CBSE 12th Standard Physics Subject Communication Systems HOT Questions 2 Mark Questions 2021
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CBSE 12th Standard Physics Subject Communication Systems HOT Questions 2 Mark Questions 2021
12th Standard CBSE
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Reg.No. :
Physics
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In the given block diagram of a receiver identify the boxes labelled as X and Y and write their functions.
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(a) Describe briefly the three factors which justify the need for translating a low frequency signal into high frequencies before transmission.
(b) Figure shows a block diagram of a detector for AM signal.
Draw the waveforms for the
(i) input AM wave at A,
(ii) output B at the rectifier, and (iii) output signal at C. -
Differentiate between (i) PAM and (ii) PPM
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An AM wave is represented by the expression: v = 5 (1 + 0.6 cos 6280 t) sin 221 x 104 t volts
(i) What are the maximum and minimum amplitudes of the AM wave.
(ii) What frequency components are contained in the modulated wave -
The antenna current of an AM transmitter is 8A when only carrier is sent but it increases to 8.93A when the carrier is sinusoidally modulated. Find the percent-age modulation index
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CBSE 12th Standard Physics Subject Communication Systems HOT Questions 2 Mark Questions 2021 Answer Keys
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X \(\rightarrow\) Intermediate Frequency (IF) stage
Y \(\rightarrow\) Power amplifier
The IF stage change the high-frequency carrier wave into lower frequency wave and amplifier enhances the strength of the signals. -
(a) There are three factors:
(i) Wider Bandwidth: High frequency provide wider bandwidth.
(ii) Distance: High-frequency wave used to carry baseband signals (low-frequency signal) over a distance of several thousand kilometres.
(iii) Cost: High frequency transmits with low cost.
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(i) Pulse Amplitude Modulation : Amplitude of the pulse varies in accordance with the modulating signal.
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(ii) (ii) Pulse Position Modulation. : Pulse position (ie) time of rise or fall of the pulse ) changes with the modulating signal.
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The AM wave equation is given by ;
v = 5(1+0.6cos6280t) sin 221 X 104 t volts ………….(i)
(i) Maximum amplitude of AM wave
= EC + maEC =5 + 0.6 X 5 = 8V
Minimum amplitude of AM wave
= Ec - maEC =5 - 0.6 X 5 = 2V
(ii) The AM wave will contain three frequency vizfc-fs fc fc+fs 336-1 336 336+1 335KHz 336Khz 337KHz -
Ps = 1/2 ma2 Pc
1.246 = 1 + \(\frac { ma^{ 2 } }{ 2 } \)
ma2/2 = 0.246
ma = (2 x 0.246)1/2 = 0.701 = 70.1%
