Samacheer Kalvi Class 12 Physics Solutions Chapter 10 Communication Systems

Part - I:

Textbook Evaluation:

I. Multiple Choice Questions:

 

Question 1. The output transducer of the communication system converts the radio signal into
(a) Sound
(b) Mechanical energy
(c) Kinetic energy
(d) None of the options
Answer: (a) Sound
In simple words: The final part of a communication system turns the electronic radio signal back into something we can understand, like sound. This lets us hear the message.

🎯 Exam Tip: Remember that transducers convert one form of energy into another; here, the output transducer changes electrical energy back into audible sound.

 

Question 2. The signal is affected by noise in a communication system
(a) At the transmitter
(b) At the modulator
(c) In the channel
(d) At the receiver
Answer: (c) In the channel
In simple words: Noise, which is unwanted disturbance, usually gets added to the signal while it travels through the channel. This channel is the path the signal takes from sender to receiver.

🎯 Exam Tip: Understanding where noise typically enters helps in designing systems to reduce its impact, often focusing on the transmission medium itself.

 

Question 3. The variation of frequency of carrier wave with respect to the amplitude of the modulating signal is called
(a) Amplitude modulation
(b) Frequency modulation
(c) Phase modulation
(d) Pulse width modulation
Answer: (b) Frequency modulation
In simple words: When the loudness (amplitude) of your message signal changes the speed (frequency) of the main carrier wave, this process is called frequency modulation (FM). It's like changing how fast a sound repeats to carry information.

🎯 Exam Tip: Clearly differentiate between amplitude, frequency, and phase modulation by focusing on which property of the carrier wave is being varied by the modulating signal.

 

Question 4. The internationally accepted frequency deviation for the purpose of FM broadcasts is
(a) 75 kHz
(b) 68 kHz
(c) 80 kHz
(d) 70 kHz
Answer: (a) 75 kHz
In simple words: For FM radio stations to broadcast clearly, there is a standard maximum change allowed in their signal's frequency, which is 75 kilohertz. This standard helps all radio receivers work correctly.

🎯 Exam Tip: Specific numerical values like international standards for frequency deviation are important to remember for factual recall questions.

 

Question 5. The frequency range of 3 MHz to 30 MHz is used for
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer: (c) Sky wave propagation
In simple words: Radio waves in the range of 3 to 30 Megahertz are typically sent up into the sky, where they bounce off a layer in the atmosphere called the ionosphere. This bouncing helps them travel long distances around the Earth.

🎯 Exam Tip: Associate different frequency ranges with their corresponding propagation modes (ground, sky, space wave) as this is a common area for questions.

II. Short Answers Questions:

 

Question 1. Give the factors that are responsible for transmission impairments.
Answer: The factors that cause problems or reduce quality during signal transmission are called impairments. These include:
Attenuation: This is when the strength of the signal gets weaker as it travels a longer distance. This happens because the signal loses energy over time.
Distortion: This means the shape of the signal changes or gets altered during transmission. The signal might not look the same as when it was sent.
Noise: This refers to unwanted or random signals that mix with the original sound or information. Noise makes the signal less clear and harder to understand.
In simple words: Transmission gets worse due to three main reasons: attenuation (signal gets weaker), distortion (signal shape changes), and noise (unwanted sounds mix in).

🎯 Exam Tip: When listing factors, provide a clear, concise definition or explanation for each to demonstrate a full understanding.

 

Question 2. Distinguish between wireline and wireless communication? Specify the range of electromagnetic waves in which it is used.
Answer: Here is a comparison between wireline and wireless communication:

The specific range of electromagnetic waves used depends on the wireless communication type. For example, radio waves are used for mobile communication, and microwaves are used for satellite communication.
In simple words: Wireline communication uses physical cables to send signals, while wireless communication sends signals through the air. Wireless works over longer distances using different radio waves.

🎯 Exam Tip: For distinction questions, using a table format helps in clearly presenting the differences and makes your answer easier to read and score.

 

Question 3. Explain centre frequency or resting frequency in frequency modulation.
Answer: In frequency modulation (FM), when there is no input signal (meaning the baseband signal's frequency is zero), the carrier wave's frequency does not change. It stays at its normal, unmodulated frequency. This specific normal frequency is called the centre frequency or resting frequency. It is the base frequency from which the carrier wave will deviate when modulated.
In simple words: The centre frequency is the normal speed of the carrier wave when no information is being added. It's like the quiet, unchanging hum before you start speaking into a microphone.

🎯 Exam Tip: Clearly define "baseband signal" and "carrier wave" when explaining modulation concepts to show a complete understanding of the terms.

 

Question 4. What does RADAR stand for?
Answer: RADAR stands for Radio Detection and Ranging. It is a system that uses radio waves to find and track distant objects. RADAR can sense, detect, and locate objects like aircraft, ships, and spacecraft by sending out radio waves and listening for the echoes.
In simple words: RADAR means Radio Detection and Ranging. It uses radio waves to find out where faraway things like planes or ships are.

🎯 Exam Tip: For acronym questions, always write out the full form first and then add a brief explanation of its function or purpose.

 

Question 5. What do you mean by the Internet of Things?
Answer: The Internet of Things (IoT) is a network that connects many different devices, allowing them to be controlled from a single device, like a mobile phone. This means everyday objects can communicate with each other and with you over the internet. A simple example is using your phone to turn on lights or adjust the temperature at home, which is called home automation.
In simple words: The Internet of Things (IoT) connects many devices to the internet so you can control them from one place. For example, you can switch on your home lights using your phone.

🎯 Exam Tip: When defining a concept like IoT, provide a clear explanation and a relevant, easy-to-understand example to illustrate its practical application.

III. Long Answers Questions:

 

Question 1. What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation: For signals to travel long distances, low-frequency information (like your voice) needs to be combined with a high-frequency radio signal. This process is called modulation, where the properties of the high-frequency carrier wave are changed according to the low-frequency input signal. It's like putting your small message on a fast, strong wave to carry it far away.
Types of Modulation:
i) Amplitude Modulation (AM):
1. In AM, the strength (amplitude) of the carrier signal is changed based on the immediate strength (amplitude) of the information signal.
2. The speed (frequency) and direction (phase) of the carrier signal stay the same.
3. AM is commonly used in radio and TV broadcasting.



Advantages of AM:
1. It's easy to transmit and receive AM signals.
2. It needs less bandwidth (frequency space) to send information.
3. The cost of setting up AM systems is low.
Disadvantages of AM:
1. AM signals often have a high level of noise, making them less clear.
2. The efficiency is low because a lot of power is used for the carrier wave itself, not just the information.
3. AM has a smaller operating range compared to other types of modulation.
ii) Frequency Modulation (FM):
1. In FM, the speed (frequency) of the carrier signal is changed based on the immediate strength (amplitude) of the information signal.
2. The strength (amplitude) and direction (phase) of the carrier signal stay the same.
3. If the amplitude of the information signal gets bigger, the carrier signal's frequency also increases.
4. Stronger signals create areas where waves are closer together (compressions), and weaker signals create areas where waves are farther apart (rarefactions).
5. When the information signal's frequency is zero (no input), the carrier wave's frequency does not change. This is called the centre frequency or resting frequency.



Advantages of FM:
1. There is a large reduction in noise, and the signal-to-noise ratio improves.
2. The operating range for FM is quite large.
3. FM transmission is very efficient.
4. FM signals have better quality, especially for music.
Disadvantages of FM:
1. FM requires a wider channel (more frequency space) for transmission.
2. It covers less area compared to AM broadcasting.
3. The equipment for FM transmitters and receivers is more complex and expensive.
iii) Phase Modulation (PM):
1. In PM, the immediate strength (amplitude) of the information signal changes the direction (phase) of the carrier signal. The strength and speed (frequency) of the carrier signal stay the same.
2. The amount of frequency shift in PM depends on both the strength and the speed of the signal.
3. When the information signal becomes positive, the carrier wave's phase shifts forward (phase lead) by an amount related to the information signal's amplitude.
4. The carrier signal can appear compressed or its frequency can seem to increase during these phase shifts.
5. When the information signal becomes negative, the carrier wave's phase shifts backward (phase lag).
6. Like FM, a phase-modulated wave also has areas of compression and rarefaction.
7. When the information signal voltage is zero, the carrier frequency remains unchanged.

Advantages:
1. The FM signal created from PM is very steady and reliable.
2. The centre frequency, which is also called the resting frequency, stays stable.
In simple words: Modulation is changing a fast radio wave to carry a slower message, like your voice. Amplitude Modulation (AM) changes the wave's strength. Frequency Modulation (FM) changes the wave's speed. Phase Modulation (PM) changes the wave's starting point. Each has pros and cons like how clear the sound is or how far it can travel.

🎯 Exam Tip: When explaining types of modulation, always include what changes (amplitude, frequency, phase) and what remains constant for the carrier wave, and illustrate with clear diagrams.

 

Question 2. Elaborate on the basic elements of a communication system with the necessary block diagram.
Answer: A communication system has several basic parts that work together to send information from one place to another. Here are the main elements:
1. Information (Baseband or input signal):
i. This is the original message or data, which could be sound (like speech or music), pictures, or computer data.
ii. This information is given to the input transducer.
2. Input transducer:
i. This device changes physical changes (like pressure, temperature, or sound) into electrical signals, or vice versa. For example, a microphone turns sound waves into electrical signals.
ii. It converts the original information into corresponding electrical signals.
iii. The electrical form of the original information is called the baseband signal.

3. Transmitter: This part is usually found at a broadcasting station and prepares the signal for sending.
a) Amplifier: The very weak signal from the transducer is made stronger here.
b) Oscillator: It creates a high-frequency carrier wave that is sent into space for long-distance travel. This wave has a lot of energy because its energy is proportional to its frequency.
c) Modulator: This component combines the input (baseband) signal with the carrier signal, creating the modulated signal.
d) Power amplifier: After modulation, this part increases the power of the electrical signal so it can travel a great distance.
4. Transmitting antenna:
i. This is where the electrical signals are converted into electromagnetic waves. These waves then travel through the air at the speed of light.
ii. Antennas are essential for sending signals over long distances.
iii. Examples include antennas for mobile phones, radio, TV broadcasting, and satellite communication.
5. Communication channel: This is the path through which the electrical signal travels from the transmitter to the receiver. The goal is to carry the signal with as little noise or distortion as possible.
Wireline Communication:
i. Uses physical mediums like wires, cables, and optical fibers.
ii. It cannot be used for very long-distance transmission.
Wireless Communication:
i. Uses free space (air) as the medium.
ii. Signals travel as electromagnetic waves.
6. Noise:
i. This refers to unwanted electrical signals that interfere with the transmitted signal.
ii. Noise reduces the quality of the signal.
iii. It can be caused by natural sources (like lightning) or human-made sources.
iv. While noise cannot be completely removed, its effects can be reduced.
7. Receiver: This part collects the transmitted signals.
i. The transmitted signals are fed into the receiver.
ii. It includes a demodulator, amplifier, and detector.
iii. The demodulator separates the original baseband signal from the carrier wave.
iv. The separated signal is then amplified and sent to the output transducer.
8. Repeaters:
i. Repeaters are a combination of a transmitter and a receiver.
ii. They are used to increase the range or distance over which signals can be sent.
iii. They receive signals, make them stronger, and then retransmit them, often on a different frequency, to their destination.
iv. A communication satellite in space acts as a repeater.
9. Output transducer:
i. This device converts the electrical signal back into its original form, such as sound (for loudspeakers), images (for picture tubes), or data (for computer monitors).
ii. Examples include loudspeakers, picture tubes, and computer monitors.
10. Attenuation:
The loss of strength of a signal as it travels through a medium.
11. Range:
i. This is the maximum distance between the source (transmitter) and the destination (receiver).
ii. Within this range, the signal is received with enough strength to be useful.
In simple words: A communication system has many parts. First, an input device turns your message into an electrical signal. A transmitter then gets this signal ready, boosts it, and puts it on a fast carrier wave. This combined signal travels through a channel (like air or cables). A receiver then catches it, takes your message off the carrier wave, and an output device turns it back into something you can understand.

🎯 Exam Tip: When drawing block diagrams, ensure all key components are present, correctly labelled, and connected with directional arrows to show the flow of information.

 

Question 3. Explain the three modes of propagation of electromagnetic waves through space.
Answer: Electromagnetic waves travel through space in three main ways, depending on their frequency:
1. Ground Wave Propagation (or) Surface Wave Propagation:
i. In this mode, electromagnetic waves travel along the Earth's surface to reach the receiver. They essentially "glide" over the ground.
ii. Both the sending and receiving antennas must be close to the Earth's surface.
iii. The size of the antenna plays a role in how well the signals are sent.
iv. Ground waves lose strength (attenuate) due to several reasons:
a) Increasing distance: The signal becomes weaker as it travels further. This weakening depends on:
(i) The power of the transmitter.
(ii) The frequency of the transmitter.
(iii) The condition of the Earth's surface.
b) Absorption of energy by the Earth: When the transmitted signal touches the Earth, it creates electrical charges in the ground, which then form a current. This uses up some of the signal's energy.
ii. The Earth can act like a leaky capacitor, which also causes the wave to lose strength.
c) Tilting of the wave:
i. As the wave travels along the Earth's curve, its front gradually tilts.
ii. When this tilt increases, the electrical field strength of the wave goes down.
iii. Eventually, the surface wave dies out due to this energy loss.
iv. Ground waves mostly have frequencies below 2 MHz.
v. This mode is used for local broadcasting, radio navigation, and communication between ships or from ship-to-shore, as well as mobile communication.
2. Skywave Propagation (or) Ionospheric Propagation:
i. Electromagnetic waves are sent upwards from an antenna at high angles. They then bounce off a charged layer in the upper atmosphere called the ionosphere and come back down to Earth.
ii. The ionosphere is full of charged particles created by the absorption of ultraviolet rays, cosmic rays, and other high-energy radiation from the sun.
iii. These charged ions act like a mirror, allowing radio waves to be reflected.
iv. This bending of radio waves back to Earth is caused by a process similar to total internal reflection.
v. The shortest distance along the ground between the transmitter and the point where the skywave is first received is called the skip distance.
vi. If the angle at which the waves are sent increases, the reception of ground waves decreases.
vii. A "skip zone" or "skip area" is a region where electromagnetic waves cannot be received by either ground wave or skywave propagation.
viii. This propagation mode typically uses a frequency range of 3 to 30 MHz.
ix. It is used for shortwave broadcast services, allowing signals to travel very long distances.
x. Extremely long-distance communication is possible with skywaves.
xi. Even a single reflection can help a signal travel approximately 4000 km.
3. Space Wave Propagation:
i. This mode involves sending and receiving information signals directly through space.
ii. It uses electromagnetic waves with very high frequencies, typically above 30 MHz.
iii. The waves travel in a straight line from the transmitter to the receiver.
iv. For this to work, the transmission towers must be tall to ensure a clear line of sight.
v. At these high frequencies, the signals do not bend around the Earth's curvature.
vi. Space waves travel with less weakening (attenuation) and less loss of signal strength.
vii. This mode is used for TV broadcasts, satellite communication, and RADAR systems.
viii. Space wave propagation offers advantages like larger bandwidth, high data rates, smaller antenna sizes, and low power consumption.
ix. The range (or distance, \( d \)) depends on the height of the antenna \( (h) \). The formula for this range is given by \( d = \sqrt{2Rh} \), where \( R \) is the radius of the Earth.
x. For example, with \( R = 6400 \mathrm{~km} \), the calculation for range involves this formula.
In simple words: Electromagnetic waves travel in three ways. Ground waves hug the Earth's surface for local areas. Skywaves bounce off the atmosphere's ionosphere, covering long distances. Space waves travel in straight lines between tall antennas, used for very high frequencies and satellite links.

🎯 Exam Tip: For each mode of propagation, clearly state the frequency range, how the waves travel, and practical applications. Mentioning the ionosphere's role in skywave propagation is key.

 

Question 4. What do you know about GPS? Write a few applications of GPS.
Answer:
GPS (Global Positioning System):
1. GPS stands for Global Positioning System. It is a satellite-based system that provides exact location and time information anywhere on or near Earth.
2. It works with the help of a network of satellites orbiting Earth.
3. GPS broadcasts a precise radio signal, similar to an ordinary radio signal.
4. GPS software translates the location data received from these signals.
5. This software identifies the satellites, their locations, and how long it took for the signals to travel from each satellite.
6. By processing data from multiple satellites, the receiver can accurately determine its own location.
Applications of GPS:
1. Fleet Vehicle Management: GPS is used to track cars, trucks, and buses, helping manage fleets efficiently.
2. Wildlife Management: It assists in counting and monitoring wild animals.
3. Engineering: Used in big projects like building tunnels and bridges for precise measurements.
4. Fisheries: Helps locate fishing zones.
5. Military and Navigation Systems: Crucial for guiding military operations and for navigation on ships and aircraft.
6. Meteorological Observation: Used to measure things like rainfall rates and wind speeds.
7. Emergency Situations: Helps to find and rescue people in difficult situations.
In simple words: GPS, or Global Positioning System, uses satellites to tell you exactly where you are on Earth and what time it is. It's used for tracking cars, helping animals, building things, and even finding fishing spots.

🎯 Exam Tip: When describing GPS, ensure you explain its core function (location and time) and then provide a diverse set of applications from different fields, showing its broad utility.

Question 5. Give the applications of ICT is mining and agriculture sectors.
Answer:
ICT in mining:
1. ICT helps improve operational efficiency in mining by using technology to manage tasks.
2. It provides audio-visual warnings to miners who are trapped underground, helping to ensure their safety.
3. It also connects remote mining sites, making communication and coordination easier. Modern mining operations benefit greatly from these digital tools.
ICT in agriculture:
1. ICT boosts food production and helps manage farms better.
2. It helps farmers use water, seeds, and fertilizers more effectively.
3. Farmers use ICT to decide the best places to plant different crops.
4. It is also used in advanced technologies like robots, temperature sensors, aerial images, and GPS for smart farming.
In simple words: In mining, ICT helps make work run smoothly, gives warnings to trapped miners, and links remote sites. In farming, it increases crop yield, optimizes resource use, helps choose planting spots, and uses new tech like robots and GPS.

🎯 Exam Tip: When listing applications, ensure you provide at least three distinct points for each sector and include a clear example for better understanding.

 

Question 6. Modulation helps to reduce the antenna size in wireless communication Explain.
Answer:
1. An antenna is needed for both sending and receiving signals in wireless communication.
2. The height of the antenna is a crucial factor for efficient wireless communication.
3. For best performance, the antenna's height (`h`) should be a multiple of \( \frac{\lambda}{4} \), where \( \lambda \) is the wavelength of the signal. So, \( h = \frac{\lambda}{4} \).
Here, \( \lambda = \frac{C}{\gamma} \), where `C` is the velocity of light and \( \gamma \) is the frequency.
4. Let's compare two baseband signals.
5. One signal is modulated, and the other is not.
6. If the original baseband signal has a frequency \( \gamma = 10 \text{ KHz} \).
7. If the modulated signal has a frequency \( \gamma = 1 \text{ MHz} \).
8. For the original baseband signal, the required antenna height (\( h_1 \)) would be:
\( h_1 = \frac{\lambda}{4} = \frac{C}{4\gamma} \)
\( = \frac{3 \times 10^8}{4 \times 10 \times 10^3} = 7.5 \text{ Km} \)
This shows that for low frequencies, a very tall antenna is needed.
For the modulated signal, the required antenna height (\( h_2 \)) would be:
\( h_2 = \frac{\lambda}{4} = \frac{C}{4\gamma} \)
\( = \frac{3 \times 10^8}{4 \times 1 \times 10^6} = 75 \text{ m} \)
9. From this example, it is clear that using modulated signals significantly reduces the required height of the antenna. This makes communication practical.
In simple words: When we send radio signals, the antenna needs to be a certain size, usually linked to the signal's wavelength. If we use a low-frequency signal, the antenna has to be very, very long (kilometers!). But if we change the signal using modulation to a much higher frequency, the antenna can become much smaller (just a few meters). This makes it easier and cheaper to build communication systems.

🎯 Exam Tip: Remember the inverse relationship between frequency and wavelength. Higher frequencies mean shorter wavelengths, which in turn means smaller antennas are required, making the system more compact and feasible.

 

Question 7. Fiber-optic communication is gaining popularity among the various transmission media - justify.
Answer:
Fiber-optic communication is quickly replacing traditional wire transmission systems. This is because light, which fiber optics uses, has a very high frequency (400 THz - 790 THz) compared to microwave radio systems. The fibers are typically made from silica glass or silicon dioxide, materials that are very common on Earth. Newer materials like chalcogenide glasses and fluoroaluminate crystalline materials are now used because they allow for longer infrared wavelengths and offer better transmission capabilities. Since fiber optic cables do not conduct electricity, they are ideal for situations where many channels need to be installed and protection from electrical and electromagnetic interference is required. This makes them highly preferred for modern communication needs.
Applications:
Optical fiber systems have many uses, including international communication, communication between cities, data links, managing plant operations and traffic control, and defense applications.
Merits:
1. Fiber optic cables are very thin and much lighter than copper cables, making them easier to handle and install.
2. This system offers a much larger bandwidth, meaning it can carry a significantly greater amount of information at once.
3. Fiber optic systems are not affected by electrical interference, ensuring clearer signal transmission.
4. Fiber optic cables are also generally cheaper to produce than copper cables, which is an important cost advantage.
In simple words: Fiber optics use light to send information, which is very fast. They are popular because they can carry a lot more data than old wires, don't get messed up by electricity, are lighter, and often cheaper. They are used for long-distance calls, internet, and even in military systems.
Demerits:
1. Fiber optic cables are more delicate and can break more easily than copper wires.
2. The technology can be expensive to set up initially.
In simple words: However, they are more fragile than copper wires and can be costly to set up at first.

🎯 Exam Tip: When justifying the popularity of fiber optics, focus on bandwidth, immunity to interference, and physical advantages. Also, briefly mention both merits and demerits to provide a balanced answer.

 

PART - II:

I. Matching Type Questions:

Question 1.

Answer:
1. d
2. c
3. b
4. a
In simple words: Match each type of communication with its typical frequency range. Wireless communication covers a very wide range, ground waves use lower frequencies, sky waves use medium frequencies, and space waves use higher frequencies.

🎯 Exam Tip: Knowing the typical frequency ranges for different propagation modes is key. Associate ground waves with low frequencies, sky waves with medium, and space/wireless with high frequencies.

 

Question 2.

Answer:
1. d
2. c
3. a
4. b
In simple words: This question matches different technologies with their main uses. GPS helps count animals, GSM is for mobile phones, ICT is used in fishing, and RADAR is for military and navigation.

🎯 Exam Tip: For matching questions, quickly recall the primary function or application of each term. If unsure, eliminate obvious mismatches first.

 

Question 3.

Answer:
1. c
2. d
3. b
4. a
In simple words: This match helps us connect common frequency types or bands with their specific values. Uplink and downlink are for satellite communication, light has extremely high frequencies, and audio is in the human hearing range.

🎯 Exam Tip: Distinguish between uplink (sending to satellite) and downlink (receiving from satellite) frequencies. Also, remember the vast difference in frequency between light waves and audio waves.

 

II. Fill in the blanks:

Question 1. _________ gives the difference between the upper and lower frequency limits of the signal.
Answer: Bandwidth
In simple words: Bandwidth tells you the total range of frequencies a signal uses, from the lowest to the highest.

🎯 Exam Tip: Clearly define bandwidth as the difference between the maximum and minimum frequencies in a signal or channel.

 

Question 2. For a frequency less than the critical frequency, skip distance is _________.
Answer: Zero
In simple words: If a radio wave's frequency is too low, it won't reflect from the ionosphere and thus will have no skip distance, meaning it won't travel far by skywave.

🎯 Exam Tip: The critical frequency is the highest frequency that can be reflected by the ionosphere for vertical incidence. Below this, reflection occurs, but if the frequency is too low, the wave might be absorbed or not propagate effectively as a skywave over distance.

 

Question 3. Multimode fibers operate at the speed of _________.
Answer: 10 Mbps
In simple words: Multimode fiber optic cables can send data at a speed of 10 Megabits per second.

🎯 Exam Tip: While 10 Mbps is a common speed for older multimode fibers, modern ones can achieve much higher speeds (up to 10 Gbps or more). If the question is from an older syllabus, 10 Mbps might be the expected answer. Clarify if higher speeds are relevant for context.

 

Question 4. Radar uses _________ for communication.
Answer: electromagnetic waves
In simple words: Radar systems use invisible electromagnetic waves, like radio waves, to find objects.

🎯 Exam Tip: Understand that radar uses radio frequency electromagnetic waves, which are part of the broader electromagnetic spectrum, to detect and range objects.

 

III. Choose the Odd man out:

Question 1.
(a) Input transducer
(b) Amplifier
(c) Oscillator
(d) Demodulator
Answer: (d) Demodulator
In simple words: Input transducer, amplifier, and oscillator are all parts of a transmitter system, helping to prepare and send the signal. A demodulator, however, is part of the receiver, used to extract the original signal.

🎯 Exam Tip: Know the distinct functions of components in a communication system. Transmitters prepare and send, while receivers pick up and decode signals.

 

Question 2.
(a) Telephone
(b) mobile
(c) intercom
(d) Cable TV
Answer: (b) Mobile
In simple words: Telephone, intercom, and Cable TV generally rely on physical wires or cables for connection. Mobile communication is different because it works wirelessly.

🎯 Exam Tip: Categorize communication systems by their medium-wireline (telephone, intercom, cable TV) vs. wireless (mobile, radio, satellite).

 

Question 3.
(a) Weather satellites
(b) Communication satellites
(c) Navigation satellites
(d) RADAR
Answer: (e) RADAR
In simple words: Weather, communication, and navigation satellites are all types of satellites orbiting Earth for various purposes. RADAR, however, is a ground-based or airborne system that detects objects using radio waves, not a satellite itself.

🎯 Exam Tip: Understand that while RADAR technology can be used with satellites (like RADAR satellites), RADAR itself is a detection system, not a type of satellite in the same way as the others listed.

 

Question 4.
(a) Total internal reflection
(b) skip zone
(c) amplitude modulation
(d) skip zone
Answer: (c) Amplitude modulation
In simple words: Total internal reflection and skip zone are concepts related to how waves travel, especially radio waves and light in optical fibers. Amplitude modulation is a method of changing a radio wave to carry information, which is a different kind of concept.

🎯 Exam Tip: Distinguish between wave propagation phenomena (total internal reflection, skip zone) and modulation techniques (amplitude modulation). The question lists "skip zone" twice, implying it's a valid related concept to wave travel.

 

IV. Choose the incorrect pair:

Question 1.
(i) Space wave propagation - LOS
(ii) Tracking cars - ICT
(iii) Counting of wild animals - GPS
(iv) Home automation using a mobile phone - IoT
Answer: (ii) Tracking cars - ICT
In simple words: While ICT (Information and Communication Technology) is a broad field, tracking cars specifically uses GPS technology, not just general ICT. All other pairs correctly link the technology to its main use.

🎯 Exam Tip: ICT is a very broad term. When specific technologies like GPS or IoT are more directly applicable, prefer them over the general ICT unless ICT is the most appropriate broad category for the given function.

 

Question 2.
(i) Input transducer - Microphone
(ii) Output transducer - Loudspeakers
(iii) Transmitter - Broadcasting station
(iv) Oscillator - Low-frequency carrier wave
Answer: (iv) Oscillator - Low-frequency carrier wave
In simple words: An oscillator in a communication system produces high-frequency carrier waves, not low-frequency ones. The other pairs correctly describe components and their functions: a microphone is an input transducer, a loudspeaker is an output transducer, and a broadcasting station acts as a transmitter.

🎯 Exam Tip: Remember that oscillators in transmitters generate high-frequency carrier waves to enable long-distance transmission, as low-frequency waves would require impractically large antennas.

 

Question 3.
(i) Bandwidth, BW - \( \gamma_1 - \gamma_2 \)
(ii) Amplitude modulation system - 10 KHz
(iii) Single side-band system - 5 KHz
(iv) Height of the Antenna - \( \frac{\lambda}{4} \)
Answer: (i) Bandwidth, BW - \( \gamma_1 - \gamma_2 \)
In simple words: Bandwidth is always calculated as the higher frequency minus the lower frequency, so it should be \( \gamma_2 - \gamma_1 \), not the other way around. The other options correctly state characteristics or common values.

🎯 Exam Tip: Always define bandwidth as the difference between the upper limit frequency and the lower limit frequency (highest minus lowest), ensuring a positive value.

 

Question 4.
(i) Amplitude - High-efficiency modulation
(ii) Baseband signal - input signal
(iii) Carrier signal - a radio signal
(iv) Resting frequency - centre frequency
Answer: (i) Amplitude - High-efficiency modulation
In simple words: Amplitude modulation (AM) is actually known for having lower efficiency compared to frequency modulation (FM) because a lot of power is wasted in transmitting the carrier wave. The other statements are correct definitions or descriptions.

🎯 Exam Tip: Recall the key characteristics of AM and FM. AM is generally less efficient due to carrier power, while FM is known for better noise immunity and efficiency. This is a common point of comparison.

 

V. Choose the correct pair:

Question 1.
(i) Attenuation - Amplitude of the transmitter
(ii) High frequency - High skip distance
(iii) Skip distance - The longest distance
(iv) RADAR - Ground wave propagation
Answer: (ii) High frequency - High skip distance
In simple words: Higher frequencies often lead to longer skip distances in skywave propagation, as they penetrate the ionosphere less or are reflected at a shallower angle. The other pairs contain incorrect associations.

🎯 Exam Tip: Remember that skip distance is the minimum distance from the transmitter where a skywave can be received. Higher frequencies typically result in a greater skip distance because they require a shallower angle of incidence for reflection from the ionosphere.

 

Question 2.
(i) Audio frequency - 200 to 2000 Hz
(ii) Carrier wave - Cosine wave
(iii) Louder signal - Compressions
(iv) Frequency shift - Rarefaction
Answer: (iii) Louder signal- Compressions
In simple words: In frequency modulation, a louder signal (higher amplitude) causes more "compressions," meaning the carrier wave's frequency changes more. This means the change in frequency is greater.

🎯 Exam Tip: In FM, the amplitude of the modulating signal is directly related to the frequency deviation. A louder (higher amplitude) signal will cause a larger frequency shift, leading to more "compressions" or denser wave patterns.

 

Question 3.
(i) Velocity of light - \( 3 \times 10^6 \text{ m/s} \)
(ii) Radius of the earth - 6800 Km
Answer: Statement (ii) Radius of the earth - 6800 Km is correct.
In simple words: The velocity of light is actually much faster, \( 3 \times 10^8 \text{ m/s} \). The radius of the Earth is approximately 6400 km or 6800 km, depending on the exact model used, making this a plausible figure.

🎯 Exam Tip: Be precise with fundamental physical constants. The speed of light is \( 3 \times 10^8 \text{ m/s} \). The Earth's average radius is approximately 6371 km, so 6800 km is a reasonable approximation in some contexts.

 

Question 4.
(i) Skywave propagation - 3 MHz to 30 MHz
(ii) Space wave propagation - 30 MHz to 40 MHz
(iii) Ground wave propagation - 20 KHz to 20 MHz
(iv) Wireless communication - 1 KHz to 2 KHz
Answer: (i) Skywave propagation - 3 MHz to 30 MHz
In simple words: Skywave propagation, which uses the ionosphere to reflect signals, works best within the frequency range of 3 MHz to 30 MHz. The other frequency ranges listed are not typical for their respective propagation modes.

🎯 Exam Tip: Memorize the typical frequency ranges associated with different radio wave propagation modes (ground wave, skywave, space wave). This is a fundamental concept in communication systems.

 

VI. Assertion and Reason:

Question 1. Assertion: Short wave bands are used for the transmission of radio waves to a large distance. Reason: Short waves are reflected by the ionosphere.
(a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
(c) Assertion is true but reason is false
(d) Assertion is false but reason is true.
Answer: (a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
In simple words: The statement that short waves travel far is true. This happens because the ionosphere, a layer in Earth's atmosphere, reflects these waves back down to Earth, allowing them to cover long distances. So, the reason correctly explains why the assertion is true.

🎯 Exam Tip: For assertion-reason questions, first check if both statements are individually true. If so, then evaluate if the reason directly explains the assertion. The ionosphere's reflective properties are key for shortwave long-distance communication.

 

Question 2. Assertion: Television broadcasting becomes weaker with increasing distance. Reason: The power transmitted from the TV transmitter varies inversely as the distance of the receiver.
(a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
(c) Assertion is true but reason is false
(d) Assertion is false but reason is true.
Answer: (c) Assertion is true but reason is false
In simple words: It is true that TV signals get weaker the farther you are from the transmitter. However, the reason given is wrong; the power does not vary inversely with just the distance. Instead, the signal strength decreases with the square of the distance. So, the explanation is not correct, even though the first statement is true.

🎯 Exam Tip: Signal strength typically follows an inverse square law (proportional to \( \frac{1}{r^2} \)), not just inversely proportional to distance (\( \frac{1}{r} \)). This distinction is crucial for understanding signal attenuation.

 

Question 3. Assertion: Optical fibre communication has immunity to cross talk. Reason: Optical interference between fibres is zero.
(a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
(c) Assertion is true but reason is false
(d) Assertion is false but reason is true.
Answer: (a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
In simple words: Optical fiber communication is indeed protected from crosstalk (signals mixing up). This is because light signals in one fiber do not cause optical interference with signals in other nearby fibers. So, the reason explains why crosstalk is not an issue.

🎯 Exam Tip: Optical fibers are excellent for secure and clear transmission because light waves do not interfere with each other when traveling in separate, insulated fibers, unlike electrical signals in copper wires which can suffer from crosstalk.

 

Question 4. Assertion: A dish antenna is highly directionals Reason: This is because a dipole antenna is omnidirectional.
(a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
(c) Assertion is true but reason is false
(d) Assertion is false but reason is true.
Answer: (b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
In simple words: A dish antenna is very good at sending and receiving signals in one specific direction, which is true. It is also true that a simple dipole antenna sends signals in all directions (omnidirectional). However, the omnidirectional nature of a dipole antenna does not explain why a dish antenna is directional. These are two separate facts about different types of antennas.

🎯 Exam Tip: Understand the characteristics of different antenna types. Dish antennas achieve directionality through their parabolic shape, focusing waves, while dipole antennas are inherently omnidirectional in a plane perpendicular to the antenna element.

 

VII. Choose the correct statement:

Question 1.
(i) A transducer is a device which converts an electrical signal into variations in a physical quantity.
(ii) Transmitting antenna radiates the radio signal into space in all directions.
(iii) The power amplifier decreases the power level of the signal
(iv) The minimum distance between the source and the destination is called a range
Answer: Statement (ii) is correct.
In simple words: A transmitting antenna does indeed send radio signals out in all directions. Statement (i) is mostly correct but transducers can convert physical quantities into electrical signals as well. Statement (iii) is wrong because a power amplifier increases the power of a signal. Statement (iv) is incorrect because "range" typically refers to the maximum distance, not the minimum.

🎯 Exam Tip: A transducer converts energy from one form to another (e.g., sound to electrical or vice-versa). A transmitting antenna radiates signals, often in specific patterns, not always omnidirectionally (though many can be approximated as such). Power amplifiers boost signal strength, and range refers to the effective coverage area.

 

Question 2.
(i) Attenuation depends on the frequency of the transmitter.
(ii) Increase in the tilt increases the electric field strength.
(iii) Transmitting and receiving antenna must be close to the earth.
(iv) The frequency of the ground waves is most less than 1 MHz.
Answer: Statements (i) and (iii) are correct
In simple words: Attenuation (signal loss) definitely changes with the frequency being used. Also, for ground wave propagation, both the sending and receiving antennas need to be close to the Earth's surface. These two statements are correct.

🎯 Exam Tip: Attenuation in ground wave propagation is frequency-dependent; higher frequencies attenuate more quickly. Ground wave antennas are positioned close to the ground, unlike skywave or space wave antennas.

 

Question 3.
(i) The higher the frequency, lower is the skip distance.
(ii) Bending of radio waves back to earth is called refraction.
(iii) The frequency range of EM wave in skywave propagation is 30 to 300 MHz.
(iv) When the angle of emission increases, the reception of ground waves decreases.
Answer: Statement (iv) is correct.
In simple words: When a radio wave is sent out at a higher angle (angle of emission increases), it means less of its energy travels along the ground. So, the reception of ground waves will go down. This statement is correct.

🎯 Exam Tip: Understand the interplay between emission angle and propagation modes. A higher emission angle favors skywave propagation, while a lower angle is better for ground wave propagation. Also, clarify that the bending of radio waves back to Earth in skywave propagation is primarily due to reflection/refraction in the ionosphere.

 

Question 4.
(i) The transducer output is very weak
(ii) The electrical equivalent of the original information is called the carrier signal.
(iii) Microphone converts electrical energy into sound energy.
(iv) Wireline communication uses free space as a communication medium.
Answer: Statement (i) is correct.
In simple words: The electrical signal produced by a transducer (like a microphone) is usually very small and needs an amplifier to boost it. This statement is true. The other statements are incorrect: the electrical equivalent of information is a baseband signal, a microphone converts sound to electrical energy, and wireline communication uses wires, not free space.

🎯 Exam Tip: Transducers produce low-level signals that require amplification. Clarify the definitions: baseband signal (electrical equivalent of information), carrier signal (high frequency for transmission), and the distinction between wireline and wireless media.

 

VIII. Choose the incorrect statement:

Question 1.
(i) Noise is the undesirable electrical signal.
(ii) It attenuates or reduces the quality of the signal
(iii) It may be man-made or natural
(iv) Noise can be completely eliminated.
Answer: Statement (iv) is correct.
In simple words: Noise is unwanted interference that reduces signal quality and can come from natural or man-made sources. However, it is impossible to get rid of noise completely; it can only be reduced. So, saying it can be completely eliminated is incorrect.

🎯 Exam Tip: While noise can be significantly reduced through various techniques and technologies, it is a fundamental characteristic of electronic systems and communication channels, meaning it can never be 100% eliminated.

 

Question 2.
(i) Operating range is quite large in FM
(ii) In FM, there is a large increase in noise
(iii) Am radio has better quality compared to FM radio.
(iv) FM requires a much under channel
Answer: Statement (ii) & (iii) are incorrect
In simple words: FM is actually known for having less noise than AM, not a large increase. Also, FM radio generally has better sound quality than AM radio. So, statements (ii) and (iii) are false.

🎯 Exam Tip: Remember the primary advantages of FM over AM: better noise immunity (so less noise, higher quality) and wider bandwidth for higher fidelity, though it requires a wider channel.

 

Question 3.
(i) Noise level is low in AM
(ii) There is a lesser landwidth in AM
(iii) In A.M the efficiency is low
(iv) Small operating range in AM
Answer: Statement (i) is incorrect
In simple words: AM (Amplitude Modulation) is generally more prone to noise compared to FM. So, the statement that noise level is low in AM is incorrect. The other statements about AM's efficiency and operating range are generally considered true.

🎯 Exam Tip: A key disadvantage of AM is its susceptibility to noise. Noise typically affects the amplitude of a signal, making AM less robust than FM, which encodes information in frequency changes.

 

Question 4.
(i) ICT in mining helps to connect remote sites.
(ii) ICT is widely used in decreasing food productivity
(iii) ICT improves operational efficiency
(iv) ICT are also used in fisheries
Answer: Statement (ii) is incorrect
In simple words: ICT (Information and Communication Technology) aims to improve efficiency and productivity in most fields, including agriculture, not decrease food productivity. So, the statement that ICT reduces food productivity is false.

🎯 Exam Tip: ICT is generally associated with advancements and improvements across various sectors, including enhancing productivity. A statement suggesting a decrease in productivity due to ICT would almost always be incorrect in this context.

IX. Choose the Best Answer:

 

Question 1. Which of the following frequencies will be suitable for beyond the horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Answer: (b) 10 MHz
In simple words: For sky wave communication that travels far, a frequency like 10 MHz is just right because the ionosphere can reflect it. Very low or very high frequencies don't work well for this type of long-distance communication.

🎯 Exam Tip: Remember that sky waves are reflected by the ionosphere, making them suitable for long-distance communication, and that the ionosphere has a specific frequency range for effective reflection.

 

Question 2. Cellular phones use radio waves in ultra-high frequency band.
(a) long-wave
(b) short wave
(c) medium wave
(d) ultra-high frequency
Answer: (d) ultra-high frequency
In simple words: Cellular phones operate using very high frequency radio waves, known as ultra-high frequency (UHF) signals. These frequencies are ideal for short-range, point-to-point communication in cellular networks.

🎯 Exam Tip: Familiarize yourself with the different frequency bands and their common applications, as this helps understand how various communication technologies work.

 

Question 3. Antenna is resistive at the resonant frequency.
(a) inductive
(b) capacitive
(c) resonant frequency
(d) resistive at the resonant frequency
Answer: (d) resistive at the resonant frequency
In simple words: An antenna works best when it is at its resonant frequency. At this specific frequency, the antenna acts like a purely resistive circuit, which means it efficiently converts electrical energy into radio waves and vice versa.

🎯 Exam Tip: Recall that at resonance, the inductive and capacitive reactances in a circuit cancel each other out, leaving only the resistance. This principle applies to antennas for maximum power transfer.

 

Question 4. In space wave propagation, the range of the propagation depends on the height (h) of the antenna given by the equation.
(a) \( \sqrt{2 \mathrm{Rh}} \)
(b) \( \sqrt{\frac{R h}{2}} \)
(c) \( \sqrt{2 R^{2} h} \)
(d) \( \frac{h}{2 R} \)
Answer: (a) \( \sqrt{2 \mathrm{Rh}} \)
In simple words: When radio signals travel directly from one antenna to another in space (space wave propagation), the distance they can cover depends on how tall the transmitting antenna is. This relationship is described by a specific formula involving the Earth's radius and the antenna's height.

🎯 Exam Tip: The formula \( d = \sqrt{2Rh} \) is crucial for calculating the maximum line-of-sight distance in space wave propagation, so ensure you remember it along with the variables it represents.

 

Question 5. Amplitude modulation is used in radio and TV broadcasting.
(a) Frequency
(b) Phase
(c) Amplitude
(d) Carrier
Answer: (c) Amplitude
In simple words: Amplitude modulation, often called AM, is a common way to send radio and TV signals. It works by changing the strength (amplitude) of a radio wave to match the sound or picture information.

🎯 Exam Tip: Differentiate between amplitude modulation (AM) and frequency modulation (FM) by understanding how each modifies the carrier wave to encode information.

 

Question 6. If a Square wave is used as the baseband signal, then phase reversal takes place.
(a) Square
(b) Sine
(c) Cosine
(d) triangular
Answer: (a) Square
In simple words: When a square wave is used as the basic signal, it causes a sudden flip in the phase of the carrier wave. This abrupt change is a characteristic feature when square waves are involved in modulation.

🎯 Exam Tip: Note the unique characteristics of different waveform types (like square, sine, triangular) and how they affect the modulation process, particularly phase changes.

 

Question 7. An oscillator is producing FM waves of frequency 2 kHz with a variation of 10 kHz. What is the modulation index?
(a) 0.67
(b) 5.00
(c) 0.20
(d) 1.5
Answer: (b) 5.00
In simple words: To find the modulation index, we divide the change in frequency (10 kHz) by the original signal frequency (2 kHz). This calculation tells us how much the frequency is being altered.

🎯 Exam Tip: Remember the formula for modulation index in FM, \( m_f = \frac{\Delta f}{f_m} \), where \( \Delta f \) is frequency deviation and \( f_m \) is modulating frequency. Pay attention to units.

 

Question 8. The best example of transducer is microphone.
(a) Loudspeaker
(b) picture tubes
(c) computer monitor
(d) microphone
Answer: (d) microphone
In simple words: A microphone is a perfect example of a transducer because it changes sound energy into electrical energy. This conversion is a key function of transducers, which change one form of energy into another.

🎯 Exam Tip: Understand that transducers are devices that convert energy from one form to another. Microphones and loudspeakers are common examples, converting sound to electrical signals and vice versa.

 

Question 9. Oscillator generates high - frequency carrier wave.
(a) Transducer
(b) Oscillator
(c) Amplifier
(d) Power amplifier
Answer: (b) Oscillator
In simple words: An oscillator is a circuit that creates a steady, high-frequency signal, which is called a carrier wave. This carrier wave is then used to carry information in communication systems.

🎯 Exam Tip: Know the role of each component in a communication system. The oscillator is fundamental for generating the high-frequency signal needed for modulation.

 

Question 10. The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(a) \( h^{1/2} \)
(b) h
(c) \( h^{3/2} \)
(d) \( h^2 \)
Answer: (a) \( h^{1/2} \)
In simple words: The furthest distance a TV signal from a tower can travel and still be picked up depends on the height of the TV tower. This distance is related to the square root of the tower's height.

🎯 Exam Tip: This relationship, \( d \propto \sqrt{h} \), is a key concept in line-of-sight communication and helps explain why taller antennas are used for wider broadcast coverage.

 

Question 11. The height of the antenna must be a multiple of \( \frac{\lambda}{4} \).
(a) \( \frac{\lambda}{4} \)
(b) \( \frac{\lambda}{8} \)
(c) \( 4\lambda \)
(d) \( 2\lambda \)
Answer: (a) \( \frac{\lambda}{4} \)
In simple words: For an antenna to work most efficiently, its height should be a specific fraction of the wavelength of the radio waves it sends or receives. This ideal length allows the antenna to resonate properly.

🎯 Exam Tip: The quarter-wavelength antenna (\( \lambda/4 \)) is a common and efficient design. Understanding this relationship is fundamental to antenna theory and design.

 

Question 12. C = \( \gamma\lambda \) en the frequency, the velocity of light, and wavelength is given by both (a) and (c).
(a) \( C = \gamma\lambda \)
(b) \( C = \frac{\lambda}{\gamma} \)
(c) \( \lambda = \frac{C}{\gamma} \)
(d) both (a) and (c)
Answer: (d) both (a) and (c)
In simple words: The speed of light (C) is equal to its frequency (\( \gamma \)) multiplied by its wavelength (\( \lambda \)). This means you can also find the wavelength by dividing the speed of light by its frequency. This is a fundamental relationship in wave physics.

🎯 Exam Tip: The wave equation \( C = f\lambda \) (where f is frequency) is crucial. Ensure you can rearrange it to solve for any variable and understand its application to electromagnetic waves.

 

Question 13. In AM, the channel bandwidth is twice the signal frequency.
(a) thrice
(b) twice
(c) equal to
(d) half
Answer: (b) twice
In simple words: For Amplitude Modulation (AM), the range of frequencies needed for a clear channel is twice the highest frequency present in the original signal. This wide range ensures that all the information is transmitted correctly.

🎯 Exam Tip: Remember that AM bandwidth is \( 2 \times f_m \), where \( f_m \) is the maximum frequency of the modulating signal. This contrasts with FM bandwidth, which is generally wider.

 

Question 14. The height of the transmitting and receiving antenna must be a multiple of \( \lambda/4 \).
(a) \( 2\lambda \)
(b) \( \lambda/4 \)
(c) \( \lambda/2 \)
(d) \( 4\lambda \)
Answer: (b) \( \lambda/4 \)
In simple words: For antennas to work effectively, their height needs to be a specific length compared to the wavelength of the signals. A common and efficient height is one-fourth of the wavelength.

🎯 Exam Tip: Understanding optimal antenna lengths, like \( \lambda/4 \), is important for practical communication system design and performance. Incorrect antenna length can lead to poor signal transmission or reception.

 

Question 15. A laser beam is used for carrying out surgery because it can be sharply focused.
(a) is highly monochromatic
(b) is highly coherent
(c) is directional
(d) can be sharply focused
Answer: (d) can be sharply focused
In simple words: Laser beams are very useful in surgery because they can be made extremely narrow and precise. This allows surgeons to target very small areas without harming surrounding tissue, making operations more accurate.

🎯 Exam Tip: Lasers are valuable tools due to their properties: monochromaticity (single color), coherence (waves in phase), directionality (travels in one direction), and the ability to be highly focused, making them suitable for precise applications like surgery.

 

Question 16. The size of the antenna plays a major role in deciding the efficiency of the radiation of the signals.
(a) shape
(b) distance
(c) direction
(d) size
Answer: (d) size
In simple words: The physical size of an antenna is very important. It directly affects how well the antenna can send out (radiate) radio signals and how far those signals can travel.

🎯 Exam Tip: Antenna size is directly related to the wavelength of the signal it transmits or receives. An antenna that is correctly sized for the wavelength will be more efficient at radiating or capturing signals.

 

Question 17. A single reflection helps the radio waves to travel a distance of approximately
(a) 40 km
(b) 400 km
(c) 4000 km
(d) 4 km
Answer: (c) 4000 km
In simple words: When a radio wave bounces off the ionosphere just once, it can cover a very long distance. This single bounce can allow the signal to travel around 4000 kilometers, making long-range communication possible.

🎯 Exam Tip: Skywave propagation, which involves reflection from the ionosphere, enables very long-distance communication, with a single hop often covering thousands of kilometers.

 

Question 18. When the angle of emission increases, the reception of ground waves decreases.
(a) increases
(b) decreases
(c) remains constant
(d) none of the options
Answer: (b) decreases
In simple words: If a radio signal is sent out at a higher angle, the strength of the ground waves that are received will go down. This is because more of the signal energy is directed towards the sky rather than along the ground.

🎯 Exam Tip: Understand the inverse relationship between the angle of emission and ground wave reception. Higher emission angles favor skywave propagation, while lower angles are better for ground waves.

 

Question 19. The principle used in the transmission of signals through an optical fiber is
(a) total internal reflection
(b) refraction
(c) dispersion
(d) interference
Answer: (a) total internal reflection
In simple words: Optical fibers work by trapping light inside them. The light bounces repeatedly off the inner walls of the fiber, staying inside until it reaches the end. This process is called total internal reflection.

🎯 Exam Tip: Total internal reflection is a fundamental concept in optics and is the core principle behind how optical fibers efficiently transmit light signals over long distances with minimal loss.

 

Question 20. The fibres are made up of both a & c.
(a) glass
(b) silicon
(c) silica glass
(d) both a & c
Answer: (d) both a & c
In simple words: Optical fibers, which carry light signals, are mainly made from a type of glass. Specifically, they often use silica glass, which is a very pure form of glass.

🎯 Exam Tip: While glass is the general material, remembering "silica glass" is more precise for optical fibers. High purity is essential for efficient light transmission.

 

Question 21. Fiber optic cables fastest transmission rate compared to any other form of transmission.
(a) Cable wires
(b) Fibers
(c) Fiber optic cables
(d) Copper wire
Answer: (c) Fiber optic cables
In simple words: Fiber optic cables are known for being the quickest way to send information. They use light pulses to carry data, which allows for much faster transmission speeds compared to traditional copper wires or other methods.

🎯 Exam Tip: Fiber optic cables offer superior bandwidth and speed compared to traditional copper cables due to their use of light signals and the principle of total internal reflection.

 

Question 22. Internet is a fast-growing technology in the field communication system.
(a) Mobile
(b) Satellite
(c) Fiber optics
(d) Internet
Answer: (d) Internet
In simple words: The Internet is a rapidly growing technology in communication. It connects people and devices worldwide, allowing for instant sharing of information and resources.

🎯 Exam Tip: Recognize the Internet's transformative role in modern communication, impacting global connectivity, data access, and various social and economic activities.

 

Question 23. The skywave propagation is suitable for radio waves of frequency
(a) from 2 MHz to 50
(b) upto 2 MHz
(c) from 2 MHz to 30 MHz
(d) from 2MHz to 20 MHz
Answer: (c) from 2 MHz to 30 MHz
In simple words: Skywave propagation works best for radio signals that fall within a certain frequency range, typically from 2 MHz to 30 MHz. Signals in this range can effectively bounce off the ionosphere and travel long distances.

🎯 Exam Tip: Knowing the typical frequency ranges for different propagation modes (ground wave, sky wave, space wave) is important for determining suitable communication methods for various applications.

 

Question 24. To store all the information available on the internet, we need over
(a) 2 billion DVDs
(b) 1 billion DVDs
(c) 10 billion DVDs
(d) 4 billion DVDs
Answer: (b) 1 billion DVDs
In simple words: The amount of information on the internet is huge. To save all of it, you would need storage space equivalent to more than one billion DVDs. This shows how vast the digital world is.

🎯 Exam Tip: This question highlights the immense scale of data on the Internet. While the exact number can change, the core idea is to appreciate the vast storage requirements of global digital information.

 

Question 25. Search engine is used to search for information on the World Wide Web.
(a) Communication
(b) Satellite
(c) E-Commerce
(d) Search engine
Answer: (d) Search engine
In simple words: A search engine is a special tool on the internet that helps us find information. You type what you are looking for, and it shows you web pages that have those words or topics.

🎯 Exam Tip: Search engines are crucial for navigating the vast amount of information on the World Wide Web, using algorithms to index and retrieve relevant content based on user queries.

 

Question 26. GPS is used for making tunnels, bridges etc.
(a) GSM
(b) TDMA
(c) GPS
(d) GPRS
Answer: (c) GPS
In simple words: The Global Positioning System (GPS) is used to accurately map locations. This is very helpful in big construction projects like building tunnels and bridges, as it ensures everything is placed in the correct spot.

🎯 Exam Tip: GPS provides precise location and timing data, making it invaluable for applications requiring high accuracy, such as civil engineering, surveying, and navigation.

 

Question 27. Fiber optic cables provide data speed of 1 Gbps for homes and businesses.
(a) 1 Gbps
(b) 2 Gbps
(c) 10 Mbps
(d) 5 Gbps
Answer: (a) 1 Gbps
In simple words: Fiber optic cables can send data at extremely high speeds, like 1 Gigabit per second (Gbps). This makes them perfect for fast internet connections in homes and businesses.

🎯 Exam Tip: Data transfer speeds are typically measured in Mbps (Megabits per second) or Gbps (Gigabits per second). Fiber optic technology is preferred for its high bandwidth and fast data rates.

 

Question 28. A modem is a device which performs
(a) modulation
(b) demodulation
(c) rectification
(d) modulation and demodulation
Answer: (d) modulation and demodulation
In simple words: A modem is a device that does two main things: it changes digital signals from a computer into analog signals for transmission (modulation), and it changes analog signals back into digital ones for the computer to understand (demodulation).

🎯 Exam Tip: The term "modem" is a portmanteau of "modulator-demodulator." Its dual function is essential for converting digital data for analog transmission over telephone lines, cable, or fiber, and vice versa.

X. Two Marks Questions:

 

Question 1. What is a communication system?
Answer: A communication system is a setup designed to send information from one place to another. This system includes all the parts needed for sending, transmitting, and receiving messages. It ensures that messages can be delivered effectively over distances.
In simple words: A communication system is a way to send information from one person or place to another.

🎯 Exam Tip: Define a communication system by highlighting its purpose (information transfer) and basic components (transmitter, channel, receiver).

 

Question 2. What is called a baseband signal?
Answer: A baseband signal is the original form of information, like your voice or data, before it is prepared for transmission. It is the electrical signal that represents the actual message. This signal contains the full range of frequencies of the original information.
In simple words: A baseband signal is the original electrical form of information before it is sent out.

🎯 Exam Tip: Distinguish the baseband signal as the raw, unmodulated information signal. It carries the actual message content, unlike a carrier wave.

 

Question 3. What is a transducer?
Answer: A transducer is a device that changes one type of physical energy into another, typically into an electrical signal, or vice versa. For example, it can turn sound waves into electrical signals or light into electrical signals. It acts as a converter for various physical quantities.
In simple words: A transducer changes one form of energy, like sound or light, into an electrical signal.

🎯 Exam Tip: When defining a transducer, emphasize its role in converting physical quantities (like pressure, temperature, sound) into corresponding electrical signals, which is vital for input and output in communication systems.

 

Question 4. Define bandwidth.
Answer: Bandwidth refers to the range of frequencies that a baseband signal or information signal uses during transmission. It includes the entire span from the lowest to the highest frequency present in the signal, such as voice, music, or video. A wider bandwidth allows more information to be transmitted.
In simple words: Bandwidth is the full range of frequencies that a signal takes up when it is sent.

🎯 Exam Tip: Bandwidth is a critical measure of a communication channel's capacity. A larger bandwidth means more data can be transmitted in a given time.

 

Question 5. Define bandwidth?
Answer: Bandwidth is the frequency range within which baseband signals or information signals (like speech, music, or images) are transmitted. It measures the difference between the highest and lowest frequencies of a signal, determining the channel's capacity. This range is essential for effective communication.
In simple words: Bandwidth is the total span of frequencies used to send information signals like voice or music.

🎯 Exam Tip: Clarify that bandwidth is the difference between the upper and lower frequency limits of a signal. It's often associated with data rate: higher bandwidth generally permits higher data rates.

 

Question 6. What is skip distance?
Answer: Skip distance is the shortest distance from a transmitter where sky waves can be received after reflecting from the ionosphere. Any area closer than this distance will not receive the skywave signal because it will have passed over. This creates a "skip zone" where no signal is received.
In simple words: Skip distance is the closest spot where you can catch a radio signal that bounced off the sky (ionosphere).

🎯 Exam Tip: Emphasize that skip distance is the minimum distance for skywave reception, occurring due to the signal passing over nearer areas before reflection back to Earth.

 

Question 7. What is skip zone or skip area?
Answer: A skip zone, also known as a skip area, is a region where radio signals are not received. This happens because the ground waves have faded out, and the sky waves have "skipped" over that area before returning to Earth. Thus, it's a dead zone for electromagnetic waves.
In simple words: A skip zone is an area where radio signals from a distant transmitter cannot be picked up.

🎯 Exam Tip: Connect the skip zone to the concept of skip distance. It is the area between the point where ground waves become too weak and the point where sky waves first land.

 

Question 8. What is called fibre optic communication?
Answer: Fiber optic communication is a method where information is sent from one place to another using light pulses. These pulses travel through tiny glass or plastic strands called optical fibers. This technology allows for very fast and clear transmission of data over long distances.
In simple words: Fiber optic communication sends information as light through thin glass cables.

🎯 Exam Tip: Key elements of fiber optic communication are light pulses and optical fibers. It's known for its high bandwidth and immunity to electromagnetic interference.

 

Question 9. What is mean by fibre optic communication?
Answer: Fiber optic communication is a modern way to transmit information by sending light signals through thin glass or plastic threads called optical fibers. This method allows data, voice, and video to travel quickly and reliably over long distances with very little loss. It relies on the principle of total internal reflection to guide the light.
In simple words: Fiber optic communication is sending data as light flashes through special thin cables made of glass.

🎯 Exam Tip: Explain that fiber optic communication leverages light as the carrier and total internal reflection for efficient transmission. Its advantages include high speed, large bandwidth, and low signal loss.

 

Question 10. Expand GPS and write a note on it.
Answer: GPS stands for Global Positioning System. It is a worldwide navigation satellite system that provides location and time information to a GPS receiver anywhere on or near the Earth where there is an unobstructed line of sight to four or more GPS satellites. This system is used for navigation, mapping, and tracking.
In simple words: GPS means Global Positioning System. It uses satellites to tell you exactly where you are and what time it is, anywhere on Earth.

🎯 Exam Tip: When describing GPS, highlight its key functions: providing accurate geolocation and precise timing, and its reliance on a network of satellites orbiting Earth.

 

Question 11. Why do we need carrier waves of very high frequency in the modulation of signals?
Answer: We use high-frequency carrier waves for several reasons:
1. They increase the operating range, allowing signals to travel further.
2. They help reduce the required antenna length, making antennas smaller and more practical.
3. They convert wideband signals into narrowband signals, making them easier to manage and transmit efficiently over communication channels. This enables easier multiplexing of signals.
In simple words: We use fast-moving carrier waves to send signals far, make antennas smaller, and fit more signals into narrow communication paths.

🎯 Exam Tip: Focus on the three main benefits of high-frequency carrier waves: increased range, reduced antenna size, and efficient channel utilization through conversion to narrowband signals.

 

Question 12. Why modulation is needed at all?
Answer: Modulation is needed to:
1. Transmit a low-frequency signal over long distances, as low-frequency signals attenuate quickly.
2. Protect the waveform of the signal from noise and interference during transmission.
3. Keep the height of the antenna small, as the antenna's size is inversely related to the signal's frequency. High-frequency signals require smaller antennas.
In simple words: Modulation is important because it helps send signals far away, keeps them safe from unwanted noise, and allows us to use smaller antennas.

🎯 Exam Tip: Remember the three main reasons for modulation: efficient long-distance transmission, improving signal-to-noise ratio, and practical antenna design considerations.

 

Question 13. Which is better for high fidelity reception FM or AM?
Answer: FM (Frequency Modulation) transmission provides higher fidelity reception compared to AM (Amplitude Modulation). This is because FM signals contain a large number of sidebands, which carry more information and are less affected by noise. The constant amplitude of FM signals also makes them more immune to amplitude noise, leading to clearer sound.
In simple words: FM radio sounds better and clearer than AM radio. This is because FM signals can carry more sound details and are not as easily disturbed by static noise.

🎯 Exam Tip: When comparing FM and AM, highlight FM's advantages in fidelity due to its resistance to noise (constant amplitude) and broader sideband content, which allows for better sound reproduction.

 

Question 14. Why is the transmission of signals through a coaxial cable not possible for frequencies greater than 20 MHz?
Answer: Transmission of signals through a coaxial cable is not suitable for frequencies greater than 20 MHz because, at these higher frequencies, the dielectric loss within the cable becomes very significant. This loss causes the signal to weaken considerably over distance, making it inefficient for long-range transmission. The energy gets absorbed and converted into heat.
In simple words: Coaxial cables don't work well for signals above 20 MHz because they lose too much energy at those high speeds.

🎯 Exam Tip: Focus on the concept of dielectric loss. Higher frequencies lead to greater energy absorption by the insulating material in coaxial cables, limiting their effective use for very high-frequency signals.

 

Question 15. Why are short wave bands used for long-distance transmission of signals?
Answer: Short wave bands are used for long-distance transmission of signals because their radio waves can be easily reflected by the ionosphere. This reflection allows the signals to bounce back to Earth over significant distances, enabling communication far beyond the line of sight. This phenomenon is known as skywave propagation.
In simple words: Short radio waves are good for long distances because they can bounce off the ionosphere, letting them travel very far around the world.

🎯 Exam Tip: The key reason for using shortwave bands for long-distance communication is their ability to be reflected by the ionosphere, enabling multiple "hops" across the globe.

 

Question 16. Why microphone is used as an input transducer?
Answer: A microphone is used as an input transducer because it converts sound variations, which are a physical quantity, into equivalent electrical signals. When you speak into a microphone, the sound waves cause a diaphragm to vibrate, and these vibrations are then changed into electrical energy. This electrical signal can then be processed and transmitted.
In simple words: A microphone is an input transducer because it changes sound into electrical signals that can be used by other devices.

🎯 Exam Tip: Remember that an input transducer's role is to convert a non-electrical signal (like sound) into an electrical one, making it compatible with electronic communication systems.

 

Question 17. What are the elements in the transmitter?
Answer: The main elements in a transmitter are:
1. Amplifier: Strengthens the weak input signal.
2. Modulator: Superimposes the information signal onto a high-frequency carrier wave.
3. Oscillator: Generates the high-frequency carrier wave.
4. Power amplifier: Boosts the power of the modulated signal for effective transmission. These components work together to prepare the signal for sending over a distance.
In simple words: A transmitter uses an amplifier, modulator, oscillator, and power amplifier to get a signal ready to be sent out.

🎯 Exam Tip: Be able to list and briefly explain the function of each component within a transmitter, understanding their sequence and contribution to signal preparation.

 

Question 18. Draw the block diagram of Transmission.
Answer:

In simple words: This picture shows how a signal moves through a transmitter. First, the sound or data goes into a transducer, then it's amplified, modulated by an oscillator, boosted by a power amplifier, and finally sent out through the communication channel.

🎯 Exam Tip: Practice drawing the block diagram of a transmitter from memory, labeling each component and indicating the direction of signal flow. This helps in understanding the overall communication process.

 

Question 19. Define range.
Answer: In communication, "range" refers to the maximum distance between a signal source (transmitter) and its destination (receiver) where the signal can still be received with enough strength. If the signal travels beyond this distance, it becomes too weak to be useful. It defines the effective coverage area of a transmission.
In simple words: Range is how far a signal can travel from where it starts to where it can still be heard or used.

🎯 Exam Tip: When defining range, specify that it is the maximum effective distance, not just any distance, and emphasize the condition of "sufficient strength" for signal usability.

 

Question 20. Mention the application of ICT is fisheries.
Answer: ICT (Information and Communication Technology) has important applications in fisheries:
1. Satellite vessel monitoring systems help identify fishing zones, allowing fishermen to find fish more efficiently and manage resources better.
2. Barcodes are used to identify the time and date of catch, species name, and quality of fish. This helps with tracking, managing stock, and ensuring food safety.
These technologies improve the efficiency, sustainability, and traceability within the fishing industry.
In simple words: In fishing, ICT uses satellites to find fish and barcodes to track fish quality, making fishing smarter and more organized.

🎯 Exam Tip: When listing applications, provide specific examples. For fisheries, focus on how ICT helps with locating fish, managing catches, and ensuring quality control.

XI. Three Marks Questions:

 

Question 1. What are the advantages of FM?
Answer:
1. FM significantly reduces noise, leading to a much better signal-noise ratio.
2. Its operating range is quite wide.
3. The transmission efficiency of FM is very high.
4. FM bandwidth covers the entire range of frequencies needed.
5. FM radio generally offers better sound quality compared to AM radio. This makes FM popular for high-quality audio broadcasts.
In simple words: FM is good because it has less noise, works over a larger area, and sends signals very efficiently, making the sound clear.

🎯 Exam Tip: When listing advantages, focus on key benefits like noise reduction, range, and quality, as these are critical aspects of communication systems.

 

Question 2. What are the disadvantages of FM?
Answer:
1. FM requires a much wider channel for transmission.
2. FM transmitters and receivers are more complex and therefore more costly to build.
3. In FM reception, the signal covers a smaller geographic area compared to AM. These factors can limit its use in certain applications.
In simple words: FM needs more space on the radio waves, its equipment costs more, and it does not cover as wide an area as AM.

🎯 Exam Tip: Remember to contrast FM's benefits with its drawbacks, especially regarding bandwidth and cost, to show a complete understanding.

 

Question 3. Give the reason why transmission of TV signals via sky wave is not possible.
Answer:
1. Television frequencies are typically in the range of 100-220 MHz. The ionosphere cannot reflect these high frequencies.
2. Therefore, skywave propagation cannot be used for TV transmission. The ionosphere reflects only lower frequencies, making skywave unsuitable for TV signals.
In simple words: TV signals use very high frequencies that bounce off the ionosphere instead of being reflected. So, we cannot use sky waves to send TV signals.

🎯 Exam Tip: Highlight the specific frequency range of TV signals and the ionosphere's inability to reflect them as the core reason.

 

Question 4. Show in diagram. the skip distance and the skip zone.
Answer:A skip zone is an area where radio waves cannot be received, found between the maximum range of ground waves and the minimum range of sky waves reflected by the ionosphere. The skip distance is the shortest distance from the transmitter at which the sky wave is first received after reflection from the ionosphere. This phenomenon depends on the wave frequency and the ionosphere's state.
In simple words: The skip zone is an empty area where no radio signals are received. The skip distance is how far away the sky wave first reaches the ground.

🎯 Exam Tip: Clearly label the transmitter, ground wave, sky wave, ionosphere, skip zone, and skip distance in your diagram for full marks. Use arrows to show wave paths.

 

Question 5. It is necessary to use satellites for long distance TV transmission. Why?
Answer:
1. TV signals have high frequencies that the ionosphere cannot reflect back to Earth.
2. Ground wave transmission is only possible for a limited range, meaning it cannot cover very long distances.
3. Because of these limitations, satellites are used for long-distance TV transmission. Satellites orbiting Earth can receive signals from one point and retransmit them to vast areas globally, overcoming the Earth's curvature.
In simple words: TV signals are too high in frequency to bounce off the atmosphere, and ground signals do not travel far. So, we use satellites to send TV signals over long distances.

🎯 Exam Tip: Emphasize the high frequency of TV signals and the Earth's curvature as key reasons why traditional ground and skywave methods are insufficient for long-distance TV broadcasting.

 

Question 6. Which is more efficient mode of transmission FM or AM?
Answer:
1. FM transmission is more efficient because all the transmitted power is useful for carrying information. This leads to better signal quality and less energy waste.
2. In AM transmission, a significant part of the power is used to transmit only the carrier wave, which does not carry information. This makes AM less efficient in terms of power utilization.
In simple words: FM is more efficient because all its power carries useful information. AM wastes power by sending the carrier wave, which does not carry information.

🎯 Exam Tip: The core difference in efficiency lies in how power is distributed; FM uses all power for information, while AM wastes some on the carrier wave.

 

Question 7. Give an account of E-commerce and search engine.
Answer:
**E-commerce:** This involves buying and selling goods and services, as well as transferring money, all done over an electronic network. It simplifies shopping and business operations by allowing transactions online.
**Search engine:** This is a web-based tool that allows users to search for information on the World Wide Web. It helps people find websites, images, videos, and other data by using keywords.
In simple words: E-commerce is shopping and money transfer online. A search engine is a tool to find information on the internet.

🎯 Exam Tip: Define each term clearly, highlighting their primary function in simple terms for full understanding.

 

Question 8. Expand GSM. Give a note on it.
Answer:
GSM stands for Global System for Mobile Communication. It is a digital mobile network widely used for mobile communication. GSM helps increase the use of the network's bandwidth and supports features like network sharing and error detection. It is a standard that allows mobile phones to work worldwide.
In simple words: GSM means Global System for Mobile Communication. It is a system that helps mobile phones work by using the network better and finding errors.

🎯 Exam Tip: Remember to provide both the full form of the acronym and a brief explanation of its purpose and key features.

 

Question 9. Mention some of the applications of optical fiber system.
Answer:
1. It is used for international communication, connecting countries across continents.
2. It enables inter-city communication, linking different cities within a country.
3. Optical fibers are used for data links, allowing fast data transfer between devices.
4. They are crucial for plant and traffic control systems, especially in industrial settings.
5. Optical fiber systems are also vital for defense applications due to their high security and resistance to interference.
In simple words: Optical fibers are used for talking between countries, connecting cities, sending data, controlling factories and traffic, and for military uses.

🎯 Exam Tip: List a variety of applications, emphasizing both large-scale communication (international, inter-city) and specific uses like data links and control systems.

 

Question 10. What are repeaters in an electronic communication system?
Answer:
1. Repeaters are devices that combine a transmitter and a receiver.
2. They are used to increase the range or distance over which signals can be sent.
3. Repeaters receive signals, amplify them, and then retransmit them at a different frequency to reach their destination. This helps signals travel further without losing strength.
In simple words: Repeaters are like signal boosters. They catch weak signals, make them strong, and send them further to cover longer distances.

🎯 Exam Tip: Emphasize that repeaters are a combination of transmitter and receiver, and their main function is to extend signal range through amplification and retransmission.

 

Question 11. State two factors by which the range of transmission of T.V. signals can be increased.
Answer:
The range of TV transmission can be increased by using:
1. Tall antennas, as a greater antenna height increases the line-of-sight distance.
2. Geostationary satellites, which can relay signals over vast distances across the Earth's surface. These satellites stay in one place relative to the Earth, making them ideal for broadcasting.
In simple words: You can send TV signals farther by using taller antennas or by sending them through satellites that stay in one place above the Earth.

🎯 Exam Tip: Focus on practical engineering solutions: antenna height for direct transmission and satellites for overcoming Earth's curvature.

 

Question 12. Compare the difference between PM and FM.
Answer:

Both Phase Modulation (PM) and Frequency Modulation (FM) are types of angle modulation, but they differ in how they encode information onto the carrier wave, impacting their bandwidth and data capacity.
In simple words: PM uses less radio space and sends information faster, while FM uses more radio space and sends information slower.

🎯 Exam Tip: When comparing, make sure each point in one column directly contrasts with a corresponding point in the other column to highlight the differences clearly.

 

Question 13. Why should transmitters broadcasting programmes use different carrier frequencies?
Answer:
1. Different audio signals from various stations can fall within the same range of frequencies.
2. To prevent interference, different transmitting stations are assigned separate slots in the radio frequency range. This ensures each station has its own unique space.
3. As a result, a single receiver can tune into these different frequencies without any confusion or overlap between programs. This allows listeners to choose specific stations clearly.
In simple words: Radio stations use different frequencies so their signals do not mix up. This way, you can clearly hear one station without another one getting in the way.

🎯 Exam Tip: The primary reason is to avoid interference and allow distinct channel selection. Mentioning 'spectral range' or 'frequency slots' shows a deeper understanding.

 

Question 14. Give one example each of a system that uses the (i) sky - wave (ii) space wave mode of propagation.
Answer:
(i) Short broadcast services commonly use skywave propagation. This method allows radio signals to travel long distances by reflecting off the ionosphere.
(ii) TV broadcasting, microwave links, and satellite communication use space wave propagation. This mode involves signals traveling in a straight line, often requiring line-of-sight.
In simple words: Short radio stations use sky waves. TV, microwave, and satellite systems use space waves.

🎯 Exam Tip: Provide clear and distinct examples for each propagation mode, as specified by the question.

 

Question 15. A transmitting antenna has a height of 40m and the height of the receiving antenna is 30m. What is the maximum distance between them for line-of-sight communication? [The radius of the earth is \( 6.4 \times 10^6 \) m]
Answer:
Given:
Height of transmitting antenna \( h_1 = 40 \) m
Height of receiving antenna \( h_2 = 30 \) m
Radius of Earth \( R = 6.4 \times 10^6 \) m
The total distance \( d \) between the transmitting and receiving antennas for line-of-sight communication is the sum of the individual distances of coverage.
The maximum line-of-sight distance \( d \) is given by the formula:
\( d = \sqrt{2Rh_1} + \sqrt{2Rh_2} \)

First, calculate each part:
\( \sqrt{2Rh_1} = \sqrt{2 \times 6.4 \times 10^6 \times 40} \)
\( \implies \sqrt{512 \times 10^6} \)
\( \implies \sqrt{512} \times 10^3 \)
\( \implies 22.627 \times 10^3 \) m
\( \implies 22.627 \) km

\( \sqrt{2Rh_2} = \sqrt{2 \times 6.4 \times 10^6 \times 30} \)
\( \implies \sqrt{384 \times 10^6} \)
\( \implies \sqrt{384} \times 10^3 \)
\( \implies 19.596 \times 10^3 \) m
\( \implies 19.596 \) km

Now, sum these distances to get the total maximum distance:
\( d = 22.627 \text{ km} + 19.596 \text{ km} \)
\( \implies d = 42.223 \) km
Therefore, the maximum distance between the antennas for line-of-sight communication is approximately \( 42.223 \) km.
In simple words: We find how far each antenna can see on its own using a special formula with the Earth's radius. Then, we add those two distances together to get the total maximum distance they can communicate directly.

🎯 Exam Tip: Remember the formula for line-of-sight distance over Earth's curvature. Ensure you correctly handle the powers of 10 and square roots in your calculations.

XII. Five Mark Questions:

 

Question 1. Explain mobile communication? Write its applications.
Answer:
Mobile communication allows people to talk to each other in different places without needing wires or cables. It sends signals over a large area, letting people communicate from anywhere, whether they are at home, in the office, or even in remote areas. It also offers a "roaming" feature, which means users can move around without losing their connection. Setting up and keeping this network running is also quite affordable.
**Applications:**
1. It helps manage vehicle fleets, allowing companies to track cars, trucks, and buses.
2. It is used in wildlife management to count wild animals.
3. Using the Internet of Things (IoT), mobile communication makes it possible to control various devices from one phone, like home automation.
4. It supports smart classrooms by providing online notes and monitoring student activities in education. Mobile technology has transformed how we connect and share information.
In simple words: Mobile communication lets us talk without wires, even far away. It is used for tracking cars, counting animals, smart homes, and helping in schools.

🎯 Exam Tip: Define mobile communication by highlighting its wireless nature and flexibility. For applications, provide varied examples from daily life, industry, and education.

 

Question 2. Give a brief account on RADAR and its applications.
Answer:
RADAR stands for Radio Detection and Ranging. It is a system that can find the angle, distance, or speed of objects that cannot be seen by the human eye. It uses electromagnetic waves for this communication. A signal is first sent out from an antenna in all directions. When this signal hits an object, it reflects back to the radar antenna. The antenna then sends this reflected radio signal to a receiver. These signals are processed to find out details about the object. The distance to the object is found by calculating how long it takes for the signal to travel to the object and come back.
**Applications:**
1. It helps to sense, find, and locate distant objects such as aircraft, ships, and spacecraft.
2. In military use, it is crucial for finding and tracking targets.
3. It is used in navigation systems like ship-borne surface search, air search, and weapon guidance systems.
4. Radar can measure rainfall rates and wind speed in weather observation.
5. It is also used to find and rescue people in emergencies.
In simple words: RADAR means Radio Detection and Ranging. It sends radio waves to find objects we cannot see, telling us their position and speed. It is used for planes, ships, military, weather, and rescue.

🎯 Exam Tip: Start with the full form of RADAR and explain its basic working principle (sending waves, receiving reflections). Provide a wide range of applications from different fields.

 

Question 3. What do you know about mobile communication? Give its applications.
Answer:
Mobile communication allows people to communicate in different locations without needing physical connections like wires or cables. It allows signals to be sent over a wide area without the need for a direct link. This system provides communication access to people in remote areas and offers a roaming facility, meaning users can move between places without losing their connection. The maintenance and installation costs for this type of network are also relatively low.
**Applications:**
1. It is used for personal communication and for connecting cellular phones, which offer high-speed voice and data services.
2. News can be transmitted across the globe in just a few seconds.
3. Mobile communication is used in smart classrooms to provide online notes and monitor student activities.
4. It is used in fleet vehicle management for tracking cars, trucks, and buses.
5. In wildlife management, it helps in counting wild animals.
6. It helps engineers in making tunnels and bridges.
In simple words: Mobile communication is talking without wires, letting us connect from far away. It helps with phones, news, schools, tracking vehicles, counting animals, and building tunnels.

🎯 Exam Tip: Ensure your explanation of mobile communication covers its wireless nature, flexibility, and roaming capability. List diverse applications across personal, social, and professional domains.

 

Question 4. What is bandwidth? Explain the bandwidth of the transmission system.
Answer:
**Bandwidth:** Bandwidth refers to the frequency range over which baseband signals or information signals, like voice, music, or pictures, are transmitted. It essentially tells us the capacity of a communication channel. The communication system depends on the specific frequency band used for a given signal. Bandwidth also represents the portion of the electromagnetic spectrum occupied by a signal. If \( \gamma_1 \) and \( \gamma_2 \) are the lower and upper-frequency limits of a signal, then Bandwidth \( BW = \gamma_2 - \gamma_1 \). This difference indicates the range of frequencies a signal uses.
**Bandwidth of a Transmission System:**
1. This is the range of frequencies needed to send a specific piece of information through a particular channel.
2. An amplitude modulation (AM) system requires a channel bandwidth of 10 kHz to transmit a 5 kHz signal.
3. A single side-band (SSB) system, however, only needs a 5 kHz channel bandwidth for a 5 kHz signal.
4. This is because, in amplitude modulation, the channel bandwidth is twice the signal frequency.
5. Reducing the channel bandwidth is important to fit more communication channels into the available electromagnetic spectrum. This optimizes the use of limited frequency resources.
In simple words: Bandwidth is how much frequency space a signal uses. For a transmission system, it is the range of frequencies needed to send information. Different systems need different amounts of bandwidth.

🎯 Exam Tip: Clearly define bandwidth as a frequency range. When explaining transmission system bandwidth, use examples like AM and SSB to illustrate how bandwidth requirements differ for various modulation techniques.

 

Question 5. Explain the process of modulation.
Answer:
Modulation is a process where a low-frequency baseband signal (input signal) is superimposed onto a high-frequency radio signal for long-distance transmission. The original information signal usually does not have enough energy to travel far by itself. This process ensures that the signal can travel long distances with less weakening. The carrier wave is a high-frequency sine wave signal, which is more suitable for propagation and more compatible with transmission systems.
A sine wave can be shown as \( e_c = E_c \sin(2\pi\gamma_ct + \phi) \), where \( E_c \) is the amplitude, \( \gamma_c \) is the frequency, and \( \phi \) is the initial phase of the carrier wave at time 't'. The characteristics of the carrier signal are modified in three main ways during modulation:
1. **Amplitude modulation (AM):** The amplitude of the carrier wave changes according to the input signal.
2. **Frequency modulation (FM):** The frequency of the carrier wave changes according to the input signal.
3. **Phase modulation (PM):** The phase of the carrier wave changes according to the input signal.
In simple words: Modulation is like attaching a small voice signal onto a powerful radio wave to send it far away. We can change the radio wave's strength, speed, or starting point to carry the voice.

🎯 Exam Tip: Begin with a clear definition of modulation and why it is needed. Then, explain the three main types of modulation (AM, FM, PM) by stating which characteristic of the carrier wave is varied.

 

Question 6. What do you know about INTERNET? Write its few applications?
Answer:
The Internet is a rapidly growing technology in communication systems, offering many tools. It provides new ways for people to interact and connect with each other. It is the largest computer network globally, connecting millions of people through various devices. The Internet has extensive uses in almost every part of life, revolutionizing how we access information and communicate. This global network continuously evolves, making communication faster and more accessible.
**Applications:**
1. **Search engine:** Tools like Google or Bing help users find information on the World Wide Web using keywords.
2. **Communication:** It helps millions of people connect through social networking, emails, instant messaging, and other online tools.
In simple words: The Internet is a huge global computer network that lets people connect and share information. It is used for searching things online and for talking to others through messages and social media.

🎯 Exam Tip: Define the Internet as a global network enabling communication. For applications, focus on common and impactful uses like information retrieval (search engines) and social interaction.