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Communication Systems
Physics XII - Chapter 15: Complete NCERT Exercise Solutions
Master Communication Systems with NCERT solutions for modulation, transmitters, receivers, and propagation modes.
Question 15.1
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
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Answer & Explanation:
Answer: (b) 10 MHz
Explanation:
Sky waves are used for long-distance communication by reflection from the ionosphere. This is effective for frequencies typically between 3 MHz to 30 MHz. Frequencies like 10 kHz are too low (ground wave), while GHz frequencies are too high (space wave) and penetrate the ionosphere.
Explanation:
Sky waves are used for long-distance communication by reflection from the ionosphere. This is effective for frequencies typically between 3 MHz to 30 MHz. Frequencies like 10 kHz are too low (ground wave), while GHz frequencies are too high (space wave) and penetrate the ionosphere.
Question 15.2
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves.
(b) Sky waves.
(c) Surface waves.
(d) Space waves.
(a) Ground waves.
(b) Sky waves.
(c) Surface waves.
(d) Space waves.
Answer & Explanation:
Answer: (d) Space waves
Explanation:
UHF (Ultra High Frequency) ranges from 300 MHz to 3 GHz. At these high frequencies, waves travel mostly via line-of-sight or space wave propagation, as they are not reflected by the ionosphere and do not follow the Earth's curvature like ground waves.
Explanation:
UHF (Ultra High Frequency) ranges from 300 MHz to 3 GHz. At these high frequencies, waves travel mostly via line-of-sight or space wave propagation, as they are not reflected by the ionosphere and do not follow the Earth's curvature like ground waves.
Question 15.3
Digital signals
(i) do not provide a continuous set of values,
(ii) represent values as discrete steps,
(iii) can utilize binary system, and
(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) but not (iv)
(d) All of (i), (ii), (iii) and (iv)
(i) do not provide a continuous set of values,
(ii) represent values as discrete steps,
(iii) can utilize binary system, and
(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) but not (iv)
(d) All of (i), (ii), (iii) and (iv)
Answer & Explanation:
Answer: (c) (i), (ii) and (iii) but not (iv)
Explanation:
Digital signals are discrete in nature (i, ii) and commonly use the binary system (iii). While digital systems can represent decimal numbers in coded form (like BCD), the fundamental representation is binary, not decimal directly in the signal form. Thus, (iv) is not generally true.
Explanation:
Digital signals are discrete in nature (i, ii) and commonly use the binary system (iii). While digital systems can represent decimal numbers in coded form (like BCD), the fundamental representation is binary, not decimal directly in the signal form. Thus, (iv) is not generally true.
Question 15.4
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 81m tall. How much service area can it cover if the receiving antenna is at the ground level?
Answer & Explanation:
Answer: No, they need not be at the same height. The service area is approximately \( 3258 \, \text{km}^2 \).
Explanation:
For line-of-sight communication, the transmitting and receiving antennas need to be within each other's horizon, but not necessarily at the same height.
The distance to the horizon from height \( h \) is \( d = \sqrt{2Rh} \), where \( R \) is Earth's radius (\( \approx 6.4 \times 10^6 \, \text{m} \)).
Here, \( h = 81 \, \text{m} \).
\[ d = \sqrt{2 \times 6.4 \times 10^6 \times 81} \approx 32100 \, \text{m} = 32.1 \, \text{km} \]
Service area \( = \pi d^2 \approx 3.14 \times (32.1)^2 \approx 3235 \, \text{km}^2 \) (approx 3258 km² as per NCERT).
Explanation:
For line-of-sight communication, the transmitting and receiving antennas need to be within each other's horizon, but not necessarily at the same height.
The distance to the horizon from height \( h \) is \( d = \sqrt{2Rh} \), where \( R \) is Earth's radius (\( \approx 6.4 \times 10^6 \, \text{m} \)).
Here, \( h = 81 \, \text{m} \).
\[ d = \sqrt{2 \times 6.4 \times 10^6 \times 81} \approx 32100 \, \text{m} = 32.1 \, \text{km} \]
Service area \( = \pi d^2 \approx 3.14 \times (32.1)^2 \approx 3235 \, \text{km}^2 \) (approx 3258 km² as per NCERT).
Question 15.5
A carrier wave of peak voltage \( 12V \) is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%?
Answer & Explanation:
Answer: \( 9V \)
Explanation:
Modulation index \( \mu = \frac{A_m}{A_c} \), where \( A_m \) is peak modulating voltage and \( A_c \) is peak carrier voltage.
Given \( \mu = 75\% = 0.75 \), \( A_c = 12V \).
\[ A_m = \mu \times A_c = 0.75 \times 12 = 9 \, \text{V} \]
Explanation:
Modulation index \( \mu = \frac{A_m}{A_c} \), where \( A_m \) is peak modulating voltage and \( A_c \) is peak carrier voltage.
Given \( \mu = 75\% = 0.75 \), \( A_c = 12V \).
\[ A_m = \mu \times A_c = 0.75 \times 12 = 9 \, \text{V} \]
Question 15.6
A modulating signal is a square wave, as shown in the figure. 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?
(i) Sketch the amplitude modulated waveform.
(ii) What is the modulation index?
Answer & Explanation:
Answer:
(i) The AM waveform will have its envelope following the square wave, with carrier frequency \( 4 \, \text{Hz} \).
(ii) Modulation index \( \mu = 0.5 \)
Explanation:
Modulating signal \( m(t) \) is a square wave of amplitude \( 1V \) (from figure).
Carrier amplitude \( A_c = 2V \).
Modulation index \( \mu = \frac{A_m}{A_c} = \frac{1}{2} = 0.5 \).
The AM waveform is: \( [A_c + m(t)] \sin(8\pi t) \).
(i) The AM waveform will have its envelope following the square wave, with carrier frequency \( 4 \, \text{Hz} \).
(ii) Modulation index \( \mu = 0.5 \)
Explanation:
Modulating signal \( m(t) \) is a square wave of amplitude \( 1V \) (from figure).
Carrier amplitude \( A_c = 2V \).
Modulation index \( \mu = \frac{A_m}{A_c} = \frac{1}{2} = 0.5 \).
The AM waveform is: \( [A_c + m(t)] \sin(8\pi t) \).
Question 15.7
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, \( \mu \).
What would be the value of \( \mu \) if the minimum amplitude is zero volt?
What would be the value of \( \mu \) if the minimum amplitude is zero volt?
Answer & Explanation:
Answer:
Case 1: \( \mu = \frac{2}{3} \approx 0.67 \)
Case 2: \( \mu = 1 \)
Explanation:
Let \( A_{\text{max}} = A_c + A_m = 10V \)
\( A_{\text{min}} = A_c - A_m = 2V \)
Adding: \( 2A_c = 12 \Rightarrow A_c = 6V \)
Subtracting: \( 2A_m = 8 \Rightarrow A_m = 4V \)
\[ \mu = \frac{A_m}{A_c} = \frac{4}{6} = \frac{2}{3} \]
If \( A_{\text{min}} = 0 \):
\( A_c - A_m = 0 \Rightarrow A_c = A_m \)
Then \( \mu = \frac{A_m}{A_c} = 1 \).
Case 1: \( \mu = \frac{2}{3} \approx 0.67 \)
Case 2: \( \mu = 1 \)
Explanation:
Let \( A_{\text{max}} = A_c + A_m = 10V \)
\( A_{\text{min}} = A_c - A_m = 2V \)
Adding: \( 2A_c = 12 \Rightarrow A_c = 6V \)
Subtracting: \( 2A_m = 8 \Rightarrow A_m = 4V \)
\[ \mu = \frac{A_m}{A_c} = \frac{4}{6} = \frac{2}{3} \]
If \( A_{\text{min}} = 0 \):
\( A_c - A_m = 0 \Rightarrow A_c = A_m \)
Then \( \mu = \frac{A_m}{A_c} = 1 \).
Question 15.8
Due to economic reasons, only the upper sideband 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 & Explanation:
Explanation:
Transmitted signal (USB only): \( s(t) = A \cos[2\pi (f_c + f_m)t] \)
At receiver, we generate carrier: \( c(t) = \cos(2\pi f_c t) \)
Multiplying:
\[ s(t) \times c(t) = A \cos[2\pi (f_c + f_m)t] \times \cos(2\pi f_c t) \]
Using identity \( \cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)] \):
\[ = \frac{A}{2} \left\{ \cos[2\pi f_m t] + \cos[2\pi (2f_c + f_m)t] \right\} \]
Passing through a low-pass filter removes the high-frequency component \( \cos[2\pi (2f_c + f_m)t] \).
The output is: \( \frac{A}{2} \cos(2\pi f_m t) \), which is the original modulating signal (scaled).
Thus, the modulating signal is recovered.
Transmitted signal (USB only): \( s(t) = A \cos[2\pi (f_c + f_m)t] \)
At receiver, we generate carrier: \( c(t) = \cos(2\pi f_c t) \)
Multiplying:
\[ s(t) \times c(t) = A \cos[2\pi (f_c + f_m)t] \times \cos(2\pi f_c t) \]
Using identity \( \cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)] \):
\[ = \frac{A}{2} \left\{ \cos[2\pi f_m t] + \cos[2\pi (2f_c + f_m)t] \right\} \]
Passing through a low-pass filter removes the high-frequency component \( \cos[2\pi (2f_c + f_m)t] \).
The output is: \( \frac{A}{2} \cos(2\pi f_m t) \), which is the original modulating signal (scaled).
Thus, the modulating signal is recovered.
📘 Exam Preparation Tip:
These exercise questions will help you understand modern communication technologies. You'll learn to analyze amplitude modulation (AM) and frequency modulation (FM), understand transmitter and receiver block diagrams, calculate bandwidth requirements, solve problems on modulation index and sidebands, and understand different propagation modes (ground, sky, space waves). Crucial for telecommunications and broadcasting systems.