12 signals,each band-limited to 5, kHz,are to | vedclass

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

Question

(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

Vedclass · Accepted Answer

$71$

Vedclass · Answer

(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 ....... $kHz$.

Similar Questions

$A$ carrier is simultaneously modulated by two sine waves with modulation indices of $0.4$ and $0.3$. The resultant modulation index will be

The amplitude of the side bands of the modulated signal, if the carrier signal and message signal amplitudes are $25 \,V$ and $5 \,V$ respectively, is (in $\,V$)

$A$ message signal of $3 \text{ kHz}$ is used to modulate a carrier signal of frequency $1 \text{ MHz}$ using amplitude modulation. The upper sideband frequency and bandwidth respectively are:

For which of the following $AM$ waves is the modulation index equal to $1$?

The maximum and minimum amplitude of an amplitude modulated wave is $16\, V$ and $8\, V$ respectively. The modulation index for this amplitude modulated wave is $x \times 10^{-2}$. The value of $x$ is .........

Vedclass Test Series

Exam Paper Generator

Online Exam Module