# Microwave Engineering Questions and Answers – Circular Waveguide

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Circular Waveguide”.

1. In TE mode of a circular waveguide, EZ=0. The wave equation is:
a) ∇2HZ+k2HZ=0
b) ∇2HZ-k2HZ=0
c) ∇2HZ-HZ=0
d) ∇2HZ+HZ=0

Explanation: In TE mode, EZ=0. Hence, when we substitute it in the wave equation, we get ∇2HZ+k2HZ=0.

2. Bessel’s differential equation for a circular waveguide is:
a) ρ2(d2R/ dρ2) + ρ(dR/dρ) + (ρ2kC2– n2) R=0
b) n2(d2R/ dρ2) + n(dR/dρ) + (ρ2kC2– n2) R=0
c) d2R/ dρ2 + dR/dρ + (ρ2kC2– n2) R=0
d) None of the mentioned

Explanation: After solving the wave equation ∇2HZ+k2HZ=0 in TE mode by making suitable assumptions and making appropriate substitutions, the final equation obtained is ρ2(d2R/ dρ2) + ρ(dR/dρ) + (ρ2kC2– n2) R=0.

3. The lowest mode of TE mode propagation in a circular waveguide is:
a) TE10 mode
b) TE00 mode
c) TE01 mode
d) TE11 mode

Explanation: A circular waveguide can support various modes of propagation. Among these, the lowest mode of propagation supported by the waveguide is TE10 mode of propagation.

4. What is the cutoff frequency for TE₁₁ mode in a circular waveguide of radius 2 cm with P’₁₁= 1.841?
a) 5.5 GHz
b) 4.3 GHz
c) 7.7 GHz
d) 8.1 GHz

Explanation: The cutoff frequency for TE11 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 4.3 GHz.

5. In a circular waveguide, if the propagation is in TE21 mode with P21=3.054, with a diameter of 60 mm, then the cutoff frequency for the mode is:
a) 5.6 GHz
b) 6.4 GHz
c) 3.5 GHz
d) 4.8 GHz

Explanation: The cutoff frequency for TE21 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 4.8 GHz.
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6. For a circular waveguide in TM11 mode of propagation with inner radius of 30mm, and the phase constant being equal to 0.3, then the wave impedance is equal to:
a) 0.16 Ω
b) 0.15 Ω
c) 0.5 Ω
d) 0.4 Ω

Explanation: For a given mode of propagation in a circular waveguide, wave impedance is given by the expression k/β, where is the intrinsic impedance of the air k is the wave number, β is the phase constant. Determining the wave number and substituting in the given equation, the wave impedance is 0.16 Ω.

7. For TM mode. The wave equation in cylindrical co ordinates is:
a) (∂2/∂ρ2+1/ρ ∂/∂ρ + 1/ρ2 (∂2/∂∅2 + kc2) =0
b) ∂2E2/∂ρ2 + 1/ρ ( ∂E/∂ρ)=0
c) ∂2E2/∂ρ2 + 1/ρ2 (∂2E2/∂∅2 ) = 0
d) None of the mentioned

Explanation: The wave propagation in a cylindrical waveguide in TM mode of propagation is governed by the equation (∂2/∂ρ2+1/ρ ∂/∂ρ + 1/ρ2 (∂2/∂∅2 + kc2) =0. This is a second order differential equation.

8. In TM mode, what is the first propagating mode?
a) TM01 mode
b) TM11 mode
c) TM12 mode
d) TM10 mode

Explanation: TM mode in a circular waveguide supports various modes of propagation. Among these modes of propagation, the first or the lowest mode of propagation is TM01 mode.

9. For TM01 mode of propagation in a circular waveguide with P01=2.405, with the inner diameter of the circular waveguide being equal to 25 mm. What is the cut off frequency for this mode of propagation?
a) 2.8 GHz
b) 6 GHz
c) 3.06 GHz
d) 4 GHz.

Explanation: The cutoff frequency for TM01 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 3.06 GHz.

10. If β is 0.3 for a circular wave guide operating in TM12 mode with P21=5.315, with the radius of the circular waveguide being equal to 25 mm, then the intrinsic impedance of the wave is:
a) 0.55 Ω
b) 0.4 Ω
c) 0.3 Ω
d) 1.2 Ω

Explanation: For a given mode of propagation in a circular waveguide, wave impedance is given by the expression k/β, where is the intrinsic impedance of the air k is the wave number, β is the phase constant. Determining the wave number and substituting in the given equation, the wave impedance is 0.55 Ω.

11. The cutoff frequencies of the first two propagating modes of a Teflon on a filled circular waveguide with a=0.5 with ∈r=2.08 is:
a) 12.19 GHz, 15.92 GHz
b) 10 GHz, 12 GHz
c) 12 GHz, 15 GHz
d) 15 GHz, 12 GHz

Explanation: The cutoff frequencies are given by the expression p*C/2πa√∈. Substituting the given values in the above expression, the cutoff frequencies are 12.19 GHz, 15.92 GHz.

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