This set of Microwave Engineering Questions and Answers for Aptitude test focuses on “Rectangular Waveguide Cavity Resonators”.
1. Microwave resonators can be constructed from open sections of waveguide.
Explanation: For resonance to occur in waveguides, a closed structure is required. They resonate between the walls of the rectangular waveguide. Also radiation loss from an open ended waveguide can be significant.
2. There is no energy stored inside a rectangular waveguide cavity resonator.
Explanation: Energy is stored in a waveguide resonator in the form of electric field and magnetic field. Power is dissipated in the metallic walls of the cavity as well as in the dielectric material that may fill the cavity.
3. A rectangular cavity supports:
a) TEM mode of resonance
b) TM mode of resonance
c) TE mode of resonance
d) TE, TM modes of resonance
Explanation: A rectangular wave guide supports both TE and TEM mode of propagation. Likewise, when a rectangular waveguide is used as resonator, it supports both TE and TM modes of resonance.
4. A waveguide is open circuited at both the ends to use it as a waveguide resonator.
Explanation: A closed cavity structure is required in order to bring resonance in the rectangular cavity. Also open ended waveguides result in radiation losses. Hence the waveguide is short circuited to form a resonator.
5. In order to obtain the resonant frequency of a rectangular waveguide, the closed cavity has to satisfy:
a) Gaussian equation
b) Helmholtz equation
c) Ampere’s law
d) None of the mentioned
Explanation: Helmholtz wave equation is considered and solved using variable separable form. Then the boundary conditions are applied to the wave equation considering the walls of the cavity. Solving this gives the expression for resonant frequency.
6. Given the dimension of the waveguide as b<a<d, no resonant mode exists for this specification of dimensions.
Explanation: For the given dimensional specification b<a<d, the dominant resonant mode (lowest resonant frequency) will be the TE101 mode, corresponding to the TE10 dominant waveguide mode in a shorted guide of length λg/2.
7. Unloaded Q of a rectangular waveguide cavity resonator:
a) Does not exist
b) Defined as the ratio of length of the waveguide to breadth of the waveguide
c) Defined as the ratio of stored energy to the power dissipated in the walls
d) None of the mentioned
Explanation: Quality factor signifies the power loss in the circuit. It is defined as the ratio of stored energy to the power dissipated in the walls. Higher the power dissipation in the walls, lower is the quality factor of the waveguide resonator.
8. Find the wave number of a rectangular cavity resonator filled with a dielectric of 2.25 and designed to operate at a frequency of 5 GHz.
Explanation: The wave number of rectangular wave resonator is 2πf√∈r/C, substituting the given values in the above equation, the wave number of the rectangular cavity resonator is 157.08.
9. The required length of the cavity resonator for l=1 mode (m=1, n=0) given that the wave number of the cavity resonator is 157.01 and the broader dimension of the waveguide is 4.755 cm:
a) 1.10 cm
b) 2.20 cm
c) 2.8 cm
d) 1.8 cm
Explanation: The required length of the cavity resonator for the given mode is given by the expression d=lπ/√(k>sup>2-(π/a)2. Substituting the given values in the equation, the required length of the waveguide is 2.20 cm.
10. If the loss tangent of a rectangular waveguide is 0.0004, then Q due to dielectric loss is:
Explanation: Q of a rectangular waveguide due to dielectric loss is given by 1/tanδ. Substituting for tanδ in the above equation, Q due to dielectric loss is 2500.
Sanfoundry Global Education & Learning Series – Microwave Engineering.
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