JEE Physics Communication System MCQs Set 04

Answer: (d) \( \frac12 \)

Explanation:
1. Modulation index \( \mu=\frac{A_m}{A_c} \).
2. \( \mu=\frac12 \) since the message amplitude is 1 V and the carrier amplitude is 2 V.

Teacher's Note:
1. Read the peak (not peak-to-peak) amplitude of each wave.
2. Modulation index above 1 causes distortion (over-modulation).

Answer: (c) 10 kHz

Explanation:
1. In AM the sidebands are at \( f_c\pm f_m \).
2. Bandwidth \( =2f_m=2\times5=10\ \text{kHz} \).

Teacher's Note:
1. The carrier frequency does not affect the bandwidth.
2. Bandwidth is the difference between the highest and lowest frequency present.

Answer: (a) A-II, B-I, C-IV, D-III

Explanation:
1. AM broadcast uses 540-1600 kHz (II).
2. FM broadcast uses 88-108 MHz (I).
3. Television uses 54-590 MHz (IV).
4. Satellite communication uses the GHz range, 3.7-4.2 GHz (III).

Teacher's Note:
1. Higher frequency bands carry more bandwidth.
2. FM is transmitted in VHF, far above the medium-wave AM band.

Answer: (d) 64 km

Explanation:
1. Maximum line of sight \( d=\sqrt{2Rh_T}+\sqrt{2Rh_R}=2\sqrt{2Rh} \).
2. \( \sqrt{2\times6.4\times10^6\times80}=\sqrt{1.024\times10^9}=32\ \text{km} \), so \( d=2\times32=64\ \text{km} \).

Teacher's Note:
1. Each antenna contributes its own horizon distance.
2. Add the two distances because both antennas have the same height.

Answer: (b) 0.4

Explanation:
1. \( \mu=\frac{V_{max}-V_{min}}{V_{max}+V_{min}}=\frac{14-6}{14+6}=\frac{8}{20}=0.4 \).

Teacher's Note:
1. The envelope maximum is Ac + Am and the minimum is Ac - Am.
2. μ is always the ratio Am/Ac.

Answer: (c) A-IV, B-III, C-II, D-I

Explanation:
1. Attenuation is loss of signal strength (IV).
2. A transducer converts energy from one form into another (III).
3. Demodulation recovers the message from the carrier (II).
4. A repeater is a receiver plus a transmitter (I).

Teacher's Note:
1. These are standard definitions from communication systems.
2. A repeater regenerates a weakened signal to extend range.

Answer: (b) A, D and E only

Explanation:
1. Carrier frequency \( f_c=\frac{1000\pi}{2\pi}=500\ \text{Hz} \) and message frequency \( f_m=\frac{4\pi}{2\pi}=2\ \text{Hz} \).
2. An AM wave contains \( f_c \), \( f_c+f_m=502 \) Hz and \( f_c-f_m=498 \) Hz: A, D and E.

Teacher's Note:
1. The message frequency itself is not present in the modulated wave.
2. The wave has a carrier and two sidebands.

Answer: (d) Statement I is correct but Statement II is incorrect

Explanation:
1. Statement I is correct: an effective antenna needs a length of about \( \lambda/4 \) or more.
2. Statement II is incorrect: in AM the amplitude of the carrier varies in proportion to the message signal (in FM the amplitude stays constant).

Teacher's Note:
1. Amplitude modulation changes the amplitude, frequency modulation changes the frequency.
2. Low-frequency signals need huge antennas, so they are modulated onto high-frequency carriers.

Answer: (b) 64 MHz

Explanation:
1. The FM broadcast band is 88-108 MHz.
2. 64 MHz lies outside this band.

Teacher's Note:
1. Know the standard bands: AM 540-1600 kHz, FM 88-108 MHz.
2. Television bands lie between 54 MHz and 590 MHz.

Answer: (c) 2 : 1

Explanation:
1. \( \mu=\frac{A_m}{A_c}\Rightarrow\mu_1=\frac XY,\ \mu_2=\frac X{2Y} \).
2. \( \mu_1:\mu_2=2:1 \).

Teacher's Note:
1. Modulation index is inversely proportional to the carrier amplitude for a fixed message.
2. Doubling the carrier amplitude halves μ.

Answer: (a) 10 V

Explanation:
1. \( A_c+A_m=120,\ A_c-A_m=80\Rightarrow A_c=100\ \text{V},\ A_m=20\ \text{V} \).
2. Each sideband has amplitude \( \frac{\mu A_c}{2}=\frac{A_m}{2}=10\ \text{V} \).

Teacher's Note:
1. The envelope maximum is Ac + Am and the minimum is Ac - Am.
2. The modulated wave has carrier plus two sidebands, each of amplitude μAc/2.