Question

Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an audio signal requires a bandwidth of 8 kHz, how many channels can be accommodated for transmission:

पर्याय

• 375 × 10⁷

• 75 × 10⁷

• 375 × 10⁸

• 75 × 10⁹

Answer 1

75 × 10⁷

Explanation: Given:

Wavelength of signal = 1000 mm = 10³ × 10^–9 = 10^–6 m

c = Speed of light = 3 × 10⁸ m/s

As   c = nλ

n = `"c"/lambda` 

= `(3xx10^8)/10^-6`

= 3 × 10¹⁴ Hz

Thus, frequency of optical signal is 3 × 10¹⁴ Hz

Given only 2% of optical source frequency is available for communication system.

= 0.75 × 10⁹ = 75 × 10⁷

∴ Available frequency = `2/100xx3xx10^14`

= 6 × 10¹² Hz

Bandwidth of communication system

= 6 × 10¹² Hz

Bandwidth of each channel to be accommodated

= 8 kHz = 8 × 10³ Hz

Thus, number system of channel

= `"Bandwidth of system"/"Bandwidth of each channel"`

n = `(6xx10^12)/(8xx10^3)`

= 0.75 × 10⁹

= 75 × 10⁷