Question

Consider a rectangular wave guide with dimensions $\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{}\mathit{c}\mathit{m}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{}\mathit{c}\mathit{m}$. What TE modes will propagate in this waveguide if the driving frequency is $\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z}$? Suppose you wanted to excite only one TE mode; what range of frequencies could you use? What are the corresponding wavelengths (in open space)?

Short answer

Answer

The frequency range is $\mathbf{0}\mathbf{.}\mathbf{657}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z}\mathbf{<}\mathit{v}\mathbf{<}\mathbf{1}\mathbf{.}\mathbf{314}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z}$, and the wavelength range is $\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{}\mathit{c}\mathit{m}\mathbf{<}\mathit{\lambda }\mathbf{<}\mathbf{4}\mathbf{.}\mathbf{56}\mathbf{}\mathit{c}\mathit{m}$. The excited TE mode that propagate are ${\mathit{f}}_{\mathbf{10}}\mathbf{,}{\mathit{f}}_{\mathbf{11}}\mathbf{,}{\mathit{f}}_{\mathbf{20}}\mathbf{,}$and ${\mathit{f}}_{\mathbf{01}}$.

Step-by-step solution

Given Information:

Given data:

The dimension of a rectangular guide wave is $\mathit{a}\mathbf{\times }\mathit{b}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{}\mathit{c}\mathit{m}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{}\mathit{c}\mathit{m}$.

The driving frequency is $\mathit{f}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z}$.

Expression for the cutoff frequency for the mode:

Write the expression for the cutoff frequency.

${\mathit{\omega }}_{\mathbf{m}\mathbf{n}}\mathbf{=}\mathit{c}\mathit{\pi }\sqrt{{\left(\frac{m}{a}\right)}^{\mathbf{2}}\mathbf{+}{\left(\frac{n}{m}\right)}^{\mathbf{2}}}$ …… (1)

Here, c is the speed of light $\left(3\times {10}^{8}m/s\right)$.

Write the lowest cutoff frequency for a given waveguide that occurs for the mode $\mathit{T}{\mathit{E}}_{\mathbf{10}}$.

${\mathit{\omega }}_{\mathbf{10}}\mathbf{=}\frac{\mathbf{c\pi }}{\mathbf{a}}$

Now, write the frequency for a given waveguide.

$$\begin{gathered} {\mathit{f}}_{\mathbf{10}}\mathbf{=}\frac{{\mathbf{\omega }}_{\mathbf{10}}}{\mathbf{2}\mathbf{\pi }} \\ {\mathit{f}}_{\mathbf{10}}\mathbf{=}\frac{\left({\displaystyle \frac{\mathrm{c\pi }}{a}}\right)}{\mathbf{2}\mathbf{\pi }} \\ {\mathit{f}}_{\mathbf{10}}\mathbf{=}\frac{\mathbf{c}}{\mathbf{2}\mathbf{a}} \end{gathered}$$ …… (2)

Determine the frequency range:

Substitute $\mathit{c}\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$and $\mathit{a}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{}\mathit{c}\mathit{m}$in equation (2).

$$\begin{gathered} {\mathit{f}}_{\mathbf{10}}\mathbf{=}\frac{\left(3\times {10}^{8}m/s\right)}{\mathbf{2}\mathbf{\times }\left(2.28cm\times {\displaystyle \frac{{10}^{-2}m}{1cm}}\right)} \\ {\mathit{f}}_{\mathbf{10}}\mathbf{=}\frac{\left(3\times {10}^{8}m/s\right)}{\mathbf{0}\mathbf{.}\mathbf{0456}\mathbf{}\mathbf{m}} \\ {\mathit{f}}_{\mathbf{10}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{657}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z} \end{gathered}$$

For 20 mode range, calculate the frequency range.

$$\begin{gathered} {\mathit{f}}_{\mathbf{20}}\mathbf{=}\mathbf{2}\frac{\mathbf{c}}{\mathbf{2}\mathbf{a}} \\ {\mathit{f}}_{\mathbf{20}}\mathbf{=}\mathbf{2}\mathbf{\times }\frac{\left(3\times {10}^{8}m/s\right)}{\mathbf{2}\mathbf{\times }\left(2.28cm\times {\displaystyle \frac{{10}^{-2}m}{1cm}}\right)} \\ {\mathit{f}}_{\mathbf{20}}\mathbf{=}\mathbf{2}\mathbf{\times }\frac{\left(3\times {10}^{8}m/s\right)}{\mathbf{0}\mathbf{.}\mathbf{0456}\mathbf{}\mathbf{m}} \\ {\mathit{f}}_{\mathbf{20}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{314}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z} \end{gathered}$$

For 30 mode range, calculate the frequency range.

$$\begin{gathered} {\mathit{f}}_{\mathbf{30}}\mathbf{=}\mathbf{3}\frac{\mathbf{c}}{\mathbf{2}\mathbf{a}} \\ {\mathit{f}}_{\mathbf{30}}\mathbf{=}\mathbf{3}\mathbf{\times }\left(2.28cm\times \frac{{10}^{-2}m}{1cm}\right) \\ {\mathit{f}}_{\mathbf{30}}\mathbf{=}\mathbf{3}\mathbf{\times }\frac{\left(3\times {10}^{8}m/s\right)}{\mathbf{0}\mathbf{.}\mathbf{0456}\mathbf{}\mathbf{m}} \\ {\mathit{f}}_{\mathbf{30}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{97}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z} \end{gathered}$$

For 01 mode range, calculate the frequency range.

$$\begin{gathered} {\mathit{f}}_{\mathbf{01}}\mathbf{=}\frac{\mathbf{c}}{\mathbf{2}\mathbf{b}} \\ {\mathit{f}}_{\mathbf{01}}\mathbf{=}\frac{\left(3\times {10}^{8}m/s\right)}{\mathbf{2}\mathbf{\times }\left(1.01cm\times {\displaystyle \frac{{10}^{-2}m}{1cm}}\right)} \\ {\mathit{f}}_{\mathbf{01}}\mathbf{=}\frac{\left(3\times {10}^{8}m/s\right)}{\mathbf{0}\mathbf{.}\mathbf{0202}\mathbf{}\mathbf{m}} \\ {\mathit{f}}_{\mathbf{01}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{49}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z} \end{gathered}$$

For 02 mode range, calculate the frequency range.

$$\begin{gathered} {\mathit{f}}_{\mathbf{02}}\mathbf{=}\mathbf{2}\frac{\mathbf{c}}{\mathbf{2}\mathbf{b}} \\ {\mathit{f}}_{\mathbf{02}}\mathbf{=}\mathbf{2}\mathbf{\times }\frac{\left(3\times {10}^{8}m/s\right)}{\mathbf{2}\mathbf{\times }\left(1.01cm\times {\displaystyle \frac{{10}^{-2}m}{1cm}}\right)} \\ {\mathit{f}}_{\mathbf{02}}\mathbf{=}\mathbf{2}\mathbf{\times }\frac{\left(3\times {10}^{8}m/s\right)}{\mathbf{0}\mathbf{.}\mathbf{0202}\mathbf{}\mathbf{m}} \\ {\mathit{f}}_{\mathbf{02}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{97}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z} \end{gathered}$$

For 11 mode range, calculate the frequency range.

$$\begin{gathered} {\mathit{V}}_{\mathbf{11}}\mathbf{=}\frac{\mathbf{c}}{\mathbf{2}}\sqrt{{\left(\frac{1}{a}\right)}^{\mathbf{2}}\mathbf{+}{\left(\frac{1}{b}\right)}^{\mathbf{2}}} \\ {\mathit{V}}_{\mathbf{11}}\mathbf{=}\frac{\left(3\times {10}^{8}m/s\right)}{\mathbf{2}}\sqrt{{\left(\frac{1}{2.28cm\times {\displaystyle \frac{{10}^{-2}m}{1cm}}}\right)}^{\mathbf{2}}\mathbf{+}{\left(\frac{1}{1.01cm\times {\displaystyle \frac{1^{-2}m}{1cm}}}\right)}^{\mathbf{2}}} \\ {\mathit{f}}_{\mathbf{11}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{62}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z} \end{gathered}$$

As the driving frequency is given as $\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z}$, the possible modes will be ${\mathit{f}}_{\mathbf{10}}\mathbf{,}{\mathit{f}}_{\mathbf{20}}\mathbf{,}{\mathit{f}}_{\mathbf{01}}$and ${\mathit{f}}_{\mathbf{11}}$.

To get only one mode, the waveguide at a frequency between ${\mathit{f}}_{\mathbf{10}}$and ${\mathit{f}}_{\mathbf{20}}$must be driven. Hence, the frequency range will be,

$\mathbf{0}\mathbf{.}\mathbf{657}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z}\mathbf{<}\mathit{v}\mathbf{<}\mathbf{1}\mathbf{.}\mathbf{314}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z}$.

Determine the wavelength range:

Using the wavelength 10 mode, the wavelength range will be,

$$\begin{gathered} {\mathit{\lambda }}_{\mathbf{10}}\mathbf{=}\frac{\mathbf{c}}{{\mathbf{f}}_{\mathbf{10}}} \\ {\mathit{\lambda }}_{\mathbf{10}}\mathbf{=}\frac{\mathbf{c}}{\left({\displaystyle \frac{c}{2a}}\right)} \\ {\mathit{\lambda }}_{\mathbf{10}}\mathbf{=}\mathbf{2}\mathit{a} \end{gathered}$$

Substitute $\mathit{a}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{80}\mathbf{}\mathit{c}\mathit{m}$in the above expression.

$$\begin{gathered} {\mathit{\lambda }}_{\mathbf{10}}\mathbf{=}\mathbf{2}\mathbf{\times }\left(2.28cm\right) \\ {\mathit{\lambda }}_{\mathbf{10}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{56}\mathbf{}\mathit{c}\mathit{m} \end{gathered}$$

Using the wavelength of 20 mode, the wavelength range will be,

$$\begin{gathered} {\mathit{\lambda }}_{\mathbf{20}}\mathbf{=}\frac{\mathbf{c}}{{\mathbf{f}}_{\mathbf{20}}} \\ {\mathit{\lambda }}_{\mathbf{20}}\mathbf{=}\frac{\mathbf{c}}{\mathbf{2}\left({\displaystyle \frac{c}{2a}}\right)} \\ {\mathit{\lambda }}_{\mathbf{20}}\mathbf{=}\mathit{a} \\ {\mathit{\lambda }}_{\mathbf{20}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{}\mathit{c}\mathit{m} \end{gathered}$$

Hence, the wavelength range will be $\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{}\mathit{c}\mathit{m}\mathbf{<}\mathit{\lambda }\mathbf{<}\mathbf{4}\mathbf{.}\mathbf{56}\mathbf{}\mathit{c}\mathit{m}$.

Therefore, the frequency range is $\mathbf{0}\mathbf{.}\mathbf{657}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z}\mathbf{<}\mathit{v}\mathbf{<}\mathbf{1}\mathbf{.}\mathbf{314}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z}$, and the wavelength range is $\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{}\mathit{c}\mathit{m}\mathbf{<}\mathit{\lambda }\mathbf{<}\mathbf{4}\mathbf{.}\mathbf{56}\mathbf{}\mathit{c}\mathit{m}$. The excited TE mode that propagate are ${\mathit{f}}_{\mathbf{10}}\mathbf{,}{\mathit{f}}_{\mathbf{11}}\mathbf{,}{\mathit{f}}_{\mathbf{20}}\mathbf{,}$and ${\mathit{f}}_{\mathbf{01}}$.