For a television network,5 times 10^5 channel | vedclass

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(A) Given:Number of channels $(N)$ = $5 	imes 10^5$Central frequency $(f_c)$ = $25 	ext{ GHz} = 25 	imes 10^9 	ext{ Hz}$Bandwidth per channel $(\D

Vedclass · Accepted Answer

$4$

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(A) Given:Number of channels $(N)$ = $5 	imes 10^5$Central frequency $(f_c)$ = $25 	ext{ GHz} = 25 	imes 10^9 	ext{ Hz}$Bandwidth per channel $(\Delta f)$ = $2 	ext{ kHz} = 2 	imes 10^3 	ext{ Hz}$The total bandwidth required for the network is $N 	imes \Delta f = (5 	imes 10^5) 	imes (2 	imes 10^3) = 10 	imes 10^8 = 10^9 	ext{ Hz}$.The percentage of the link used is given by the ratio of the total bandwidth required to the central frequency (total available bandwidth capacity):Percentage = $\frac{	ext{Total Bandwidth Required}}{	ext{Central Frequency}} 	imes 100$Percentage = $\frac{10^9 	ext{ Hz}}{25 	imes 10^9 	ext{ Hz}} 	imes 100$Percentage = $\frac{1}{25} 	imes 100 = 4 \%$.Therefore,$4 \%$ of the link is used for the network.

For a television network,$5 \times 10^5$ channels are granted. If the central frequency of the microwave link is $25 \text{ GHz}$ and the allotted bandwidth for each channel is $2 \text{ kHz}$,then what percentage of the link is used for the network (in $\%$)?

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