Class 12 - Physics - Communication Systems

**Question1.**

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:

Correct option: - ** 10 KHz**.

10KHz cannot be radiated because of size of antenna would be large. 1GHz and 1000GHz are very high frequency,

wave which will penetrate the atmosphere and so cannot be transmitted.

**Question2.**

Frequencies in the UHF range normally propagate by means of:

(a) Ground waves. (b) Sky waves.

(c) Surface waves. (d) Space waves.

Answer:

Correct option: - (d) ** Space waves**.

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

**Question 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:

Correct option: - __(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.

**Question 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 81m tall. How much service area can it cover if the receiving antenna is at the ground level?

Answer:

The requirement for line-of-sight communication is that the signals emitted by the transmitted antenna should be received by the receiving antenna directly.

For this transmitting antenna should not necessarily have the same height as that of receiving antenna.

Given:

Height of the given antenna h = 81cm

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

d = √ (2Rh)

= √ (2 x 6.4 x 10^{6} x 81)

=3.2 x 10^{4} m

Therefore, the service area covered by TV transmitting antenna is

A = ∏ (d)^{ 2}

= (3.14) x (3.2 x 10^{4} x 3.2 x 10^{4})

=3258 x 10^{6} km^{2}

**Question 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:

Given:

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:

m= (A_{m} / A_{c})

Therefore, A_{m} = (m A_{c})

= (0.75 x 12)

=9V

**Question 6.**

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

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

(i) Sketch the amplitude modulated waveform

(ii) What is the 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:

ω_{m }= (2∏/T_{m})

=2∏ rads^{-1} (i)

The angular frequency of the carrier signal is calculated as:

ω_{c} = 8∏ rads^{-1} (ii)

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

ω_{c} = 4 ω_{m}

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

- Modulation index, m (A
_{m}/ A_{c})

= (1/2) = 0.5

**Question 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 zero volts?

Answer:

Given:

Maximum Amplitude E_{max} = 10V

Minimum Amplitude E_{min} =2V

Modulation index, μ = (E_{S}/E_{C})

As E_{C} = ((E_{max} + E_{min})/ (2)) and E_{S} = ((E_{max} - E_{min})/ (2))

Therefore, μ = (E_{max} - E_{min})/ (E_{max} - E_{min})

= (10 -2)/ (10+2)

= (8/12)

= (2/3)

When E_{min} = 0,

μ = (10-0)/ (10+0)

= (10/10)

μ =1

E_{min} = 0, μ = 1, irrespective of the value of E_{max}.

**Question 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 ω_{c} = frequency of carrier wave.

ω_{s} = frequency of the signal waves.

Signal received at the receiving station, V = V_{1} cos (ω_{c} + ω_{s}) t

Instantaneous voltage of the carrier wave, V_{in} = V_{c} cos ω_{c}t

Therefore, V V_{in} = (V_{1} cos (ω_{c} + ω_{s}) t) (V_{c} cos ω_{c}t)

= V_{1} V_{c }[cos (ω_{c} + ω_{s}) t. cos ω_{c}t]

= (V_{1} V_{c})/ (2) [2 cos (ω_{c} + ω_{s}) t. cos ω_{c}t]

= (V_{1} V_{c})/ (2) [{(cos (ω_{c} + ω_{s}) t + cos ω_{c}t}) + (cos (ω_{c} + ω_{s}) t - cos ω_{c}t})

= (V_{1} V_{c})/ (2) [cos {(2 ω_{c} + ω_{s}) t + cos ω_{s}t}]

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,

((V_{1} V_{c})/ (2) (cos ω_{s}t)) which is the signal frequency.

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