CBSE Class 12 Physics Important Questions Communication system

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COMMUNICATION SYSTEMS
CLASS 12 PHYSICS : DIFFICULT TOPICS AND IMPORTANT QUESTIONS

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? [1 Mark]

Answer:
1. Wavelength λ = c/f = (3 × 10⁸)/(5 × 10⁸) = 0.6 m.
2. Length of a half wave dipole L = λ/2 = 0.6/2 = 0.3 m.

Teacher's Note:
a) The dipole length is half the wavelength, not the full wavelength.

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. [1 Mark]
(a) Transducer
(b) Repeater

Answer: Device X is (a) Transducer, and device Y is (b) Repeater.

Teacher's Note:
a) A repeater receives a signal and retransmits it, so it combines a receiver with a transmitter.

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

Answer:
1. First block: Transmitter.
2. Middle block: Medium (channel) through which the signal travels.
3. Last block: Receiver.

Teacher's Note:
a) The transmitter converts the message into a suitable signal for transmission.
b) The medium carries the signal from the transmitter to the receiver.
c) The receiver recovers the original message from the signal it receives.

5. Calculate the length of a half wave dipole antenna at (a) 1 MHz (b) 100 MHz (c) 1000 MHz. What conclusion you draw from the results? [3 Marks]

Answer:
1. Length of a half wave dipole L = λ/2 = c/(2f).
2. (a) At 1 MHz: λ = 300 m, so L = 150 m.
3. (b) At 100 MHz: λ = 3 m, so L = 1.5 m.
4. (c) At 1000 MHz: λ = 0.3 m, so L = 15 cm.
5. Conclusion: the length of the antenna decreases as the frequency increases, since L is inversely proportional to f.

Teacher's Note:
a) Frequency and wavelength are inversely related through c = fλ.
b) Convert MHz to Hz before using c = 3 × 10⁸ m/s.
c) Higher frequency antennas are much smaller, which is why mobile phone antennas are short.

6. A TV tower has a height of 300m. What is the maximum distance upto which this TV transmission can be received? [1 Mark]

Answer:
1. Maximum distance d = √(2Rh), with R = 6.4 × 10⁶ m and h = 300 m.
2. d = √(2 × 6.4 × 10⁶ × 300) = √(3.84 × 10⁹) ≈ 6.2 × 10⁴ m = 62 km.

Teacher's Note:
a) Convert the radius of the earth into metres before substituting.

7. How does the effective power radiated by an antenna vary with wavelength? [1 Mark]

Answer: The effective power radiated by an antenna is inversely proportional to the square of the wavelength, P ∝ 1/λ².

Teacher's Note:
a) The radiated power depends on the ratio of antenna length to wavelength, so a shorter wavelength radiates more power for the same length.

8. Why ground wave propagation is not suitable for high frequency? (OR) Why is ground wave propagation restricted to frequency up to 1500 kHz? [1 Mark]

Answer: Ground wave propagation is restricted to frequencies up to about 1500 kHz because radio waves of higher frequency are strongly absorbed by the ground.

Teacher's Note:
a) The printed hint says 1500 MHz, but the correct limit is 1500 kHz, which is 1.5 MHz.

9. Why are signals not significantly absorbed by ionosphere in satellite communication? [1 Mark]

Answer: Satellite communication uses microwave carriers of very high frequency, which are much higher than the critical frequency of the ionosphere, so they pass through the ionosphere and are not significantly absorbed.

Teacher's Note:
a) The printed hint calls the carrier an HF carrier, but HF is only 3 to 30 MHz; satellite links use microwaves.

10. How many geostationary satellites are required to provide communication link over the entire globe and how should they be parked? [1 Mark]

Answer: Three geostationary satellites are required, placed in the same orbit above the equator at equal angular separations of 120°.

Teacher's Note:
a) A geostationary satellite stays over one point of the equator at a height of about 36000 km.

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. [2 Marks]

Answer:
1. Carrier frequency ν_c = 1/(2π√(LC)) = 1/(2π√(50 × 10^-6 × 1 × 10^-9)).
2. ν_c ≈ 712 kHz.
3. Upper side frequency = ν_c + ν_m = 712 + 10 = 722 kHz.
4. Lower side frequency = ν_c - ν_m = 712 - 10 = 702 kHz.
5. Range of AM wave = 702 kHz to 722 kHz, so its bandwidth is 20 kHz.

Teacher's Note:
a) The bandwidth of an AM wave is twice the highest audio frequency.
b) Convert μH and nF into SI units before calculating the square root.

12. What is the population covered by the transmission, if the average Population density around the tower is 1200km^-2? [2 Marks]

Answer:
1. Radius of coverage d = √(2Rh) = √(2 × 6.4 × 10⁶ × 160) ≈ 45 km, taking the tower height h = 160 m.
2. Area covered = π d² = 3.14 × 45² ≈ 6359 km².
3. Population covered = area × population density = 6359 × 1200 ≈ 76,30,800.

Teacher's Note:
a) The printed hint shows a tower height of 160 m, which is not stated in the question text, so this height is used here.
b) The printed key gives 7,63,020, which is a tenfold arithmetic error; 6359 × 1200 is 76,30,800.

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). [2 Marks]

Answer:
1. Maximum LOS distance d = √(2Rh_t) + √(2Rh_r).
2. √(2 × 6.4 × 10⁶ × 36) ≈ 21.5 km.
3. √(2 × 6.4 × 10⁶ × 49) ≈ 25.0 km.
4. d ≈ 21.5 + 25.0 = 46.5 km.

Teacher's Note:
a) Each antenna contributes its own line of sight distance, and the two are added.
b) Use the radius of the earth in metres to keep the units consistent.

14. Derive an expression for covering range of TV transmission tower [2 Marks]

Answer:
1. Let R be the radius of the earth, h the height of the tower, and d the distance from the foot of the tower to the tangent point on the earth's surface.
2. Since the tangent is perpendicular to the radius at the tangent point, (R + h)² = R² + d².
3. d² = 2Rh + h².
4. As h is much smaller than R, h² is negligible, so d = √(2Rh).

Teacher's Note:
a) The tangent to the earth at the point of contact makes a right angle with the radius.
b) Dropping the h² term is valid only because the tower is short compared with the earth's radius.

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. [3 Marks]

Answer:
1. Space wave propagation is the transmission of radio waves directly from the transmitting antenna to the receiving antenna through the lower atmosphere, along the line of sight.
2. Line of sight (LOS) communication and satellite communication use space wave propagation.
3. Let h₁ and h₂ be the heights of the transmitting and receiving antennae, with h₁ + h₂ = h.
4. Maximum range d_m = √(2Rh₁) + √(2Rh₂) = √(2R)(√h₁ + √h₂).
5. Put h₁ = x, so h₂ = h - x, and d_m = √(2R)(√x + √(h - x)).
6. Differentiate: dd_m/dx = √(2R) [ 1/(2√x) - 1/(2√(h - x)) ].
7. Set dd_m/dx = 0, so √x = √(h - x), which gives x = h - x, that is, x = h/2.
8. Hence h₁ = h₂ = h/2. The second derivative is negative, so this value gives the maximum range.

Teacher's Note:
a) The printed key writes d1 = √2Rh2 and d2 = √2Rh2; the first should read d1 = √(2Rh1).
b) Use the constraint h1 + h2 = h before differentiating.
c) The square-root sum is largest when the two terms are equal.

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, Nmax = 10¹² m^-3 [3 Marks]

Answer:
1. Maximum space wave range d = √(2Rh) = √(2 × 6400 × 10³ × 300) ≈ 62 km.
2. The receiver is at 100 km, which is beyond 62 km, so direct space wave communication is not possible for either signal.
3. Critical frequency f_c = 9√N_max = 9 × √(10¹²) = 9 × 10⁶ Hz = 9 MHz.
4. Signal (i) at 5 MHz is below 9 MHz, so it is reflected by the ionosphere and comes via sky wave propagation.
5. Signal (ii) at 100 MHz is above 9 MHz, so it passes through the ionosphere and comes via a satellite transponder.

Teacher's Note:
a) A signal below the critical frequency is reflected back by the ionosphere.
b) Compare the distance with the maximum space wave range before deciding the mode.
c) The printed key uses 6.2 × 10⁴ without the unit; it is 62 km, that is, 6.2 × 10⁴ m.

17. By what percentage will the transmission range of a TV tower be affected when the height of the tower is increased by 21%. ? [3 Marks]

Answer:
1. Transmission range d = √(2Rh), so d is proportional to √h.
2. New height h' = h + (21/100)h = 1.21h.
3. d'/d = √(h'/h) = √1.21 = 1.1.
4. Increase in range = (d' - d)/d × 100% = (1.1 - 1) × 100% = 10%.

Teacher's Note:
a) The range depends on the square root of the height, so a 21% rise in height gives only a 10% rise in range.
b) The printed key says 1.1% in one line, but the ratio 1.1 means a 10% increase.
c) Always take the ratio of the new range to the old range before converting it to a percentage.

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'? [2 Marks]

Answer:
1. Modulation index m_a = E_m/E_c.
2. Maximum amplitude a = E_c + E_m, and minimum amplitude b = E_c - E_m.
3. Adding and subtracting: E_c = (a + b)/2 and E_m = (a - b)/2.
4. m_a = \( \frac{a - b}{a + b} \).

Teacher's Note:
a) The printed key writes Ec = a+b/2; the correct value is (a + b)/2.
b) Modulation index is always between 0 and 1 for a properly modulated wave.

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% ? [2 Marks]

Answer:
1. Modulation index m_a = E_m/E_c, so E_m = m_a × E_c.
2. E_m = 0.80 × 20 V = 16 V.

Teacher's Note:
a) Convert the percentage into a decimal, 80% = 0.80, before multiplying.
b) The modulating peak voltage must stay below the carrier peak voltage for undistorted modulation.

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. [2 Marks]

Answer:
1. (i) Modulation index m_a = E_m/E_c = 8/20 = 0.4.
2. (ii) Side band frequencies are f_c + f_m and f_c - f_m.
3. Upper side band = 1000 kHz + 10 kHz = 1010 kHz.
4. Lower side band = 1000 kHz - 10 kHz = 990 kHz.

Teacher's Note:
a) Both side bands are equally spaced from the carrier by the message frequency.
b) Keep the units of frequency the same before adding, 1 MHz equals 1000 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 Marks]

Answer:
1. Condition for good detection: the reactance of the capacitor at the carrier frequency must be much less than R.
2. Reactance X_C = 1/(2πf_cC) = 1/(2π × 100 × 10³ × 10 × 10^-12) ≈ 1.59 × 10⁵ Ω = 159 kΩ.
3. Since 159 kΩ is much greater than R = 1 kΩ, the circuit is not a good detector.
4. For good detection, C must be much greater than 1/(2πf_cR) = 1/(2π × 100 × 10³ × 1 × 10³) ≈ 1.59 nF.
5. A suitable value is C = 1 μF, which is much greater than 1.59 nF.

Teacher's Note:
a) The printed key writes "C much greater than 1/1.59 x 10^-9", which is wrong; the threshold is 1/(2πfR), that is 1.59 nF.
b) The printed key writes "fared"; the unit is farad.
c) A larger capacitor passes the carrier easily while the resistor keeps the audio signal.

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

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

Here d₁=√2Rh₂ and d₂= √2Rh₂

For maximum range,

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

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.

Solution:

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

h’ = h + 21\100h = 1.21h

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

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