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Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed signal is
  • a)
    115.2 kbps
  • b)
    28.8 kbps
  • c)
    57.6 kbps
  • d)
    38.4 kbps
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, ar...
Analog signals, having bandwidth 1200 Hz, 600 Hz and 600 Hz have 2400, 1200 samples/sec at Nyquist rate. Hence 48000 sample/sec
bit rate  = 48000 sample/sec x 12 = 57.6 kbps
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Most Upvoted Answer
Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, ar...
To determine the bit rate for the multiplexed signal, we need to consider the bandwidth and the number of bits used to represent each sample for each analog signal.

Given:
Bandwidth of the first analog signal = 1200 Hz
Bandwidth of the second analog signal = 600 Hz
Bandwidth of the third analog signal = 600 Hz
Number of bits used to represent each sample = 12 bits

Step 1: Calculate the Nyquist rate for each analog signal
The Nyquist rate is twice the bandwidth of the signal. So, we can calculate the Nyquist rates as follows:

Nyquist rate for the first analog signal = 2 * 1200 Hz = 2400 samples per second
Nyquist rate for the second analog signal = 2 * 600 Hz = 1200 samples per second
Nyquist rate for the third analog signal = 2 * 600 Hz = 1200 samples per second

Step 2: Determine the total number of samples per second for all three signals
Since the three analog signals are time division multiplexed, we need to add up the number of samples per second for each signal.

Total number of samples per second = Nyquist rate of first signal + Nyquist rate of second signal + Nyquist rate of third signal
= 2400 samples per second + 1200 samples per second + 1200 samples per second
= 4800 samples per second

Step 3: Calculate the bit rate for the multiplexed signal
The bit rate is given by the formula:

Bit rate = Total number of samples per second * Number of bits per sample

Bit rate = 4800 samples per second * 12 bits per sample
= 57,600 bits per second
= 57.6 kbps

Therefore, the correct answer is option 'C' - 57.6 kbps.
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