2nd PUC Physics Question Bank Chapter 15 Communication Systems

You can Download Chapter 15 Communication Systems Questions and Answers, Notes, 2nd PUC Physics Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

2nd PUC Physics Question Bank Chapter 15 Communication Systems

2nd PUC Physics Chapter 15 Communication Systems NCERT Text Book Questions and Answers

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:
At frequencies above 40 MHz, the communication is limited to line of sight paths. Hence, options (a) and (b) are correct.

Question 2.
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves.
(b) Skywaves.
(c) Surface waves.
(d) Space waves.
Answer:
In the UHF range, the transmission is done by means of space waves. Hence, option (d) is correct.

Question 3.
Digital signals
(i) do not provide a continuous set of values,
(ii) represent values as discrete steps,
(iii) can utilize a 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:
For digital signals, all four options are true. Hence the option (d) is correct.

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 81m tall. How much service area can it cover if the receiving antenna is at the ground level?
Answer:
For the line of sight communication, the receiving antenna must intercept the transmitted signal. If the signal is to be received. beyond the horizon, the receiving antenna must be high enough to intercept the transmitted signal. Therefore, it is not necessary for a transmitting antenna to be at the same high as that of the receiving antenna.
Here, h = 81m; R = 6.4 x 10⁶ m If d is the radius of the circle within which the transmission can be observed, then

Question 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:
Here, E_c = 12V, m_a = 75% = 0.75
Now, modulation factor,

Question 6.
A modulating signal is a square wave, as shown in Fin. 15.14.

The carrier wave is given by c (t) = 2sin (8 π t) volts.
(i) Sketch – the amplitude modulated waveform
Answer:
(i) Here, c (t) = 2 sin 8 n t
It follows that the carrier is a sine wave having a period of T = 2π/8π = 0.25 s and amplitude of E_C = ±2V
On the other hand, the modulating signal is a square wave of period 1s and amplitude of E = ±1V.
Therefore, the maximum and minimum amplitude of the modulated waveform will be

Hence the amplitude modulated waveform will be as

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 μ. What would be the value of p if the minimum amplitude is zero volts?
Answer:

Question 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:
When only the upper sideband (w_e + w_m) is transmitted, the transmitted signal may be represented as
e = E_t cos (w_e + w_m )t   ………………(1)
where is the amplitude of the transmitted signal Let the carrier wave available at the receiving station be
e =E_c cos w_ct  ……………(2)
where,
E_c is the amplitude of the carrier wave Multiplying the respective sides of equations (1) and (2).

If this signal is passed through a filter circuit, which acts as a block for the higher frequency (2w_e + w_m) and allows the low frequency to pass through it, then the modulating signal given by,

can be recovered at the receiving station

2nd PUC Physics Chapter 15 Communication Systems Additional Questions and Answers

Question 1.
The audio signal can not be transmitted directly into space. Why?
Answer:

• The length of the antenna required is so large (500m) that is practically impossible to set up it.
• The energy radiated from the antenna in ‘ audio frequency range is practically zero
• The audio signals transmitted from the different broadcasting stations get inseparably mixed up.

Question 2.
600 Hz modulating voltage fed into FM generator produces a frequency deviation of 3.36 kHz. Find the modulation index.
Answer:
Modulating frequency, FM = 600 Hz Frequency deviation,

Question 3.
Why is an FM signal less susceptible to noise than an AM signal?
Answer:
In FM transmission, the message signal is in the form of frequency variations of carrier waves. During the modulation process, the noise gets amplitude modulated. Since it brings about variations only in the amplitude of carrier waves, it does no harm to the message signal. For this reason, an FM signal is less susceptible to noise than an AM signal.

Question 4.
The ozone layer is present in
(A) Troposphere
(B) stratosphere
(C) Ionosphere
(D) Mesosphere
Answer:
(B) stratosphere

Question 5.
What is the cause of the greenhouse effect?
(A) IR rays
(B) UV rays
(C) x-rays
(D) Radio waves
Answer:
(A) IR rays

Question 6.
Communication on the ground is through electromagnetic waves of wavelength
(A) Larger than 600m
(b) Between 200 and 600m
(C) Between 1 and 5m
(D) Between 10^-13 and 0.1m
Answer:
(D) Between 10^-13 and 0.1m

Question 7.
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^3/2
(C) h
(D) h²
Answer:
(D) h²

Question 8.
The output from a LASER is monochromatic. It means that it is
(A) Directional
(B) Polarised
(C) Narrow beam
(D) Single-frequency
Answer:
(D) Single-frequency

Question 9.
Losses in optical fiber are caused by
(A) Impurities in glass
(B) Imperfect transparency of the glass
(C) Its extremely small area of cross-section
(D) Stepped index structure
Answer:
(A) Impurities in glass

Question 10.
The number of geostationary satellites needed for uninterrupted global coverage is
(A) 3
(B) 4
(C) 1
(D) 2
Answer:
(A) 3