Electromagnetic Theory Questions and Answers – Cut-off Frequency and Wavelength

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This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Cut off Frequency and Wavelength”.

1. The real part of the propagation constant is the
a) Attenuation constant
b) Phase constant
c) Permittivity
d) Permeability
View Answer

Answer: a
Explanation: The propagation constant is given by γ = α + jβ. Here the real part is the attenuation constant and the imaginary part is the phase constant.
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2. The phase constant of a wave is given by
a) ω√(LC)
b) ω√(L/C)
c) ω√(C/L)
d) ω√(1/LC)
View Answer

Answer: a
Explanation: The phase constant of a wave in a transmission line is given by β = ω√(LC), where L and C are the specifications of the line.

3. 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
View Answer

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.
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4. The cut off frequency of the TE01 mode will be
a) mc/2a
b) mc/2b
c) nc/2a
d) nc/2b
View Answer

Answer: d
Explanation: The cut off frequency consists of modes m and n. For m = 0, the dimension b will be considered. Thus the frequency is nc/2b, where c is the speed of the light.

5. The condition which will satisfy the dimensions of the waveguide is
a) a = b
b) a > b
c) a < b
d) ab = 0
View Answer

Answer: b
Explanation: The dimensions a and b represent the broad wall and the side wall dimensions respectively. The broad wall will be greater than the side wall. Thus the condition a>b is true.
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6. The cut off wavelength of the TE10 mode having a broad wall dimension of 5cm is
a) 0.1
b) 1
c) 10
d) 0.01
View Answer

Answer: a
Explanation: The cut off wavelength of the waveguide is given by λc = 2a/m. on substituting for a = 0.05 and m = 1, we get λc = 2 x 0.05/1 = 0.1 units.

7. The broad wall dimension of a waveguide having a cut off frequency of 7.5 GHz is
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: b
Explanation: The cut off frequency and the broad wall dimension are related by fc = mc/2a. On substituting for m = 1 and fc = 7.5 GHz, we get a = 0.02 or 2 cm.
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8. The sin θ in the waveguide refers to the ratio of the
a) Frequency to wavelength
b) Wavelength to frequency
c) Cut off frequency to frequency
d) Frequency to cut off frequency
View Answer

Answer: c
Explanation: The ratio of the cut off frequency to the frequency at any point gives the sin θ in a waveguide.

9. Is the transmission of a frequency 5 GHz possible in waveguides?
a) Yes
b) No
View Answer

Answer: a
Explanation: The cut off frequency for waveguide operation is 6 GHz. Thus a wave of 5 GHz is not possible for transmission in a waveguide.
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10. The dimension for a waveguide in dominant mode with a cut off wavelength of 2 units is
a) 2
b) 4
c) 6
d) 8
View Answer

Answer: b
Explanation: The cut off wavelength of a waveguide is given by λc = 2a/m. For the dominant mode, m = 1. Given that λc = 2, thus we get a = 4 units.

11. The waveguides are used in a transmission line for
a) Increasing transmission coefficient
b) Increasing reflection coefficient
c) Decreasing transmission coefficient
d) Decreasing reflection coefficient
View Answer

Answer: a
Explanation: The waveguides are used to increase the transmission efficiency of the waves travelling through it.

12. The attenuation coefficient of the wave having a resistance of 15 ohm in a 50 ohm line is
a) 15
b) 1.5
c) 0.15
d) 0.015
View Answer

Answer: c
Explanation: The attenuation coefficient of a wave with a resistance of R in a line of characteristic impedance Zo is α = R/2Zo. On substituting for R = 15 and Zo = 50, we get α = 15/(2 x 50) = 0.15 units.

Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
To practice all areas of Electromagnetic Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.

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