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The correct OPTION is (a) Tropospheric Scattering

The best I can explain: Signal SCATTERS in forward DIRECTION above the radio horizon and it travels BEYOND the LOS. The signal strength is decreased as only a small amount of it is forward scattered and so high power amplifiers are equipped at the receivers.

Right option is (a) Above 30MHz

Easiest explanation: Tropospheric scatter is also known as forward scatter propagation. It OCCURS at a FREQUENCY above 30MHz. It belongs to the UHF and MICROWAVE range.

Right answer is (a) Change in the refractivity with respect to height

For explanation: Change in the refractivity with respect to height is known as the gradient of REFRACTIVE INDEX. RADIO horizon VARIES with the refractive index in the atmosphere.

Right choice is (a) \(\frac{77.6}{T}P+4810 \frac{E}{T} \)

Explanation: The expression used to calculate refractivity is given by

\(N=\frac{77.6}{T} P+4810 \frac{e}{T} \)

Where p is the total PRESSURE in millibars, e is the PARTIAL pressure of WATER vapor and T is the absolute TEMPERATURE.

Correct choice is (b) \(K=\frac{1}{1-\frac{r_e}{r_c}} \)

To explain I WOULD say: EFFECTIVE RADIUS factor k in terms of radius of curvature rcand radius of earth re is

\(k=\frac{1}{1-\frac{r_e}{r_c}} \)

The radius of curvature is four times the radius of earth. rc=4re

The effective radius factor of earth is 4/3 times actual radius of earth.

Right choice is (c) Infinity

For explanation I would say: Effective radius FACTOR k in terms of radius of curvature rc and radius of earth re is

\(k=\frac{1}{1-\frac{r_e}{r_c}}=\frac{1}{0}=∞.\)

The correct choice is (a) True

The best explanation: The TROPOSPHERIC SCATTERING OCCURS at frequency above 30MHz. Under low frequencies it falls below 30MHz and sometimes THUS leads to the Ionospheric propagation.

The CORRECT choice is (a) \(\frac{dϵ_r}{dH}=2n \frac{dn}{dH}\)

EXPLANATION: REFRACTIVE index is \(n=n=\sqrt{ϵ_r μ}= n=\sqrt{ϵ_r} (μ=1)\)

DIFFERENTIATING with respective heights treating n, ϵ_r as dependent variables

\(\frac{dn}{dH}=\frac{1}{2} \frac{1}{\sqrt{ϵ_r}} \frac{dϵ_r}{dH}\)

\(\frac{dϵ_r}{dH}=2n \frac{dn}{dH}\)

The correct answer is (a) rc=4re

For explanation: The RADIUS of curvature is FOUR times the radius of earth. rc=4re

The EFFECTIVE radius factor of earth is 4/3 times actual radius of earth.

The correct answer is (a) lower atmospheric LAYERS

The explanation is: DUCT propagation occurs at lower atmospheric layers at 15 – 50m above the earth surface. It occurs due to temperature inversion layer. The waves follow the CURVATURE of earth up to 1000km long.

Right OPTION is (a) \(\FRAC{dμ}{dh}\)is -ve

Best EXPLANATION: The height v/s the refractive index graph of duct propagation is as follows

To determine the presence of duct propagation (T.I.L) and \(\frac{dμ}{dh}\)is -ve.

The correct answer is (b) 4/3

Easy explanation: EFFECTIVE RADIUS factor k in terms of radius of curvature rcand radius of EARTH re is

\(k=\frac{1}{1-\frac{r_e}{r_c}} \)

The radius of curvature is four times the radius of earth. rc=4re

⇨ K=1/(1-(1/4)) =4/3

The effective radius factor of earth is 4/3 times actual radius of earth.

Correct option is (a) Temperature Inversion layer

To explain I WOULD SAY: Duct propagation is due to temperature inversion layer. Normal atmospheric temperature reduces at 6.5°C/km where as to provide duct wave propagation temperature must increase within 15-50m from earth surface. This layer OCCURS due to super refraction and at lower atmospheric layers.

The correct choice is (a) \(λ=2.5h_d \SQRT{∆μ×10^{-6}} \)

To ELABORATE: The wavelength of the EM signal propagation in the T.I.L for DUCT propagation is

\(λ=2.5h_d \sqrt{∆μ×10^{-6}} \)

Here ∆μ is CHANGE in the refraction index in HEIGHT hd.

Right CHOICE is (d) Air pressure, water pressure and temperature

Easy EXPLANATION: The expression USED to calculate refractivity is given by

\(N=\frac{77.6}{T} P+4810 \frac{e}{T}\)

Where p is the total pressure in millibars, e is the PARTIAL pressure of water vapor and T is the absolute temperature.

Correct CHOICE is (a) MULTIPATH propagation, TURBULENCES in atmosphere

To explain I would say: Rapid fading occurs due to the multipath propagation. As the TURBULENT conditions changes constantly, the path LENGTH and signal strength levels also change. The turbulence in the atmosphere gives rise to daily and seasonal variations in signal strength so results in long-term fading.

Correct choice is (a) increases

To elaborate: The signal take-off ANGLE determines the HEIGHT of the SCATTER volume. As the signal take-off angle increases, the height of scatter volume also increases. A low take-off produces low scatter volume.

Correct ANSWER is (a) True

Explanation: TROPOSPHERIC scatter is also known as forward scatter PROPAGATION, occurs at a frequency above 30MHz. It belongs to the UHF and microwave RANGE. Its scattering occurs above radio horizon and travels beyond the LOS.

The correct option is (a) N=(n-1)×10^6

For EXPLANATION: REFRACTIVITY is to observe the CHANGE in refractive index due to the smallest change in the relative DIELECTRIC constant.

N=(n-1)×10^6

Correct answer is (a) Super refraction

To elaborate: Around 50m of height from EARTH surface temperature increases with height, at this level the EM waves tend to REFRACT continuously than to reflect into ionosphere. This is termed as super refraction. REFLECTION and diffraction don’t affect the temperature inversion much.

Correct option is (a) tropospheric scattering

The explanation: The main cause of the tropospheric scattering is the turbulences in the ATMOSPHERE. When it MEETS the turbulences, then there will be an abrupt change in VELOCITY leading to the tropospheric scattering.The tropospheric scattering occurs at frequency above 30MHz.

The correct answer is (b) False

The explanation: The radio horizon and LOS become equal when k=1. LOS depends on the HEIGHT of the transmitting and receiving antennas and effective earth’s radius factor k. STANDARD value of k is 4/3.

Right answer is (a) True

Easy explanation: Expression for radio HORIZON in km is \(d = 3.56(\SQRT{h_t}+\sqrt{h_r}),\) Expression for the LOS DISTANCE is \(d = 4.12(\sqrt{h_t}+\sqrt{h_r})\) in km. Under same height conditions, radio horizon is less than LOS. But at effective RADIUS FACTOR k=1, both will be equal.

Right option is (d) On height of transmitting and RECEIVING antenna and effective earths radius FACTOR k

To explain: LOS (LINE of sight) is the distance covered by a direct space wave from transmitting to receiving antenna. The LS distance depends on the effective earth’s radius factor k and height of the transmitting and receiving antennas. The LOS distance is given by \(d = 4.12(\sqrt{h_t}+\sqrt{h_r})\) in km.

Right answer is (a) 8.24√h

For explanation I WOULD say: GIVEN heights of transmitting and RECEIVING ANTENNA are equal ht = hr = h

Expression for the LOS distance is \(d = 4.12(\sqrt{h_t}+\sqrt{h_r})\) in km.

⇨ D=4.12 (2√h) = 8.24√h.

Correct option is (a) 1

The BEST I can explain: DISTANCE between transmitter and receiver is \(d=d_t+d_r=\sqrt{2r_e h_t}+\sqrt{2r_e h_r} \)

And re=k*6370km, k is the effective radius factor.

Expression for RADIO horizon in KM is \(d = 3.56(\sqrt{h_t}+\sqrt{h_r}),\) LOS depends on k also so d = \(4.12(\sqrt{h_t}+\sqrt{h_r}).\) At k=1 both will be EQUAL.

The correct answer is (a) \(4.12(\SQRT{h_t}+\sqrt{h_r})\)

The explanation is: LOS is thedistance COVERED by a direct SPACE WAVE from TRANSMITTING to receiving antenna.Expression for the LOS distance is \(d = 4.12(\sqrt{h_t}+\sqrt{h_r})\) in km.

Expression for the Radio horizon distance is \(d =3.56(\sqrt{h_t}+\sqrt{h_r})\) in km.

Correct option is (a) Distance covered by a direct space wave from transmitting to RECEIVING antenna

Easy EXPLANATION: LOS (Line of sight) is defined for space wave propagation. It is the distance covered by a direct space wave from transmitting to receiving antenna. It depends on the height of the transmitting and receiving ANTENNAS and effective EARTH’s radius factor k.