Communication systems practice · Mixed difficulty

Option C ($\frac{\lambda}{4}$) is correct because the minimum size of an antenna for efficient EM signal radiation is a quarter of the wavelength.

Option A (Plate-modulated class C amplifier) is the most likely output stage for a television transmitter due to its efficiency and ability to handle large RF power.

Option C (Electromagnetic waves of frequencies higher than about 30 MHz cannot penetrate the ionosphere) is correct because electromagnetic waves with frequencies above the critical frequency can penetrate the ionosphere.

Option B (Multi-mode step index fiber) is correct because multi-mode step index fibers are generally less expensive and simpler to construct.

Option B (1%) is correct because the typical difference in refractive index between the core and cladding in step-index fibers is around 1% to ensure effective total internal reflection.

Option A (${d^2}/2R$) is correct because it represents the relationship between the height of the antenna and the range of transmission.

Option C (100) is correct because the wavelength of the signal transmitted by the antenna is four times the height of the antenna.

Option A (990) is correct because the sidebands in amplitude modulation are the carrier frequency plus and minus the message signal frequency.

Option A (377 W) is correct because the characteristic impedance of free space is approximately 377 ohms.

Option D (6) is correct because the modulating index is the coefficient of the sine function inside the argument of the main sine function.

Option C (55 km) is correct because the distance up to which the broadcast can be received is given by $d = \sqrt{2Rh}$, where R is the radius of the Earth and h is the height of the tower.

Option B ($10^{-4} {~cm}$) is correct because the diameter of an optical fiber is typically around 125 micrometers, which is equal to $1.25 \times 10^{-2}$ mm or $1.25 \times 10^{-4}$ cm, approximately $10^{-4}$ cm.

Option D (Transmitter, receiver and communication channel) is correct because these are the three essential components for any communication system to function.

Option A (Its use avoids receiver complexity) is correct because AM receivers are simpler and cheaper to build compared to other modulation techniques like FM.

Option A (1490, 1510) is correct because the lower and upper sideband frequencies are calculated by subtracting and adding the modulating frequency to the carrier frequency, respectively.

Option D (All of the above) is correct because PAM, PPM, PDM, and PCM are all common pulse modulation systems used for analog signals.

Option A (7.143) is incorrect. The carrier swing is twice the frequency deviation, so it should be 2*50kHz = 100kHz. The question asks for the carrier swing, not the ratio of frequency deviation to modulation frequency. The correct answer should be 100kHz.

Option B is correct. The diameter of an optical fiber is typically around ${10^{-4}} m$ (100 micrometers).

Option D (Inversely proportional to the square perpendicular-distance from transmitter to antenna) is correct because the intensity of a wave decreases with the square of the distance.

The correct answer is C ($6.9 \times {10^8}\mathrm{m}{\mathrm{s}^{ - 1}}$). The phase velocity of an electromagnetic wave in a medium is calculated using the refractive index, which depends on the electron density and frequency.

Option C (Modulation) is correct because modulation is the process of superimposing a signal frequency onto a carrier wave.

Option B ($5.925-6.425 \mathrm{GHz}$) is correct because it is the standard uplink frequency band used in satellite communication.

Option C (2000) is correct because optical fibers can carry approximately 2000 telephonic messages at an instant with minimal intensity loss.

Option A (Superimposing the audio frequency signal over a carrier wave) is correct because modulation is the process of varying the characteristics of a carrier wave in accordance with the instantaneous amplitude of the message signal.

Option D is correct because audio signals have low frequencies, requiring impractically large antenna lengths for efficient transmission.

Option B ($\omega_c, \omega_c+\omega_m$ and $\omega_c-\omega_m$) is correct because amplitude modulation generates the carrier frequency and sideband frequencies.

Option D (800-950 MHz) is correct because mobile phones typically operate within this frequency range to facilitate wireless communication.

Option B (${h^{1/2}}$) is correct because the range R is proportional to the square root of the transmitter height h, as given by the formula $R = \sqrt{2hr}$.

Option B is correct because optical fibers are not susceptible to electromagnetic interference due to the use of light for signal transmission.

Option D ($104 \, \mathrm{km}$) is correct because it is calculated using the formula for the maximum distance between antennas in line of sight.