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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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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)`

  • In optical communication system operating at 1200nm, only 2% of the source frequency is available for TV t ransmission having a bandwidth of 5 MHz. the number of TV channels that can be transmitted is

    A
    2 million
    B
    10 million
    C
    0.1 million
    D
    1 million