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**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 log _{10} (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 log

Input power is P

Thus, the output power is 101 mW

Ans.

Radius of earth = 6.4 x 10

= 16000 m = 16 km

Area covered A = Ï€(d)

= 3.14 x 16 x 16 = 803.84 km

(ii) At a height of h

= 16 x 10

= 33.9 x 10

= 33.9 km

Area covered =Ï€(d)

= 3.14 x 33.9 x 33.9

= 3608.52 km

Therefore, percentage increase in area

Ans.

where, R is the radius of the earth (approximately 6400 km). h

d

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 h

Assume the height of transmitting antenna or receiving antenna in order to cover the entire surface of earth through communication is h

As, space wave frequency is used, so Î» < <h

On a certain day it is observed that signals of frequencies higher than 5MHz are not received by reflection from the F

Estimate the maximum electron densities of the F

Ans.

where N

For layer F

So, 5 x 10

this implies

Then for layer F

So, 8 x 10

this implies

Suggest ways to minimise cost of radiation without compromising on information.

Ans.

(Ï‰

(Ï‰

Side band frequencies are generally close to the carrier frequency.

Only side band frequencies contain information in amplitude modulated signal, [only (Ï‰

Here, the total radiated power is due to energy carried by Ï‰

For reduction of cost of radiation without compromising on information Ï‰c can be left and transmitting the frequencies (Ï‰

**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 _{o}e^{â€“Î±x}, where I_{o} is the intensity at x = 0 and Î± 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 log_{10} (I/I_{0}). 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 = I

where I

If the intensity is reduced by 75% that means,

I = 25% of I

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, Î± is the attenuation constant expressed in dB/km. If x is the distance travelled by signal, then

....(1)

where, I

Here 50% intensity reduced by distance 50 km So, I = 50% of I

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

âˆ´ The attenuation for given optical fibre Î± = 0.0602 dB/km

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.

Velocity of waves= 3 x 10

Time to reach a receiver = 4.04 ms = 4.04 x 10

Let the height of satellite is h

Radius of earth = 6400 km

Size of transmitting antenna = h

^{}

=6.06 x 10^{5} = 606 km

Using Pythagoras theorem,

d^{2 }= x^{2} - h^{2}_{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 voltageAns.**

It is observed from the above diagram that

Maximum voltage V

Minimum voltage V

(iii) Peak value of information voltage.

(ii) Is the plot symmetrical about Ï‰c? Comment especially about plot in region Ï‰ < Ï‰

(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 Ï‰

(iii) If more modulating signals are present then there will be more crowding in the modulation signal in the region Ï‰ <Ï‰

(iv) To accommodate more signals, we should increase bandwidth and frequency carrier waves Ï‰

(i) R = 1 kâ„¦, C = 0.01ÂµF

(ii) R = 10 kâ„¦, C = 0.01ÂµF

(iii) R = 10 kâ„¦, C = 0.1ÂµF

Ans.

According to the problem, carrier wave frequency f

= 20 x 10

Bandwidth required for modulation is

2f

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 = 1kÎ© x 0.01 Î¼F = 10

Here, condition is satisfied.

Hence it can be demodulated

(ii) RC = 10 kÎ© x 0.01 Î¼F = 10

Here condition is satisfied.

Hence, it can be demodulated.

(iii) RC = 10kÎ© x 1 Î¼F = 10

Here, condition so this cannot be demodulated.

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