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 Bihar Board Class 12 Physics Solutions Chapter 15 Communication Systems

Bihar Board Class 12 Physics Communication Systems Textbook 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:
(b) i.e., 10 MHz.
10 kHz frequencies cannot be radiated because of the antenna size required for their propagation while the frequencies 1 GHz and 1000 GHz will penetrate and thus cannot reach to the receiver beyond the horizon.

Question 2.
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves
(b) Sky waves
(c) Surface waves
(d) Space waves.
Answer:
(d) i.e., space waves i.e., propagation of e.m. waves in ultra ugh frequency range is possible by space waves.

Question 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).
Answer:
(c) Digital signals represented by 0 and 1 don’t provide continuous set of values, (ii) they represent values in discrete stps. (iii) utilize only binary system.

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 it can cover if the receiving antenna is at the ground level?
Answer:
No, it is not necessary for a transmitting antenna to be at the same height as that of the receiving antenna for a line of sight communication. :
Here, h = height of TV antenna = 81 m.
R = radius of earth = 6400 km = 6.4 × 106 m.
Let d be the radius of the service area it can cover i.e., the radius of the circle within which the transmission can be observed.
Using the relation, d = 2hR−−−−√ , we get
d = 2×81×6.4×106−−−−−−−−−−−−−−√m
= 9 × 8 × 103 × 0.2−−−√ m
Let A = service area in which transmission can be received from this antenna = ?
∴ Using the formula,
A = πd2, we get
A = π × (72×103×0.2−−−√)2
= π × 81 × 64 × 106 × 0.2 m2
= 3258 × 106  × 0.2 m2 = 3258 km2.

Question 5.
A carrier wave of peak voltage 12 V 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, Ec = amplitude of the carrier = 12 V.
Em = peak voltage i.e., amplitude of the modulating signal = ?
ma = modulation index = 75% = 75100
Using the relation, ma = EmEc, we get
Em= ma × Ec
= 0. 75 × 12 = 9 V.

Question 6.
A modulating signal is a square wave as shown in fig.:

The carrier wave is given by c(t) = 2 sin (8 πt) volts.
(i) Sketch the amplitude modulated waveform
(ii) What is the modulation index?
Answer:
(i) The sketch of the amplitude modulated waveform is shown below:

(ii) Here, the carrier wave
C(t) = 2 sin (8 πt) (volts)
Ec= amplitude of carrier wave = 2V
Em= amplitude of the modulating signal = IV
ma= modulation index of A.M.
Waveform = ?
We know that, ma = EmEc = 12 = 12 = 0.5

Question 7.
For an amplitude modulated wave, the maximum amplitude is found to be 10V while the minimum amplitude is found 1 be 2V. Determine the modulation index, μ.
What would be the value of μ if the minimum amplitude is zero volt?
Answer:
Modulation index μ =  amplitude of modulating wave  amplitude of unmodulated 
μ = VmVc
Modulating index increases as modulating frequency fm decrease As mf is the ratio of two frequencies, it is measured in radians.

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 sign it at the receiver station.
Answer:
Let ec and em be the instantaneous.
Voltages or carried and modulating wavesterpngtively and ar, given by
ec = vc sin wct ……….. (1)
em = vm sin wmt ……….. (2)
For modulated wave
e = V sin wct
Where V = amplitude of modulated Wave

(∵m=Vm Vc)
Vc sin wct m Vc [12cos(wc−wm)t−12cos(wc+wm)t]
V = Vc sin wct + mvc2 cos (wc−wm)t−mvc2cos(wc+wm)t ……….. (3)
Eq (3) represents the simplitude modulated wave recived at the receiving station from the transmitting station. If only upper sideband
of AM wave is transmitted, then V = sin wct – mvc2 cos (Wc + Wm)t
To recover the modulating signal at the receiving station, the modulated wave is fet to a circuit or a device known as detector It demolutates the modulating signal from the modulated carrier wave i.e, it recovers the modulated signal from the molulated carrier wave i.e. rectifying the input signal. Thus either the top or bottom half of the wave is removed, a suitable filter removes the raido frequency of the rectified wave and a signal is produced which allows the envelope of one-half of the wave. Here, upper side band is transmitted so the out
put follows this half and is of the form of the modulating wave. The waveforms are as shown below.

Aliter:
Let A’ = Ac cos Wc)t se the carrier signal.
Let, for simplicity, the received signal be
A1cos (wc> + wa)t
the carrier Ac cos wct is available at the receiving station. By multiplying the two signals, we get
A1 Ac (wc + wa)t cos wct
= A1Ac2 [cos (2wc> + wm) t + cos wmt]
If this signal is passed through a low-pass filter, we can record the modulating signal A1Ac2 cos wmt.