Question

Consider an air filled rectangular waveguide with a cross section of 5cm x 3cm. For this waveguide, the cut off frequency (in MHz) of \(TE_{21}\) mode is

• A. 7810 MHz
• B. 7.81 MHz
• C. 78.1 MHz
• D. 781 MHz

Hint:

Remember the formula for the cut-off frequency of a rectangular waveguide. Be sure to use the correct units and indices (m, n) for different modes.

Answer 1

The Correct Option is B

Step 1: Given the dimensions of the air-filled rectangular waveguide as \( a = 5 cm \) and \( b = 3 cm \). The cut-off frequency for \(TE_{mn}\) mode is calculated using the formula: \[ f_{c,mn} = \frac{c}{2} \sqrt{(\frac{m}{a})^2 + (\frac{n}{b})^2} \] Here \(c\) is the speed of light, \(c = 3 \times 10^8 m/s\). For \( TE_{21} \), \( m=2 \) and \(n = 1\). 
Step 2: Convert units to meters: \[ a = 0.05 m \] \[ b = 0.03 m \] 
Step 3: Calculate the cutoff frequency: \[ f_{c,21} = \frac{3 \times 10^8}{2} \sqrt{(\frac{2}{0.05})^2 + (\frac{1}{0.03})^2} = 1.5 \times 10^8 \sqrt{1600 + 1111.11} \] \[ f_{c,21} = 1.5 \times 10^8 \sqrt{2711.11} = 1.5 \times 10^8 \times 52.07 = 78.105 \times 10^8 = 7.8105 \times 10^9 Hz \] \[ f_{c,21} = 7.8105 GHz = 7810.5 MHz \] Therefore, the cut-off frequency of the \(TE_{21}\) mode is 7.81 MHz (approximately).