NCERT Exemplars: Communication Systems- 2 - JEE PDF Download

SHORT ANSWER TYPE QUESTIONS

Q.21. Figure shows a communication system. What is the output power when input signal is of 1.01 mW ? (gain in dB = 10 log₁₀ (P_o/P_i ).
Ans.
According to the problem, Loss suffered in path of transmission = 2 dR/km
And the distance travelled by the signal or path length is 5 km.
So, total loss suffered in 5 km = -2 x 5 = -10 dB
Total amplifier gain = 10 dB + 20 dB - 30 dB
Overall gain in signal = 30 - 10 - 20 dB
Here, it is given that gain in dB = 10 log₁₀ (P₀/P_i)

Input power is P_i =1.01 mW and P₀ is the output power

Thus, the output power is 101 mW

Q.22. A TV transmission tower antenna is at a height of 20 m. How much service area can it cover if the receiving antenna is (i) at ground level, (ii) at a height of 25 m? Calculate the percentage increase in area covered in case (ii) relative to case (i).
Ans. In this problem, height of antenna h_T = 20 m
Radius of earth = 6.4 x 10⁶ m

= 16000 m = 16 km
Area covered A = π(d)²
= 3.14 x 16 x 16 = 803.84 km²
(ii) At a height of h_R = 25 m from ground level


= 16 x 10³ + 17.9 x 10³
= 33.9 x 10³ m
= 33.9 km
Area covered =π(d)² 
= 3.14 x 33.9 x 33.9
= 3608.52 km²
Therefore, percentage increase in area


Q.23. If the whole earth is to be connected by LOS communication using space waves (no restriction of antenna size or tower height), what is the minimum number of antennas required? Calculate the tower height of these antennas in terms of earths radius?
Ans.
Key concept: Distance or range of transmission tower, d_T =√2Rh_T
where, R is the radius of the earth (approximately 6400 km). h_T is the height of transmission tower, .
d_T is also called the radio horizon of the transmitting antenna.
Let us consider the figure given below to solve this problem.
Assume the height of transmitting antenna or receiving antenna in order to cover the entire surface of earth through communication is h₁, i.e. h_T = h_R and radius of earth is R. If d_M is the line-of-sight distance between the transmission and receiving antennas, then maximum distance
Assume the height of transmitting antenna or receiving antenna in order to cover the entire surface of earth through communication is h_t i.e. h_T - h_R and radius of earth is R. If d_M is the Line-of-sight distance between the transmission and receiving antennas,.then maximum distance


As, space wave frequency is used, so λ < <h_T, hence only lower height is to be taken into consideration, In three dimensions of earth, 6 antenna towers of each of height h_T = R would be used to cover entire surface of earth with communication programme.

Q.24. The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be f_max = 9(N_max)^1/2, where N_max is the maximum electron density at that layer of the ionosphere.
On a certain day it is observed that signals of frequencies higher than 5MHz are not received by reflection from the F₁ layer of the ionosphere while signals of frequencies higher than 8 MHz are not received by reflection from the F₂ layer of the ionosphere.
Estimate the maximum electron densities of the F₁ and F₂ layers on that day.
Ans. According to the problem, the maximum frequency for reflection of sky waves is f_max  = 9(N_max)^1/2.
where N_max is a maximum electron density
For layer F₁, f_max = 5 MHz
So, 5 x 10⁶ = 9 (N_max)^1/2
this implies
Then for layer F₂, f_max =  8 MHz
So, 8 x 10⁶ = 9 (N_max)^1/2
this implies

Q.25. On radiating (sending out) an AM modulated signal, the total radiated power is due to energy carried by ω_c, ω_c – ω_m & ω_c + ω_m.
Suggest ways to minimise cost of radiation without compromising on information.
Ans.
Key concept: Side band frequencies. The AM wave contains three frequencies ω_c, (ω_c + ω_m) and (ω_c – ω_m), ω_c is called carrier frequency, (ω_c + ω_m) and ( ω_c – ω_m) are called side band frequencies.
(ω_c + ω_m) = Upper side band (USB) frequency
(ω_c – ω_m) =Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency.
Only side band frequencies contain information in amplitude modulated signal, [only (ω_c+ ω_m) and (ω_c + ω_m)].
Here, the total radiated power is due to energy carried by ω_c, (ω_c – ω_m) and (ω_c+ω_m)
For reduction of cost of radiation without compromising on information ωc can be left and transmitting the frequencies (ω_c + ω_m), (ω_c – ω_m) or both (ω_c + ω_m) and (ω_c – ω_m).

LONG ANSWER TYPE QUESTIONS

Q.26. (i) The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I = I_oe^–αx, where I_o is the intensity at x = 0 and α is the attenuation constant.
Show that the intensity reduces by 75 per cent after a distance of
(ii) Attenuation of a signal can be expressed in decibel (dB) according to the relation dB = 10 log₁₀ (I/I₀). What is the attenuation in dB/km for an optical fibre in which the intensity falls by 50 per cent over a distance of 50 km?
Ans.
(a) According to the problem, the intensity of a light pulse travelling along a communication channel is given by I = I₀e^-ax
where I_o is the intensity at x - 0 and α is the attenuation constant.
If the intensity is reduced by 75% that means,
I = 25% of I₀
So we can write this as
By using live above relation mentioned in the problem,

Taking natural log on both sides, we get

Therefore, at  distance  the intensity is reduced to 75% of initial intensity,
(b) Here, α is the attenuation constant expressed in dB/km. If x is the distance travelled by signal, then
     ....(1)
where, I₀ is the initial intensity.
Here 50% intensity reduced by distance 50 km So, I = 50% of I₀

Substituting the value of x in Eq. (i),


∴ The attenuation for given optical fibre α = 0.0602 dB/km

Q.27. A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above earth’s surface.
Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be?
Ans. Let the receiver is at point A and source is at B.
Velocity of waves= 3 x 10⁸ m/s
Time to reach a receiver = 4.04 ms = 4.04 x 10^-3 s
Let the height of satellite is h_s = 600 km
Radius of earth = 6400 km
Size of transmitting antenna = h_T

=6.06 x 10⁵ = 606 km
Using Pythagoras theorem,
d² = x² - h²_s = (606)² - (600)² = 7236
or d = 85.06 km
So, the distance between source and receiver = 2d
= 2 x 85.06= 170 km
The maximum distance covered

size of antenna or


Q.28. An amplitude modulated wave is as shown in Figure  Calculate
(i) the percentage modulation,
(ii) peak carrier voltage and,
(iii) peak value of information voltage

Ans.
It is observed from the above diagram that
Maximum voltage V_max = 100/2 = 50 V
Minimum voltage V_{min =} 20/2 = 10V




(iii) Peak value of information voltage.


Q.29. (i) Draw the plot of amplitude versus ‘ ω’ for an amplitude modulated wave whose carrier wave (ω_c) is carrying two modulating signals, ω₁ and ω₂ ( ω₂ > ω₁).
(ii) Is the plot symmetrical about ωc? Comment especially about plot in region ω < ω_c.
(iii) Extrapolate and predict the problems one can expect if more waves are to be modulated.
(iv) Suggest solutions to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth?
Ans.
(i)
(ii) In the plotted graph shown, we note that frequency spectrum is not symmetrical about ω_c. Crowding of spectrum is present for ω < ω_c.
(iii) If more modulating signals are present then there will be more crowding in the modulation signal in the region ω <ω_c. That will result more chances of mixing of signal.
(iv) To accommodate more signals, we should increase bandwidth and frequency carrier waves ω_c. This shows that large carrier frequency enables to carry more information (i.e., more ω_m) and the same will in turn increase bandwidth.

Q.30. An audio signal is modulated by a carrier wave of 20MHz such that the bandwidth required for modulation is 3kHz. Could this wave be demodulated by a diode detector which has the values of R and C as
(i) R = 1 kΩ, C = 0.01µF
(ii) R = 10 kΩ, C = 0.01µF
(iii) R = 10 kΩ, C = 0.1µF
Ans.
According to the problem, carrier wave frequency f_c = 20MHz
= 20 x 10⁶ Hz
Bandwidth required for modulation is
2f_m = 3kHz = 3 x 10³ Hz

Demodulation by a diode is possible if the condition  is satisfied
Thus      ..(i)
and   ..(ii)
Now, gain through all the options of R and C one by one, we get
(i) RC = 1kΩ x 0.01 μF = 10³ Ω x (0.01 x 10^-6F) = 10^-5 s
Here, condition  is satisfied.
Hence it can be demodulated
(ii) RC = 10 kΩ x 0.01 μF = 10⁴Ω x 10^-8 F = 10^-4s
Here condition  is satisfied.
Hence, it can be demodulated.
(iii) RC = 10kΩ x 1 μF = 10⁴Ω x 10^-12F = 10^-8s
Here, condition  so this cannot be demodulated.