Explore topic-wise InterviewSolutions in .

Right CHOICE is (C) Minimum DISTANCE at which wave returns back at the critical angle

To explain I would SAY: The minimum distance at which the wave returns back at the critical angle is CALLED skip distance.

Correct CHOICE is (a) 15

The BEST explanation: Take-off angle for the flat EARTH surface is β = 90 – θi = 90 – 75 = 15

The correct option is (a) \(d=2h\sqrt{[\frac{f_{MUF}^2}{f_c^2}-1]}\)

EASY explanation: Relation between skip DISTANCE d, virtual height d and MUF FMUF is given by

skip distance \(d=2h\sqrt{[\frac{f_{MUF}^2}{f_c^2}-1]}\)

Correct option is (a) SINGLE hop distance

To ELABORATE: If wave TRAVELS from the TRANSMITTER to receiver without touching the ground is called the single hop distance. If it touches the ground in between, then it is called multi hop PROPAGATION.

Right choice is (c) 4.5MHz

To elaborate: \(f_c=9\sqrt{N \,MAX}\) where Nmax= electron DENSITY and angle of INCIDENCE is ἰ = 0 (given)

∴ \(f_c = 9\sqrt{25×10^{10}} = 4.5MHz\)

⇨ And FMUF = fc secθi = fc = 4.5MHz

Right OPTION is (a) MUF is equal to critical frequency

Explanation: When a ray is INCIDENT NORMALLY in an Ionosphere region, θi=0

⇨ fMUF=fc secθi=fc

Therefore, MUF is equal to critical frequency.

The correct answer is (B) 35.26km

The best explanation: Take-off angle for the flat EARTH SURFACE is β = 90 – θi

θi=90-β=90-10=80

skip distance \(d=2h\sqrt{[(secθ_i)^2-1]}=2h\sqrt{[(sec10)^2-1]}=35.26km \)

The CORRECT answer is (c) Multi hop multi layer

Best EXPLANATION: Multi HOPPING means the wave TOUCHES the ground once or more than once while travelling from transmitter to receiver. If it propagates through the multiple layers then it is CALLED multi-hop multi layer.In single hopping the wave doesn’t touches the ground.

Correct answer is (C) 5.4MHz

To explain I WOULD say: \(f_c=9\sqrt{N \,max}\) where Nmax= electron density and angle of incidence is ἰ = 0 (GIVEN)

∴ \(f_c =9\sqrt{36×10^{10}} = 5.4MHz\)

The CORRECT option is (b) 0

To explain: REFRACTIVE index \(n=\sqrt{(1-\frac{81N_{max}}{F^2})}\)

Critical frequency\(f_c=9\sqrt{N \,max} \)

When \(f_c=f_{MUF}, n=\sqrt{(1-\frac{81N_{max}}{f_{MUF}^2})}=1-1=0\)

The correct choice is (a) True

For explanation: The difference OCCURS due to the exchange of energy between wave and the ELECTRONS present in the ionosphere which changes the velocity of propagation of wave. So the difference between VIRTUAL and actual HEIGHT are influenced by the ELECTRON distribution.

Right choice is (c) Multi hop

Best explanation: When the RECEIVER is beyond the skip distance, single hop is prevented from reaching the receiver. It requires more than one hop to travel to the desired receiver. So Multi Hop PROPAGATION with single or multi layer is USED.

Right option is (a) critical frequency

The explanation: When a wave is incident normally then the acceptable HIGHEST frequency at which SIGNAL can be RETURNED is the critical frequency. Beyond critical frequency, wave is penetrated into ANOTHER REGION. When the frequency is greater than critical frequency, still some part is returned back by varying the angle of incidence and this frequency is called MUF.

Right option is (d) fMUF = FC secθi

Easiest explanation: The fMUF in TERMS of critical FREQUENCY is given by fMUF = fc secθi for a given ANGLE of incidence between TWO locations.

Correct choice is (a) β = 90 – θi – 57.3d/2R

To EXPLAIN I would say: The take-off angle for the CURVED EARTH SURFACE is β = 90 – θi – 57.3d/2R

Take-off angle for the flat earth surface is β = 90 – θi

Right option is (b) LUF

The best explanation: The frequency below which the entire power GETS absorbed is called as LUF (Lowest usable frequency). The maximum possible frequency for which the wave is reflected back for a GIVEN distance of propagation in the ionosphere LAYER is called as maximum usable frequency (MUF). The frequency at which there is optimum return of the wave is the optimum frequency.

The correct CHOICE is (a) MUF

The explanation is: ACCORDING to the secant LAW, fMUF=fc secθi

So MUF is GREATER than or equal to the critical FREQUENCY depending on the secθi

The correct option is (b) LUF

Explanation: When the FREQUENCY is LOWERED below the lowest usable frequency, the signal gets absorbed. At low frequencies the noise LEVEL is also more and signal reception will be also DIFFICULT.

Right choice is (B) 0.11×10^12/cm^3

For explanation I WOULD say: \(f_c=9\sqrt{N\, max} \)

\(N_{max}=\FRAC{f_c^2}{81}=\frac{(3×10^6)^2}{81}=0.11×10^{12}/cm^3\)

Correct ANSWER is (a) Actual height

Best explanation: The height at a point above the earth’s surface at which the WAVE bends down to the earth is called actual height or true height. Virtual height is greater than actual height. Skip DISTANCE is the distance the wave travels from the transmitter to the RECEIVER without touching the ground.

The CORRECT option is (a) True

For explanation: The SKIP zone is the region between the point where the ground wave RECEPTION BECOMES null and the sky wave returns for the first time. It ENTIRELY depends on the ground wave coverage and the skip distance.

Correct ANSWER is (a) 6.05MHz

Easy EXPLANATION: \(f_{MUF}=f_c \sqrt{((\frac{d}{2H})^2+1)}=6×10^6×\sqrt{((\frac{25}{2×100})^2+1)}=6.05MHz\)

The correct option is (b) False

The explanation is: The UPPER ray is weaker than the LOWER ray in terms of energy contents and it spreads more in the electron DENSITY region compared to the lower ray. So the lower ray is MOSTLY preferred for communication.

Correct ANSWER is (c) 34.64km

Easiest explanation: VIRTUAL height \(h=\frac{d/2}{\SQRT{secθ_i^2-1}}=34.64km\)

Correct option is (a) True

Easy explanation: The height to which a short pulse of ENERGY is sent vertically UPWARD and traveling with the speed of light would reach taking the same two-way travel TIME as does original pulse reflected from ionosphere LAYER is called the VIRTUAL height.

Correct answer is (c) Square ROOT

To elaborate: For a regular LAYER, the CRITICAL FREQUENCY is proportional to the square root ofNmax electron density. \(f_c=9\sqrt{N \,max} \)

Correct answer is (b) 0.919

The best I can EXPLAIN: Refractive INDEX \(n=\SQRT{(1-\frac{81N}{f^2})}=\sqrt{(1-\frac{81×49×10^{10}}{(16×10^{6})^2})}=\sqrt{(1-0.155)}=\sqrt{0.845}\)

n=0.919

The correct ANSWER is (a) True

Explanation: Maximum usable frequency (MUF) is defined as the maximum possible frequency for which the WAVE is reflected BACK for a given distance of propagation in the ionosphere layer. If the wave exceeds MUF then it’s not reflected back and is transmitted to other UPPER LAYERS and signal is lost.

Correct choice is (b) False

To explain I would SAY: For fMUF ≤ fc, the WAVE will REFLECTS BACK irrespective of the angle of incidence. For fMUF ≥ fc, the wave will reflects back depending of the angle of incidence (which should be SMALL).

Correct answer is (b) 7.57MHz

Explanation: REFRACTIVE index \(N=\sqrt{(1-\FRAC{81N_{max}}{f^2})}\)

⇨ \(N_{max}=\frac{(1-n^2)f^2}{81}=\frac{(1-0.54^2)(9×10^6)^2}{81}=0.708 ×10^{12}/cm^3\)

⇨ CRITICAL frequency \(f_c=9\sqrt{N \,max} =9\sqrt{0.708 ×10^{12}}=7.57Mhz\)