Communication System Question 7
Question: Consider an optical communication system operating at $ \lambda \approx 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?
Options:
A) $ 4.8\times 10^{8} $
B) 48
C) $ 6.2\times 10^{8} $
D) $ 4.8\times 10^{5} $
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Answer:
Correct Answer: A
Solution:
[a] Optical source frequency $ f=\frac{c}{\lambda } $
$ =3\times 10^{8}/(800\times {{10}^{-9}})=3.8\times 10^{14}Hz $ Bandwidth of channel (1% of above) $ =3.8\times 10^{12}Hz $ Number of channels = (Total bandwidth of channel)/ (Bandwidth needed per channel)
$ \Rightarrow $ Number of channels for audio signal $ =(3.8\times 10^{12})/(8\times 10^{3})=4.8\times 10^{8} $
BETA
