Communication Systems

10 MHz frequency will be suitable for beyond the horizon communication using skywaves.

10 KHz cannot be transmitted because the size of the antenna will be large for this apparently lower frequency.1 and 1000 GHz are very high frequency wave which will penetrate and so cannot be transmitted.

The signals having Ultra High Frequency(UHF) can propagate through Space waves. These can neither travel along the trajectory of ground as shown in green nor can propagate by reflection in the ionosphere as shown in figure in red. The UHF signal propagates as shown in Blue in the figure.

Digital signal obviously use the binary signals i.e 0 and 1 only and no other signal other than these. It is also a signal that represents discontinuous values and it cannot utilize decimal values (0 to 9) which are continuous.

No, it is not necessary for a transmitting antenna and the receiving antenna to be at the same height. Because in this specific case of transmission there is no physical obstruction between the transmitter and the receiver.

(b) Height of the given antenna, h = 81 m

Radius of earth, R = 6.4 × 10⁶ m

For range, d = (2Rh)^1/2, the service area of the antenna is given by the relation:

A = π d²

= π (2Rh)

= 3.14 × 2 × 6.4 × 10⁶ × 81 m²

= 3255.55 × 10⁶ m²

~ 3256 km²

[As 1km = 10³ m , so 1km² = 10⁶ m²]

Amplitude of the carrier wave, A_c = 12 V

Modulation index, m = 75% = 0.75 (unitless as a ratio)

Amplitude of the modulating wave = A_m

We know for modulation index:

m = A_m/A_c

⇒ A_m = mA_c

⇒ A_m = (0.75 × 12) V = 9V

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) = 2sin(8πt) volts

∴ 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:

(t)_m = 2π/T_m

= 2π rad s^-1 (as T_m = 1)……………….(i)

The angular frequency of the carrier signal is calculated as:

(t)_m = 8π rad s^-1 ……….(ii)

From equations (i) and (ii), we get:

(t)_m = 4(t)_c

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

(ii) Modulation index,

m = A_m/A_c = 1/2 = 0.5

Maximum amplitude, A_max = 10 V

Minimum amplitude, A_min = 2 V

Modulation index μ is given by the relation:

⇒ μ = 8/12

∴ μ = 0.67

(b)

If A_min= 0

Then μ = A_max/A_max = 1

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

Signal received at the receiving station, V = V₁cos(ω_c + ω_s)t

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

V.V_in = V_1.V_c{cos(ω_c + ω_s)t. cos ω_ct}

= ( V_1.V_c)/2 {2cos(ω_c + ω_s)t. cosω_ct}

= ( V_1.V_c)/2 [cos{(ω_c + ω_s)t + ω_ct} + cos {{(ω_c + ω_s)t - ω_ct }

= (V_1.V_c)/2 [cos{(2ω_c + ω_s)t} + cos ω_st ]

As the receiving centre allows only high frequency signals to pass through it. Obstructs the low frequency signal ω_s Thus, at the receiving station, one can record the modulating signal (V₁V_c/2).cosω_st , which is the signal frequency.