JEE Physics Communication System MCQs Set 15

Answer: (c) \( [A_c+m(t)]\sin\omega_ct \)

Explanation:
1. In amplitude modulation the carrier amplitude is varied in step with the signal, so \( c_m(t)=[A_c+m(t)]\sin\omega_ct \).

Teacher's Note:
1. Only the amplitude of the carrier changes; its frequency stays \( \omega_c \).

Answer: (b) 4.2 MHz

Explanation:
1. A video (TV picture) signal needs a bandwidth of about 4.2 MHz.

Teacher's Note:
1. Audio needs about 20 kHz; video needs about 4.2 MHz.

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

Explanation:
1. Modulation allows practical antenna size, multiplexing and a smaller fractional bandwidth.

Teacher's Note:
1. Radio waves travel at the speed of light with or without modulation, so there is no time lag to reduce.

Answer: (c) 50%

Explanation:
1. \( m=\dfrac{E_m}{E_c}=\dfrac{10}{20}=0.5=50\% \).

Teacher's Note:
1. The modulation index is the ratio of the two amplitudes.

Answer: (a) 8 kHz

Explanation:
1. The channel width is twice the highest modulating frequency.

Teacher's Note:
1. \( 2\times4000=8000 \) Hz \( =8 \) kHz.

Answer: (b) 70.1%

Explanation:
1. \( I_t=I_c\sqrt{1+\dfrac{m^2}{2}} \Rightarrow \left(\dfrac{8.93}{8}\right)^2=1.246 \).
2. \( \dfrac{m^2}2=0.246\Rightarrow m=0.70 \), i.e. 70.1%.

Teacher's Note:
1. Power rises by the factor \( 1+m^2/2 \) on modulation.

Answer: (b) 88 to 108 MHz

Explanation:
1. Commercial FM broadcasting uses the VHF band of 88 to 108 MHz.

Teacher's Note:
1. AM broadcasting is in the kHz range (530 to 1710 kHz).

Answer: (a) frequency varies with time

Explanation:
1. In FM the instantaneous frequency of the carrier follows the signal while the amplitude stays constant.

Teacher's Note:
1. Constant amplitude is why FM is less noisy than AM.

Answer: (a) fidelity

Explanation:
1. Fidelity is the ability of a receiver to reproduce all signal frequencies equally.

Teacher's Note:
1. Sensitivity is the ability to receive weak signals.

Answer: (b) space waves

Explanation:
1. Line-of-sight links (TV, microwave relays) use space waves of very high frequency.

Teacher's Note:
1. Ground waves follow the earth; sky waves reflect from the ionosphere.

Answer: (d) microwave

Explanation:
1. Microwaves are used in satellite and radar communication.

Teacher's Note:
1. Radio and microwave bands carry most telecommunication.

Answer: (a) 80 km

Explanation:
1. \( d=\sqrt{2Rh}=\sqrt{2\times6.4\times10^6\times500}=\sqrt{6.4\times10^9} \) m.
2. \( d\approx8\times10^4 \) m \( =80 \) km.

Teacher's Note:
1. Range of line-of-sight communication is \( \sqrt{2Rh} \).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Sky waves are reflected from the ionosphere over long distances, so they are useful for long-range links.
2. The ionosphere changes with time, which makes sky wave signals less stable, so R explains A.

Teacher's Note:
1. Ground waves do not depend on the ionosphere.
2. Ionisation depends on solar radiation.

Answer: (b) 12.8 m

Explanation:
1. \( d^2=2Rh\Rightarrow h=\dfrac{(12.8)^2}{2\times6400}=0.0128 \) km.
2. \( h=12.8 \) m.

Teacher's Note:
1. Use the same units for \( d \) and \( R \).

Answer: (a) Ground wave propagation is for high frequency

Explanation:
1. Ground waves are attenuated quickly at high frequency, so ground wave propagation is used for low frequencies (up to a few MHz).

Teacher's Note:
1. Sky waves: 3 to 30 MHz; space waves: above 30 MHz.

Answer: (c) \( \lambda/4 \)

Explanation:
1. An efficient antenna must be at least of the order of \( \lambda/4 \).

Teacher's Note:
1. Long wavelengths need very large antennas, hence modulation.

Answer: (a) 60.288 lakh

Explanation:
1. Area covered \( =\pi d^2=2\pi Rh=2\pi\times6.4\times10^6\times150=6.03\times10^9 \) m\(^2\) \( =6029 \) km\(^2\).
2. Population \( =6029\times10^3\approx60.3 \) lakh.

Teacher's Note:
1. Area covered is \( \pi d^2 \) with \( d^2=2Rh \).

Answer: (a) \( \mu_1>\mu_2 \)

Explanation:
1. Total internal reflection at the core-cladding boundary needs the core to be denser, \( \mu_1>\mu_2 \).

Teacher's Note:
1. Light then stays in the core.

Answer: (b) between 50 \( \Omega \) and 70 \( \Omega \)

Explanation:
1. Coaxial cables for communication have a characteristic impedance of about 50 to 70 \( \Omega \).

Teacher's Note:
1. Typical values are 50 \( \Omega \) and 75 \( \Omega \).

Answer: (c) \( f_c=9\sqrt{N_{max}} \)

Explanation:
1. The critical frequency is \( f_c=9\sqrt{N_{max}} \) (Hz, with \( N \) in m\(^{-3}\)).

Teacher's Note:
1. \( f_c\propto\sqrt N \).