Communication Systems

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New answer posted

2 weeks ago

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V
Vishal Baghel

Contributor-Level 10

According to question, we can write

  d = 2 h R d 2 d 1 = h 2 h 1 h 2 = ( d 2 d 1 ) 2 h 1 = ( 2 1 ) 2 * 1 2 5 = 5 0 0 m

Increment in height of tower = h2 – h1 = 500 – 125 = 375 m

New answer posted

3 weeks ago

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V
Vishal Baghel

Contributor-Level 10

Low pass filter will allow low frequency signal to pass while high pass filter allow high frequency to pass through

New answer posted

3 weeks ago

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V
Vishal Baghel

Contributor-Level 10

λ > h λ > 4 0 0 m

New answer posted

3 weeks ago

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V
Vishal Baghel

Contributor-Level 10

µ = A? /A? = 0.5 {A? = 20 volt, A? = 40 volt}
m (t) = A? sin ω? t {ω? = 2π*10? }
c (t) = A? sin ω? t {ω? = 2π*10*10³}
C? (t) = (A? + A? sin ω? t) sin ω? t ⇒ A? {1+ µsin ω? t} sin ω? t

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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a month ago

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New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Amplitude of sidebands = µAc/2
Here µ = Am/Ac = 5/15 = 1/3
Amplitude = (1/3) (15)/2 = 2.5 V
a/10 = 2.5 => a = 25
b/10 = 2.5 => b = 25
a/b = 1

New answer posted

a month ago

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Vishal Baghel

Contributor-Level 10

Band Width = 2 * n * the highest modulation frequency

n = 9 0 k H z 2 * 5 k H z = 9

New answer posted

a month ago

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Vishal Baghel

Contributor-Level 10

Sound level is given in dB = 10 log  ( P 0 P i ) o r 1 0 l o g ( l l 0 )

As sound level decreases 5 dB every km,

So, in 20 km sound level will decrease by 20 * 5 = 100 dB.

Δ β = β 2 β 1 = 1 0 l o g ( l 2 l 1 )

1 0 0 = 1 0 l o g ( l 2 l 1 )

1 0 1 0 = l 2 l 1 l 2 = 1 0 1 0 l 1

P 2 = 1 0 1 0 P 1

P 2 = 1 0 1 0 * ( 0 . 1 * 1 0 3 ) P 2 = 1 0 8 W = 1 0 x W

 x = 8

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Side band frequencies are 1 MHz ±  1 kHz

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