Consider an optical communication system operating at λ$$=800nm$$. Suppose only $$1$$ % of optical frequency is available for channel bandwidth for optical communication. If number of channels can be accommodated for transmitting audio signals required a bandwidth of $$5$$ kHz is y×$$107$$, then y is

Question

Consider an optical communication system operating at λ$$=800nm$$. Suppose only $$1$$ % of optical frequency is available for channel bandwidth for optical communication. If number of channels can be accommodated for transmitting audio signals required a bandwidth of $$5$$ kHz is y×$$107$$,  then y is

Infinity Learn · Accepted Answer

Bandwidth of channel  $$=$$ $$5$$ kHz $$=$$ $$5000$$ HZ  Optical wave length  $$=$$ λ$$=800nm$$ $$=8$$×$$10$$−$$7m$$  Optical frequency  $$=n=C$$λ $$=$$ $$3$$×$$1088$$×$$10$$−$$7=0.375$$×$$1015$$ Hz  $$1$$ % of frequency is available, so  n $$=$$ $$1$$ % of  $$0.375$$×$$1015$$                                                                  $$=$$ $$=375$$×$$1010$$ Hz    No of channels   $$=Total$$ bandwidthBandwidth per channel  $$=375$$×$$10105000=75$$×$$107$$

Modulation

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Consider an optical communication system operating at λ=800nm. Suppose only 1 % of optical frequency is available for channel bandwidth for optical communication. If number of channels can be accommodated for transmitting audio signals required a bandwidth of 5 kHz is y×107, then y is

Bandwidth of channel = 5 kHz = 5000 HZ Optical wave length = λ=800nm =8×10−7m Optical frequency =n=Cλ = 3×1088×10−7=0.375×1015 Hz 1 % of frequency is available, so n = 1 % of 0.375×1015 = =375×1010 Hz No of channels =Total bandwidthBandwidth per channel =375×10105000=75×107

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