CBSE Class 12 Physics Communication Systems Assignment Set 01

Question. Why are microwaves used in radar systems?
Answer: Microwaves are utilized in radar because they have very short wavelengths, typically in the millimeter range. Having a small wavelength makes it possible to transmit a highly directional beam signal, which is essential for radar targeting and detection.
In simple words: Radar needs a sharp, directed beam, which is only possible with very short wavelengths. Since microwaves have tiny wavelengths of just a few millimeters, they are perfect for radar.

Exam Tip: Mention both the requirement of a directional beam and the millimeter-range wavelength of microwaves to score full marks.

Question. Why cannot sky waves be used for the transmission of television signals?
Answer: TV signals operate at high frequencies ranging from 100 to 200 MHz. The ionosphere is unable to reflect radio waves that exceed a frequency of approximately 40 MHz back to Earth. Consequently, sky wave propagation cannot be employed for transmitting television signals, as these waves pass right through the ionospheric layer.
In simple words: Television signals have very high frequencies (100 - 200 MHz). Since the ionosphere can only bounce back signals below 40 MHz, TV signals cannot travel as sky waves.

Exam Tip: Clearly state the frequency range of TV signals (100 - 200 MHz) and the critical limit of the ionosphere (around 40 MHz) to justify why sky waves fail.

Question. What characteristics should a diode detector possess for proper detection?
Answer: For efficient and accurate detection, a diode detector must satisfy the following requirements:
(a) It must exhibit high rectification efficiency.
(b) It should impose a negligible loading effect on the preceding stage of the circuit.
(c) It must introduce minimal distortion to the signal.
In simple words: A good diode detector needs to be highly efficient at converting AC to DC, must not overload the previous circuit, and should keep the signal clear without distorting it.

Exam Tip: List all three key traits - high efficiency, low loading effect, and minimal distortion - to obtain full credit in short-answer questions.

Question. What is the velocity factor of a cable, and how is it related to the dielectric constant of the medium?
Answer: The velocity factor (VF) of a cable represents the ratio of the speed of light within the cable's dielectric medium to its speed in a vacuum. As light or electromagnetic waves travel through a physical medium, their speed decreases. The velocity in a medium is calculated using the formula: \[ v = \frac{c}{\sqrt{k}} \] Here, \( c \) is the speed of light in vacuum (\( 3 \times 10^8 \text{ m/s} \)) and \( k \) is the dielectric constant of the medium. The velocity factor is given by: \[ \text{VF} = \frac{v}{c} = \frac{1}{\sqrt{k}} \] Typically, the velocity factor for transmission lines ranges from 0.6 to 0.9.
In simple words: The velocity factor is how fast a signal travels inside a cable compared to its speed in empty space. It gets slower because of the cable's material, and is usually between 60% and 90% of the speed of light.

Exam Tip: Remember the formula \( \text{VF} = \frac{1}{\sqrt{k}} \) and state that its value is always less than 1, typically between 0.6 and 0.9.

Question. Why is delta modulation considered convenient to use?
Answer: Delta modulation is highly convenient because it utilizes straightforward techniques for encoding and decoding pulses. In its simplest form, this method transmits only a single bit per sample (indicating either a positive or negative step, or a 'non-zero' change), making the overall system design simple and easy to implement.
In simple words: Delta modulation is easy to use because it sends only one bit of information for each sample to show if the signal went up or down.

Exam Tip: Highlight that delta modulation requires only one bit per sample, which significantly simplifies the transmitter and receiver design.

Question. Which transmission media are used for Audio Frequency (AF), Ultra High Frequency (UHF), and optical frequency ranges?
Answer: For transmitting signals in the audio frequency (AF) and ultra-high frequency (UHF) bands, parallel two-wire lines and coaxial cables are commonly used. On the other hand, optical fibers are used for transmitting signals at optical frequencies.
In simple words: Normal wires and coaxial cables are used for lower sound and radio frequencies, while glass optical fibers are used to carry light signals.

Exam Tip: Match the transmission medium correctly with its frequency range: two-wire/coaxial for AF/UHF, and optical fibers for light frequencies.

Question. How are two different frequencies mixed in communication systems?
Answer: The mixing of two different frequencies is achieved by passing the signals through a device that exhibits non-linear impedance, which generates new frequencies representing the sum and difference of the original signals.
In simple words: To mix two different frequencies together, we pass them through a special component whose resistance changes in a non-linear way.

Exam Tip: State that a non-linear device or impedance is essential to produce the sum and difference frequencies during mixing.

Question. What is population inversion in a laser medium, and why is it necessary?
Answer: Population inversion is a state where a greater number of atoms in the laser medium occupy a higher excited energy level compared to the lower ground state. Under normal conditions, more atoms reside in the ground state. This unnatural distribution is essential because it ensures that incident light stimulates emission rather than being absorbed, allowing for signal amplification.
In simple words: Normally, most atoms are at rest. For a laser to work, we must force more atoms into an excited, energetic state than the resting state. This active state is called population inversion.

Exam Tip: Define population inversion as \( N_2 > N_1 \) (more atoms in the excited state than ground state) and mention its role in achieving stimulated emission.

Question. What are the different types of lasers based on the state of the active medium?
Answer: Lasers can be classified into four primary categories depending on the nature of their medium:
(a) Solid-state lasers
(b) Semiconductor lasers
(c) Liquid lasers
(d) Gas lasers
In simple words: Lasers are grouped by what is inside them to make the light: solids, semiconductors, liquids, or gases.

Exam Tip: List all four types clearly when asked to classify lasers based on their medium.

Question. State some important applications of lasers.
Answer: Lasers are widely used in various fields, with key applications including:
(a) Precision medical procedures, such as laser surgery.
(b) High-speed printing technology, such as laser printing.
(c) Long-distance data transmission in optical communication systems.
In simple words: Lasers are used in medical operations, printing paper quickly, and sending data through optical fibers.

Exam Tip: Be ready with at least three diverse applications from medicine, daily technology, and communication.

Question. What are the three primary systems that constitute a telecommunication network?
Answer: A standard telecommunication network is comprised of three essential systems:
(1) Transmission Systems
(2) Switching Systems
(3) Signaling Systems
In simple words: A phone or data network needs three parts: one to carry the signal, one to connect the correct lines, and one to send control signals.

Exam Tip: Remember these three core systems - transmission, switching, and signaling - as they form the backbone of any communication network.