Question

The number of propagating modes through an air filled rectangular waveguide of dimensions, \( a = 5 \text{ cm} \) and \( b = 2.5 \text{ cm} \) at 8 GHz frequency is ________.

• A. Five
• B. Four
• C. Three
• D. Six

Hint:

To find the number of modes in a waveguide, calculate cut-off frequencies of all \( TE_{mn} \) modes and count those with \( f_c<f \). Dominant mode in rectangular waveguide is always \( TE_{10} \).

Answer 1

The Correct Option is A

For a rectangular waveguide, the cut-off frequency for the \( \text{TE}_{mn} \) mode is:
\[ f_{c_{mn}} = \frac{1}{2} \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2} \cdot c \]
where:
\( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
\( a = 5 \, \text{cm} = 0.05 \, \text{m} \)
\( b = 2.5 \, \text{cm} = 0.025 \, \text{m} \)
\( m, n \) are integers (\( m \geq 0, n \geq 0, m \text{ or } n \neq 0 \))

We calculate \( f_c \) for different \( m,n \) values and check which modes satisfy \( f_c < 8 \, \text{GHz} \)

TE₁₀:
\[ f_c = \frac{c}{2a} = \frac{3 \times 10^8}{2 \times 0.05} = 3 \, \text{GHz} \]
TE₀₁:
\[ f_c = \frac{c}{2b} = \frac{3 \times 10^8}{2 \times 0.025} = 6 \, \text{GHz} \]
TE₂₀:
\[ f_c = \frac{c \cdot 2}{2a} = \frac{6 \times 10^8}{2 \times 0.05} = 6 \, \text{GHz} \]
TE₁₁:
\[ f_c = \frac{c}{2} \sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2} = \frac{3 \times 10^8}{2} \sqrt{(20)^2 + (40)^2} = \frac{3 \times 10^8}{2} \cdot \sqrt{2000} \approx 6.7 \, \text{GHz} \]
TE₀₂:
\[ f_c = \frac{c \cdot 2}{2b} = \frac{6 \times 10^8}{2 \times 0.025} = 12 \, \text{GHz} > 8 \, \text{GHz (Not allowed)} \]
Valid modes with \( f_c < 8 \, \text{GHz} \):
- TE₁₀: 3 GHz
- TE₀₁: 6 GHz
- TE₂₀: 6 GHz
- TE₁₁: ~6.7 GHz
- TE₂₁: Can be computed but may also fall within range

Final Answer: Five modes