Chapter 12, Problem 7P

(a)

To determine

Find the value of the dimensions of an air-filled rectangular waveguide.

Answer to Problem 7P

The value of the dimensions $a\times b$ of an air-filled rectangular waveguide is $3\text{cm}\times 1.25\text{cm}$.

Explanation of Solution

Calculation:

Given that the cutoff frequency $f_c$ of ${\text{TE}}_{10}$ mode is $5\text{GHz}$ and ${\text{TE}}_{01}$ mode is $12\text{GHz}$.

For ${\text{TE}}_{10}$ mode:

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

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

Therefore the Equation (1) becomes,

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

Rearrange the above equation.

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

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $5\text{GHz}$ for $f_c$ in above Equation.

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

For ${\text{TE}}_{01}$ mode:

Write the expression to calculate the cutoff frequency for the ${\text{TE}}_{01}$ mode.

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

Here,

$b$ is the inner dimension of the waveguide.

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

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

Rearrange the above equation.

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

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $12\text{GHz}$ for $f_c$ in above Equation.

$\begin{array}{c}b=\frac{3\times {10}^8\text{m}/\text{s}}{2\left(12\text{GHz}\right)}\\ =\frac{3\times {10}^8\text{m}/\text{s}}{2\left(12\times {10}^9\text{1}/\text{s}\right)}\\ =1.25\times {10}^{-2}\text{m}\\ =1.25\text{cm}\end{array}$

Conclusion:

Thus, the value of the dimensions $a\times b$ of an air-filled rectangular waveguide is $3\text{cm}\times 1.25\text{cm}$.

(b)

To determine

Find the value of the cutoff frequencies of the next three higher TE modes in an air-filled rectangular waveguide.

Answer to Problem 7P

The value of the cutoff frequencies $\left(f_c\right)$ of the next three higher TE modes ${\text{TE}}_{20}$, ${\text{TE}}_{01}$ and ${\text{TE}}_{11}$ modes are $10\text{GHz}$, $12\text{GHz}$ and $13\text{GHz}$ respectively.

Explanation of Solution

Calculation:

From part (a), the dimension of the waveguide is in form of $a>b$.

The condition $a>b$ can also be expressed as $\frac{1}{a}<\frac{1}{b}$.

Write the general expression to calculate the cutoff frequency for the waveguide.

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

Substitute $c$ for $u^{\prime }$ in above equation.

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

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$, $3\text{cm}$ for $a$ and $1.25\text{cm}$ for $b$ in above Equation.

$\begin{array}{c}{f}_c=\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{2}\sqrt{{\left(\frac{m}{3\text{cm}}\right)}^2+{\left(\frac{n}{1.25\text{cm}}\right)}^2}\\ =\left(1.5\times {10}^8\right)\sqrt{{\left(\frac{m}{3\times {10}^{-2}}\right)}^2+{\left(\frac{n}{1.25\times {10}^{-2}}\right)}^2}\text{1}/\text{s}\left\{\because 1\text{c}={10}^{-2}\right\}\end{array}$

$f_c=\left(1.5\times {10}^8\right)\sqrt{{\left(\frac{m}{0.03}\right)}^2+{\left(\frac{n}{0.0125}\right)}^2}\text{Hz}\left\{\because 1\text{Hz}=\frac{1}{1\text{s}}\right\}$        (3)

For ${\text{TE}}_{10}$ mode:

The integers are $m=1$ and $n=0$.

Therefore, the Equation (3) becomes,

$\begin{array}{c}{f}_c=\left(1.5\times {10}^8\right)\sqrt{{\left(\frac{1}{0.03}\right)}^2+{\left(\frac{0}{0.0125}\right)}^2}\text{Hz}\\ =\left(1.5\times {10}^8\right)\left(\sqrt{1111.11+0}\right)\text{Hz}\\ =5\times {10}^9\text{Hz}\\ =5\text{GHz}\end{array}$

For ${\text{TE}}_{20}$ mode:

The integers are $m=2$ and $n=0$.

Therefore, the Equation (3) becomes,

$\begin{array}{c}{f}_c=\left(1.5\times {10}^8\right)\sqrt{{\left(\frac{2}{0.03}\right)}^2+{\left(\frac{0}{0.0125}\right)}^2}\text{Hz}\\ =\left(1.5\times {10}^8\right)\left(\sqrt{4444.44+0}\right)\text{Hz}\\ =10\times {10}^9\text{Hz}\\ =10\text{GHz}\end{array}$

For ${\text{TE}}_{30}$ mode:

The integers are $m=3$ and $n=0$.

Therefore, the Equation (3) becomes,

$\begin{array}{c}{f}_c=\left(1.5\times {10}^8\right)\sqrt{{\left(\frac{3}{0.03}\right)}^2+{\left(\frac{0}{0.0125}\right)}^2}\text{Hz}\\ =\left(1.5\times {10}^8\right)\left(\sqrt{10000+0}\right)\text{Hz}\\ =15\times {10}^9\text{Hz}\\ =15\text{GHz}\end{array}$

For ${\text{TE}}_{40}$ mode:

The integers are $m=4$ and $n=0$.

Therefore, the Equation (3) becomes,

$\begin{array}{c}{f}_c=\left(1.5\times {10}^8\right)\sqrt{{\left(\frac{4}{0.03}\right)}^2+{\left(\frac{0}{0.0125}\right)}^2}\text{Hz}\\ =\left(1.5\times {10}^8\right)\left(\sqrt{17777.778+0}\right)\text{Hz}\\ =20\times {10}^9\text{Hz}\\ =20\text{GHz}\end{array}$

For ${\text{TE}}_{01}$ mode:

The integers are $m=0$ and $n=1$.

Therefore, the Equation (3) becomes,

$\begin{array}{c}{f}_c=\left(1.5\times {10}^8\right)\sqrt{{\left(\frac{0}{0.03}\right)}^2+{\left(\frac{1}{0.0125}\right)}^2}\text{Hz}\\ =\left(1.5\times {10}^8\right)\left(\sqrt{0+6400}\right)\text{Hz}\\ =12\times {10}^9\text{Hz}\\ =12\text{GHz}\end{array}$

For ${\text{TE}}_{02}$ mode:

The integers are $m=0$ and $n=2$.

Therefore, the Equation (3) becomes,

$\begin{array}{c}{f}_c=\left(1.5\times {10}^8\right)\sqrt{{\left(\frac{0}{0.03}\right)}^2+{\left(\frac{2}{0.0125}\right)}^2}\text{Hz}\\ =\left(1.5\times {10}^8\right)\left(\sqrt{0+25600}\right)\text{Hz}\\ =24\times {10}^9\text{Hz}\\ =24\text{GHz}\end{array}$

For ${\text{TE}}_{11}$ mode:

The integers are $m=1$ and $n=1$.

Therefore, the Equation (3) becomes,

$\begin{array}{c}{f}_c=\left(1.5\times {10}^8\right)\sqrt{{\left(\frac{1}{0.03}\right)}^2+{\left(\frac{1}{0.0125}\right)}^2}\text{Hz}\\ =\left(1.5\times {10}^8\right)\left(\sqrt{1111.11+6400}\right)\text{Hz}\\ =13\times {10}^9\text{Hz}\\ =13\text{GHz}\end{array}$

For ${\text{TE}}_{21}$ mode:

The integers are $m=2$ and $n=1$.

Therefore, the Equation (3) becomes,

$\begin{array}{c}{f}_c=\left(1.5\times {10}^8\right)\sqrt{{\left(\frac{2}{0.03}\right)}^2+{\left(\frac{1}{0.0125}\right)}^2}\text{Hz}\\ =\left(1.5\times {10}^8\right)\left(\sqrt{4444.44+6400}\right)\text{Hz}\\ =15.62\times {10}^9\text{Hz}\\ =15.62\text{GHz}\end{array}$

Therefore, the next three higher TE modes are ${\text{TE}}_{20}$, ${\text{TE}}_{01}$ and ${\text{TE}}_{11}$ modes.

Conclusion:

Thus, the value of the cutoff frequencies $\left(f_c\right)$ of the next three higher TE modes ${\text{TE}}_{20}$, ${\text{TE}}_{01}$ and ${\text{TE}}_{11}$ modes are $10\text{GHz}$, $12\text{GHz}$ and $13\text{GHz}$ respectively.

(c)

To determine

Find the value of the cutoff frequency for the ${\text{TE}}_{11}$ mode in an air-filled rectangular waveguide.

Answer to Problem 7P

The value of the cutoff frequency $\left(f_c\right)$ for the ${\text{TE}}_{11}$ mode in an air-filled rectangular waveguide is $8.67\text{GHz}$.

Explanation of Solution

Calculation:

For ${\text{TE}}_{11}$ mode, the integers are,

$\begin{array}{l}m=1\\ n=1\end{array}$

Write the general expression to calculate the cutoff frequency for the waveguide.

$f_c=\frac{u^{\prime }}{2}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$        (4)

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 (4).

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

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$, $2.25$ for ${\epsilon }_r$, $1$ for $m$ and $n$, $3\text{cm}$ for $a$ and $1.25\text{cm}$ for $b$ in above Equation.

$\begin{array}{c}{f}_c=\frac{3\times {10}^8\text{m}/\text{s}}{2\sqrt{2.25}}\sqrt{{\left(\frac{1}{3\text{cm}}\right)}^2+{\left(\frac{1}{1.25\text{cm}}\right)}^2}\\ =\left(\frac{1.5\times {10}^8}{\sqrt{2.25}}\right)\sqrt{{\left(\frac{1}{3\times {\text{10}}^{-2}}\right)}^2+{\left(\frac{1}{1.25\times {\text{10}}^{-2}}\right)}^2}\text{1}/\text{s}\left\{\because 1\text{c}={10}^{-2}\right\}\\ =8.67\times {\text{10}}^9\text{Hz}\left\{\because 1\text{Hz}=\frac{1}{1\text{s}}\right\}\\ =8.67\text{GHz}\end{array}$

Conclusion:

Thus, the value of the cutoff frequency $\left(f_c\right)$ for the ${\text{TE}}_{11}$ mode in an air-filled rectangular waveguide is $8.67\text{GHz}$.