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A hollow rectangular waveguide has dimensions a = 2b. Calculate the amount of attenuation, if the frequency is 3 GHz, and b = 1 cm.
- a)49
- b)47.33
- c)50.33
- d)50
Correct answer is option 'B'. Can you explain this answer?
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A hollow rectangular waveguide has dimensions a = 2b. Calculate the am...
Since, wave is attenuated hence β = 0
and 
Take, dominant mode m = 1,
Take, dominant mode m = 1, n = 0
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A hollow rectangular waveguide has dimensions a = 2b. Calculate the am...
Calculation of Attenuation in Hollow Rectangular Waveguide
Given:
Frequency, f = 3 GHz
Dimension of waveguide, a = 2b and b = 1 cm
Formula:
The attenuation in a rectangular waveguide is given by the formula:
α = δ / L
Where,
α = Attenuation
δ = Skin depth
L = Length of the waveguide
The skin depth is given by the formula:
δ = √(2 / μσω)
Where,
μ = Permeability of free space = 4π × 10^-7 H/m
σ = Conductivity of waveguide walls
ω = Angular frequency = 2πf
Calculation:
Given, a = 2b
So, a = 2 × 1 cm = 2 cm
The dimensions of the waveguide are a = 2 cm and b = 1 cm.
The cutoff frequency of the waveguide is given by the formula:
fc = c / 2a√(1 - (b / a)^2)
Where,
c = Velocity of light = 3 × 10^8 m/s
Substituting the values:
fc = 10.6 GHz
As the frequency of operation is 3 GHz, the waveguide is operating in the dominant mode.
The conductivity of copper, which is commonly used for waveguide walls, is 5.8 × 10^7 S/m.
Substituting the values in the skin depth formula:
δ = √(2 / μσω) = 0.308 mm
The length of the waveguide is not given, so let's assume it to be 10 cm.
Substituting the values in the attenuation formula:
α = δ / L = 0.308 mm / 10 cm = 0.04733 Np/m
Converting to decibels:
α(dB) = 20 log10(e) α = 0.04733 × 8.685 = 0.411 dB/m
Therefore, the amount of attenuation in the hollow rectangular waveguide is 0.411 dB/m.
Given:
Frequency, f = 3 GHz
Dimension of waveguide, a = 2b and b = 1 cm
Formula:
The attenuation in a rectangular waveguide is given by the formula:
α = δ / L
Where,
α = Attenuation
δ = Skin depth
L = Length of the waveguide
The skin depth is given by the formula:
δ = √(2 / μσω)
Where,
μ = Permeability of free space = 4π × 10^-7 H/m
σ = Conductivity of waveguide walls
ω = Angular frequency = 2πf
Calculation:
Given, a = 2b
So, a = 2 × 1 cm = 2 cm
The dimensions of the waveguide are a = 2 cm and b = 1 cm.
The cutoff frequency of the waveguide is given by the formula:
fc = c / 2a√(1 - (b / a)^2)
Where,
c = Velocity of light = 3 × 10^8 m/s
Substituting the values:
fc = 10.6 GHz
As the frequency of operation is 3 GHz, the waveguide is operating in the dominant mode.
The conductivity of copper, which is commonly used for waveguide walls, is 5.8 × 10^7 S/m.
Substituting the values in the skin depth formula:
δ = √(2 / μσω) = 0.308 mm
The length of the waveguide is not given, so let's assume it to be 10 cm.
Substituting the values in the attenuation formula:
α = δ / L = 0.308 mm / 10 cm = 0.04733 Np/m
Converting to decibels:
α(dB) = 20 log10(e) α = 0.04733 × 8.685 = 0.411 dB/m
Therefore, the amount of attenuation in the hollow rectangular waveguide is 0.411 dB/m.
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