Communication Systems Ch 15 Physics

 NCERT Solution For Class 12 Science

Exercises

15.1. Which of the following frequencies will be suitable for beyond-the-horizon communication using sky    waves ?

(a) 10 kHz

(b) 10 MHz

(c) 1 GHz

(d) 1000 GHz

Answer : (b) 10 MHz

Frequency of 10 kHZ cannot be used because it will require very large transmitting antenna. Also frequencies of 1 GHz and 1000 GHz cannot be used because these will penetrate through the ionosphere; the ionosphere cannot reflect these high frequencies.

15.2. Frequencies in the UHF range normally propagate by means of :

(a) Ground waves.

(b) Sky waves

(c) Surface waves.

(d) Space waves.

Answer : (d) Space waves

Frequencies in the UHF range (0.3 GHz to 3 GHz) normally propagate by means of space waves.

15.3. Digital signals

(i) do not provide a continuous set of values,

(ii) represent values as discrete steps,

(iii) can utilize binary system, and

(iv) can utilize decimal as well as binary systems.

Which of the above statements are true ?

(a) (i) and (ii) only

(b) (ii) and (iii) only

(c) (i), (ii) and (iii) but not (iv)

(d) All of (i), (ii), (iii) and (iv)

Answer: (i) , (ii) and (iii) but not (iv)

It is because digital signals cannot utilise decimal system which represents continuous set of values.

15.4. Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication ? A TV transmitting antenna is 81 m tall. How much service area can it cover if the receiving antenna is at the ground level ?

Answer :

No. However, for line-of-sight (LOS) communication, the receiving antenna must be able to directly intercept the signals radiated by the transmitting antenna. 

Here, h = 81 m; Radius of earth, R = 6.4 X 10^6 m

∴ Radius (d) of the area covered by TV transmitting antenna is 

d = √2Rh = √2 ✖ 6.4 X 10^6 X 81 = 3.2 X 10^4 m

∴ Service area covered by TV transmitting antenna is 

A = πd^2 = π X (3.2 X 10^4)^2 = 3258 X 10^6 m^2 = 3258 km^2

15.5. A carrier wave of peak voltage 12V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75% ?

Answer:

15.6. A modulating signal is a square wave, as shown in Fig. 15.14. 

Figure 15.14

The carrier wave is given by c(t) = 2 sin(8πt) volts.

(i) Sketch is the amplitude modulated waveform

(ii) What is the modulation index ?

Answer:

(i) Here, c(t) = 2 sin(8πt)

It is clear that carrier is a sine wave having a period T = 2π/8π = 0.25s and amplitude Ec = 士 2V.

The modulating signal is a square wave of period 1s and amplitude of Es = 士 1V.

∴ Maximum amplitude of modulated wave is

emax = Ec + Es = 士 (2+1) = 士 3 V

Minimum amplitude of modulated wave is

emin = Ec - Es = 士 (2-1) = 士 1 V

Therefore, the amplitude modulated wave will be as shown below.

(ii) Modulation index, ma = Es / Ec = 1 V / 2 V = 0.5

15.7. For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2V. Determine the modulation index, μ.

What would be the value of μ if the minimum amplitude is zero volt ?

Answer :

15.8. Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.

Answer:

Let ωc and ωs be the angular frequency of carrier waves and signal wave respectively. When only the upper sideband (ωc + ωs) is transmitted, then the signal available at the receiving station may be respresented as :

e = Et cos (ωc + ωs) t

where Et = Amplitude of transmitted signal