Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an signal requires a bandwith of 8 kHz, how many channels can be accommodated for transmission:
Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an signal requires a bandwith of 8 kHz, how many channels can be accommodated for transmission:
Option 1 -
375 * 107
Option 2 -
75 * 107
Option 3 -
375 * 108
Option 4 -
75 * 109
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According to question, we can write
Increment in height of tower = h2 – h1 = 500 – 125 = 375 m
Low pass filter will allow low frequency signal to pass while high pass filter allow high frequency to pass through
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
Kindly consider the following figure
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