CBSE Class 12 Physics Important Questions Communication system

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COMMUNICATION SYSTEMS

DIFFICULT TOPICS

1. Sky waves Propagation

2. Space wave Propagation

3. Amplitude modulation

4. Satellite communication

5. Band width of signals

6. Expression for covering range of TV transmission tower.

QUESTIONS

ELEMENTS OF COMMUNICATION SYSTEMS

1. What should be the length of dipole antenna for a carrier wave of 5 x 108 Hz? 

Ans: L = λ\2 = c\v x 2 = 3 x 10⁸ / 5 x 10⁸ x 2 = 0.3m.

2. A device X can convert one form of energy into another. Another device Y can be regarded as a combination of a transmitter and a receiver. Name the devices X and Y.

(a) Transducer (b) Repeater

3. Complete the following block diagram depicting the essential elements of a basic communication system. 

ANS:TRANSMITTER,MEDIUM AND RECIEVER

5. Calculate the length of a half wave dipole antenna at

(a) 1 MHz (b) 100 MHz (c) 1000MHz

What conclusion you draw from the results?

Hint: Length of dipole antenna, L = λ/ 2

(a) 150m (b) 1.5m (c) 15 cm

6. A TV tower has a height of 300m. What is the maximum distance upto which this TV transmission can be received?

Ans: d = √2Rh = √ 2 x 6400 x 1000 x 300 = 62km

7. How does the effective power radiated by an antenna vary with wavelength? 

8. Why ground wave propagation is not suitable for high frequency? (OR)Why is ground wave propagation restricted to frequency up to 1500 kHz?

Hint: It is because radio waves having frequency greater than 1500MHz are strongly absorbed by the ground.

9. Why are signals not significantly absorbed by ionosphere in satellite communication?

Hint: It is because satellite communication employs HF carrier i.e. microwaves

10. How many geostationary satellites are required to provide communication link over the entire globe and how should they be parked? 

11. The tuned circuit of oscillator in a single AM transmitter employs 50 uH coil and 1nF capacitor. The oscillator output is modulated by audio frequency up to 10KHz. Determine the range of AM wave.

Hint: υ_c = 1/2π√LC ; USF = υ_c + υ_m ; LSF = υ_c – υ_m

12. What is the population covered by the transmission, if the average Population density around the tower is 1200km‐2? 

Hint: d = √2Rh=√2×6.4×10³ ×16⁰×10^‐3 =45km Range 2d=2×45=90km

Population covered=area × population density=1200×6359= 763020

13. A transmitting antenna at the top of tower has a height of 36m and the height of the receiving antenna is 49m. What is the maximum distance between them, for the satisfactory communication in the LOS mode? (Radius of the earth =6400km). 

Hint: Using d= √2Rht + √2Rhr we get =46.5km

14. Derive an expression for covering range of TV transmission tower 

15.  What is space wave propagation? Which two communication methods make use of this mode of propagation? If the sum of the heights of transmitting and receiving antennae in line of sight of communication is fixed at h, show that the range is maximum when the two antennae have a height h/2 each.

Ans: Satellite communication and line of sight (LOS) communication make use of space waves.

Here d₁=√2Rh₂ and d₂= √2Rh₂

For maximum range,

D_m=√2Rh₁ + √2Rh₂

where d_m =d₁ + d₂= d

Given h₁ + h₂ = h

Let h₁ = x then h₂ = h‐x

Then d_m = √2Rx + √2R(h‐x) ,

dd_m /dx = √R/2x ‐ √R/2(h‐x) = 0

i.e., 1/2x = 1/2(h‐x) i.e., x = h/2

=> h1 = h2 = h/2.

16. A ground receiver station is receiving signals at (i) 5 MHz and (ii) 100 MHz, transmitted from a ground transmitter at a height of 300 m located at a distance of 100km. Identify whether the signals are coming via space wave or sky wave propagation or satellite transponder. Radius of earth = 6400 km; Maximum electron density in ionosphere, N_max = 10¹²m^‐3 

Solution:

Maximum coverage range of transmitting antenna, d = √2Reh

Therefore d = √2 x 6400 x 10³ x 300 = 6.2 x 10⁴

The receiving station (situated at 100 km) is out of coverage range of transmitting antenna, so space wave communication is not possible, in both cases (i) and (ii) The critical frequency (or maximum frequency) of ionospheric propagation is fc = 9(N_max)^1/2 = 9 x √10¹²= 9 x 10⁶ Hz = 9 MHz Signal (i) of 5MHz (< 9 MHz) is coming via ionosphere mode or sky wave propagation, while signal (ii) of 100MHz is coming via satellite mode.

17. By what percentage will the transmission range of a TV tower be affected when the height of the tower is increased by 21%. ? 

Solution:

Transmission range of TV tower = d = √2hR If the height is increased by 21%, new height

h’ = h + 21\100h = 1.21h

If d’ is the new average range, then d’/d =√h’ / √h = 1.1% increase in range Δd\ d x 100% = (d’ – d \ d)

x 100% = (d’/ d ‐1) x100% = (1.1 – 1) x 100% = 10%

18. What would be the modulation index for an amplitude modulated wave for which the maximum amplitude is ‘a’ while the minimum amplitude is ‘b’? 

Ans: Modulation index, a_m = E_m/E_c … (1)

Maximum amplitude of modulated wave a=E_c + E_m.....(2)

Minimum amplitude of modulated wave b = E_c ‐ E_m …(3)

From (2) and (3), Ec = a+b/2, Em = a‐b/2

From (1), modulation index, a_m = E_m/E_c = (a‐b)/2 / (a+b)/2 = a‐b/ a+b

19. A carrier wave of peak voltage 20 V 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 80% ?

Hint: Modulation index, m_a = E_m / E_c

E_m=m_a x E_c= 0.80 x 20 V = 16 V

20. A message signal of frequency 10 kHz and peak value of 8 volts is used to modulate a carrier of frequency 1MHz and peak voltage of 20 volts. Calculate: (i) Modulation index (ii) The side bands produced.

Solution: (i) Modulation index, m_a = E_m / E_c = 8/20 = 0.4

(ii) Side bands frequencies = f_c ± fm

Thus the side bands are at 1010KHz and 990 kHz.

21. An amplitude modulation diode detector, the output circuit consists of resistance R = 1kΩ and capacitance C = 10pf. It is desired to detect a carrier signal of 100 kHz by it. Explain whether it is a good detector or not? If not what value of capacitance would you suggest? (3)

Solution: The satisfactory condition for demodulation is that reactance at carrier frequency must be much less than R.

Reactance = 1 / ώC = 1 / 2πfCC = 1/ 2 x 3.14 x 100 x 10³ x 10 x 10^‐12

= 1.59 x 105 Ω = 159 kΩ

This is much greater than the given resistance, so it is not a good detector. For detection, the condition is 1 / 2πfCC<< R = C >> 1 / 1.59 x 10^‐9 fared or C >> 1.59 nF.

Thus for proper detection the capacitance of output circuit must be much greater than 1.59 nF. The suitable capacitance is 1μF.

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