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Q.

Consider an optical communication system operating at λ ~800 nm. Suppose, only 1 % of the optical source frequency is the available channel bandwidth for optical communication. Number of channels accommodated for transmitting audio signals requiring a bandwidth of 8 kHz are 4.68×10p. Find the value of p.

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Detailed Solution

Optical source frequency f=cλ=3×108/(800×10−9)=3.8×1014Hz Bandwidth of channel (1% of above )=3.75×1012Hz Number of channels = (Total bandwidth of channel) / (Bandwidth needed per channel) ⇒Number of channels for audio signal =(3.75×1012)/(8×103)≃4.68×108

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