CBSE Class 12 Physics HOTs Communication Systems Set 03

Communication Systems

Question. Why cannot audio signals be transmitted directly into space without modulation?
Answer: Direct transmission of audio signals without modulation is not feasible due to three primary reasons:
(1) The required length of the transmitting antenna would be extremely large (e.g., around 5000 meters for audio frequencies), making its construction practically impossible.
(2) The electromagnetic energy radiated by an antenna at audio frequencies is virtually zero, leading to extremely poor transmission efficiency.
(3) If multiple stations broadcast audio signals directly, their frequencies would overlap, causing them to get completely and inseparably mixed.
In simple words: We cannot send sound signals directly because the antenna would need to be miles long, the signal would carry almost no energy, and all different stations would get jumbled up together.

Exam Tip: This is a highly important exam question. Always list these three factors: antenna height requirement, low radiation energy at low frequencies, and signal mixing/interference.

Question. What is pulse modulation, and how does it work?
Answer: Pulse modulation is a technique where continuous analog waveforms are sampled at periodic, regular intervals. Instead of transmitting the entire continuous wave, the signal's information is transmitted only at specific sampling moments, along with necessary synchronization pulses. This process transforms a continuous message into a series of discontinuous pulses using specialized modulation methods.
In simple words: Pulse modulation takes quick snapshots of a continuous signal at regular times and sends only those snapshots as a series of separate pulses.

Exam Tip: Define pulse modulation by emphasizing "sampling at regular intervals" and the transmission of "discontinuous pulses."

Question. What does the term "channel" mean in communication systems?
Answer: In communication systems, a channel refers to the specific range of frequencies allocated for a particular transmission from a broadcaster or transmitter. For instance, a telephone channel serves as the physical or frequency-domain link connecting a transmitter to a receiver.
In simple words: A channel is the specific pathway or range of frequencies set aside for a sender and a receiver to talk to each other.

Exam Tip: Define a channel as an allocated frequency band or physical medium that connects a transmitter and a receiver.

Question. Why does the electrical conductivity of the Earth's atmosphere increase with altitude?
Answer: Atmospheric pressure falls as altitude increases. Energetic cosmic rays and other radiations from space ionize gas atoms upon entering the upper atmosphere. As these rays descend toward the Earth's surface, their ionizing capability drops sharply because there are fewer collisions with gas atoms in the denser air near the ground. Consequently, the concentration of free ions and electrons is much higher at greater altitudes, which causes the electrical conductivity of the atmosphere to increase with height.
In simple words: High-energy space rays hit and ionize the thin air high up, creating lots of charged particles. As they get closer to the ground, they lose energy, so the air high above is a much better conductor of electricity than the air near us.

Exam Tip: Explain this by linking low atmospheric pressure at high altitudes to the higher rate of ionization caused by incoming cosmic rays.

Question. Explain the propagation of light through an optical fiber. Define numerical aperture and state how it depends on core diameter.
Answer: In an optical fiber, the refractive index of the core (\( \mu_1 \)) is designed to be slightly larger than that of the outer glass cladding (\( \mu_2 \)). Light travels through the fiber via a sequence of successive total internal reflections at the boundary between the core and cladding. For total internal reflection to happen, the light must enter the fiber core at an angle \( \theta \) that lies within the acceptance angle \( \theta_c \). The numerical aperture (NA) is defined as: \[ \text{NA} = \sin \theta_c = \sqrt{\mu_1^2 - \mu_2^2} \] The numerical aperture is also dependent on the diameter of the core; specifically, it decreases as the core diameter is reduced, and increases when the core diameter is made larger.
In simple words: An optical fiber has an inner core that bends light more than the outer cladding. This traps the light inside, making it bounce down the fiber. The numerical aperture shows how much light the fiber can take in, and this capacity gets smaller if the core is made thinner.

Exam Tip: Write down the expression \( \text{NA} = \sqrt{\mu_1^2 - \mu_2^2} \) clearly, and remember that \( \mu_1 > \mu_2 \) is the fundamental condition for total internal reflection inside the fiber.

Question. Define the following terms:
(i) Pulse Amplitude Modulation (PAM)
(ii) Pulse Position Modulation (PPM)
Answer:
(i) **Pulse Amplitude Modulation (PAM):** In this scheme, the amplitude of each individual pulse is varied in direct proportion to the instantaneous amplitude of the modulating (message) signal.
(ii) **Pulse Position Modulation (PPM):** In this scheme, the positions in time (i.e., the relative starting or arrival time of each pulse) are shifted in accordance with the amplitude of the modulating signal, while the amplitude and width of the pulses remain constant.
In simple words:
(i) Pulse Amplitude Modulation means changing how tall the pulses are to match the signal.
(ii) Pulse Position Modulation means shifting the timing of the pulses back and forth to match the signal.

Exam Tip: For PAM, the height of the pulse changes. For PPM, the timing of the pulse changes while its height and width stay the same.

Question. Why are coaxial cables not used for transmitting signals at extremely high microwave frequencies?
Answer: At extremely high frequencies (such as microwave ranges), the dielectric loss within the insulating material of a coaxial cable increases drastically, leading to severe signal attenuation and power loss.
In simple words: Coaxial cables are not used for ultra-high frequencies because the plastic insulation inside them starts absorbing too much of the signal's energy as heat.

Exam Tip: Mention "dielectric loss" as the main reason for high-frequency attenuation in coaxial cables.

Question. How does the effective power radiated by an antenna vary with wavelength?
Answer: The effective power radiated by an antenna is inversely proportional to the square of the wavelength (\( \lambda \)) of the transmitted signal: \[ P \propto \frac{1}{\lambda^2} \] Thus, shorter wavelengths (or higher frequencies) result in significantly greater radiated power.
In simple words: The power sent out by an antenna goes up quickly as the wavelength gets shorter.

Exam Tip: Remember that \( P \propto \frac{1}{\lambda^2} \). A smaller wavelength means more power is radiated into space.

Question. What should be the length of a dipole antenna for a carrier wave of frequency \( 5 \times 10^8 \text{ Hz} \)?
Answer: The length \( L \) of a dipole (half-wave) antenna is given by: \[ L = \frac{\lambda}{2} = \frac{c}{2\nu} \] Given: Frequency, \( \nu = 5 \times 10^8 \text{ Hz} \) Speed of light, \( c = 3 \times 10^8 \text{ m/s} \) Substituting these values: \[ L = \frac{3 \times 10^8}{2 \times 5 \times 10^8} \] \[ \implies L = \frac{3}{10} = 0.3 \text{ m} = 30 \text{ cm} \] Thus, the length of the dipole antenna should be 0.3 meters.
In simple words: To find the antenna length, we divide the speed of light by twice the frequency. For this frequency, the antenna needs to be 0.3 meters (or 30 centimeters) long.

Exam Tip: Use the standard relation \( L = \frac{\lambda}{2} \) for a half-wave dipole, and make sure to convert your final answer into standard units (meters or centimeters).

Question. If the coverage range of a transmission is to be doubled, how many times must the height of the transmitting antenna be increased?
Answer: The maximum range of transmission from an antenna of height \( h \) is given by: \[ d = \sqrt{2hR} \] If the new range is to be \( d' = 2d \), then: \[ d' = \sqrt{2h'R} \] \[ \implies 2d = \sqrt{2h'R} \] Squaring both sides: \[ 4d^2 = 2h'R \] Since \( d^2 = 2hR \): \[ 4(2hR) = 2h'R \implies h' = 4h \] Thus, the height of the antenna must be increased to four times its original height.
In simple words: To double the distance a signal can travel, we have to make the antenna four times taller.

Exam Tip: Remember that range \( d \propto \sqrt{h} \). Doubling the range always requires quadrupling the height of the antenna.

Question. A TV tower has a height of 100 m. How much population is covered by the TV broadcast if the average population density around the tower is \( 1000 \text{ km}^{-2} \)? (Take \( R_E = 6400 \text{ km} \))
Answer: Given: Height of the TV tower, \( h = 100 \text{ m} = 0.1 \text{ km} \) Radius of the Earth, \( R = 6400 \text{ km} \) Population density = \( 1000 \text{ km}^{-2} \) The transmission range \( d \) of the antenna is calculated as: \[ d = \sqrt{2hR} \] \[ d = \sqrt{2 \times 0.1 \times 6400} \] \[ \implies d = \sqrt{1280} \text{ km} \] The area covered by the broadcast is: \[ A = \pi d^2 = 3.14 \times 1280 = 3919.2 \text{ km}^2 \] The total population covered is: \[ \text{Population Covered} = A \times \text{Population Density} \] \[ \implies \text{Population Covered} = 3919.2 \times 1000 = 3,919,200 \] Thus, the broadcast covers a population of 3,919,200 people.
In simple words: The tower's signal covers a circular area of 3,919.2 square kilometers. Since there are 1,000 people per square kilometer, the broadcast reaches 3,919,200 people.

Exam Tip: Always make sure to write down step-by-step calculations for range, area, and population coverage to ensure you get full partial marks even if a calculation error occurs.