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The cut off frequency of the dominant mode in a TE wave in the line having a and b as 2.5 cm and 1 cm respectively is
- a)4.5 GHz
- b)5 GHz
- c)5.5 GHz
- d)6 GHz
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The cut off frequency of the dominant mode in a TE wave in the line ha...
Answer: d
Explanation: The dominant mode in TE is TE10. The cut off frequency will be mc/2a, where m = 1 and a = 0.025 are given. On substituting, we get the frequency as 1 x 3 x 108/2 x 0.025 = 6 GHz.
Explanation: The dominant mode in TE is TE10. The cut off frequency will be mc/2a, where m = 1 and a = 0.025 are given. On substituting, we get the frequency as 1 x 3 x 108/2 x 0.025 = 6 GHz.
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The cut off frequency of the dominant mode in a TE wave in the line ha...
Cut-off frequency of TE wave in a transmission line:
A transmission line is a two-conductor structure used to guide high-frequency signals from one point to another. A waveguide is a type of transmission line that is used to guide electromagnetic waves. The cut-off frequency of a waveguide is the frequency below which the wave will not propagate. In a waveguide, the dominant mode is the mode with the lowest cut-off frequency.
TE wave in a transmission line:
A TE (Transverse Electric) wave is a type of electromagnetic wave in which the electric field is transverse to the direction of propagation. In a transmission line, the TE wave is characterized by the transverse electric field and the absence of a longitudinal magnetic field.
Calculation of cut-off frequency of dominant mode in a TE wave in a transmission line:
The cut-off frequency of the dominant mode in a TE wave in a transmission line can be calculated using the following formula:
fc = c / 2a√(1-(b/a)^2)
where
fc = cut-off frequency
c = speed of light in vacuum
a = width of the waveguide
b = height of the waveguide
Given,
a = 2.5 cm = 0.025 m
b = 1 cm = 0.01 m
Substituting the values in the formula, we get:
fc = 3 x 10^8 / (2 x 0.025)√(1-(0.01/0.025)^2)
fc = 6 GHz
Therefore, the correct answer is option D, i.e., 6 GHz.
A transmission line is a two-conductor structure used to guide high-frequency signals from one point to another. A waveguide is a type of transmission line that is used to guide electromagnetic waves. The cut-off frequency of a waveguide is the frequency below which the wave will not propagate. In a waveguide, the dominant mode is the mode with the lowest cut-off frequency.
TE wave in a transmission line:
A TE (Transverse Electric) wave is a type of electromagnetic wave in which the electric field is transverse to the direction of propagation. In a transmission line, the TE wave is characterized by the transverse electric field and the absence of a longitudinal magnetic field.
Calculation of cut-off frequency of dominant mode in a TE wave in a transmission line:
The cut-off frequency of the dominant mode in a TE wave in a transmission line can be calculated using the following formula:
fc = c / 2a√(1-(b/a)^2)
where
fc = cut-off frequency
c = speed of light in vacuum
a = width of the waveguide
b = height of the waveguide
Given,
a = 2.5 cm = 0.025 m
b = 1 cm = 0.01 m
Substituting the values in the formula, we get:
fc = 3 x 10^8 / (2 x 0.025)√(1-(0.01/0.025)^2)
fc = 6 GHz
Therefore, the correct answer is option D, i.e., 6 GHz.
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The cut off frequency of the dominant mode in a TE wave in the line having a and b as 2.5 cm and 1 cm respectively isa)4.5 GHzb)5 GHzc)5.5 GHzd)6 GHzCorrect answer is option 'D'. Can you explain this answer?
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