A) 4.8 ´ 108
B) 48
C) 6.2 ´ 108
D) 4.8 ´ 105
Correct Answer: A
Solution :
Optical source frequency \[f=\frac{c}{\lambda }\] = 3 ´ 108/(800 ´ 10?9) = 3.8 ´ 1014 Hz Bandwidth of channel (1% of above) = 3.8 ´ 1012 Hz Number of channels =(Total bandwidth of channel)/ (Bandwidth needed per channel) Number of channels for audio signal \[=(3.8\times {{10}^{12}})/(8\times {{10}^{3}})\] ~ 4.8 ´ 108
You need to login to perform this action.
You will be redirected in
3 sec
