JEE Physics Communication System MCQs Set 08

Answer: (d) should be less than (or at most equal to) 100%

Explanation:
1. When the modulation factor exceeds 1 the envelope of the wave no longer follows the message (over-modulation), and the message is distorted.
2. So the degree of modulation is kept below 100%.

Teacher's Note:
1. Over-modulation makes the carrier amplitude cross zero and the envelope is no longer a copy of the signal.
2. The undistorted limit is \( m_a\le 1 \).

Answer: (c) small operating range

Explanation:
1. Amplitude modulation suffers from noisy reception, low efficiency, small operating range and poor audio quality.
2. Of the options, only the small operating range is a limitation; the others are advantages.

Teacher's Note:
1. Noise changes amplitude, so AM picks it up easily.
2. Most of the power is in the carrier, which carries no information, hence the low efficiency.

Answer: (d) Troposphere

Explanation:
1. In the troposphere (the lowest layer) the temperature falls steadily with height.
2. In the stratosphere it rises (ozone absorbs ultraviolet), and in the ionosphere it rises again.

Teacher's Note:
1. Weather occurs in the troposphere, where air cools with altitude (about 6.5 degrees per km).
2. The ionosphere starts at roughly 65 km and is the layer used for sky-wave reflection.

Answer: (c) \( Q=\frac{\omega L}{R} \)

Explanation:
1. Q is the ratio of the reactance of the inductor at resonance to the resistance: \( Q=\frac{\omega L}{R} \).

Teacher's Note:
1. A high Q means a sharp resonance, which is needed for good selectivity.
2. Check the units: \( \omega L \) and R are both in ohms, so Q is dimensionless; L/R has the unit of time.

Answer: (a) Amplifier

Explanation:
1. A transducer converts one form of energy into another (sound to electrical, electrical to sound, light to electrical).
2. An amplifier only increases the strength of an electrical signal and does not change its form of energy, so it is not a transducer.

Teacher's Note:
1. The microphone converts sound into an electrical signal; the loudspeaker does the reverse.
2. An amplifier takes energy from a DC supply to increase the signal amplitude.

Answer: (b) one carrier and two side-band frequencies

Explanation:
1. For a single-tone message the AM wave has the carrier plus an upper and a lower side band.

Teacher's Note:
1. The carrier is \( f_c \) and the side bands are at \( f_c\pm f_m \).
2. A message made of many tones produces many side-band pairs.

Answer: (d) 1490 and 1510

Explanation:
1. \( f_{LSB}=f_c-f_m=1500-10=1490\ \text{kHz} \), \( f_{USB}=f_c+f_m=1510\ \text{kHz} \).
2. The percentage of modulation and the amplitude do not affect the side-band frequencies.

Teacher's Note:
1. Side bands are always at \( f_c\pm f_m \).
2. The modulation index fixes only the side-band amplitudes.

Answer: (c) twice the audio (signal) frequency

Explanation:
1. \( BW=(f_c+f_m)-(f_c-f_m)=2f_m \).

Teacher's Note:
1. Only the two side bands determine the bandwidth, and the carrier does not enter into it.
2. This is why radio channels are spaced by about 10 kHz for 5 kHz audio.

Answer: (d) 10

Explanation:
1. \( BW=2f_m=2\times5=10\ \text{kHz} \).

Teacher's Note:
1. Bandwidth depends only on the modulating frequency.
2. The carrier frequency (100 kHz) is a distractor.

Answer: (a) 9.72

Explanation:
1. \( P_t=P_c\left(1+\frac{m^2}{2}\right)=9\left(1+\frac{0.16}{2}\right)=9\times1.08=9.72\ \text{kW} \)

Teacher's Note:
1. Unmodulated power is the carrier power.
2. Total power = carrier power times (1 + m²/2).

Answer: (d) 8

Explanation:
1. The channel width is twice the highest modulating frequency: \( 2\times4000=8000\ \text{Hz}=8\ \text{kHz} \).

Teacher's Note:
1. Bandwidth is set by the highest signal frequency.
2. The lowest frequency (200 Hz) and the carrier frequency do not matter.

Answer: (a) 1.28

Explanation:
1. \( d=\sqrt{2Rh} \)
2. \( A=\pi d^2=2\pi Rh=2\pi(6.4\times10^6)(100)=1.28\pi\times10^{9}\ \text{m}^2=1.28\pi\times10^3\ \text{km}^2 \)
3. So n = 1.28.

Teacher's Note:
1. The coverage is a circle of radius d, the radio horizon.
2. Convert m² to km² by dividing by 10⁶.

Answer: (b) 12.8 m

Explanation:
1. \( d=\sqrt{2Rh}\Rightarrow h=\frac{d^2}{2R}=\frac{(12.8\times10^3)^2}{2\times6.4\times10^6}=12.8\ \text{m} \)

Teacher's Note:
1. Work in metres: d = 12800 m, R = 6.4 × 10⁶ m.
2. Height grows as the square of the range.

Answer: (d) to reduce the time lag between transmission and reception of the information signal

Explanation:
1. Modulation raises the frequency, so a short antenna suffices, the fractional bandwidth is small and selectivity is easier.
2. It does not change the time lag: the signal still travels at the speed of light.

Teacher's Note:
1. Time lag depends only on distance and the speed of the wave.
2. Antenna size, bandwidth and selectivity all improve with a high carrier frequency.

Answer: (b) \( C_m(t)=A(1+\sin\omega_m t)\sin\omega_c t \) and 1

Explanation:
1. \( C_m=(A_c+A_m\sin\omega_m t)\sin\omega_c t=A\,(1+\sin\omega_m t)\sin\omega_c t \)
2. The modulation index is \( A_m/A_c=A/A=1 \).

Teacher's Note:
1. The message sits inside the bracket (it multiplies the carrier amplitude) and the carrier is the outside sine.
2. Equal amplitudes mean 100% modulation.

Answer: (a) In amplitude modulation the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

Explanation:
1. AM varies the carrier amplitude with the message amplitude.
2. FM varies the carrier frequency with the message amplitude, so the other three statements are incorrect.

Teacher's Note:
1. Mnemonic: the name of the modulation is the carrier parameter that changes.
2. In FM the amplitude of the carrier stays constant.

Answer: (b) \( \omega_m \)

Explanation:
1. The modulated wave contains \( \omega_c \) and the side bands \( \omega_c\pm\omega_m \).
2. The signal frequency \( \omega_m \) itself is not present.

Teacher's Note:
1. The baseband signal appears only after detection.
2. Only the sum and difference frequencies appear after multiplication.

Answer: (c) \( 2\times10^{5} \)

Explanation:
1. Available bandwidth = 10% of 10 GHz = \( 10^{9}\ \text{Hz} \).
2. \( N=\frac{10^9}{5\times10^3}=2\times10^5 \)

Teacher's Note:
1. Divide the usable bandwidth by the bandwidth of one channel.
2. Check units: 10 GHz is 10¹⁰ Hz.

Answer: (b) 2000 kHz

Explanation:
1. Each station occupies a band of \( 2f_m=500\ \text{kHz} \), so the carriers must differ by at least 500 kHz.
2. Allowed frequencies are 2000 kHz or below, or 3000 kHz or above. Of the options only 2000 kHz qualifies; 2750, 2250 and 2900 kHz are too close to 2500 kHz.

Teacher's Note:
1. Two stations overlap if their carriers are closer than the bandwidth.
2. Find the carrier first: 250 kHz is 10% of 2500 kHz.

Answer: (c) 0.6

Explanation:
1. \( m=\frac{V_{max}-V_{min}}{V_{max}+V_{min}}=\frac{160-40}{160+40}=0.6 \).

Teacher's Note:
1. Alternatively, \( A_m=160-100=60\ \text{V} \) and \( m=60/100 \).
2. Both approaches give the same answer.

Answer: (d) 1500 nm

Explanation:
1. Silica fibre has its lowest loss near 1300 nm to 1550 nm, so modern optical-fibre networks use infrared light of about 1500 nm.

Teacher's Note:
1. Loss in glass is minimum in the near infrared.
2. 600 nm and 900 nm lose much more, and 2400 nm is absorbed strongly in silica.

Answer: (b) \( 0.25;\ 2\times10^{8}\ \text{Hz} \)

Explanation:
1. \( m=\frac{100}{400}=0.25 \).
2. Side bands are at \( f_c\pm f_m \), so their difference is \( 2f_m=2\times10^{8}\ \text{Hz} \).

Teacher's Note:
1. The modulation index is message amplitude divided by carrier amplitude, which cannot be 4 for undistorted AM.
2. The difference of the side bands equals the bandwidth, 2fm.