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In a communication system, operating at 1200 mm, only 2% of the source frequency is available for T.V. transmission , having a bandwidth of 5 MHz. The number of T.V. channels that can be transmitted isA. 2 millionB. 10 millionC. 0.1 millionD. 1 million |
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Answer» Correct Answer - D Optical source frequency, `v=(c)/(lambda)` or `v=(3xx10^(8)xxms^(-1))/(1200xx10^(-9)m)=2.5xx10^(14)Hz` Bandwidth of channel=2% of the source frequency `=(2)/(100)xx2.5xx10^(14)Hz=5xx10^(12)Hz`. Number of channels `=("Total bandwidth of channel")/("Bandwidth needed per channel")` `=(5xx10^(12)Hz)/(5xx10^(6)Hz)=10^(6)`=1 million |
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