\\ =3.991\times {10}^9{\text{s}}^{-1}\\ =3.991\text{GHz}\left\{\begin{array}{l}\because 1\text{Hz}=\frac{1}{1\text{s}},\\ 1\text{G}={10}^9\end{array}\right\}\end{array}$

Write the expression to calculate the phase constant of the TE waveguide.

$\beta ={\beta }^{\prime }\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}$

For dielectric medium, the above equation becomes,

$\beta =\frac{2\pi f\sqrt{{\epsilon }_r}}{c}\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}$        (2)

Here,

$f_c$ is the cutoff frequency and

$f$ is the operating frequency.

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$, $7.5\text{GHz}$ for $f$, $2.26$ for ${\epsilon }_r$ and $3.991\text{GHz}$ for $f_c$ in Equation (2).

$\begin{array}{c}\beta =\frac{2\pi \left(7.5\text{GHz}\right)\sqrt{2.26}}{\left(3\times {10}^8\text{m}/\text{s}\right)}\sqrt{1-{\left(\frac{3.991\text{GHz}}{7.5\text{GHz}}\right)}^2}\\ =\frac{2\pi \left(7.5\times {\text{10}}^9\right)\sqrt{2.26}}{\left(3\times {10}^8\right)}\sqrt{1-0.2832}\\ =199.94\text{rad}/\text{m}\end{array}$

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 $2.26$ for ${\epsilon }_r$ in above Equation.

$\begin{array}{c}{\eta }^{\prime }=\frac{377}{\sqrt{2.26}}\Omega \\ =250.78\Omega \end{array}$

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

${\alpha }_d=\frac{\sigma {\eta }^{\prime }}{2\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}$        (3)

Substitute $250.78\Omega$ for ${\eta }^{\prime }$, ${10}^{-4}\text{S}/\text{m}$ for $\sigma$, $7.5\text{GHz}$ for $f$ and $3.991\text{GHz}$ for $f_c$ in Equation (3).

$\begin{array}{c}{\alpha }_d=\frac{\left({10}^{-4}\text{S}/\text{m}\right)\left(250.78\Omega \right)}{2\sqrt{1-{\left(\frac{3.991\text{GHz}}{7.5\text{GHz}}\right)}^2}}\\ =\frac{\left({10}^{-4}\right)\left(250.78\right)\Omega \cdot \text{S}\cdot {\text{m}}^{-1}}{2\sqrt{1-0.2832}}\\ =1.481\times {10}^{-2}\Omega \cdot {\Omega }^{-1}\cdot {\text{m}}^{-1}\left\{\because 1\text{S}=1{\Omega }^{-1}\right\}\\ =1.481\times {10}^{-2}\text{Np}/\text{m}\end{array}$

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

${\alpha }_c=\frac{2R_s}{b{\eta }^{\prime }\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}\left(0.5+\frac{b}{a}{\left(\frac{f_c}{f}\right)}^2\right)$        (4)

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 ${\mu }_o$, $7.5\text{GHz}$ for $f$ and $1.1\times {10}^7\text{S}/\text{m}$ for ${\sigma }_c$ in above equation.

$\begin{array}{c}{R}_s=\sqrt{\frac{\pi \left(7.5\text{GHz}\right)\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)}{1.1\times {10}^7\text{S}/\text{m}}}\\ =\sqrt{\frac{\pi \left(7.5\times {10}^9\right)\left(4\pi \times {10}^{-7}\right){\text{s}}^{-1}\cdot \text{H}\cdot {\text{m}}^{-1}}{1.1\times {10}^7\text{S}/\text{m}}}\left\{\begin{array}{l}\because 1\text{G}={10}^9,\\ 1\text{Hz}=\frac{1}{1\text{s}}\end{array}\right\}\\ =\sqrt{\frac{29608.8132\Omega \cdot \text{s}\cdot {\text{s}}^{-1}}{1.1\times {10}^7{\Omega }^{-1}}}\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s},\\ 1\text{S}=\frac{1}{1\Omega }\end{array}\right\}\\ =\sqrt{2.6917\times {10}^{-3}{\Omega }^2}\end{array}$

Simplify the above Equation.

$R_s=0.0519\Omega$

Substitute $0.0519\Omega$ for $R_s$, $2.5\text{cm}$ for $a$, $1.5\text{cm}$ for $b$, $250.78\Omega$ for ${\eta }^{\prime }$, $7.5\text{GHz}$ for $f$ and $3.991\text{GHz}$ for $f_c$ in Equation (4).

$\begin{array}{c}{\alpha }_c=\frac{2\left(0.0519\Omega \right)}{\left(1.5\text{cm}\right)\left(250.78\Omega \right)\sqrt{1-{\left(\frac{3.991\text{GHz}}{7.5\text{GHz}}\right)}^2}}\left(0.5+\left(\frac{1.5\text{cm}}{2.5\text{cm}}\right){\left(\frac{3.991\text{GHz}}{7.5\text{GHz}}\right)}^2\right)\\ =\frac{0.1038}{\left(1.5\times {10}^{-2}\right)\left(250.78\right)\sqrt{1-0.2831}}\left(0.5+\left(0.6\right)\left(0.2831\right)\right)\\ =0.02183\text{Np}/\text{m}\end{array}$

Write the expression to calculate the phase velocity for the waveguide.

$u_p=\frac{u^{\prime }}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}$

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

$u_p=\frac{c}{\sqrt{{\epsilon }_r}\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$, $7.5\text{GHz}$ for $f$, $2.26$ for ${\epsilon }_r$ and $3.991\text{GHz}$ for $f_c$ in above equation.

$\begin{array}{c}{u}_p=\frac{3\times {10}^8\text{m}/\text{s}}{\sqrt{2.26}\sqrt{1-{\left(\frac{3.991\text{GHz}}{7.5\text{GHz}}\right)}^2}}\\ =\frac{3\times {10}^8}{\sqrt{2.26}\sqrt{1-0.2832}}\text{m}/\text{s}\\ =2.357\times {10}^8\text{m}/\text{s}\end{array}$

Write the expression to calculate the group velocity for the waveguide.

$u_g=u^{\prime }\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}$

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

$u_g=\frac{c}{\sqrt{{\epsilon }_r}}\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$, $7.5\text{GHz}$ for $f$, $2.26$ for ${\epsilon }_r$ and $3.991\text{GHz}$ for $f_c$ in above equation.

$\begin{array}{c}{u}_g=\frac{3\times {10}^8\text{m}/\text{s}}{\sqrt{2.26}}\sqrt{1-{\left(\frac{3.991\text{GHz}}{7.5\text{GHz}}\right)}^2}\\ =\left(\frac{3\times {10}^8}{\sqrt{2.26}}\right)\sqrt{1-0.2832}\text{m}/\text{s}\\ =1.689\times {10}^8\text{m}/\text{s}\end{array}$

Write the expression to calculate the wavelength of the waveguide.

${\lambda }_c=\frac{u^{\prime }}{f_c}$

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

$\begin{array}{c}{\lambda }_c=\frac{\left(\frac{c}{\sqrt{{\epsilon }_r}}\right)}{f_c}\\ =\frac{c}{f_c\sqrt{{\epsilon }_r}}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$, $2.26$ for ${\epsilon }_r$ and $3.991\text{GHz}$ for $f_c$ in above equation.

$\begin{array}{c}{\lambda }_c=\frac{3\times {10}^8\text{m}/\text{s}}{\left(3.991\text{GHz}\right)\sqrt{2.26}}\\ =\frac{3\times {10}^8\text{m}/\text{s}}{\left(3.991\times {10}^9\text{1}/\text{s}\right)\sqrt{2.26}}\\ =5\times {10}^{-2}\text{m}\\ =5\text{cm}\end{array}$

Conclusion:

Thus, the value of the phase constant $\left(\beta \right)$, attenuation constant due to the dielectric losses $\left({\alpha }_d\right)$, attenuation constant due to the conduction losses $\left({\alpha }_c\right)$, phase velocity $\left(u_p\right)$, group velocity $\left(u_g\right)$ and the wavelength $\left({\lambda }_c\right)$ of a rectangular waveguide is $199.94\text{rad}/\text{m}$, $1.481\times {10}^{-2}\text{Np}/\text{m}$, $0.02183\text{Np}/\text{m}$, $2.357\times {10}^8\text{m}/\text{s}$, $1.689\times {10}^8\text{m}/\text{s}$ and $5\text{cm}$ respectively.