Consider an optical communication system oper | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The frequency of the optical source is given by $f = c / \lambda = (3 	imes 10^{8} \, m/s) / (800 	imes 10^{-9} \, m) = 3.75 	imes 10^{14} \, H

Vedclass · Accepted Answer

$4.8 	imes 10^{8}$

Vedclass · Answer

(A) The frequency of the optical source is given by $f = c / \lambda = (3 	imes 10^{8} \, m/s) / (800 	imes 10^{-9} \, m) = 3.75 	imes 10^{14} \, Hz$.The available bandwidth for communication is $1\%$ of the source frequency:$\Delta f = 0.01 	imes 3.75 	imes 10^{14} \, Hz = 3.75 	imes 10^{12} \, Hz$.The number of channels that can be accommodated for a bandwidth of $8 \, kHz$ $(8 	imes 10^{3} \, Hz)$ is:$N = \Delta f / 	ext{bandwidth per channel} = (3.75 	imes 10^{12} \, Hz) / (8 	imes 10^{3} \, Hz) \approx 4.6875 	imes 10^{8} \approx 4.8 	imes 10^{8}$.

Consider an optical communication system operating at a wavelength of $\lambda \sim 800 \, nm$. If only $1\%$ of the optical source frequency is available for optical communication,how many channels can be accommodated for a bandwidth of $8 \, kHz$?

Similar Questions

$A$ modulated signal $C_m(t)$ has the form $C_m(t) = 30 \sin(300\pi t) + 10 \cos(200\pi t) - 10 \cos(400\pi t)$. The carrier frequency $f_c$,the modulating frequency $f_m$,and the modulation index $\mu$ are respectively given by:

The size of the antenna required for a carrier wave of frequency $3 \text{ MHz}$ is: (in $\text{ m}$)

$A$ transmitting antenna has a height of $320\, m$ and that of the receiving antenna is $2000\, m$. The maximum distance between them for satisfactory communication in line-of-sight mode is $d$. The value of $d$ is $\ldots \ldots \, km$.

The process of retrieval of information from the carrier wave is known as:

Consider telecommunication through optical fibres. Which of the following statements is $NOT$ true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module