Question

$12$ Signals each band limited to $5kHz$ are to be transmitted by the frequency-division multiplexer. If an AM-SSB modulation guard band $1kHz$ is used then the bandwidth of the multiplexed signal is
$\left( a \right)$ $101kHz$
$\left( b \right)$ $99kHz$
$\left( c \right)$ $84kHz$
$\left( d \right)$ $71kHz$

Answer 1

Hint First of all we will calculate the total signal bandwidth and for this, we will have it by the product of the total number of signals and the band. And as we know there are $11$ guard bands required between the $12$ signals. So by using this we will calculate the total bandwidth.
Formula used:
Total signal bandwidth will be given by the formula,
$B{W_{Total}} = Total{\text{ }}Signals \times {\text{ Band}}$
Here,
$B{W_{Total}}$, will be the total bandwidth.

Complete Step By Step Solution
So the total signal bandwidth will be given by
$B{W_{Total}} = Total{\text{ }}Signals \times {\text{ Band}}$
Now, we will substitute the values, and we get
$ \Rightarrow B{W_{Total}} = 12 \times 5$
And after solving the equation, we will get
$ \Rightarrow B{W_{Total}} = 60Hz$
Since the total of $11$ the guard, the band are required between $12$ signals
So thee guard bandwidth will be,
Guard bandwidth $ = 11 \times 1kHz$
Therefore, on adding the above equation, we get
Guard bandwidth $ = 11kHz$
Now we will calculate the total bandwidth
So,
Total bandwidth $ = \left( {60 + 11} \right)1kHz$
Therefore, on solving we get
Total bandwidth$ = 711kHz$
Therefore, $71kHz$will be the bandwidth of the multiplexed signal.

So, option $D$ is correct.

Additional information DSBSC is AM without the transporter - so we're involving a transmission capacity twice that of the first sign, however without the transporter we can at this point don't utilize a basic demodulator/indicator. We'll have to either re-infuse a transporter (i.e., utilize a BFO/coordinated finder) or use envelope identification or comparable. The bit of leeway here is that we're presently putting the entirety of the transmitter power into the sidebands, and not squandering any on the transporter (which has no data content).

Note AM is both sidebands, every one of which involves a similar data transfer capacity as the first adjusting signal, and incorporates a full transporter. At $100\% $ balance, we have completely $66\% $ of the absolute force going into the transporter - and obviously, we don't average $100\% $ balance. So AM is the easiest (both to create and demodulate), however takes up the greatest transmission capacity and is reasonably power wasteful, at any rate regarding putting the accessible force behind the genuine data you're attempting to communicate.