Chapter 7, Problem 7.1P

Find the cutoff frequency for the first eight modes of WR430.

To determine

The cutoff frequency of eight modes of $\text{WR430}$ waveguide.

Answer to Problem 7.1P

The cutoff frequency of the eight modes of $\text{WR430}$ the waveguide are $1.374\text{ GHz}$ , $2.747\text{ GHz}$ , $2.747\text{ GHz}$ , $3.07\text{ GHz}$ , $3.07\text{ GHz}$ , $3.885\text{ GHz}$ , $3.885\text{ GHz}$ and $4.121\text{ GHz}$ .

Explanation of Solution

Given:

The dimensions of the waveguide are $a=4.3\text{ in}$ and $b=2.15\text{ in}$ .

Concept used:

The cutoff frequency of the waveguide with dimensions $a$ and $b$ is shown below.

  $f_{{\text{c}}_{mn}}=\frac{c}{2}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$ ..... (1)

Here, $m$ and $n$ represents the half-wave variations.

Calculation:

Convert the dimensions of the waveguide in meters as

  $\begin{array}{c}a=\frac{4.3\text{ in}}{39.37}\left(\frac{1\text{ m}}{1\text{ in}}\right)\\ \approx 0.109\text{ m}\end{array}$

The value of $b$ in meters is calculated below.

  $\begin{array}{c}b=\frac{\text{2}\text{.15 in}}{39.37}\left(\frac{1\text{ m}}{1\text{ in}}\right)\\ \approx 0.0546\text{ m}\end{array}$

The first mode of the waveguide is ${\text{TE}}_{10}$ and its cutoff frequency is calculated as,

  $\begin{array}{c}{f}_{{\text{c}}_{10}}=\frac{c}{2}\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{0}{b}\right)}^2}\\ =\frac{c}{2a}\end{array}$

Substitute $3\times {10}^8$ for $c$ and $0.109$ for $a$ in the above equation.

  $\begin{array}{c}{f}_{{\text{c}}_{10}}=\frac{3\times {10}^8}{2\left(0.109\right)}\\ =13.74\times {10}^8\\ \approx 1.374\times {10}^9\end{array}$

Therefore, the cutoff frequency for ${\text{TE}}_{10}$ mode is $1.374\text{ GHz}$ .

The second mode of the waveguide is ${\text{TE}}_{01}$ and its cutoff frequency is calculated as,

  $\begin{array}{c}{f}_{{\text{c}}_{01}}=\frac{c}{2}\sqrt{{\left(\frac{0}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}\\ =\frac{c}{2b}\end{array}$

Substitute $3\times {10}^8$ for $c$ and $0.0546$ for $a$ in the above equation.

  $\begin{array}{c}{f}_{{\text{c}}_{01}}=\frac{3\times {10}^8}{2\left(0.0546\right)}\\ =27.47\times {10}^8\\ \approx 2.747\times {10}^9\end{array}$

Therefore, the cutoff frequency for ${\text{TE}}_{01}$ mode is $2.747\text{ GHz}$ .

The third mode of the waveguide is ${\text{TE}}_{20}$ and its cutoff frequency is calculated as,

  $\begin{array}{c}{f}_{{\text{c}}_{20}}=\frac{c}{2}\sqrt{{\left(\frac{2}{a}\right)}^2+{\left(\frac{0}{b}\right)}^2}\\ =\frac{c}{a}\end{array}$

Substitute $3\times {10}^8$ for $c$ and $0.109$ for $a$ in the above equation.

  $\begin{array}{c}{f}_{{\text{c}}_{20}}=\frac{3\times {10}^8}{0.109}\\ =27.47\times {10}^8\\ \approx 2.747\times {10}^9\end{array}$

Therefore, the cutoff frequency for ${\text{TE}}_{20}$ mode is $2.747\text{ GHz}$ .

The fourth mode of the waveguide is ${\text{TE}}_{11}$ and its cutoff frequency is calculated as,

  $\begin{array}{c}{f}_{{\text{c}}_{11}}=\frac{c}{2}\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}\\ =\frac{c}{2}\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}\end{array}$

Substitute $3\times {10}^8$ for $c$ , $0.109$ for $a$ and $0.0546$ for $b$ in the above equation.

  $\begin{array}{c}{f}_{{\text{c}}_{11}}=\frac{c}{2}\sqrt{\frac{1}{{\left(0.109\right)}^2}+\frac{1}{{\left(0.0546\right)}^2}}\\ =30.7\times {10}^8\\ \approx 3.07\times {10}^9\end{array}$

Therefore, the cutoff frequency for ${\text{TE}}_{11}$ mode is $3.07\text{ GHz}$ .

The fifth mode of the waveguide is ${\text{TM}}_{11}$ and its cutoff frequency is calculated as,

  $\begin{array}{c}{f}_{{\text{c}}_{11}}=\frac{c}{2}\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}\\ =\frac{c}{2}\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}\end{array}$

Substitute $3\times {10}^8$ for $c$ , $0.109$ for $a$ and $0.0546$ for $b$ in the above equation.

  $\begin{array}{c}{f}_{{\text{c}}_{11}}=\frac{c}{2}\sqrt{\frac{1}{{\left(0.109\right)}^2}+\frac{1}{{\left(0.0546\right)}^2}}\\ =30.7\times {10}^8\\ \approx 3.07\times {10}^9\end{array}$

Therefore, the cutoff frequency for ${\text{TM}}_{11}$ mode is $3.07\text{ GHz}$ .

Next mode of the waveguide is ${\text{TE}}_{21}$ and its cutoff frequency is calculated as,

  $\begin{array}{c}{f}_{{\text{c}}_{21}}=\frac{c}{2}\sqrt{{\left(\frac{2}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}\\ =\frac{c}{2}\sqrt{\frac{4}{a^2}+\frac{1}{b^2}}\end{array}$

Substitute $3\times {10}^8$ for $c$ , $0.109$ for $a$ and $0.0546$ for $b$ in the above equation.

  $\begin{array}{c}{f}_{{\text{c}}_{21}}=\frac{c}{2}\sqrt{\frac{4}{{\left(0.109\right)}^2}+\frac{1}{{\left(0.0546\right)}^2}}\\ =38.85\times {10}^8\\ \approx 3.885\times {10}^9\end{array}$

Therefore, the cutoff frequency for ${\text{TE}}_{21}$ mode is $3.885\text{ GHz}$ .

Next mode of the waveguide is ${\text{TM}}_{21}$ and its cutoff frequency is calculated as,

  $\begin{array}{c}{f}_{{\text{c}}_{21}}=\frac{c}{2}\sqrt{{\left(\frac{2}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}\\ =\frac{c}{2}\sqrt{\frac{4}{a^2}+\frac{1}{b^2}}\end{array}$

Substitute $3\times {10}^8$ for $c$ , $0.109$ for $a$ and $0.0546$ for $b$ in the above equation.

  $\begin{array}{c}{f}_{{\text{c}}_{21}}=\frac{c}{2}\sqrt{\frac{4}{{\left(0.109\right)}^2}+\frac{1}{{\left(0.0546\right)}^2}}\\ =38.85\times {10}^8\\ \approx 3.885\times {10}^9\end{array}$

Therefore, the cutoff frequency for ${\text{TM}}_{21}$ mode is $3.885\text{ GHz}$ .

The eighth modeof the waveguide is ${\text{TE}}_{30}$ and its cutoff frequency is calculated as,

  $\begin{array}{c}{f}_{{\text{c}}_{30}}=\frac{c}{2}\sqrt{{\left(\frac{3}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}\\ =\frac{3c}{2a}\end{array}$

Substitute $3\times {10}^8$ for $c$ , and $0.109$ for $a$ in the above equation.

  $\begin{array}{c}{f}_{{\text{c}}_{30}}=\frac{3\left(3\times {10}^8\right)}{2\left(0.109\right)}\\ =41.21\times {10}^8\\ \approx 4.121\times {10}^9\end{array}$

Therefore, the cutoff frequency for ${\text{TE}}_{30}$ mode is $4.121\text{ GHz}$ .

Conclusion:

Thus, the cutoff frequency of the eight modes of $\text{WR430}$ the waveguide are $1.734\text{ GHz}$ , $2.747\text{ GHz}$ , $2.747\text{ GHz}$ , $3.07\text{ GHz}$ , $3.07\text{ GHz}$ , $3.885\text{ GHz}$ , $3.885\text{ GHz}$ and $4.121\text{ GHz}$ .