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.

Question

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.

Infinity Learn · Accepted Answer

Optical source frequency ​$$f=c$$λ$$=3$$×$$108/(800$$×$$10$$−$$9)=3.8$$×$$1014Hz$$​  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 $$=(3.75$$×$$1012)/(8$$×$$103)$$≃$$4.68$$×$$108$$

Modulation

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Modulation

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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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