In a communication system operating at a wave | vedclass

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(C) Given: Wavelength $\lambda = 800 \,nm = 800 	imes 10^{-9} \,m$. Speed of light $c = 3 	imes 10^8 \,m/s$. Frequency of the source $f = c / \lambd

Vedclass · Accepted Answer

$\frac{1}{16} 	imes 10^7$

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(C) Given: Wavelength $\lambda = 800 \,nm = 800 	imes 10^{-9} \,m$. Speed of light $c = 3 	imes 10^8 \,m/s$. Frequency of the source $f = c / \lambda = (3 	imes 10^8) / (800 	imes 10^{-9}) = 3.75 	imes 10^{14} \,Hz$. Available signal bandwidth is $1 \%$ of the source frequency: Bandwidth $= 0.01 	imes 3.75 	imes 10^{14} = 3.75 	imes 10^{12} \,Hz$. Bandwidth required for one $TV$ channel $= 6 \,MHz = 6 	imes 10^6 \,Hz$. Number of channels $= 	ext{Total bandwidth} / 	ext{Bandwidth per channel} = (3.75 	imes 10^{12}) / (6 	imes 10^6) = 0.625 	imes 10^6 = 6.25 	imes 10^5$. Expressing this in the form of the options: $6.25 	imes 10^5 = (1/16) 	imes 10^7 = 0.0625 	imes 10^7$.

In a communication system operating at a wavelength of $800 \,nm$, only $1 \%$ of the source frequency is available as signal bandwidth. The number of channels accommodated for transmitting $TV$ signals of bandwidth $6 \,MHz$ is

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