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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?

Text Solution

AI Generated Solution

To solve the problem step by step, we will follow the given information and apply the relevant formulas. ### Step 1: Understand the given data - Wavelength (λ) = 800 nm = \(800 \times 10^{-9}\) m - Bandwidth required for video TV signal = 4.5 MHz = \(4.5 \times 10^{6}\) Hz - Only 1% of the optical source frequency is available for channel bandwidth. ### Step 2: Calculate the frequency of the optical source ...
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Knowledge Check

  • In a communication system, operating at lamda=700nm, only 1% of the optical source frequency is the available channel band width. How many channels can be accommodated for transmitting the signal requiring an approximate band width of 4.5 MeV ?

    A
    `8.5xx10^(5)`
    B
    `9.5xx10^(6)`
    C
    `9.5xx10^(5)`
    D
    `8.5xx10^(4)`

  • Consider an optical communication system operating at lambda=800 nm . Suppose, only 1% of the optical source frequency is the available channel bandwidth for optical communication. How many channels can be accommondated for transmitting audio signals requiring a bandwidth of 8kHz

    A
    `4.8xx10^(8)`
    B
    `48`
    C
    `6.2xx10^(8)`
    D
    `4.8xx10^(5)`

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    A
    2 million
    B
    10 million
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    D
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