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The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of ...

 

Q. The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of inner dimension 0.28 inch×0.14 inch is __. (rounded off to two decimal places).

Given Information

  • Waveguide dimensions = 0.28 inch × 0.14 inch
  • Wider dimension (a) = 0.28 inch
  • Smaller dimension (b) = 0.14 inch
  • Medium = Air
  • Dominant mode = TE10
  • Speed of light (c) = 3 × 108 m/s

Step 1: Cutoff Frequency Formula

The cutoff frequency of a rectangular waveguide is:

fc = (c / 2) √[(m/a)2 + (n/b)2]

For the dominant TE10 mode:

m = 1, n = 0

fc = c / 2a

Step 2: Convert Inch into Meter

We know that 1 inch = 0.0254 meter.

a = 0.28 × 0.0254

a = 0.007112 m

Step 3: Calculate Cutoff Frequency

Substitute the values into the formula:

fc = c / 2a

= (3 × 108) / (2 × 0.007112)

= 21.0911 × 109 Hz

= 21.0911 GHz

Final Answer

21.09 GHz

Cutoff frequency of the dominant TE10 mode

Important Note

For the dominant TE10 mode, the cutoff frequency depends only on the wider dimension (a), not on the smaller dimension (b).

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