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(D) The bandwidth of each signal is $5\, kHz$. Since there are $12$ signals,the total signal bandwidth is $12 	imes 5\, kHz = 60\, kHz$.For $N$ signa

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$71$

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(D) The bandwidth of each signal is $5\, kHz$. Since there are $12$ signals,the total signal bandwidth is $12 	imes 5\, kHz = 60\, kHz$.For $N$ signals,the number of guard bands required between them is $N-1$.Here,$N = 12$,so the number of guard bands required is $12 - 1 = 11$.Each guard band has a bandwidth of $1\, kHz$.Therefore,the total guard bandwidth is $11 	imes 1\, kHz = 11\, kHz$.The total bandwidth of the multiplexed signal is the sum of the signal bandwidths and the guard bandwidths:Total Bandwidth $= 60\, kHz + 11\, kHz = 71\, kHz$.

$12$ signals,each band-limited to $5\, kHz$,are to be transmitted by frequency-division multiplexing. If an $AM-SSB$ modulation guard band of $1\, kHz$ is used,then the bandwidth of the multiplexed signal is ....... $kHz$.

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