Chapter 12, Problem 18P

(a)

To determine

Find the value of the phase constant, phase velocity, group velocity and intrinsic impedance of the air-filled rectangular waveguide.

Answer to Problem 18P

The value of the phase constant $\left(\beta \right)$, phase velocity $\left(u_p\right)$, group velocity $\left(u_g\right)$ and intrinsic impedance $\left({\eta }_{\text{TE}}\right)$ of the air-filled rectangular waveguide is $122.32\frac{\text{rad}}{\text{m}}$, $3.185\times {10}^8\frac{\text{m}}{\text{s}}$, $2.826\times {10}^8\frac{\text{m}}{\text{s}}$ and $400.21\Omega$ respectively.

Explanation of Solution

Calculation:

Given dimensions $a\times b$ for the waveguide is $7.2\text{cm}\times 3.4\text{cm}$, since $a>b$, the dominant mode is ${\text{TE}}_{10}$ mode.

Write the expression to calculate the cutoff frequency.

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

Here,

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

$a$ is the inner dimension of the waveguide.

Given waveguide is air-filled, therefore the phase velocity $u^{\prime }$ is the speed of light in vacuum $c$, which is $3\times {10}^8\frac{\text{m}}{\text{s}}$.

Substitute $c$ for $u^{\prime }$ in Equation (1).

$f_c=\frac{c}{2a}$

Substitute $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$ and $7.2\text{cm}$ for $a$ in above Equation.

$\begin{array}{c}{f}_c=\frac{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}{2\left(7.2\text{cm}\right)}\\ =\frac{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}{2\left(7.2\times {10}^{-2}\right)\text{m}}\left\{\because 1\text{c}={10}^{-2}\right\}\\ =2.083\times {10}^9\frac{\text{1}}{\text{s}}\\ =2.083\text{GHz}\left\{\begin{array}{l}\because 1\text{Hz}=\frac{1}{1\text{s}},\\ 1\text{G}={10}^9\end{array}\right\}\end{array}$

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

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

Here,

$\omega$ is the angular frequency,

$f_c$ is the cutoff frequency and

$f$ is the operating frequency.

Substitute $c$ for $u^{\prime }$ in Equation (2).

$\begin{array}{c}\beta =\frac{\omega }{c}\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}\\ =\frac{2\pi f}{c}\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}\left\{\because \omega =2\pi f\right\}\end{array}$

Substitute $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$, $2.083\text{GHz}$ for $f_c$ and $6.2\text{GHz}$ for $f$ in above Equation.

$\begin{array}{c}\beta =\frac{2\pi \left(6.2\text{GHz}\right)}{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}\sqrt{1-{\left(\frac{2.083\text{GHz}}{6.2\text{GHz}}\right)}^2}\\ =\frac{2\pi \left(6.2\times {10}^9\right)}{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}\sqrt{0.8871}\text{Hz}\left\{\because 1\text{G}={10}^9\right\}\\ =\frac{\left(3.6691\times {10}^{10}\right)}{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}\frac{1}{\text{s}}\left\{\because 1\text{Hz}=\frac{1}{1\text{s}}\right\}\\ =122.32\frac{\text{rad}}{\text{m}}\end{array}$

Write the expression to calculate the phase velocity.

$u_p=\frac{u^{\prime }}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}$        (3)

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

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

Substitute $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$, $2.083\text{GHz}$ for $f_c$ and $6.2\text{GHz}$ for $f$ in above Equation.

$\begin{array}{c}{u}_p=\frac{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}{\sqrt{1-{\left(\frac{2.083\text{GHz}}{6.2\text{GHz}}\right)}^2}}\\ =\frac{\left(3\times {10}^8\right)}{\sqrt{0.8871}}\frac{\text{m}}{\text{s}}\\ =3.185\times {10}^8\frac{\text{m}}{\text{s}}\end{array}$

Write the expression to calculate the group velocity.

$u_g=u^{\prime }\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}$        (4)

Substitute $c$ for $u^{\prime }$ in Equation (4).

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

Substitute $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$, $2.083\text{GHz}$ for $f_c$ and $6.2\text{GHz}$ for $f$ in above Equation.

$\begin{array}{c}{u}_g=\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)\sqrt{1-{\left(\frac{2.083\text{GHz}}{6.2\text{GHz}}\right)}^2}\\ =\left(3\times {10}^8\right)\sqrt{0.8871}\frac{\text{m}}{\text{s}}\\ =2.826\times {10}^8\frac{\text{m}}{\text{s}}\end{array}$

Write the expression to calculate the intrinsic impedance for the free space.

${\eta }_{\text{TE}}=\frac{{\eta }^{\prime }}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}$        (5)

Here,

${\eta }^{\prime }$ is the intrinsic impedance of the free space which is $377\Omega$.

Substitute $377\Omega$ for ${\eta }^{\prime }$, $6.2\text{GHz}$ for $f$ and $2.083\text{GHz}$ for $f_c$ in Equation (5).

$\begin{array}{c}{\eta }_{\text{TE}}=\frac{377\Omega }{\sqrt{1-{\left(\frac{2.083\text{GHz}}{6.2\text{GHz}}\right)}^2}}\\ =\frac{377\Omega }{\sqrt{0.8871}}\\ =400.21\Omega \end{array}$

Conclusion:

Thus, the value of the phase constant $\left(\beta \right)$, phase velocity $\left(u_p\right)$, group velocity $\left(u_g\right)$ and intrinsic impedance $\left({\eta }_{\text{TE}}\right)$ of the air-filled rectangular waveguide is $122.32\frac{\text{rad}}{\text{m}}$, $3.185\times {10}^8\frac{\text{m}}{\text{s}}$, $2.826\times {10}^8\frac{\text{m}}{\text{s}}$ and $400.21\Omega$ respectively.

(b)

To determine

Find the value of the phase constant, phase velocity, group velocity and intrinsic impedance of the rectangular waveguide with a material.

Answer to Problem 18P

The value of the phase constant $\left(\beta \right)$, phase velocity $\left(u_p\right)$, group velocity $\left(u_g\right)$ and intrinsic impedance $\left({\eta }_{\text{TE}}\right)$ of the rectangular waveguide with a material is $189.83\frac{\text{rad}}{\text{m}}$, $2.052\times {10}^8\frac{\text{m}}{\text{s}}$, $1.949\times {10}^8\frac{\text{m}}{\text{s}}$ and $257.88\Omega$ respectively.

Explanation of Solution

Calculation:

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,

${\epsilon }_r$ 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 $2.25$ for ${\epsilon }_r$, $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$ and $7.2\text{cm}$ for $a$ in above Equation.

$\begin{array}{c}{f}_c=\frac{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}{2\left(7.2\text{cm}\right)\sqrt{2.25}}\\ =\frac{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}{2\sqrt{2.25}\left(7.2\times {10}^{-2}\right)\text{m}}\left\{\because 1\text{c}={10}^{-2}\right\}\\ =1.389\times {10}^9\frac{\text{1}}{\text{s}}\\ =1.389\text{GHz}\left\{\begin{array}{l}\because 1\text{Hz}=\frac{1}{1\text{s}},\\ 1\text{G}={10}^9\end{array}\right\}\end{array}$

Substitute $\frac{c}{\sqrt{{\epsilon }_r}}$ for $u^{\prime }$ in Equation (2).

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

Substitute $2.25$ for ${\epsilon }_r$, $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$, $1.389\text{GHz}$ for $f_c$ and $6.2\text{GHz}$ for $f$ in above Equation.

$\begin{array}{c}\beta =\frac{2\pi \left(6.2\text{GHz}\right)\sqrt{2.25}}{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}\sqrt{1-{\left(\frac{1.389\text{GHz}}{6.2\text{GHz}}\right)}^2}\\ =\frac{2\pi \sqrt{2.25}\left(6.2\times {10}^9\right)}{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}\sqrt{0.9498}\text{Hz}\left\{\because 1\text{G}={10}^9\right\}\\ =\frac{\left(5.6948\times {10}^{10}\right)}{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}\frac{1}{\text{s}}\left\{\because 1\text{Hz}=\frac{1}{1\text{s}}\right\}\\ =189.83\frac{\text{rad}}{\text{m}}\end{array}$

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

$\begin{array}{c}{u}_p=\frac{\left(\frac{c}{\sqrt{{\epsilon }_r}}\right)}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}\\ =\frac{c}{\left(\sqrt{{\epsilon }_r}\right)\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}\end{array}$

Substitute $2.25$ for ${\epsilon }_r$, $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$, $1.389\text{GHz}$ for $f_c$ and $6.2\text{GHz}$ for $f$ in above Equation.

$\begin{array}{c}{u}_p=\frac{\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}{\left(\sqrt{2.25}\right)\sqrt{1-{\left(\frac{1.389\text{GHz}}{6.2\text{GHz}}\right)}^2}}\\ =\frac{\left(3\times {10}^8\right)}{\left(\sqrt{2.25}\right)\sqrt{0.9498}}\frac{\text{m}}{\text{s}}\\ =2.052\times {10}^8\frac{\text{m}}{\text{s}}\end{array}$

Substitute $\frac{c}{\sqrt{{\epsilon }_r}}$ for $u^{\prime }$ in Equation (4).

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

Substitute $2.25$ for ${\epsilon }_r$, $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$, $1.389\text{GHz}$ for $f_c$ and $6.2\text{GHz}$ for $f$ in above Equation.

$\begin{array}{c}{u}_g=\left(\frac{3\times {10}^8\frac{\text{m}}{\text{s}}}{\sqrt{2.25}}\right)\sqrt{1-{\left(\frac{1.389\text{GHz}}{6.2\text{GHz}}\right)}^2}\\ =\left(\frac{3\times {10}^8}{\sqrt{2.25}}\right)\sqrt{0.9498}\frac{\text{m}}{\text{s}}\\ =1.949\times {10}^8\frac{\text{m}}{\text{s}}\end{array}$

Write the expression to calculate the intrinsic impedance for the medium.

${\eta }_{\text{TE}}=\frac{\sqrt{\frac{\mu }{\epsilon }}}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}$

${\eta }_{\text{TE}}=\frac{\sqrt{\frac{{\mu }_o}{2.25{\epsilon }_o}}}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}\left\{\begin{array}{l}\because \epsilon =2.25{\epsilon }_o\\ \mu ={\mu }_o\end{array}\right\}$        (6)

Here,

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

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

Substitute $4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}$ for ${\mu }_o$, $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_o$, $1.389\text{GHz}$ for $f_c$ and $6.2\text{GHz}$ for $f$ in Equation (6).

$\begin{array}{c}{\eta }_{\text{TE}}=\frac{\sqrt{\frac{\left(4\pi \times {10}^{-7}\frac{\text{H}}{\text{m}}\right)}{2.25\left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)}}}{\sqrt{1-{\left(\frac{1.389\text{GHz}}{6.2\text{GHz}}\right)}^2}}\left\{\because \epsilon =2.25{\epsilon }_o\right\}\\ =\frac{\left(\sqrt{\frac{\left(4\pi \times {10}^{-7}\right)\text{H}}{\left(19.9215\times {10}^{-12}\right)\text{F}}}\right)}{\sqrt{0.9498}}\\ =\frac{\left(\sqrt{\frac{\left(4\pi \times {10}^{-7}\right)\Omega \cdot \text{s}}{\left(19.9215\times {10}^{-12}\right)\frac{\text{s}}{\Omega }}}\right)}{0.9746}\left\{\begin{array}{l}\because 1\text{H}=1\Omega \cdot 1\text{s,}\\ 1\text{F}=\frac{1\text{s}}{1\Omega }\end{array}\right\}\end{array}$

Simplify the above Equation.

$\begin{array}{c}{\eta }_{\text{TE}}=\frac{\sqrt{6.3079\times {10}^4{\Omega }^2}}{0.9746}\\ =\frac{251.1562}{0.9746}\Omega \\ =257.88\Omega \end{array}$

Conclusion:

Thus, the value of the phase constant $\left(\beta \right)$, phase velocity $\left(u_p\right)$, group velocity $\left(u_g\right)$ and intrinsic impedance $\left({\eta }_{\text{TE}}\right)$ of the rectangular waveguide with a material is $189.83\frac{\text{rad}}{\text{m}}$, $2.052\times {10}^8\frac{\text{m}}{\text{s}}$, $1.949\times {10}^8\frac{\text{m}}{\text{s}}$ and $257.88\Omega$ respectively.