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A rectangular waveguide, in dominant TE mode, has dimensions 10 cm x 15 cm. The cut off frequency is
- a)10 GHz
- b)1 GHz
- c)15 GHz
- d)25 GHz
Correct answer is option 'B'. Can you explain this answer?
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A rectangular waveguide, in dominant TE mode, has dimensions 10cmx 15c...
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A rectangular waveguide, in dominant TE mode, has dimensions 10cmx 15c...
Explanation:
A rectangular waveguide supports different modes of propagation. The dominant mode is the mode that has the lowest cutoff frequency. The cutoff frequency is the frequency below which the waveguide cannot support any mode of propagation.
Formula:
The cutoff frequency of a rectangular waveguide is given by the formula:
fc = (c/2π) * √((m/a)² + (n/b)²)
where,
c = speed of light in free space
a = width of the waveguide
b = height of the waveguide
m, n = mode numbers
Given:
Width of waveguide, a = 10 cm
Height of waveguide, b = 15 cm
Dominant mode, TE
Solution:
For the dominant TE mode, m = 1, n = 0 or m = 0, n = 1.
Using the formula, we can calculate the cutoff frequency for both modes:
For m = 1, n = 0:
fc = (c/2π) * √((m/a)² + (n/b)²)
fc = (3 x 10^8 / 2π) * √((1/0.1)² + (0/0.15)²)
fc = 4.77 GHz
For m = 0, n = 1:
fc = (c/2π) * √((m/a)² + (n/b)²)
fc = (3 x 10^8 / 2π) * √((0/0.1)² + (1/0.15)²)
fc = 6.32 GHz
Therefore, the lowest cutoff frequency is 4.77 GHz, which is option B.
A rectangular waveguide supports different modes of propagation. The dominant mode is the mode that has the lowest cutoff frequency. The cutoff frequency is the frequency below which the waveguide cannot support any mode of propagation.
Formula:
The cutoff frequency of a rectangular waveguide is given by the formula:
fc = (c/2π) * √((m/a)² + (n/b)²)
where,
c = speed of light in free space
a = width of the waveguide
b = height of the waveguide
m, n = mode numbers
Given:
Width of waveguide, a = 10 cm
Height of waveguide, b = 15 cm
Dominant mode, TE
Solution:
For the dominant TE mode, m = 1, n = 0 or m = 0, n = 1.
Using the formula, we can calculate the cutoff frequency for both modes:
For m = 1, n = 0:
fc = (c/2π) * √((m/a)² + (n/b)²)
fc = (3 x 10^8 / 2π) * √((1/0.1)² + (0/0.15)²)
fc = 4.77 GHz
For m = 0, n = 1:
fc = (c/2π) * √((m/a)² + (n/b)²)
fc = (3 x 10^8 / 2π) * √((0/0.1)² + (1/0.15)²)
fc = 6.32 GHz
Therefore, the lowest cutoff frequency is 4.77 GHz, which is option B.
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A rectangular waveguide, in dominant TE mode, has dimensions 10cmx 15cm. The cut off frequency isa)10 GHzb)1 GHzc)15 GHzd)25 GHzCorrect answer is option 'B'. Can you explain this answer?
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