Question

Find the cutoff frequency of dominant mode in an air-filled circular waveguide having inner radius of 1 cm which is excited in dominant mode at 12 GHz.
1. 10 GHz
2. 3.5 GHz
3. 6.5 GHz
4. 8.791 GHz

Answer 1

Correct Answer - Option 4 : 8.791 GHz

Concept:

For the circular waveguide dominant mode is TE₁₁

For the dominant mode, TE₁₁ cutoff frequency is given by

\(f_{c}= \frac{1.841c}{2\pi a}~Hz\)

Where,

c = velocity of light = 3×10⁸ m/s

a = radius of the circular waveguide in meter

Calculations:

Given

a = 1 cm = 0.01 m

The cutoff frequency is given by

\(f_{c}= \frac{1.841c}{2\pi a}\)

\(f_c=\frac{1.841\times 3\times10^8}{2\times \pi \times 0.01}\)

f_c = 8.791 GHz

The next higher mode of the circular waveguide is TM₀₁ 

The cutoff frequency f_c for TM₀₁ is given by

\(f_{c}= \frac{2.405c}{2\pi a}~Hz\)