Question

Consider an air filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm. The increasing order of the cut-off frequencies for different modes is:
A. TE\(_1_0\),
B. TE\(_0_1\),
C. TE\(_2_0\),
D. TE\(_1_1\).
Choose the correct answer from the options given below:

• A. A, B, C, D
• B. A, C, B, D
• C. B, A, D, C
• D. C, B, D, A

Hint:

For a standard waveguide where \(a>b\), the TE$_{10}$ mode is always dominant (lowest \(f_c\)). Since usually \(a \approx 2b\), the next mode is often TE$_{20}$ (\(2/a\)), followed by TE$_{01}$ (\(1/b\)). Higher-order modes like TE$_{11}$ will have even higher cutoff frequencies.

Answer 1

The Correct Option is B

Step 1: Recall the formula for the cutoff frequency in a rectangular waveguide.

\[ f_{c,mn} = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2} \] where \(c\) is the speed of light, \(a\) is the wide dimension, and \(b\) is the narrow dimension. Here, \(a=2.286\) cm and \(b=1.016\) cm. Note that \(a \approx 2.25b\).

Step 2: Calculate the cutoff frequency for each mode proportionally.

We only need the relative order, so we can ignore the constant \(\frac{c}{2}\) and compare the values inside the square root. Let's use \(a=2.25, b=1\) for simplicity.

• A. TE\(_{10}\): \(\sqrt{\left(\frac{1}{a}\right)^2 + 0} = \frac{1}{a} = \frac{1}{2.286} \approx 0.437\)
• B. TE\(_{01}\): \(\sqrt{0 + \left(\frac{1}{b}\right)^2} = \frac{1}{b} = \frac{1}{1.016} \approx 0.984\)
• C. TE\(_{20}\): \(\sqrt{\left(\frac{2}{a}\right)^2 + 0} = \frac{2}{a} = \frac{2}{2.286} \approx 0.875\)
• D. TE\(_{11}\): \(\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2} = \sqrt{(0.437)^2 + (0.984)^2} = \sqrt{0.191 + 0.968} = \sqrt{1.159} \approx 1.077\)

Step 3: Order the values.

The calculated proportional values are:

\(0.437\) (A) \(<\) \(0.875\) (C) \(<\) \(0.984\) (B) \(<\) \(1.077\) (D)

Thus, the correct increasing order of cutoff frequencies is:

A, C, B, D