Consider an optical communication system operating at λ$$=600nm$$. Suppose, only $$1$$% of the optical source frequency is the available channel bandwidth for optical communication. Number of channels that can be accommodated for transmitting audio signals requiring a bandwidth of $$8kHz$$ are $$_____$$ ×$$108$$

Question

Consider an optical communication system operating at λ$$=600nm$$. Suppose, only $$1$$% of the optical source frequency is the available channel bandwidth for optical communication. Number of channels that can be accommodated for transmitting audio signals requiring a bandwidth of $$8kHz$$ are $$_____$$ ×$$108$$

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Optical source frequency  $$f=c$$λ$$=3$$×$$108/(600$$×$$10$$−$$9)=5$$×$$1014Hz$$ Bandwidth of channel  (1$$% of $$above) $$=1$$%of $$5$$×$$1014=1100$$×$$5$$×$$1014=5$$×$$1012HzNumber$$ of channels $$=$$ (Total$$ bandwidth of $$channel)/$$ $$(Bandwidth$$ needed per $$channel) So, Number of channels for audio $$signal=(5$$×$$1012)/(8$$×$$103)=6.25$$×$$108$$

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Consider an optical communication system operating at λ=600nm. Suppose, only 1% of the optical source frequency is the available channel bandwidth for optical communication. Number of channels that can be accommodated for transmitting audio signals requiring a bandwidth of 8kHz are _____ ×108

Optical source frequency f=cλ=3×108/(600×10−9)=5×1014Hz Bandwidth of channel (1% of above) =1%of 5×1014=1100×5×1014=5×1012HzNumber of channels = (Total bandwidth of channel)/ (Bandwidth needed per channel) So, Number of channels for audio signal=(5×1012)/(8×103)=6.25×108

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