CBSE Class 12 Physics Communication Systems Assignment Set 02

Question- Encoding of signal is required for

(a) modulation at transmitting end

(b) modulation at receiving end

(c) demodulation at receiving end

(d) demodulation at transmitting end

Answer-(a)

Question- Range of communication is extended by using

(a) transmitter

(b) transducer

(c) processor

(d) repeater

Answer-(d)

Question- For transmission of e.m.wave of audible frequency, these waves are superimposed with waves of

(a) frequency less than 20 Hz

(b) frequency less than 10 KHz.

(c) frequency in the audible range.

(d) radio-frequency.

Answer-(d)

Question-In which of the following remote sensing technique is not used?

(a) Forest density

(b) Pollution

(c) Wetland mapping

(d) Medical treatment

Answer-(d)

Question- A radio station has two channels. One is AM at 1020 kHz and the other FM at 89.5 MHz. For good results you will use

(a) longer antenna for the AM channel and shorter for the FM

(b) shorter antenna for the AM channel and longer for the FM

(c) same length antenna will work for both

(d) information given is not enough to say which one to use for which

Answer-(a)

Question- For transmission of TV- signal, sound-part is

(a) amplitude modulated

(b) frequency modulated

(c) phase modulated

(d) pulse modulated

Answer-(b)

Question- Wave obtained on superimposition of audible frequency e.m. wave is known as

(a) carrier wave

(b) high frequency wave

(c) modulating wave

(d) modulated wave

Answer-(d)

Question- In sky-wave propagation, skip-distance depends on

(a) frequency of e.m. waves transmitted

(b) critical frequency of the layer

(c) height of layer above earth’s surface

(d) all of the above

Answer-(d)

Question-Which mode of communication is most suitable for carrier wave of frequencies around 100 MHz?

(a) Satellite

(b) Ground wave

(c) Line of sight

(d) Ionospheric

Answer-(c)

Question- The layer of atmosphere which contains water vapour is

(a) stratosphere

(b) mesospshere

(c) troposphere

(d) ionosphere

Answer-(c)

Question- Space wave communication is limited

(a) to the line of sight distance

(b) by earth’s curvature

(c) either (a) or (b)

(d) both (a) and (b)

Answer-(d)

Question-The waves used in telecommunication are

(a) IR

(b) UV

(c) Microwave

(d) Cosmic rays

Answer-(c)

36. Critical frequency that gets reflected back from ionosphere is

(a) same for all layers of the ionosphere

(b) different for different layers of the ionosphere

(c) not dependent on layers of the ionosphere

(d) None of these

Answer-(d)

Question-The losses in transmission lines are

(a) radiation losses only

(b) conductor heating only

(c) dielectric heating only

(d) all of these

Answer-(d)

Question- Which of the following is drawback of amplitude modulation?

(a) low efficiency

(b) noise reception

(c) operating range is small

(d) all of these

Answer-(d)

Question-Basic components of transmitter are

(a) message signal generator and antenna

(b) modulator and antenna

(c) signal generator and modulator and antenna

(d) message signal generator, modulator and antenna

Answer-(d)

Question-Which of the following device is fully duplex?

(a) Mobile phone

(b) Walky-talky

(c) Loud speaker

(d) Radio

Answer-(a)

Question- In Laser communication there is

(a) low loss of signal

(b) loss of signal

(c) no signal security

(d) low band width

Answer-(a)

Question-Citizen’s band ratio is the application of

(a) amplitude modulation

(b) frequency modulation

(c) phase modulation

(d) None of these

Answer-(a)

Question-A geosynchronous satellite is

(a) located at a height of 34860 km to ensure global coverage

(b) appears stationary over a place on earth’s magnetic pole

(c) not really stationary at all, but orbits the earth within 24 hours.

(d) always at fixed location in space and simply spins about its own axis

Answer-(c)

Question-During ground wave propagation the transmitted waves gets attenuated because

(a) earth surface absorbs the waves

(b) frequency of the waves are too low

(c) energy content of these waves are high

(d) earth surface offers resistance.

Answer-(d)

Short Answer Type Questions
 

Question. State the suitable mode of communication propagation for the following signal frequencies:
(i) 5 MHz
(ii) 100 MHz
Answer:
(i) **For a 5 MHz signal:** Since 5 MHz is less than the critical frequency (\( f_c \)) of the ionosphere, sky wave propagation (ionospheric propagation) is the appropriate mode of communication.
(ii) **For a 100 MHz signal:** Since 100 MHz is much greater than the critical frequency (\( f_c \)) of the ionospheric layers, the ionosphere cannot reflect it. Therefore, satellite communication or line-of-sight space wave propagation must be used.
In simple words:
(i) A 5 MHz signal is low enough to bounce off the ionosphere, so we can send it as a sky wave.
(ii) A 100 MHz signal is too strong and passes right through the ionosphere, so we have to use satellite communication to relay it.

Exam Tip: Use the condition \( f < f_c \) for sky wave and \( f > f_c \) for satellite/space wave to explain your reasoning.

Question. Draw the block diagram of a basic sinusoidal AM transmitter and receiver system.
Answer: Below are the block diagrams representing the essential transmitter and receiver systems in a sinusoidal AM communication network:

In simple words: A transmitter takes sound from a microphone, modulates it onto a carrier wave, amplifies it, and sends it from an antenna. A receiver catches the wave with an antenna, tunes and amplifies it, extracts the sound with a demodulator, amplifies the audio, and plays it through a speaker.

Exam Tip: Draw each block in the correct sequence, showing the signal flow with arrows from left to right. Make sure to label the antennas and transducers (microphone and loudspeaker) clearly.

Question. The maximum peak-to-peak voltage of an AM wave is 16 mV and the minimum peak-to-peak voltage is 4 mV. Calculate the modulation factor.
Answer: Given: Maximum peak-to-peak voltage = 16 mV Minimum peak-to-peak voltage = 4 mV First, we determine the maximum and minimum amplitudes of the AM wave: \[ V_{\max} = \frac{16}{2} = 8 \text{ mV} \] \[ V_{\min} = \frac{4}{2} = 2 \text{ mV} \] The modulation factor (\( m_a \)) is calculated using the formula: \[ m_a = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}} \] Substituting the values: \[ m_a = \frac{8 - 2}{8 + 2} \] \[ m_a = \frac{6}{10} = 0.6 \] Thus, the modulation factor of the AM wave is 0.6 (or 60%).
In simple words: To find the modulation index, we first halve the peak-to-peak voltages to get the maximum and minimum heights of the wave (8 mV and 2 mV). Then we divide their difference by their sum, which gives 0.6.

Exam Tip: Make sure to divide the peak-to-peak values by 2 first to obtain the correct single-peak voltages (\( V_{\max} \) and \( V_{\min} \)) before applying the formula.

Question. An AM wave is represented by the expression: \( v = 5(1 + 0.6 \cos 6280 t) \sin 211 \times 10^4 t \) volts.
(i) What are the maximum and minimum amplitudes of the AM wave?
(ii) What frequency components are contained in the modulated wave?
Answer: Comparing the given wave equation with the standard AM wave equation: \[ v = E_c (1 + m_a \cos \omega_s t) \sin \omega_c t \] We get: Carrier amplitude, \( E_c = 5 \text{ V} \) Modulation index, \( m_a = 0.6 \) Modulating angular frequency, \( \omega_s = 6280 \text{ rad/s} \) Carrier angular frequency, \( \omega_c = 211 \times 10^4 \text{ rad/s} \) Let's calculate the frequencies:
- Modulating signal frequency: \[ f_s = \frac{\omega_s}{2\pi} = \frac{6280}{2 \times 3.14} = 1000 \text{ Hz} = 1 \text{ kHz} \]
- Carrier frequency: \[ f_c = \frac{\omega_c}{2\pi} = \frac{211 \times 10^4}{2 \times 3.14} \approx 3.36 \times 10^5 \text{ Hz} = 336 \text{ kHz} \]
(i) **Maximum and minimum amplitudes:**
- Maximum amplitude of the AM wave: \[ V_{\max} = E_c + m_a E_c = 5 + 0.6 \times 5 = 8 \text{ V} \]
- Minimum amplitude of the AM wave: \[ V_{\min} = E_c - m_a E_c = 5 - 0.6 \times 5 = 2 \text{ V} \]
(ii) **Frequency components contained in the modulated wave:** The AM wave contains three frequencies: the carrier frequency (\( f_c \)), the lower sideband frequency (\( f_c - f_s \)), and the upper sideband frequency (\( f_c + f_s \)):
- Lower sideband: \( f_c - f_s = 336 - 1 = 335 \text{ kHz} \)
- Carrier frequency: \( f_c = 336 \text{ kHz} \)
- Upper sideband: \( f_c + f_s = 336 + 1 = 337 \text{ kHz} \)
In simple words:
(i) The wave's height rises to a maximum of 8 V and dips to a minimum of 2 V.
(ii) The broadcast signal is made of three key frequencies: 335 kHz, 336 kHz, and 337 kHz.

Exam Tip: Identify \( E_c \) and \( m_a \) directly from the given expression. Make sure to divide \( \omega_s \) and \( \omega_c \) by \( 2\pi \) to convert angular frequencies to cyclic frequencies in Hz/kHz.

Question. An audio signal of frequency 1 kHz is used to amplitude-modulate a carrier of frequency 500 kHz.
(i) Determine the sideband frequencies.
(ii) Calculate the bandwidth required for this transmission.
Answer: Given: Modulating signal frequency, \( f_s = 1 \text{ kHz} \) Carrier frequency, \( f_c = 500 \text{ kHz} \)
(i) **Sideband frequencies:** The AM sidebands are given by:
- Upper Sideband Frequency (USB): \[ f_c + f_s = 500 \text{ kHz} + 1 \text{ kHz} = 501 \text{ kHz} \]
- Lower Sideband Frequency (LSB): \[ f_c - f_s = 500 \text{ kHz} - 1 \text{ kHz} = 499 \text{ kHz} \]
(ii) **Bandwidth required:** The channel bandwidth required is the difference between the upper and lower sideband frequencies: \[ \text{Bandwidth} = \text{USB} - \text{LSB} = 501 \text{ kHz} - 499 \text{ kHz} = 2 \text{ kHz} \] *(Alternatively, \( \text{Bandwidth} = 2 f_s = 2 \times 1 \text{ kHz} = 2 \text{ kHz} \))* So the bandwidth ranges from 499 kHz to 501 kHz.
In simple words:
(i) The two new frequency bands created are 499 kHz and 501 kHz.
(ii) The total width of frequency space needed for this station to broadcast is 2 kHz.

Exam Tip: The sidebands are located at \( f_c \pm f_s \), and the total bandwidth is always twice the frequency of the modulating signal (\( 2 f_s \)).

Question. The antenna current of an AM transmitter is 8 A when only the carrier is sent, but it increases to 8.93 A when the carrier is sinusoidally modulated by a single audio signal. Find the percentage modulation index.
Answer: Given: Unmodulated carrier antenna current, \( I_c = 8 \text{ A} \) Total antenna current after modulation, \( I_T = 8.93 \text{ A} \) The relation between the modulated total current (\( I_T \)) and unmodulated carrier current (\( I_c \)) is given by: \[ I_T = I_c \sqrt{1 + \frac{m_a^2}{2}} \] Squaring both sides and rewriting the ratio: \[ \left(\frac{I_T}{I_c}\right)^2 = 1 + \frac{m_a^2}{2} \] Substituting the given currents: \[ \left(\frac{8.93}{8}\right)^2 = 1 + \frac{m_a^2}{2} \] \[ \implies 1.246 = 1 + \frac{m_a^2}{2} \] \[ \implies \frac{m_a^2}{2} = 1.246 - 1 = 0.246 \] \[ \implies m_a^2 = 2 \times 0.246 = 0.492 \] \[ \implies m_a = \sqrt{0.492} \approx 0.701 \] Thus, the percentage modulation index is: \[ m_a = 70.1\% \]
In simple words: When we modulate the signal, the antenna current increases from 8 A to 8.93 A. By squaring the ratio of these currents, we find that the modulation index is about 0.701, which is 70.1%.

Exam Tip: Use the correct formula relating modulated and unmodulated currents: \( I_T = I_c \sqrt{1 + \frac{m_a^2}{2}} \), and solve systematically for \( m_a \).

Question. Define the modulation index for a frequency-modulated (FM) wave.
Answer: In frequency modulation (FM), the modulation index (\( m_f \)) is defined as the ratio of the maximum frequency deviation of the carrier wave to the frequency of the modulating signal: \[ m_f = \frac{\text{Maximum frequency deviation}}{\text{Modulating signal frequency}} \]
In simple words: The FM modulation index tells us how much the carrier's frequency shifts compared to the frequency of the audio or message signal itself.

Exam Tip: Do not confuse the AM modulation index (which depends on amplitude) with the FM modulation index (which depends on frequency deviation and modulating frequency).

Question. A TV transmitting antenna has a height of 160 m.
(i) Find its coverage range.
(ii) What should be the new height of the antenna to double its coverage range? (Take \( R = 6400 \text{ km} \))
Answer: Given: Initial height of the antenna, \( h_1 = 160 \text{ m} \) Radius of the Earth, \( R = 6400 \text{ km} = 6400 \times 10^3 \text{ m} \)
(i) **Coverage range:** The initial coverage range \( d_1 \) is given by: \[ d_1 = \sqrt{2 R h_1} \] Substituting the values: \[ d_1 = \sqrt{2 \times 6400 \times 10^3 \times 160} \] \[ d_1 = \sqrt{2048 \times 10^6} \approx 45,255 \text{ m} = 45.255 \text{ km} \]
(ii) **New height to double the coverage range:** To double the coverage range (\( d_2 = 2 d_1 \)), the height of the antenna must be increased to four times its initial value: \[ h_2 = 4 h_1 \] \[ \implies h_2 = 4 \times 160 \text{ m} = 640 \text{ m} \] Thus, the new height of the antenna must be 640 meters (corresponding to an increase of \( 640 - 160 = 480 \text{ m} \)).
In simple words:
(i) With a 160-meter tower, the signal can reach about 45.26 kilometers away.
(ii) To make the signal reach twice as far (90.52 km), the tower needs to be four times taller, which is 640 meters high.

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

Question. A TV tower has a height of 110 m. How much population is covered by the TV broadcast if the average population density around the tower is \( 1000 \text{ km}^{-2} \)? (Take the radius of the Earth, \( R = 6.4 \times 10^6 \text{ m} \))
Answer: Given: Height of the TV tower, \( h = 110 \text{ m} \) Radius of the Earth, \( R = 6.4 \times 10^6 \text{ m} = 6400 \text{ km} \) Population density = \( 1000 \text{ km}^{-2} \) The transmission radius (range) of the area covered by the TV broadcast is: \[ d = \sqrt{2Rh} \] \[ d = \sqrt{2 \times (6.4 \times 10^6 \text{ m}) \times 110 \text{ m}} \] \[ d = \sqrt{1.408 \times 10^9} \approx 37,500 \text{ m} = 37.5 \text{ km} \] The area covered by the broadcast is: \[ A = \pi d^2 \] \[ A = 3.14 \times (37.5)^2 \approx 4415.6 \text{ km}^2 \] Therefore, the population covered is: \[ \text{Population Covered} = \text{Area} \times \text{Population Density} \] \[ \text{Population Covered} = 4415.6 \text{ km}^2 \times 1000 \text{ km}^{-2} \approx 4.4 \times 10^6 \] Thus, the television broadcast covers a population of approximately \( 4.4 \times 10^6 \) people.
In simple words: A 110-meter tower can broadcast signals over a circle with a radius of 37.5 kilometers. With 1,000 people living per square kilometer, the total area has about 4.4 million people in it.

Exam Tip: Ensure you keep the units consistent: convert the radius \( d \) of the covered area from meters to kilometers before calculating the area if the population density is given in \( \text{km}^{-2} \).

Question. A microwave telephone link operating at a central frequency of 10 GHz has been established. If only 2% of this frequency is available for microwave communication channels, how many telephone channels can be simultaneously granted if each channel is allotted a bandwidth of 8 kHz?
Answer: Given: Central frequency of the link = \( 10 \text{ GHz} = 10 \times 10^9 \text{ Hz} \) Percentage of frequency available = 2% Bandwidth allocated per telephone channel = \( 8 \text{ kHz} = 8 \times 10^3 \text{ Hz} \) First, we calculate the total available bandwidth: \[ \text{Available Bandwidth} = 2\% \text{ of } 10 \text{ GHz} \] \[ \text{Available Bandwidth} = \frac{2}{100} \times 10 \times 10^9 \text{ Hz} = 0.2 \text{ GHz} = 2 \times 10^8 \text{ Hz} \] Next, we find the number of channels that can be simultaneously accommodated: \[ \text{Number of Channels} = \frac{\text{Total Available Bandwidth}}{\text{Bandwidth per Channel}} \] \[ \text{Number of Channels} = \frac{2 \times 10^8 \text{ Hz}}{8 \times 10^3 \text{ Hz}} \] \[ \implies \text{Number of Channels} = 0.25 \times 10^5 = 2.5 \times 10^4 = 25,000 \] Thus, 25,000 telephone channels can be simultaneously granted.
In simple words: We have a total frequency bandwidth of 10 GHz, but we can only use 2% of it, which is 0.2 GHz (or 200 million Hz). Since each phone line needs 8,000 Hz, we can fit exactly 25,000 phone lines at the same time.

Exam Tip: Express all frequencies in standard Hz before division to avoid power-of-ten mistakes during calculations.

Question. You are given three semiconductors A, B, and C with respective band gaps of 3 eV, 2 eV, and 1 eV, to be used in a photodetector to detect a signal of wavelength \( \lambda = 1400 \text{ nm} \). Select the suitable semiconductor and justify your choice.
Answer: Given: Wavelength of the signal, \( \lambda = 1400 \text{ nm} = 1400 \times 10^{-9} \text{ m} \) Band gap of A, \( E_{gA} = 3 \text{ eV} \) Band gap of B, \( E_{gB} = 2 \text{ eV} \) Band gap of C, \( E_{gC} = 1 \text{ eV} \) The energy \( E \) of the incident photon of wavelength \( \lambda \) is given by: \[ E = \frac{hc}{\lambda} \] Substituting the values (\( h = 6.63 \times 10^{-34} \text{ J}\cdot\text{s} \), \( c = 3 \times 10^8 \text{ m/s} \)): \[ E = \frac{1.989 \times 10^{-25}}{1400 \times 10^{-9}} \text{ J} \approx 1.42 \times 10^{-19} \text{ J} \] To convert this energy into electron-volts (eV), we divide by the charge of an electron (\( 1.6 \times 10^{-19} \text{ C} \)): \[ E = \frac{1.42 \times 10^{-19}}{1.6 \times 10^{-19}} \text{ eV} \approx 0.89 \text{ eV} \approx 1 \text{ eV} \] For a photodetector to successfully detect light, the energy of the incident photon must be greater than or equal to the band gap of the semiconductor material (\( E \ge E_g \)):
- For semiconductor A: \( 1 \text{ eV} < 3 \text{ eV} \) (Not suitable)
- For semiconductor B: \( 1 \text{ eV} < 2 \text{ eV} \) (Not suitable)
- For semiconductor C: \( 1 \text{ eV} \ge 1 \text{ eV} \) (Suitable) Therefore, only semiconductor C is suitable for this photodetector.
In simple words: For a material to detect light, the light must carry more energy than the material's internal barrier (its band gap). The light has an energy of about 1 eV, so only material C (which has a 1 eV barrier) can detect it.

Exam Tip: State the core condition for detection clearly: \( E \ge E_g \). Calculate \( E \) in Joules first, convert it to eV, and then compare it directly to each semiconductor's band gap.

Question. The critical frequency of an ionospheric layer is 10 MHz. Calculate the maximum electron density of this layer.
Answer: The critical frequency (\( f_c \)) of the ionosphere and its maximum electron density (\( N_{\max} \)) are related by the formula: \[ f_c = 9 \sqrt{N_{\max}} \] Squaring both sides to solve for \( N_{\max} \): \[ N_{\max} = \frac{f_c^2}{81} \] Given: Critical frequency, \( f_c = 10 \text{ MHz} = 10 \times 10^6 \text{ Hz} = 10^7 \text{ Hz} \) Substituting \( f_c \) into the equation: \[ N_{\max} = \frac{(10^7)^2}{81} \] \[ N_{\max} = \frac{10^{14}}{81} \approx 1.23 \times 10^{12} \text{ m}^{-3} \] Thus, the maximum electron density of the ionospheric layer is \( 1.23 \times 10^{12} \text{ m}^{-3} \).
In simple words: The critical frequency depends on the density of free electrons in the sky. For a frequency of 10 MHz, the calculation shows there are about \( 1.23 \times 10^{12} \) free electrons in every cubic meter of that ionosphere layer.

Exam Tip: Remember the formula \( f_c = 9 \sqrt{N_{\max}} \) where \( f_c \) must be in Hz (not MHz) and \( N_{\max} \) is in \( \text{m}^{-3} \).