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For an optical communication system, operating at `lambda = 800 nm`, only `1 %` the optical source frequency is the available channel band width. How many channels can be accomodated for transmitting video T.V. signal requiring an approximate band width of 4.5 MHz? |
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Answer» Optical source frequency, `v = c/lambda = ( 3xx 10^8)/(800 xx 10^(-9)) = 3.75 xx 10^(14) Hz` Band width of channel = 1% of source frequency =` 1/100 xx 3.75 xx 10^(14) = 3.75 xx 10^(12) Hz` Number of channels for video T.V. signal `= (3.75 xx 10^(12))/(4.5 xx 10^(6)) = 8.3 xx 10^(5) `. |
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