A TDM link has 24 signals and each channel is sampled 8000 times/sec. Each sample is represented by 8 binary bits and it contains an additional bit for synchronization. The total bit rate made from the TDM link is 

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UGC NET Paper 2: Electronic Science Dec 2019 Official Paper
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  1. 1600 K bits/sec
  2. 1728 K bits/sec
  3. 1826 K bits/sec
  4. 2056 K bits/sec

Answer (Detailed Solution Below)

Option 2 : 1728 K bits/sec
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Detailed Solution

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Concept:

Time Division Multiplexing (TDM) provides the time-sharing of a common channel by a large number of users.

The data sources are sampled at a Nyquist rate or higher.

Proper operation depends on synchronization between the input and the output communications.

F2 S.B Madhu 08.04.20 D5

Calculation:

Given sampling rate of channel each = 8000 Hz = fch

Since 24 such signals are Multiplexed, we get a total pulse per second (frequency) of the multiplied signal as:

fs = nfch

Putting on the respective values:

fs = 24 × 8000

fs = 192,000 = 192 KHz

Now each sample is represented by 8 bits and contains an additional bit for synchronization (n).

The total no. of bits per sample will be:

n = 8 + 1 = 9 bits

∴ The total bit rate of TDM link is given as:

Rb = nfs

Rb = 9 × 192 Kbps

Rb = 1728 Kilo-bits/sec
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