Antennas Questions and Answers – Radio Wave Propagation – Field Strength

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This set of Antennas Multiple Choice Questions & Answers (MCQs) focuses on “Radio Wave Propagation – Field strength”.

1. The field strength due to space wave propagation is E=____
a) \(\frac{4πh_t h_r}{λd^2}E_0\)
b) \(\frac{2πh_t h_r}{λd^2}E_0\)
c) \(\frac{4πh_t h_r}{λd}E_0\)
d) \(\frac{4πh_t h_r}{λd^4}E_0\)
View Answer

Answer: a
Explanation: Space wave propagation reflects frequencies above 30MHz. The field strength due to space wave propagation is \(E=\frac{2E_0}{d} sin⁡ \frac{2πh_t h_r}{λd} ≈ \frac{4πh_t h_r}{λd^2}E_0\)
Here E0 is the field strength due to LOS at a unit distance which depends on transmits power, ht and hr is the height of transmitting and receiving antenna and d is the distance between two antennas.
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2. The path difference given by two-ray model is ___
a) \(\frac{4πh_t h_r}{λd^2}\)
b) \(\frac{4πh_t h_r}{d}\)

c) \(\frac{4h_t h_r}{d}\)

d) \(\frac{2h_t h_r}{d}\)

View Answer

Answer: d
Explanation: The path difference given by two-ray model is \(\frac{2h_t h_r}{d}.\) and its phase difference is given as \(∅=\frac{2π}{λ} ∆x=\frac{4πh_t h_r}{d}.\)

3. The field strength of the wave at a unit distance from transmitting antenna depends on _____
a) only the power radiated by the transmitting antenna
b) only on the power received by the receiving antenna
c) only on directivity of antenna in vertical and horizontal planes
d) both on the power radiated by the transmitting antenna & directivity in vertical and horizontal planes
View Answer

Answer: d
Explanation: The field strength E0 of the wave at a unit distance from transmitting antenna depends on both on the power radiated by the transmitting antenna & directivity in vertical and horizontal planes. The overall field strength is calculated from this. The field strength of the surface wave of flat earth is given by E=AE0/d.
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4. What is the condition of roughness of earth for earth to be electrically smooth?
a) R < 0.1
b) R > 10
c) R > 0.1
d) R > 1
View Answer

Answer: a
Explanation: Earth is classified as electrically smooth and electrically rough.
If the roughness of earth R < 0.1, then it is electrically smooth.
If the roughness of earth R > 10, then it is electrically rough.

5. Power density of the receiving antenna is given by _________
a) \(P_D=\frac{P_t G_t}{4πd^2}\)
b) \(P_D=\frac{P_t G_t}{(4πd/λ)^2}\)
c) \(P_D=\frac{P_t G_t G_r}{(4πd/λ)^2}\)
d) \(P_D=\frac{P_t G_t}{(4πd)^2}\)
View Answer

Answer: a
Explanation: Power density is the power per unit area at a distance d from the transmitter and is given by \(P_D=\frac{P_t G_t}{4πd^2}\) where Pt is the transmitter power and Gt is the transmitter antenna gain.
Power received by the antenna is given by \(P_r=\frac{P_t G_t G_r}{(4πd/λ)^2}\)
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6. Spatial attenuation coefficient is given by ____
a) (4πd/λ)2
b) (4πλ/d)2
c) 4πλ/d
d) 4πd/λ
View Answer

Answer: a
Explanation: Spatial attenuation coefficient reflects the decrease in the power density due to the spherical spread of the wave. It is also known as the loss factor. Expression for loss factor is given by Ls=(4πd/λ)2

7. The field strength of the surface wave of flat earth is given by ____
a) E=AE0/d
b) E=AE0
c) E=dE0/A
d) E=λE0/A
View Answer

Answer: a
Explanation: The field strength of the surface wave of flat earth is given by E=AE0/d, here A is the ground attenuation factor, d is the distance and E0 is the field strength per unit distance. The field strength of the wave at a unit distance from transmitting antenna depends on both on the power radiated by the transmitting antenna & directivity in vertical and horizontal planes
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8. Dissipation factor for the dielectric is Df= ____
a) \(1.8×10^6 \frac{σ}{f(MHz)}\)
b) \(1.8 \frac{σ}{f(MHz)}\)
c) \(1.8×10^6\frac{σ}{f(GHz)}\)
d) \(1.8×10^3 \frac{σ}{f(MHz)}\)
View Answer

Answer: a
Explanation: Dissipation factor for the dielectric is Df = \(1.8×10^6 \frac{σ}{f(MHz)}\), σ is the conductivity of earth. This is used to calculate the phase constant for the surface wave given by \(b=tan^{-1}(\frac{ϵ+1}{D_f})\)

9. The effective radius of earth is how many times the actual radius of earth at standard atmospheric conditions?
a) 3/4
b) 4/3
c) 1/4
d) 1/3
View Answer

Answer: a
Explanation: Effective radius of earth is the equivalent radius of earth used to correct atmospheric refraction as the refractive index varies linearly with height. It is 4/3 times the geometrical radius of the earth.
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10. What is the equivalent radius of the earth when the ray appears to be a straight line over a flat earth surface?
a) 0
b) Infinity
c) Actual radius
d) 4/3 times actual radius
View Answer

Answer: b
Explanation: When the ray appears to be a straight line over a flat earth surface horizontally, its refractive index is not changing with respect to height. \(\frac{dM}{dh}=0\)
Effective radius is k times the actual radius and \(k=0.048/\frac{dM}{dh}=∞\)

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter