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I am new to the concept of Time Division Multiplexing in school, after reading a few books I came over this question.

With the help of sketches explain Time Division Multiplexing (TDM) in terms of sources with the same bit rate as well as different bit rates. How can the existing analog lines be used? What are T1 and E1 circuits? show how the bit rate of T1 is 1.544 Mbps and that of E1 is 2.048 Mbps but yet all have the same sampling period of 12 msec. Use either one of them to explain how it works using TDM (using either T1 or E1).

I would like anybody with any knowledge on how to solve the question to please help me out here am stuck.

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closed as not a real question by PeterJ, Keelan, Leon Heller, Dave Tweed, W5VO May 17 '13 at 13:44

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  • \$\begingroup\$ I got something, the answer to how bit rate of T1 is 1.544 Mbps is that a T1 network bundles 24 64kps, hence would have 24*64kps = 1.544Mbps \$\endgroup\$ – James Okpe George May 13 '13 at 10:46
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A T1 circuit combines 24 "channels" of 64 kbps each by interleaving 8 bits at a time from each channel into one "frame" of data. It also adds 8 kbps (one bit per frame) of "overhead", used for finding the frame and channel boundaries, which brings the total data rate to 1.544 Mbps.

Similarly, an E1 circuit combines 32 channels for a total data rate of 2.048 Mbps (one of the channels is used for the overhead signal).

The question is incorrect in one respect: the sample period (frame period) is 1/8000 Hz, or 125 µs.

The 64 kbps in each channel is a stream of 8-bit PCM samples taken at a sample rate of 8000 sps, representing the audio signal on an ordinary analog phone line.

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  • \$\begingroup\$ okay so the 125us is gotten from the inverse of the sampling rate \$\endgroup\$ – James Okpe George May 13 '13 at 21:52

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