# NCERT Solutions for Class 12 Physics Chapter 15 - Communication System

## NCERT Solutions for Class 12 Physics Chapter 15 Free PDF Download

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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

Sol. (b) Due to large antenna size, 10 kHz frequencies cannot be radiated, 1 GHz and 1000 GHz will be penetrated. So, option (b) is correct.

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

(a) ground waves

(b) sky waves

(c) surface waves

(d) space waves

Sol. (d) The frequencies in UHF range normally propagate by means of space waves. The high frequency space does not bend with ground but they are ideal for frequency modulation.

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)

(b) (ii) and (iii)

(c) (i),(ii) and (iii)

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

Sol. (c) A digital signal is a discontinuous function of time in contrast to an analog signal. The digital signals can be stored as digital data and cannot be transmitted along the telephone lines. Digital signal cannot utilize decimal signals.

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?

Sol. Given, antenna height, h = 81 m

Radius of earth, R = 6.4 × 106 m
No, it is not necessary the two antennas may not be at the same height for line of sight communication.

Area = πd2

$$\because\space\text{Range, d =}\sqrt{2hR}\\\therefore\space\text{Service area =} \pi × 2hR\\=\frac{22}{7}×2×81×6.4×10^{6}$$

= 3258.5 × 106 m2s

= 3258.5 km2

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%?

Sol. Given, peak voltage, V0 = 12V

$$\text{Modulation index,}\space\mu = 75\% =\frac{75}{100}\\\text{Modulation index}(\mu)=\frac{\text{Peak voltage of modulating signal(V}_m)}{\text{Peakvoltage(V}_0)}\\\text{So, peak voltage of modulating signal,}\\\text{V}_{m}=\mu×\text{Peak voltage}\\=\frac{75}{100}×12=9\text{V}$$

$$\text{Modulation index,}\space\mu = 75\% =\frac{75}{100}\\\text{Modulation index}(\mu)=\frac{\text{Peak voltage of modulating signal(V}_m)}{\text{Peakvoltage(V}_0)}\\\text{So, peak voltage of modulating signal,}\\\text{V}_{m}=\mu×\text{Peak voltage}\\=\frac{75}{100}×12=9\text{V}$$

6. A modulating signal is a square wave as shown in figure. The carrier wave is given by c(t) = 2 sin (8 πt) volt.

(a) Sketch the amplitude modulated waveform.

(b) What is the modulation index ?

Sol. Given, the equation of carrier wave

c (t) = 2 sin (8pt) …(i)

(a) According to the figure,
Amplitude of modulating signal

Am = 1V

Amplitude of carrier wave AC = 2V
[By Eq. (i)]
Tm = 1s (From figure)
ωm = 2Tπm= 2π1= 2p rad/s …(ii)
From equation (i),
c(t) = 2 sin 8pt
= AC sin wCt
So wc = 8p
From Eq. (ii),
wc = 4 wm
Amplitude of modulated wave A = Am + AC = 2 + 1 = 3V
The figure of the amplitude modulated waveform is shown here.

$$\omega_m=\frac{2\pi}{\text{T}_{m}}=\frac{2\pi}{1}=2\pi\text{rad/s}\qquad \text{…(ii)}$$

From equation (i),

c(t) = 2 sin 8πt

= AC sin ωCt

So ωc = 8p

From Eq. (ii),

ωc = 4 ωm

Amplitude of modulated wave A = Am + AC = 2 + 1 = 3V

The figure of the amplitude modulated waveform is shown here.

$$\text{(b) Modulation index, }\mu =\frac{\text{A}_m}{\text{A}_C}=\frac{1}{2}=0.5$$

7. For an amplitude modulated wave, the maximum amplitude is found to be 10V 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 0V?

Sol. Given, maximum amplitude Amax = 10V

Minimum amplitude Amin = 2V

Let AC and Am be the amplitudes of carrier wave and signal wave respectively

∴ Amax = AC + Am = 10 …(i)

and Amin = AC – Am = 2 …(ii)

Adding the equations (i) and (ii), we get

2 AC = 12

or AC = 6V

and Am = 10 – 6 = 4V

$$\text{Modulation index,}\space\mu=\frac{A_m}{A_c}=\frac{4}{6}=\frac{2}{3}$$

When the minimum amplitude is zero, then i.e., Amin = 0

AC + Am = 10 …(iii)

AC – Am = 0 …(iv)

By solving equations (iii) and (iv), we get

2Am = 10

or Am = 5

and

AC = 5

$$\text{Modulation index,}\mu=\frac{A_m}{A_c}=\frac{5}{5}=1$$

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.

Sol. Let ωc and ωm be the angular frequency of carrier waves and signal waves respectively.

Let the signal received at the receiving station be

e = E1 cos (ωc + wm )t …(i)

Let the instantaneous voltage of carrier wave

ec = Ec cos wc t …(ii)

is available at receiving station.

Multiplying equations (i) and (ii), we get e × ec = E1Ec cos ωc t cos (ωc + ωm)t

$$\text{E}=\frac{\text{E}_{1}\text{E}_{c}}{2}.2\text{cos}\space\omega_ct.\text{cos}(\omega_c+\omega_m)t\\\text{(Let e × e}_c \text{= E)}\\=\frac{\text{E}_1\text{E}_c}{2}.[\text{cos}(\omega_c+\omega_c+\omega_m)t+\text{cos}(\omega_c+\omega_m-\omega_c)t]\\\lbrack\because \space2\text{cos A cos B}=\text{cos (A +B) + cos (A – B)}\rbrack\\=\frac{\text{E}_1\text{E}_c}{2}[\text{cos}(2\omega_c + \omega_m)t + \text{cos} \omega_mt]$$

Now, at the receiving end as the signal passes through filter, it will pass the high frequency
(2 ωc + ωm) but obstruct the frequency ωm . So, we can record the modulating signal

$$\frac{\text{E}_1\text{E}_2}{2}\text{cos}\omega_m t\text{which is a signal of angular frequency}\space\omega_m.$$