JEE Physics Communication System MCQs Set 03

Answer: (c) modulation

Explanation:
1. Varying the amplitude, frequency or phase of a high-frequency carrier according to the message signal is called modulation.

Teacher's Note:
1. Modulation lets low-frequency information ride on a high-frequency carrier.
2. Detection (demodulation) is the reverse process.

Answer: (c) three

Explanation:
1. A sinusoidal carrier has three characteristics that can be varied: amplitude, frequency and phase.
2. So three types of modulation are possible: amplitude (AM), frequency (FM) and phase (PM) modulation.

Teacher's Note:
1. Pulse modulation of digital signals is based on the same three ideas.
2. A carrier is described by A sin(ωt + φ).

Answer: (c) both the strength and the quality of the signal

Explanation:
1. Modulation index m = Am/Ac sets the sideband power, which is the information-carrying part, so it affects the strength of the transmitted signal.
2. Too small an m gives weak signals, while m > 1 distorts the signal, so it also determines the quality.

Teacher's Note:
1. Sideband power fraction is m²/(2 + m²).
2. The best choice is m close to 1 without exceeding it.

Answer: (b) should be less than 100%

Explanation:
1. If m exceeds 1 (over-modulation) the envelope of the AM wave no longer follows the signal and the information is distorted.
2. So the degree of modulation should not exceed 100%.

Teacher's Note:
1. Degree of modulation = m x 100%.
2. Ideal AM uses a modulation index between 0 and 1.

Answer: (c) \( f_c,\ f_c+f_s,\ f_c-f_s \)

Explanation:
1. Multiplying the carrier \( A_c\sin\omega_ct \) by the factor \( (1+m\sin\omega_mt) \) gives terms at the carrier frequency and at the sum and difference frequencies.
2. So the wave contains fc, fc + fs (upper sideband) and fc - fs (lower sideband).

Teacher's Note:
1. The product of two sines gives the sum and difference frequencies.
2. The bandwidth of an AM wave is 2fs.

Answer: (c) \( P_c=\frac{E_c^2}{2R} \)

Explanation:
1. Average power of a sinusoidal voltage of peak value Ec across R is \( P_c=\frac{(E_c/\sqrt2)^2}{R}=\frac{E_c^2}{2R} \).

Teacher's Note:
1. Use the rms value Ec/√2 for the power.
2. Total AM power is Pc(1 + m²/2).

Answer: (c) Small operating range

Explanation:
1. AM suffers from noise (it affects the amplitude), low efficiency (most power is in the carrier) and a small operating range.
2. Clear reception, high efficiency and good audio quality are not features of AM.

Teacher's Note:
1. FM is less noisy and gives better audio quality than AM.
2. AM transmitters waste power in the carrier.

Answer: (c) Amplitude modulation

Explanation:
1. Amplitude modulation is used for the standard (medium-wave and short-wave) radio broadcasts of All India Radio.
2. FM is used for the VHF band broadcasts and for TV sound.

Teacher's Note:
1. Radio broadcasts in the medium wave band, 540 to 1600 kHz, are AM.
2. FM broadcasts operate at 88 to 108 MHz.

Answer: (c) the transverse nature of e.m. waves

Explanation:
1. Polarisation means restricting the vibration to one direction perpendicular to the direction of propagation.
2. This is possible only for transverse waves, and electromagnetic waves are transverse.

Teacher's Note:
1. Longitudinal waves such as sound cannot be polarised.
2. Reflection and refraction occur for all waves.

Answer: (d) 7.5 × 10⁷ m/s

Explanation:
1. \( v=\frac{c}{\sqrt{\varepsilon_r\mu_r}}=\frac{3\times10^{8}}{\sqrt{16}}=7.5\times10^{7}\ \text{m/s} \).

Teacher's Note:
1. Refractive index n = √(εr μr).
2. Speed in the medium is c/n.

Answer: (a) 20 × 10¹² m⁻³

Explanation:
1. \( f_c=9\sqrt{N}\Rightarrow N=\left(\frac{f_c}{9}\right)^2=\left(\frac{40\times10^{6}}{9}\right)^2=1.98\times10^{13}\ \text{m}^{-3} \), i.e. about 20 x 10¹² m⁻³.

Teacher's Note:
1. The critical frequency is the highest frequency reflected for vertical incidence.
2. fc = 9 √Nmax with fc in Hz.

Answer: (a) 6 Re

Explanation:
1. A geostationary satellite has orbital radius about 42,000 km, i.e. a height of about 36,000 km above the surface.
2. With Re = 6400 km this is about 5.6 Re, which is close to 6 Re.

Teacher's Note:
1. The time period of such a satellite is 24 hours.
2. It stays above the same point on the equator.

Answer: (d) All of them

Explanation:
1. Ground (surface) waves are used for low frequencies, sky waves (ionospheric reflection) for medium and short waves, and space waves (line of sight) for VHF and above.
2. All three modes can carry radio waves from place to place.

Teacher's Note:
1. Each mode suits a different frequency range.
2. Choice of the mode depends on frequency and distance.

Answer: (b) 10⁴ Hz to 10¹¹ Hz

Explanation:
1. Radio and microwave communication in space uses frequencies from about 10 kHz to 100 GHz.
2. That is 10⁴ Hz to 10¹¹ Hz.

Teacher's Note:
1. Audio frequencies (below 20 kHz) are too low for radiation.
2. Above 10¹¹ Hz absorption by the atmosphere becomes severe.

Answer: (c) both of them

Explanation:
1. Ground waves are attenuated more at higher frequency and travel farther when the transmitter power is larger.
2. Hence the maximum range depends on both frequency and power.

Teacher's Note:
1. Ground wave range falls rapidly above about 1.5 MHz.
2. More power gives more range at a given frequency.

Answer: (c) UHF

Explanation:
1. Space (line-of-sight) waves are used for frequencies above about 30 MHz.
2. The range given in the question that is normally used is UHF (300 to 3000 MHz), which is typical for TV and mobile communication.

Teacher's Note:
1. Space wave transmission is line-of-sight.
2. Higher frequencies are better for space waves because they are not reflected by the ionosphere.

Answer: (c) modulation

Explanation:
1. Impressing the information signal on a carrier wave is modulation.
2. Detection is the recovery of the signal at the receiver.

Teacher's Note:
1. Modulation is done at the transmitter, detection at the receiver.
2. The carrier makes long-distance radiation possible.

Answer: (d) 50,000 cycles/s

Explanation:
1. The carrier frequency must be much larger than the signal frequency (at least about 10 times) so the carrier cycles outline the signal faithfully.
2. Among the options only 50,000 cycles/s satisfies this.

Teacher's Note:
1. fc must be much greater than fm.
2. A smaller or equal carrier cannot be modulated properly.

Answer: (d) 80 km

Explanation:
1. \( d=\sqrt{2Rh}=\sqrt{2\times6.4\times10^{6}\times500}=\sqrt{6.4\times10^{9}}=8\times10^{4}\ \text{m} \) = 80 km.

Teacher's Note:
1. The radar range on the ground is limited by line of sight to the horizon.
2. Power and frequency do not affect this geometric limit.

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

Explanation:
1. Modulation allows a high-frequency carrier to be used, which gives a smaller antenna, a smaller fractional bandwidth and the possibility of tuning to different stations (selectivity).
2. It does not change the speed at which the signal travels, so it does not reduce the time lag; statement (a) is not correct.

Teacher's Note:
1. Signals travel at the speed of light with or without modulation.
2. Higher carrier frequency means shorter wavelength and smaller antennas.

Answer: (d) 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. In AM the carrier amplitude follows the signal amplitude, which is statement (d).
2. In FM the carrier frequency (not its amplitude) varies with the signal amplitude, so (a), (b) and (c) are wrong.

Teacher's Note:
1. AM varies the amplitude; FM varies the frequency.
2. In FM the frequency deviation is proportional to the signal amplitude.

Answer: (b) ωm

Explanation:
1. The modulated wave contains the carrier ωc and the side frequencies ωc + ωm and ωc - ωm.
2. The signal frequency ωm itself is absent from the modulated wave.

Teacher's Note:
1. AM shifts the signal spectrum up around the carrier.
2. The audio frequency is recovered only after detection.

Answer: (d) 2 × 10⁵

Explanation:
1. Usable bandwidth \( =10\%\times10\times10^{9}=10^{9}\ \text{Hz} \).
2. \( \text{Number of channels}=\frac{10^{9}}{5\times10^{3}}=2\times10^{5} \).

Teacher's Note:
1. Divide the usable bandwidth by the bandwidth of one channel.
2. 10% of 10 GHz is 1 GHz.

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

Explanation:
1. Sky waves are reflected by the ionosphere, so they can travel very long distances. The assertion is true.
2. The ionosphere's electron density changes with time of day and season, so the reflection of sky waves changes and the signal fades. The reason is true, and it explains why sky waves are less stable than ground waves.

Teacher's Note:
1. Sky wave signals depend on ionospheric conditions, which change continuously.
2. Ground waves follow the surface and depend much less on atmospheric conditions.