Question

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

• A. 4.8 $\times {10}^8$
• B. 48
• C. 6.2 $\times {10}^8$
• D. 4.8 $\times {10}^5$

Answer 1

The correct option is A 4.8 $\times {10}^8$

Optical source frequency f = $\frac{c}{\lambda }$
= $\frac{3\times {10}^8}{800\times {10}^{-9}}=3.8\times {10}^{14}Hz$
Bandidth of channel (1 % of above) = 3.8 $\times$ ${10}^{12}$ Hz
Number of channel = ($\frac{Totalbandwidthofchannel}{Bandwidthneededperchannel}$)
(a) Number of channels for audio signal
= $\frac{(3.8\times {10}^{12})}{8\times {10}^3}$ ~4.8 $\times {10}^8$