Communication Systems

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Question
CBSEENPH12039851

Two sources of sound placed close to each other, are emitting progressive waves given by
y1 = 4sin 600π t and y2 = 5 sin 608πt

An observer located near these two sources of sound will hear

  •  4 beats per second  with intensity ratio 25: 16 between waxing and waning

  • 8 beats per second with intensity ratio 25:16 between  waxing and waning

  • 8 beats per second with intensity ratio 81:1 between waxing and waning

  • 4 beats per second with intensity ratio 81:1 between waxing and waning

Solution

D.

4 beats per second with intensity ratio 81:1 between waxing and waning

Given, y1 = 4sin 600π t and y2 = 5 sin 608πt
Comparing with general equation
y = a sin 2πft
 we get, f2 = 300 Hz and f2 304 Hz
Number of beats = f2-f1 = 4s-1
straight I subscript max over straight I subscript min space equals space open parentheses fraction numerator a subscript 1 plus a subscript 2 over denominator a subscript 1 minus a subscript 2 end fraction close parentheses squared
space equals space open parentheses fraction numerator 4 plus 5 over denominator 4 minus 5 end fraction close parentheses squared space equals space 81 over 1

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Question
CBSEENPH12048011

A hemisphere is positively charged uniformly. The electric field at a point on a diameter away from the centre is directed

  • perpendicular to the diameter

  • parallel to the diameter

  • at an angle tilted towards the diameter

  • at an angle tilted away from the diameter

Solution

A.

perpendicular to the diameter

When the point is on the diameter and away from the centre of hemisphere which is charged uniformly and positively, the component of electric field intensity parallel to the diameter cancel out. So the electric field is perpendicular to the diameter.

Question
CBSEENPH12047766

A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of

  • 1000

  • 10000

  • 10

  • 100

Solution

D.

100

Let intensity of sound be I and I'. Loudness of sound initially

β1 = 10 log IIoβ2 = 10 log I'Io

β2 - β1 = 20

 20 = 20 log I'I

∴ I' = 100 I

Question
CBSEENPH12040162

A source emits electromagnetic waves of wavelength 3m. One beam reaches the observer directly and other after reflection from a water surface, travelling 1.5 m extra distance and with an intensity reduced to 1/4 as compared to intensity due to the direct beam alone. The resultant intensity will be.

  • (1/4) fold

  • (3/4) fold

  • (5/4) fold

  • (9/4) fold

Solution

D.

(9/4) fold

We know that a phase change of π occurs when the reflection takes place at the boundary of the denser medium. This is equivalent to a path difference of λ/2.

Therefore, Total phase difference = π -π = 0

Thus, the two waves superimpose in phase.

Resultant,

Resultant amplitude = I + I4 = 32IResultant amplitude = 32I2 = 94 I = 94 fold 

Question
CBSEENPH12047774

What maximum frequency can be reflected from ionosphere?

  • 5 MHz

  • 6 GHz

  • 5 KHz

  • 500 MHz

Solution

A.

5 MHz

The ionosphere extends from a height of 80 km to 300 km. The refractive index of ionosphere is less than its free space value. That is, it behaves as a rarer medium. As we go deep into the ionosphere, the refractive index keeps on decreasing. The bending of beam (away from the normal) will continue till it reaches critical angle after which it will be reflected back. The different points on earth receive signals reflected from different depths of the ionosphere. There is a critical frequency f, (S to 100 MHz) beyond which the waves cross the ionosphere and do not return back to earth.

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