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Communication Systems

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Question 1
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A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it. 

  • 10.62 MHz

  • 10.62 kHz

  • 5.31 mHz

  • 5.31 kHz

Solution

B.

10.62 kHz

The frequency is given as
straight f space equals space fraction numerator 1 over denominator 2 πμτ end fraction straight tau space equals space RC therefore space straight f space equals open parentheses space fraction numerator 1 over denominator 2 πRC end fraction close parentheses
μ =0.6
R = 100 k = 100 x1000 Ω
c = 250 pico farad = 250 x 10-12 F
So,
straight f space equals space fraction numerator 1 over denominator 2 straight pi space straight x space 0.6 space straight x space 1000 space straight x 100 space straight x 250 space straight x 10 to the power of negative 12 end exponent end fraction rightwards double arrow straight f equals space fraction numerator 1 over denominator 9.42 space straight x 1 to the power of negative 5 end exponent end fraction space equals space 0.1061 space straight x space 10 to the power of negative 5 end exponent straight f space equals space 10.61 space kHz

Question 2
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A radar has a power of 1 Kw and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6.4 x 106 m) is

  • 80 km

  • 16 km

  • 40km

  • 64 km

Solution

A.

80 km

Maximum distance on earth where object can be detected is d, then

(h + R)2 = d2 + R2

⇒ d2 = h2 + 2Rh
Since, h<<R,
⇒d2 =2hR
straight d equals square root of 2 left parenthesis 500 right parenthesis left parenthesis 6.4 space straight x 10 to the power of 6 right parenthesis end root equals 80 space km

Question 3
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A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are:

  • 2005 kHz, and 1995 kHz

  • 2005 kHz, 2000 kHz and 1995 kHz

  • 2000 kHz and 1995 kHz

  • 2 MHz only

Solution

C.

2000 kHz and 1995 kHz

Frequency associated with AM are
fc - fm, f, fc + fm
according to the question
fc = 2 MHz = 2000 kHz

fm = 5 kHz
Thus, frequency of the resultant signal is are carrier frequency fc = 2000 kHz, LSB frequency  fc-fm = 2000 kHz-5kHz = 1995 kHz and USB frequency fc+fm = 2005 kHz
Question 4
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A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of

  • 1000

  • 10000

  • 10

  • 100

Solution

D.

100

straight B subscript 1 space equals space 10 space log space open parentheses straight I over straight I subscript 0 close parentheses straight B subscript 2 space equals space log space space open parentheses fraction numerator straight I apostrophe over denominator straight I subscript 0 end fraction close parentheses given space straight B subscript 2 minus straight B subscript 1 20 space equals space straight l 0 space log space space open parentheses fraction numerator straight I apostrophe over denominator straight I end fraction close parentheses straight I apostrophe space equals space 100
Question 5
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A train is moving on a straight track with speed 20 ms-1 .It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms-1 ) close to :

  • 6%

  • 12%

  • 18%

  • 24%

Solution

B.

12%

Apparent frequency heard by the person before crossing the train.
straight f subscript 1 space equals space open parentheses fraction numerator straight c over denominator straight c minus straight v subscript straight s end fraction close parentheses space straight f subscript straight o space equals space open parentheses fraction numerator 320 over denominator 320 minus 20 end fraction close parentheses space 1000
Similarly, apparent frequency heard, after crossing the trainsstraight f subscript 2 space equals space open parentheses fraction numerator straight c over denominator space straight c plus straight V subscript straight s end fraction close parentheses straight f subscript straight o space equals space open parentheses fraction numerator 320 over denominator 320 plus 20 end fraction close parentheses 1000 left square bracket straight c space equals speed space of space sound right square bracket increment straight f space equals space straight f subscript 1 minus straight f subscript 2 space equals space open parentheses fraction numerator 2 cv over denominator straight c squared minus straight v subscript straight s superscript 2 end fraction close parentheses straight x space straight f subscript straight o or space fraction numerator increment straight f over denominator straight f subscript straight o end fraction straight x 100 space equals space open parentheses fraction numerator 2 cv over denominator straight c squared minus straight v subscript straight s superscript 2 end fraction close parentheses straight x 100 space equals space fraction numerator 2 space straight x space 320 space straight x space 20 over denominator 300 space straight x space 340 end fraction straight x 100 equals space fraction numerator 2 space space straight x space 32 space straight x space 20 over denominator 3 space straight x 34 end fraction space equals space 12.54 space equals space 12 percent sign

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