GSEB Class 12 Physics Solutions Chapter 15 Communication Systems

 

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 Hz
Answer: (b) 10 MHz
In simple words: Sky waves are used for long-distance communication because they bounce off the ionosphere. Frequencies around 10 MHz are best for this; lower frequencies (like 10 kHz) cannot be sent effectively by antennas, and higher frequencies (like 1 GHz) pass straight through the ionosphere.

🎯 Exam Tip: Understand the characteristics of different frequency ranges (LF, MF, HF, UHF) and how they interact with the atmosphere (ground, sky, space waves) for effective communication. Pay attention to antenna size requirements and ionospheric penetration/reflection.

 

Question 2. Frequencies in the UHF range normally propagate by means of:
(a) Ground waves.
(b) Sky waves.
(c) Space waves.

Service	Frequency bands	Comments
Standard AM broadcast	540-1600 kHz
FM broadcast	88-108 MHz
Television	54-72 MHz	VHF (very high frequencies)
	76-88 MHz	TV
	174-216 MHz	UHF (ultra high frequencies)
	420-890 MHz	TV
Cellular Mobile Radio	896-901 MHz	Mobile to base station
	840-935 MHz	Base station to mobile
Satellite Communication	5.925-6.425 GHz	Uplink
	3.7-4.2 GHz	Downlink

Answer: The provided text doesn't explicitly select an option for Question 2 in the standard MCQ format, but gives a table of frequency bands and services. Based on common knowledge in physics, UHF range frequencies typically propagate using space waves.
In simple words: Ultra High Frequency (UHF) signals, like those used for TV and cellular mobile radio, travel mainly through space waves. This means they move in a straight line from the transmitting antenna to the receiving antenna. The table above shows the specific frequency ranges for various communication services, including UHF for television.

🎯 Exam Tip: While the question lists MCQ options, the provided answer is a table detailing frequency bands for various services. For such cases, focus on extracting relevant information from the table that broadly relates to the question. Note that UHF typically uses space wave propagation, which is implicitly supported by its use in line-of-sight applications like TV broadcasting mentioned in the table.

 

Question 3. Digital signals
(i) do not provide a continuous set of values
(ii) represent values as discrete steps
(iii) can utilize binary system
(iv) can utilize decimal as well as binary systems
Which of the above statements are true?
(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (i), (ii) and (iii) only
(d) All of (i), (ii), (iii) and (iv).
Answer: (c) (i), (ii) and (iii) only
In simple words: Digital signals are not smooth; they use specific, separate values, like steps. They mostly use a binary system (0s and 1s). The decimal system, which has a continuous range of values, is not used for digital signals in their core form.

🎯 Exam Tip: Remember that digital signals are discrete and typically binary, contrasting with analog signals which are continuous. Understanding these fundamental differences is crucial for distinguishing between digital and analog communication systems.

 

Question 4. Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81 m tall. How much service area can it cover if the receiving antenna is at the ground level?
Answer: No. The service area will be \( A = \pi d_T^2 \). Using the formula, \( A = \frac{22}{7} \times 162 \times 6.4 \times 10^6 = 3258 \text{ km}^2 \).
In simple words: For line-of-sight communication, antennas do not need to be at the same height. The transmitting antenna just needs to be high enough for its signal to reach the receiving antenna without obstacles. For a 81-meter tall TV antenna, the area it can cover is about 3258 square kilometers.

🎯 Exam Tip: For line-of-sight (LOS) communication, the range (d) is typically calculated using \( d = \sqrt{2Rh} \), where R is the Earth's radius and h is the antenna height. The service area is then \( A = \pi d^2 = \pi (2Rh) \). Ensure to use consistent units for calculation (e.g., all in meters or all in kilometers).

 

Question 5. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%, if a carrier wave with a peak voltage of 12V is used to transmit a message signal?
Answer: The modulation index is `\( \mu = 0.75 \)` and is given by `\( \mu = \frac{A_m}{A_C} \)`. With a carrier voltage `\( A_C = 12 \text{V} \)`, the modulating signal voltage `\( A_m = 0.75 \times 12 = 9 \text{V} \)`.
In simple words: To get a 75% modulation, if the carrier wave has a peak voltage of 12V, the modulating signal needs to have a peak voltage of 9V. This calculation ensures the signal strength changes appropriately with the carrier.

🎯 Exam Tip: The modulation index (\(\mu\)) is a key parameter in amplitude modulation, representing the depth of modulation. It's calculated as the ratio of modulating signal amplitude (\(A_m\)) to carrier wave amplitude (\(A_c\)). Remember that `\(\mu\)` should ideally be less than or equal to 1 for distortion-free transmission.

 

Question 6. A modulating signal is a square wave, as shown in the figure.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वर्गाकार मॉडुलन सिग्नल को दर्शाता है। इसमें सिग्नल का आयाम (वोल्टेज) समय के साथ दो निश्चित स्तरों (जैसे 1 वोल्ट और 0 वोल्ट) के बीच बदलता रहता है, जो एक वर्ग तरंग का आकार बनाता है। यह ग्राफ दिखाता है कि सिग्नल 0 से 1 सेकंड तक 1 वोल्ट पर है, फिर 1 से 2 सेकंड तक 0 वोल्ट पर है, और फिर वापस 1 वोल्ट पर आता है, जिससे एक आयताकार या वर्गाकार पल्स पैटर्न बनता है।
The carrier wave is given by `\( c(t) = 2\sin(8\pi t) \)` volts.
i. Sketch the amplitude modulated waveform
ii. What is the modulation index?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक आयाम-मॉडुलित (AM) तरंगरूप को दर्शाता है। इसमें, वाहक तरंग का आयाम (ऊंचाई) एक वर्ग तरंग मॉडुलन सिग्नल के अनुसार बदलता है। जब मॉडुलन सिग्नल उच्च होता है (जैसे 1 वोल्ट), तो वाहक तरंग का आयाम बढ़ जाता है, और जब मॉडुलन सिग्नल निम्न होता है (जैसे 0 वोल्ट), तो वाहक तरंग का आयाम घट जाता है। इससे एक लिफाफा बनता है जो मॉडुलन सिग्नल के आकार का अनुसरण करता है।
(b) The modulation index, `\( \mu = 0.5 \)`
In simple words: The first diagram shows a simple square wave used to change the carrier signal. The carrier signal itself is a sine wave with a peak voltage of 2V. When these are combined in amplitude modulation, the final waveform will have its peak strength changing according to the square wave pattern, making the envelope look like a square wave. The modulation index for this process is calculated as 0.5.

🎯 Exam Tip: When sketching AM waveforms, remember that the envelope of the modulated wave should follow the shape of the modulating signal. The modulation index indicates how much the carrier amplitude changes; for a square wave modulating signal, the carrier amplitude will shift between two distinct levels.

 

Question 7. For an amplitude modulated wave, the maximum amplitude is found to be 10V while the minimum amplitude is found to be 2V. Determine the modulation index. What would be the value of it if the minimum amplitude is zero volt?
Answer: An AM wave is typically expressed as `\( (A_c + A_m \sin(\omega_m t)) \cos(\omega_c t) \)`. The maximum amplitude, `\( M_1 = A_c + A_m \)`, is 10V. The minimum amplitude, `\( M_2 = A_c - A_m \)`, is 2V.
`\( m = \frac{A_m}{A_c} = \frac{M_1 - M_2}{M_1 + M_2} = \frac{10 - 2}{10 + 2} = \frac{8}{12} = \frac{2}{3} \)`
If `\( M_2 = 0 \)`, then `\( m = 1 \)` (which indicates 100% modulation), regardless of the value of `\( M_1 \)`.
In simple words: If an AM wave swings between a high of 10V and a low of 2V, its modulation index is 2/3. If the lowest point of the wave was 0V instead, it means the modulation is complete or 100%, so the modulation index would be 1.

🎯 Exam Tip: The modulation index can be calculated directly from the maximum (`\( M_1 \)`) and minimum (`\( M_2 \)`) amplitudes of the AM wave using the formula `\( m = \frac{M_1 - M_2}{M_1 + M_2} \)`. Remember that `\( m=1 \)` corresponds to 100% modulation, where the minimum amplitude `\( M_2 \)` is zero.

 

Question 8. Due to economic reasons, only the upper side band of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.
Answer: For simplicity, let the received signal be `\( A_1 \cos(\omega_c + \omega_m)t \)`. The carrier `\( A_c \cos(\omega_c t) \)` is also available at the receiving station. By multiplying these two signals, we get:
`\( A_1 A_c \cos(\omega_c + \omega_m)t \cos(\omega_c t) \)`
`\( = \frac{A_1 A_c}{2} [\cos((2\omega_c + \omega_m)t) + \cos(\omega_m t)] \)`
If this combined signal is then passed through a low-pass filter, the high-frequency term `\( \cos((2\omega_c + \omega_m)t) \)` is removed. What remains is the modulating signal:
`\( \frac{A_1 A_c}{2} \cos(\omega_m t) \)`
In simple words: If we send only one part of an AM signal (the upper side band) but have the original carrier signal at the receiver, we can get back the original message. We do this by multiplying the received signal with the recreated carrier. Then, using a filter that removes high-frequency parts, we are left with only the low-frequency message signal.

🎯 Exam Tip: This process describes synchronous detection (or coherent detection), where a locally generated carrier is multiplied with the received sideband to recover the modulating signal. The use of a low-pass filter is essential to separate the desired low-frequency message from unwanted high-frequency components.

GSEB Class 12 Physics Communication Systems Additional Important Questions And Answers

 

Question 1. How will you classify communication systems?
Answer: Communication systems can be classified based on the nature of the signal source, the method of communication, the type of modulation used, and the kind of channel through which information travels.
In simple words: We can sort communication systems by what kind of information they send, how they send it, if they change the signal (modulate), and what path the signal takes (channel).

🎯 Exam Tip: When classifying communication systems, think about the key components: the signal itself (analog/digital), the transmission method (wired/wireless), and how the signal is prepared for transmission (modulation techniques).

 

Question 2. What are the types of channels used for transmission?
Answer:
i. Space communication (e.g., Broadcasting, microwave mobile communication)
ii. Line communication (e.g., Two wire lines, co-axial cables, fiber optical cables)
In simple words: Signals can travel in two main ways: through open space (like radio waves or satellite signals) or through physical lines (like phone wires, cable TV wires, or fiber optic cables).

🎯 Exam Tip: Distinguish between guided (line) and unguided (space) transmission media. Guided media offer more security and less interference over short distances, while unguided media are essential for broadcasting and mobile communication.

 

Question 3. What is the length of antenna required to transmit wave of frequency 40 Hz and 40 MHz?
Answer: The minimum length of antenna required is usually `\( \frac{\lambda}{4} \)`.
i. If `\( \nu = 40 \text{ Hz} \)`, then `\( \frac{\lambda}{4} = \frac{c}{4\nu} = \frac{3 \times 10^8}{4 \times 40} = 1.875 \times 10^6 \text{ m} \)`
ii. If `\( \nu = 40 \text{ MHz} \)`, then `\( \frac{\lambda}{4} = \frac{c}{4\nu} = \frac{3 \times 10^8}{4 \times 40 \times 10^6} = 1.875 \text{ m} \)`
In simple words: The ideal size for an antenna is usually one-quarter of the wavelength of the signal it sends. For a very low frequency of 40 Hz, an antenna would need to be extremely long (over 1.8 million meters). For a much higher frequency like 40 MHz, the antenna would be a practical size (around 1.875 meters).

🎯 Exam Tip: Remember the inverse relationship between frequency and wavelength (`\( c = \nu \lambda \)`). Antenna length is directly proportional to wavelength. This explains why very low frequencies require impractically large antennas, limiting their use in general broadcasting.

 

Question 4. Identify the sound that can travel a longer distance - siren from a factory or horn of a car. Why?
Answer: A siren from a factory travels a longer distance. This is because it has high intensity.
In simple words: A factory siren can be heard from much farther away than a car horn. This is because the factory siren produces sound with a much greater power, or intensity.

🎯 Exam Tip: Sound intensity plays a critical role in how far a sound wave can travel and be detected. Higher intensity sounds, meaning more energy per unit area, can overcome attenuation effects and propagate over greater distances.

 

Question 5. Mention the factors on which power of electromagnetic wave transmitted depends.
Answer: The power of an electromagnetic wave transmitted is related to the antenna's length and the wave's wavelength. The power is proportional to `\( \left(\frac{l}{\lambda}\right)^2 \)`, where `\( l \)` is the antenna length and `\( \lambda \)` is the wavelength.
In simple words: How powerful an electromagnetic wave is when sent out depends on the size of the antenna and the length of the wave itself. Specifically, if the antenna length compared to the wavelength is bigger, more power can be sent.

🎯 Exam Tip: The efficiency of an antenna in radiating power is significantly influenced by its dimensions relative to the wavelength. Antennas that are a significant fraction of a wavelength (like `\( \lambda/2 \)` or `\( \lambda/4 \)`) are more efficient radiators.

 

Question 6. Which range of wave is more reliable to avoid intermixing - shorter or longer wavelength?
Answer: Longer wavelength.
In simple words: To avoid signals getting mixed up, using longer wavelengths is generally more reliable. Longer waves are less prone to certain types of interference.

🎯 Exam Tip: While shorter wavelengths can carry more information (higher bandwidth), longer wavelengths often provide better propagation characteristics, less susceptibility to certain environmental interferences, and better diffraction around obstacles, which can contribute to reliability.

 

Question 7. Is there any change in the frequency or phase due to amplitude modulation?
Answer: No change in either the frequency or the phase occurs due to amplitude modulation.
In simple words: When you use amplitude modulation (AM), only the strength (amplitude) of the signal changes. The speed (frequency) and starting point (phase) of the wave stay exactly the same.

🎯 Exam Tip: Amplitude Modulation (AM) varies the carrier wave's amplitude in proportion to the modulating signal. Contrast this with Frequency Modulation (FM) and Phase Modulation (PM), which vary the carrier's frequency and phase, respectively.

 

Question 8. Which physical quantity of wave is varied in AM, FM and PM?
Answer: In Amplitude Modulation (AM), the physical quantity of the carrier wave that changes is its amplitude. In Frequency Modulation (FM), the physical quantity of the carrier wave that changes is its frequency. In Phase Modulation (PM), the physical quantity of the carrier wave that changes is its phase.
In simple words: In AM radio, the loudness of the wave is changed. In FM radio, the pitch or speed of the wave is changed. In PM, the starting point of the wave is changed.

🎯 Exam Tip: This question highlights the core difference between the three primary types of analog modulation: AM, FM, and PM. Knowing which parameter (amplitude, frequency, or phase) is modulated is fundamental to understanding each type.

 

Question 9. What are the advantages and limitations of AM and FM?
Answer:

	Advantages	Limitations
AM	Wireless transmission is possible, uses a simple circuit, and has two sidebands.	Suffers from low efficiency, small operating range, noisy receptions, and interference effects.
FM	More resistant to noise, has a large number of sidebands, uses a high carrier frequency (e.g., for television broadcasts), and allows for more economical space wave propagation.	Requires wider bandwidth, uses more complex circuits, and has a smaller area of reception for a single transmitter.

In simple words: AM is easy to use and send wirelessly but often sounds noisy and doesn't reach very far. FM is much clearer and better against noise, can carry more information, and uses higher frequencies, but it needs more complex equipment and covers a smaller area per station.

🎯 Exam Tip: When comparing AM and FM, remember that FM's primary advantage is its superior noise immunity due to its frequency varying, while AM's simplicity is its main strength. Bandwidth requirements and circuit complexity also differ significantly.

 

Question 10. What is ground wave propagation?
Answer: Ground wave propagation occurs when radio waves travel along the Earth's surface. These waves follow the Earth's curvature and are effective for carrier frequencies up to 2 MHz, typically used for AM radio. Ground waves must be vertically polarized to prevent short-circuiting electrical equipment. As a ground wave travels, it induces currents in the ground, causing it to lose energy through absorption. This lost energy is partially replenished by energy diffracted downwards from the upper part of the wave front.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दर्शाता है कि कैसे भू-तरंगें (Ground waves) पृथ्वी की सतह के साथ-साथ फैलती हैं। एक ट्रांसमीटर (T) से निकलने वाली तरंगें पृथ्वी की वक्रता का अनुसरण करती हैं। जैसे-जैसे तरंग आगे बढ़ती है, इसका तरंगमुख (wavefront) धीरे-धीरे पृथ्वी की ओर झुकता जाता है, जिसे विवर्तन (diffraction) कहते हैं। यह झुकाव तरंग को पृथ्वी की सतह के पास रहने में मदद करता है, जिससे वह दूर तक जा पाती है।
Additionally, ground waves attenuate due to diffraction, where the wavefront gradually tilts over as it propagates. This tilting increases short-circuiting of the wave's electric component, leading to a reduction in field strength. Eventually, at a certain distance from the antenna, the wave weakens and dies off. The maximum range of such a transmitter depends on its frequency and power, with ground waves being effective only at very low frequencies (VLF).
In simple words: Ground waves are radio signals that travel along the Earth's surface, following its curve. They are good for frequencies up to 2 MHz, like AM radio. They lose energy as they travel because the ground absorbs some of it, and they get weaker over long distances, especially at higher frequencies.

🎯 Exam Tip: Key features of ground wave propagation include its ability to follow Earth's curvature, its effectiveness at lower frequencies (up to 2 MHz), and the attenuation caused by ground absorption and diffraction. Remember that vertical polarization is necessary to minimize energy loss.

 

Question 11. Ground waves are not sustained for a long range communication. Why?
Answer: Ground waves are not sustained for long-range communication because they experience damping (energy loss) due to the Earth's surface.
In simple words: Ground waves cannot travel very far because the Earth's surface absorbs their energy, making them weaker over long distances.

🎯 Exam Tip: The primary reason for the limited range of ground waves is energy dissipation. Emphasize "damping" or "absorption" by the Earth's surface as the key factor when explaining this limitation.

 

Question 12. What is the range of frequencies used in ground wave propagation? Why?
Answer: Ground wave propagation primarily uses Very Low Frequencies (VLF). This is because the attenuation (weakening) of surface waves increases very rapidly as the frequency increases.
In simple words: Ground waves use very low frequencies because signals that move along the ground lose energy much faster when their frequency is high. So, lower frequencies travel better along the ground.

🎯 Exam Tip: Connect the frequency range (VLF) to the propagation characteristics. The rapid attenuation of higher frequencies over the Earth's surface is a crucial reason why ground waves are limited to lower frequency bands.

 

Question 13. How can we overcome this limitation?
Answer: This limitation can be overcome by changing to space wave communication.
In simple words: To send signals farther than ground waves allow, we can switch to space wave communication, which sends signals directly through the air or space.

🎯 Exam Tip: Recognize that different propagation modes are suited for different ranges and frequencies. Switching to space waves (line-of-sight) or sky waves (ionospheric reflection) can extend communication range beyond ground wave limitations.

 

Question 14. What is the basic requirement of space wave communication?
Answer: The basic requirement for space wave communication is the presence of both a transmitting and a receiving antenna.
In simple words: For space wave communication, you just need a sender antenna and a receiver antenna to pass the signal directly between them.

🎯 Exam Tip: Space wave communication, also known as line-of-sight (LOS) propagation, relies on a direct path between antennas. The presence of both antennas is fundamental, along with ensuring they are within each other's line of sight.

 

Question 15. Why is communication also known as line of sight communication?
Answer: The transmitting and receiving antennas must be "on sight," meaning they need a direct, unobstructed path between them.
In simple words: This type of communication is called "line-of-sight" because the sending and receiving antennas must be able to "see" each other directly, without anything blocking the signal.

🎯 Exam Tip: The term "line-of-sight" directly implies the need for an unobstructed visual (or quasi-visual) path between the transmitting and receiving antennas. This is critical for high-frequency signals that do not diffract significantly around obstacles.

 

Question 16. Why are repeaters needed in line of sight communication?
Answer: Repeaters are needed in line-of-sight communication to compensate for the loss of energy that occurs during signal propagation.
In simple words: Signals get weaker as they travel. Repeaters act like boosters, catching the weak signal and sending it out strong again, making sure it reaches far destinations in line-of-sight communication.

🎯 Exam Tip: Attenuation is a significant factor in all communication systems. Repeaters (or relays) are essential to extend the range of line-of-sight systems by amplifying and retransmitting the signal, overcoming the limitations imposed by Earth's curvature and signal loss.

 

Question 17. What are the limitations of space wave propagation?
Answer: This method requires repeaters to extend the range and needs suitable antenna length (height) for effective transmission and reception.
In simple words: Space wave communication has limits because signals can't go through the Earth, so you need tall antennas and sometimes signal boosters (repeaters) to cover long distances.

🎯 Exam Tip: The main limitations of space wave propagation are the line-of-sight restriction (due to Earth's curvature) and the need for repeaters to extend range, which increases infrastructure cost and complexity. Antenna height is crucial for maximizing the line-of-sight distance.

 

Question 18. What is sky wave communication?
Answer: Sky wave communication, also known as ionospheric communication, involves directing high-frequency electromagnetic waves towards the ionosphere, which then reflects these waves back to Earth.
In simple words: Sky wave communication is when radio signals are sent up to a layer in the sky called the ionosphere. This layer bounces the signals back down to Earth, allowing them to travel very long distances, even over the horizon.

🎯 Exam Tip: Key aspects of sky wave communication are the use of the ionosphere as a reflector and its suitability for long-distance communication (beyond the horizon). The frequency of the wave is critical, as only certain frequencies are reflected by the ionosphere.

 

Question 19. Which space transmission technology makes use of total internal reflection?
Answer: Ionospheric transmission makes use of total internal reflection.
In simple words: The way radio waves bounce off the ionosphere, similar to how light bounces off a mirror, is an example of total internal reflection in space communication.

🎯 Exam Tip: Relate the concept of total internal reflection, commonly seen in optics (e.g., fiber optics), to the reflection of radio waves from the ionosphere. This reflection is crucial for sky wave propagation.

 

Question 20. Is sky wave propagation possible on moon? Why?
Answer: No. Sky wave propagation is not possible on the Moon because the Moon has no ionosphere.
In simple words: Sky wave communication cannot happen on the Moon because it lacks an ionosphere, which is the atmospheric layer needed to bounce radio signals.

🎯 Exam Tip: The presence of an ionosphere is a prerequisite for sky wave propagation. Understanding this physical requirement helps explain why such communication modes are planet-specific.

 

Question 21. Can all frequencies be transmitted using sky wave propagation?
Answer: No. Only frequencies below a specific critical value can be transmitted using sky wave propagation.
In simple words: Not all radio frequencies can be sent using sky waves. Only frequencies lower than a certain limit (the critical frequency) can bounce off the ionosphere; higher frequencies just pass through it.

🎯 Exam Tip: The critical frequency is the highest frequency that can be reflected vertically by the ionosphere. Frequencies above this limit, even at an angle, tend to penetrate the ionosphere rather than being reflected back to Earth, thus limiting sky wave propagation.

 

Question 22. How does sky wave propagation depend on refractive index of atmosphere?
Answer: The refractive index of the ionosphere decreases below that of free space due to the change in the velocity of electrons within it. This change in refractive index causes electromagnetic waves to undergo total internal reflection, reflecting them back to Earth.
In simple words: The way radio waves bend and bounce in the sky depends on how the ionosphere changes the speed of light. Because of free electrons, the ionosphere acts like a special mirror, bending waves back down through total internal reflection.

🎯 Exam Tip: The ionosphere's refractive index is a function of electron density and wave frequency. A lower refractive index in the ionosphere causes the electromagnetic waves to bend away from the normal, eventually leading to total internal reflection if the angle of incidence is sufficient.

 

Question 23. Through which atmospheric layer does the propagation take place in ground, space and sky communications?
Answer:
Ground wave – Troposphere
Space wave - Troposphere
Sky wave – Ionosphere
In simple words: Ground waves and space waves travel through the lowest part of the atmosphere, called the troposphere. Sky waves, however, use a higher layer called the ionosphere to bounce signals over long distances.

🎯 Exam Tip: Clearly associate each propagation mode with its corresponding atmospheric layer. Troposphere for ground and space waves, and Ionosphere for sky waves. This distinction is fundamental to understanding radio wave propagation.

 

Question 24. Which type of space communication has maximum range of transmission?
Answer: Satellite communication has the maximum range of transmission.
In simple words: Satellite communication can send signals the farthest because it uses satellites orbiting Earth to relay signals over huge distances.

🎯 Exam Tip: Satellite communication excels in global coverage due to its use of geostationary or low-Earth orbit satellites, which can relay signals across continents and oceans, making it the mode with the maximum transmission range.

 

Question 25. Compare the principle applied for each type of communication.
Answer:
Ground wave - Wireless propagation along the Earth's surface.
Space wave - Line of sight propagation.
Sky wave - Total internal reflection by the ionosphere.
In simple words: Ground waves follow the Earth's surface. Space waves travel in a straight line, needing a clear view between sender and receiver. Sky waves bounce off a high layer in the atmosphere (ionosphere) like a mirror.

🎯 Exam Tip: Understand the distinct physical principle behind each communication type: surface conduction for ground waves, direct propagation for space waves, and atmospheric reflection for sky waves. This clarifies their applications and limitations.

 

Question 26. What is the range of frequency used in each case?
Answer:
Ground wave - Below 2 MHz
Space wave - Greater than 30 MHz
Sky wave - Below 10 MHz
In simple words: Ground waves work for low frequencies, under 2 MHz. Space waves are for high frequencies, above 30 MHz. Sky waves are used for medium frequencies, under 10 MHz.

🎯 Exam Tip: Memorize the typical frequency ranges associated with each propagation mode. These ranges are determined by how electromagnetic waves interact with the Earth's surface and atmospheric layers.

 

Question 27. Point out the limitations and uses in each case.
Answer:
Ground wave – Limitation: Damping effect (energy loss); Use: Wireless communication over short to medium distances.
Space wave - Limitation: Finite curvature of Earth (line of sight limited); Use: Line-of-sight communication.
Sky wave - Limitation: Critical frequency (only frequencies below it are reflected); Use: Long distance coverage.
In simple words: Ground waves get weaker quickly and are for shorter wireless chats. Space waves need a clear path because the Earth is round, used for direct line-of-sight. Sky waves only bounce certain frequencies and are for very long distance communication.

🎯 Exam Tip: For each communication type, be able to state both a key limitation and a primary application. This demonstrates a comprehensive understanding of their practical relevance.

 

Question 28. Name the type of channel used in telephone, cable TV and high speed internet connections.
Answer:
Telephone - Two wire
Cable TV - Coaxial cable
Internet - Space (satellite).
In simple words: Regular phone calls use two simple wires. Cable TV uses a special coaxial cable. High-speed internet can often use satellites to send signals through space.

🎯 Exam Tip: Be familiar with common transmission channels and their applications. Different communication needs (bandwidth, distance, interference) dictate the choice of cable or wireless channel.

 

Question 29. Which is the cheapest mode of line communication?
Answer: The two-wire system is the cheapest mode of line communication.
In simple words: Using two simple wires is the least expensive way to send signals through a physical line.

🎯 Exam Tip: Cost-effectiveness is a practical consideration in communication system design. Two-wire systems, despite their limitations in bandwidth and interference, remain a low-cost option for basic line communication.

 

Question 30. What are the merits and demerits of two wire communication?
Answer: Merits include the ability for signals to travel kilometers without needing amplification, and that both digital and analogue signals can be sent cheaply. Demerits include attenuation of the signal and susceptibility to interference.
In simple words: Two-wire communication is good because signals can travel far without getting boosted, and it's cheap for both digital and analog messages. But, signals can get weaker, and they can easily pick up unwanted noise or interference.

🎯 Exam Tip: When discussing advantages and disadvantages, focus on key aspects like cost, signal integrity (attenuation, interference), and signal types supported (analog/digital) for a balanced answer.

 

Question 31. Why twisted wires are preferred?
Answer: Twisted wires are preferred to reduce interference from electromagnetic radiations.
In simple words: Wires are twisted together to help stop unwanted electrical noise and interference from messing with the signal.

🎯 Exam Tip: The twisting of wires helps to cancel out electromagnetic interference (EMI) by ensuring that any induced noise voltage in one wire is nearly cancelled by an opposing voltage in the adjacent twisted wire. This enhances signal integrity.

 

Question 32. Under which condition does maximum power transmission occur through two wire lines?
Answer: Maximum power transmission occurs through two-wire lines when the impedance of the detecting device at the receiver (load) is matched (i.e., equal) to the characteristic impedance of the two-wire system.
In simple words: For the most power to pass through two wires, the electrical resistance of the device receiving the signal must exactly match the natural electrical resistance of the wires themselves.

🎯 Exam Tip: Impedance matching is a critical concept in electrical engineering. It ensures that maximum power is transferred from source to load and minimizes signal reflections, leading to efficient system operation.

 

Question 33. What kind of cable is used to connect a VCR to a TV?
Answer: A coaxial cable is used to connect a VCR to a TV.
In simple words: To hook up an old VCR player to a TV, you typically use a coaxial cable.

🎯 Exam Tip: Coaxial cables are widely used for video and RF signals due to their shielding properties, which help maintain signal quality over moderate distances and reduce external interference.

 

Question 34. Draw the figure of a coaxial cable.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समाक्षीय केबल (coaxial cable) की आंतरिक संरचना को दर्शाता है। इसमें सबसे अंदर एक केंद्रीय चालक (Central conductor) होता है, जो आमतौर पर तांबे का तार होता है। इस केंद्रीय चालक को इन्सुलेशन (Insulation) की एक परत से घेरा जाता है। इन्सुलेशन के ऊपर एक धात्विक शील्ड (Shield) होती है, जो बाहरी हस्तक्षेप से बचाती है। सबसे बाहरी परत को जैकेट (Jacket) कहते हैं, जो केबल को भौतिक क्षति से बचाता है।
In simple words: A coaxial cable has a center wire for the signal, surrounded by an insulating layer. Around that is a metal shield to block interference, and finally, an outer protective jacket.

🎯 Exam Tip: When describing or drawing a coaxial cable, ensure to label its four main components: central conductor, dielectric insulator, braided shield, and outer jacket. Each component serves a specific function in signal transmission and protection.

 

Question 35. What is the structure of a coaxial cable?
Answer: Coaxial cables are shielded; an outer conductor surrounds an insulated inner wire, and this outer conductor is always grounded.
In simple words: A coaxial cable is built with a central wire for the signal, then a layer of insulation, followed by a metal shield that wraps around it, and this shield is always connected to the ground for safety and to reduce noise.

🎯 Exam Tip: The shielded design of a coaxial cable, with its grounded outer conductor, is crucial for minimizing electromagnetic interference and signal leakage, making it suitable for high-frequency applications.

 

Question 36. Which type of material is suitable to use as spacer in coaxial cable? Why?
Answer: A solid dielectric material is suitable to use as a spacer in a coaxial cable. It acts as insulation.
In simple words: Inside a coaxial cable, a solid plastic-like material is used to keep the central wire separated from the outer shield. This material acts as an insulator, stopping electricity from flowing where it shouldn't.

🎯 Exam Tip: The dielectric material in a coaxial cable is crucial for maintaining a consistent impedance, minimizing signal loss, and providing electrical insulation between the inner and outer conductors. Its properties (dielectric constant, loss tangent) affect cable performance.

 

Question 37. What are the merits of coaxial cables?
Answer: Coaxial cables do not suffer from radiation problems and can be used for microwaves.
In simple words: Coaxial cables are good because they don't leak much signal energy and can handle very fast signals, like those used for microwaves.

🎯 Exam Tip: The primary advantages of coaxial cables are their reduced radiation loss (due to shielding), good bandwidth, and suitability for higher frequency signals (including microwaves), making them versatile for various applications.

 

Question 38. For establishing a communication between a transmitting and receiving station, a physical medium is used.
(a) Name the two principal classes of communication based on the physical medium used for propagation.
(b) Construct a table showing advantages and one practical application each for the two-wire, coaxial cable and optic fiber communication.
(c) In cable TV transmission, the channel in the UHF band usually carries relatively more noise compared to the VHF band. Justify.
Answer:
(a) The two principal classes of communication based on the physical medium used for propagation are Line communication and Space communication.
(b)

Communication	Advantages	Practical Application
Two-wire	Less expense	Telephone system
Co-axial	Less interference	Cable TV
Optic fibre	Low transmission loss	Carry computer signals

(c) At higher frequencies, such as those in the UHF band, radiation loss is higher, which contributes to more noise compared to the VHF band.
In simple words: (a) Communication can happen through physical lines (like wires) or through open space. (b) Simple two-wire systems are cheap for phones. Coaxial cables reduce noise for cable TV. Fiber optic cables send signals with very little loss, good for computer data. (c) UHF signals on cable TV have more noise because higher frequency signals tend to lose more energy as radiation.

🎯 Exam Tip: This question covers different aspects of communication channels. For part (b), remember to provide a distinct advantage and a specific application for each cable type. For part (c), recall that radiation loss generally increases with frequency, which is why UHF signals often experience more noise.

 

Question 39. Schematic diagram for three types of satellite orbits are shown below and named as

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पृथ्वी के चारों ओर तीन अलग-अलग प्रकार की उपग्रह कक्षाओं को दर्शाता है। कक्षा A एक भू-स्थिर (Geostationary) कक्षा का प्रतिनिधित्व करती है, जो भूमध्य रेखा के ऊपर स्थित है। कक्षा B एक ध्रुवीय (Polar) कक्षा है जो पृथ्वी के ध्रुवों के ऊपर से गुजरती है। कक्षा C एक निम्न पृथ्वी कक्षा (Low Earth Orbit) है, जो पृथ्वी की सतह के काफी करीब है।
(a) Identify the polar orbit and give its approximate height from earth.
(b) Give the criteria for selecting the frequency of EM wave to be used in photographs from satellites.
(c) A satellite TV company attempts to use 25,000 kHz for up-linking a signal to a satellite. Say whether they have selected an apt frequency. Justify.
Answer:
(a) Orbit C represents a polar orbit. Its approximate height from Earth is about 1000 km.
(b) i. The nature of the atmosphere.
ii. The reluctance of the object.
(c) No. The company has not selected an appropriate frequency. Frequencies below 20 MHz will undergo total internal reflection at the ionosphere, preventing them from reaching the satellite.
In simple words: (a) Orbit C is a polar orbit, around 1000 km high. (b) When choosing frequencies for satellite photos, we need to consider how the atmosphere affects the waves and how the objects on Earth reflect them. (c) A frequency of 25,000 kHz (25 MHz) is too low for sending signals to satellites. Signals below 20 MHz would bounce off the ionosphere and not reach the satellite in space.

🎯 Exam Tip: For satellite communication questions, remember the characteristics of different orbits (geostationary, polar, LEO) and their typical altitudes. Crucially, understand the frequency window for satellite communication: frequencies must be high enough to penetrate the ionosphere (above ~20-30 MHz) but not so high that they are excessively absorbed by atmospheric gases.

 

Question 40. The following diagrams represent some of the modulated signals.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र तीन अलग-अलग प्रकार के मॉडुलित संकेतों को दर्शाता है। चित्र (i) में, वाहक तरंग का आयाम समय के साथ बदल रहा है, जो आयाम मॉडुलन (AM) का संकेत देता है। चित्र (iii) में, वाहक तरंग की आवृत्ति समय के साथ बदल रही है, जबकि आयाम स्थिर है, जो आवृत्ति मॉडुलन (FM) का संकेत देता है। चित्र (ii) भी आयाम मॉडुलन का एक प्रकार दर्शाता है, जहाँ एक पल्स-जैसे सिग्नल वाहक के आयाम को मॉडुलित करता है।
Which among is following correct
(a) i only
(b) ii only
(c) iii only
(d) both i and ii
Answer: (d) both i and ii
In simple words: Diagrams (i) and (ii) correctly show amplitude-modulated signals, where the strength of the main wave changes according to the message. Diagram (iii) shows frequency modulation, where the speed of the wave changes, not its strength.

🎯 Exam Tip: To identify modulated signals, look at how the carrier wave's characteristics (amplitude, frequency, or phase) change. In amplitude modulation, the envelope of the high-frequency carrier signal traces the shape of the low-frequency modulating signal.

 

Question 41. (a) Name the two distinct message categories.
(b) Explain them with examples.
Answer:
(a) In electronic communication systems, we use:
i. Analog signals
ii. Digital signals
(b) i. An analog signal is continuous in both amplitude and time variables. For example, a speech signal converted by a microphone or an ECG (electrocardiogram) signal.
ii. A digital signal is discrete in both amplitude and time. Here, the signal is subjected to time-sampling and amplitude quantization. Examples include a digital video stream or data files.
In simple words: (a) Messages come in two main types: analog and digital. (b) Analog signals are smooth and continuous, like your voice when picked up by a mic. Digital signals are like steps, taking only specific values at specific times, such as computer files or online videos.

🎯 Exam Tip: Understanding the difference between analog (continuous) and digital (discrete, quantized) signals is fundamental to communication systems. Be prepared to provide clear examples for each, highlighting their defining characteristics.

 

Question 42. List the various types of communications according to
(a) nature of information
(b) mode of transmission
(c) transmission channel
(d) types of modulation
Answer:
(a) Speech, picture, fax, data transmission
(b) Analog and digital communication
(c) i. Space communication
ii. Line communication
(d) i. Sinusoidal waves - AM, FM, PM
ii. Pulsed carrier waves - PAM, PTM, PPM, PNM, PCM
In simple words: We can classify communication by: (a) what kind of message is sent (like voice, pictures, or data); (b) if it's analog (smooth) or digital (steps); (c) how it travels (through air or wires); and (d) how the signal is changed (like AM, FM, or pulse types).

🎯 Exam Tip: This question provides a comprehensive overview of communication system classifications. Ensure you can list examples for each category, as this demonstrates a broad understanding of the field.

 

Question 43. Explain the necessity of modulation.
Answer: A signal from a source will usually be too weak to be transmitted over a long distance through a channel. To overcome this, the low-frequency (long-wave) signal is suitably combined with a high-frequency (short-wave) carrier wave. During this combination, a property of the carrier (like amplitude, frequency, or phase) is varied in proportion to the signal. This process is called modulation.
In simple words: Modulation is needed because normal message signals are too weak and low-frequency to travel far. We mix them with a strong, high-frequency carrier wave, letting the message change a property of the carrier. This makes the signal strong enough and suitable for long-distance transmission.

🎯 Exam Tip: The key reasons for modulation are: reducing antenna size, avoiding signal mixing (multiplexing), increasing transmission range, and improving signal-to-noise ratio. Emphasize that low-frequency signals are inefficient for direct long-distance transmission.

 

Question 44. (a) What is meant by demodulation?
(b) What is its necessity?
(c) What are the different types of demodulation?
Answer:
(a) Demodulation, also known as detection, is the process of extracting the original information (message signal) from a modulated wave.
(b) Demodulation is an essential process for the information to be recovered and utilized at the receiving end.
(c) Different types of detectors are used for demodulation, depending on the type of modulation. Examples include optical detectors, diode detectors, etc.
In simple words: (a) Demodulation is like unpacking a message from its carrier wave. (b) It's necessary so the receiver can get the original message and understand it. (c) There are different ways to unpack messages, like using special devices such as diode detectors, depending on how the message was packed (modulated).

🎯 Exam Tip: Demodulation is the reverse process of modulation. While modulation prepares a signal for transmission, demodulation retrieves the original information at the receiver. Different modulation schemes (AM, FM, PM) require specific demodulation techniques.

 

Question 45. (a) What does the figure represent?
(b) What is the function of 'C'?
(c) What is the function of 'R'?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सरल आयाम मॉडुलन (AM) विमॉडुलक (demodulator) या डिटेक्टर सर्किट को दर्शाता है। इसमें एक डायोड (जो एक तरह से करंट को एक ही दिशा में बहने देता है) और एक संधारित्र (C) तथा एक प्रतिरोधक (R) शामिल हैं। एएम इनपुट (AM input) सिग्नल डायोड से गुजरता है, फिर संधारित्र और प्रतिरोधक के माध्यम से संसाधित होता है, जिससे अंततः आउटपुट पर मूल संदेश सिग्नल (O/P voltage) प्राप्त होता है।
Answer:
(a) This figure represents an AM demodulator (or detector) circuit.
(b) The capacitor (C) charges to the peak voltage of the rectified AM signal and then discharges through the resistor (R). It functions as a capacitor filter, smoothing the rectified wave. The diode rectifies the incoming AM signal. The filtered, rectified wave then has the radio frequency (RF) carrier component removed, revealing the original signal.
(c) The voltage across the resistor (R) is the envelope of the modulated wave, which represents the original signal.
In simple words: (a) The drawing shows a circuit that takes an AM radio signal and pulls out the original sound or message. (b) The capacitor (C) smooths out the incoming signal after it's been straightened by the diode, acting like a filter to remove the fast radio wave part. (c) The resistor (R) then allows us to measure the slower, original message signal.

🎯 Exam Tip: A diode detector (AM demodulator) typically consists of a diode for rectification and an RC filter (capacitor C and resistor R) to extract the envelope of the modulated wave, which is the original message signal. Understand how each component contributes to this process.

 

Question 46. Why is Space wave communication called troposphere wave propagation or LOS?
Answer: Space wave communication is called troposphere wave propagation or Line of Sight (LOS) because it takes place under line-of-sight conditions, meaning the waves travel directly through the troposphere (the lowest part of the atmosphere) from the transmitting to the receiving antenna.
In simple words: Space wave communication is called "troposphere wave propagation" because signals travel through the troposphere, and it's also called "line-of-sight" because the sending and receiving antennas must have a clear, direct path between them.

🎯 Exam Tip: Line-of-sight (LOS) propagation is the defining characteristic of space waves, as they travel in a straight path. The troposphere is the atmospheric layer through which this direct propagation occurs, linking these terms together.

 

Question 47. Name the four areas in which space technology finds application?

Answer: Space technology is used in four main areas. These are studying weather (meteorology), understanding climate patterns (climatology), exploring the oceans (oceanography), and looking at areas near the sea (coastal studies).
In simple words: Space technology helps us study weather, climate, oceans, and coastal areas.

🎯 Exam Tip: Listing specific applications of space technology correctly is key to scoring.

 

Question 48. “For long TV transmission we need satellites “. Give reason.

Answer: We need satellites for sending TV signals over long distances. This is because the signal from a TV camera uses about 64 Hz bandwidth. Many things can weaken or change a signal sent directly, meaning it might not get to where it needs to go. Satellites help relay these signals over vast areas.
In simple words: Satellites are needed for long-distance TV because normal signals get weak and cannot reach far.

🎯 Exam Tip: Understanding why direct transmission fails over long distances for TV and how satellites solve this is important.

 

Question 49. How Kepler's III law plays an important role in satellite communication?

Answer: Kepler's Third Law is very important for satellite communication. It helps design steady orbits for satellites. This law states that the time a satellite takes to go around the Earth, squared, is directly proportional to the cube of its average distance from the Earth. This relationship helps engineers place satellites correctly for good communication.
In simple words: Kepler's Third Law helps design satellite orbits by connecting how long a satellite takes to circle Earth to its distance from Earth.

🎯 Exam Tip: Knowing Kepler's Third Law and its application in satellite orbit design is crucial.

 

Question 50.
(a) What do you understand by synchronous satellite?
(b) Why are such satellites used for world wide communications?

Answer:
(a) A synchronous satellite is a special type of satellite. It takes exactly 24 hours to go around the Earth once.
(b) These satellites are used for worldwide communication because they are always available, very dependable, and can cover a very large area on Earth.
In simple words: (a) A synchronous satellite takes 24 hours to orbit Earth. (b) They are used for global talks because they are always there, work well, and cover big areas.

🎯 Exam Tip: Define synchronous satellites by their orbital period and explain their advantages for global communication.