Wireless & Mobile Communications Multiple Choice Questions on “Diffraction”.
1. Diffraction occurs when radio path between Tx. And Rx. Is obstructed by ____________
a) Surface having sharp irregularities
b) Smooth irregularities
c) Rough surface
d) All types of surfaces
Answer: a
Clarification: Diffraction occurs when radio path between transmitter and receiver is obstructed by a surface that has sharp irregularities (edges). The secondary waves resulting from the obstructing surface are present throughout the space and even behind the obstacle.
2. At high frequencies, diffraction does not depends on ___________
a) Geometry of the object
b) Distance between Tx and Rx
c) Amplitude of incident wave
d) Polarization of incident wave
Answer: b
Clarification: At high frequency, diffraction depends on the geometry of the object, as well as the amplitude, phase, and polarization of the incident wave at the point of diffraction. It gives rise to a bending of waves even when line of sight does not exist between transmitter and receiver.
3. Diffraction allows radio signals to propagate around ________
a) Continuous surface
b) Smooth surface
c) Curved surface of Earth
d) Does not allow propagation
Answer: c
Clarification: Diffraction allows radio signals to propagate around the curved surface of the Earth. Signals can propagate beyond the horizon and to propagate behind obstruction. It is the slight bending of light as it passes around the edge of an object.
4. Which principle explains the phenomenon of diffraction?
a) Principle of Simultaneity
b) Pascal’s Principle
c) Archimedes’ Principle
d) Huygen’s principle
Answer: d
Clarification: The phenomenon of diffraction can be explained by Huygen’s principle. It states that all points on a wavefront can be considered as point sources for the production of secondary wavelets. And these wavelets combine to produce a new wavefront in direction of propagation.
5. Diffraction is caused by propagation of secondary wavelets into _______
a) Bright region
b) Shadowed region
c) Smooth region
d) Large region
Answer: b
Clarification: Diffraction is caused due to propagation of secondary wavelets into a shadowed region. The field strength in the shadowed region is the vector sum of the electric field components of all the secondary wavelets in the space around the obstacle.
6. Difference between the direct path and the diffracted path is called _______
a) Average length
b) Radio path length
c) Excess path length
d) Wavelength
Answer: c
Clarification: Excess path length denoted by ∆, is the difference between the direct path and the diffracted path. It is calculated with the help of Fresnel zone geometry.
7. The phase difference between a direct line of sight path and diffracted path is function of _______
a) Height and position of obstruction
b) Only height
c) Operating frequency
d) Polarization
Answer: a
Clarification: The phase difference between a direct line of sight path and diffracted path is a function of height and position of the diffraction. It is also a function of transmitter and receiver location.
8. Which of the following explains the concept of diffraction loss?
a) Principle of Simultaneity
b) Pascal’s Principle
c) Fresnel zone
d) Archimedes’ Principle
Answer: c
Clarification: The concept of diffraction loss is a function of the path difference around an obstruction. It can be explained by Fresnel zones. Fresnel zones represent successive regions where secondary waves have a path length from Tx to Rx which are nλ/2 greater than total path length.
9. In mobile communication system, diffraction loss occurs due to ______
a) Dielectric medium
b) Obstruction
c) Electric field
d) Operating frequency
Answer: b
Clarification: Diffraction loss occurs from the blockage of secondary waves such that only a portion of the energy is diffracted around an obstacle. An obstruction causes a blockage of energy from source some of the Fresnel zones, allowing only some of the transmitted energy to reach the receiver.
10. For predicting the field strength in a given service area, it is essential to estimate ______
a) Polarization
b) Magnetic field
c) Height of transmitter
d) Signal attenuation
Answer: d
Clarification: Estimating the signal attenuation caused by diffraction of radio waves over hills and buildings is essential in predicting the field strength in a given service area. In practice, prediction is a process of theoretical approximation modified by necessary empirical corrections.
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