Answer: (b)

10 MHz

For beyond-the-horizon communication, it is necessary for the signal waves to travel a large distance. 10 KHz signals cannot be radiated efficiently because of the antenna size. The high energy signal waves (1GHz − 1000 GHz) penetrate the ionosphere. 10 MHz frequencies get reflected easily from the ionosphere. Hence, signal waves of such frequencies are suitable for beyond-the-horizon communication.

 Answer: (d)

Space waves

Owing to its high frequency, an ultra high frequency (UHF) wave can neither travel along the trajectory of the ground nor can it get reflected by the ionosphere. The signals having UHF are propagated through line-of-sight communication, which is nothing but space wave propagation.

 Answer: (C)

A digital signal uses the binary (0 and 1) system for transferring message signals. Such a system cannot utilise the decimal system (which corresponds to analogue signals). Digital signals represent discontinuous values.

Answer

Line-of-sight communication means that there is no physical obstruction between the transmitter and the receiver. In such communications it is not necessary for the transmitting and receiving antennas to be at the same height.

Height of the given antenna, h = 81 m

Radius of earth, R = 6.4 × 10⁶ m

For range, d = (2Rh)^½, the service area of the antenna is given by the relation:

A = πd²

= π (2Rh)

= 3.14 × 2 × 6.4 × 10⁶ × 81

= 3255.55 × 10⁶ m²

= 3255.55

∼ 3256 km²

Answer

Amplitude of the carrier wave, A_c = 12 V

Modulation index, m = 75% = 0.75

Amplitude of the modulating wave = A_m

Using the relation for modulation index:

Answer

It can be observed from the given modulating signal that the amplitude of the modulating signal, A_m = 1 V

It is given that the carrier wave c (t) = 2 sin (8πt)

∴Amplitude of the carrier wave, A_c = 2 V

Time period of the modulating signal T_m = 1 s

The angular frequency of the modulating signal is calculated as:

The angular frequency of the carrier signal is calculated as:

From equations (i) and (ii), we get:

The amplitude modulated waveform of the modulating signal is shown in the following figure.

(ii)Modulation index, 

Answer

Maximum amplitude, A_max = 10 V

Minimum amplitude, A_min = 2 V

Modulation index μ, is given by the relation:

Answer

Let ω_c and ω_s be the respective frequencies of the carrier and signal waves.

Signal received at the receiving station, V = V₁ cos (ω_c + ω_s)t

Instantaneous voltage of the carrier wave, V_in = V_c cos ω_ct

At the receiving station, the low-pass filter allows only high frequency signals to pass through it. It obstructs the low frequency signal ω_s. Thus, at the receiving station, one can record the modulating signal , which is the signal frequency.