A bandwidth of 6 , text{MHz} is available for | vedclass

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(D) The bandwidth required for a single $A.M.$ station is twice the maximum frequency of the modulating audio signal.$	ext{Bandwidth per station} = 2

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(D) The bandwidth required for a single $A.M.$ station is twice the maximum frequency of the modulating audio signal.$	ext{Bandwidth per station} = 2 	imes f_m = 2 	imes 6 \, 	ext{kHz} = 12 \, 	ext{kHz}$.The total available bandwidth is $6 \, 	ext{MHz} = 6000 \, 	ext{kHz}$.The number of stations $N$ that can be broadcasted simultaneously is given by the ratio of the total bandwidth to the bandwidth per station:$N = \frac{	ext{Total Bandwidth}}{	ext{Bandwidth per station}} = \frac{6000 \, 	ext{kHz}}{12 \, 	ext{kHz}} = 500$.Therefore,$500$ stations can be broadcasted simultaneously.

$A$. Optical Fiber Communication	$P$. Ultrasound
$B$. Radar	$Q$. Infrared Light
$C$. Sonar	$R$. Microwaves
$D$. Mobile Phones	$S$. Radio Waves

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