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

Options

A. 101 kHz
B. 99 kHz
C. 84kHz
D. 71 kHz

Explanation

To solve the problem, we need to calculate the total bandwidth required for transmitting 12 signals using frequency-division multiplexing (FDM) with a guard band. Each signal is band-limited to 5 kHz, and we have a guard band of 1 kHz between each pair of signals.

### Step-by-Step Solution:

1. **Determine the bandwidth for one signal:**
Each signal is band-limited to 5 kHz. Therefore, the bandwidth required for one signal is:
\[ \text{Bandwidth of one signal} = 5 \text{ kHz} \]

2. **Calculate the total bandwidth for 12 signals:**
Since there are 12 signals, the total bandwidth required for these signals without considering the guard bands is:
\[ \text{Total bandwidth for 12 signals} = 12 \times 5 \text{ kHz} = 60 \text{ kHz} \]

3. **Determine the number of guard bands required:**
A guard band is required between each pair of adjacent signals. For 12 signals, the number of guard bands needed is one less than the number of signals:
\[ \text{Number of guard bands} = 12 - 1 = 11 \]

4. **Calculate the total bandwidth of the guard bands:**
Each guard band is 1 kHz wide. Therefore, the total bandwidth for the guard bands is:
\[ \text{Total bandwidth for guard bands} = 11 \times 1 \text{ kHz} = 11 \text{ kHz} \]

5. **Calculate the total bandwidth of the multiplexed signal:**
Finally, we add the total bandwidth of the signals and the total bandwidth of the guard bands to find the total bandwidth of the multiplexed signal:
\[ \text{Total bandwidth of multiplexed signal} = 60 \text{ kHz} + 11 \text{ kHz} = 71 \text{ kHz} \]

### Final Answer:
The bandwidth of the multiplexed signal is **71 kHz**.

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