In time-division multiplexing, how do we make sure that there are no dropped bits?

Consider this example. Let's say we have a 4-to-1 multiplexer. Each input line is 200 bit/s. The unit of transmission is 8 bits, that is, each time slot has 8 bits, and therefore each frame is 32 bits.

So now the selector is at the first input, and because a time slot is 8 bits, the selector must stay on the first input until 8 bits are transmitted to the output. And after the first input has transmitted 8 bits, the selector will then switch to the second input.

But before the selector switch to the second input, the second input was already transmitting at the same rate, that is, although the line is open, it is still changing its state between high and low at the same rate. So therefore, before the selector switch to second input, 8 bits were already lost. And before it switches to third input, 16 bits were lost, and before the last input, 24 bits were lost, and so on.

What am I missing?

  • \$\begingroup\$ You're missing the precise allocation of time slots. All transmitters must have a common clock so that they know when exactly is their time to transmit. \$\endgroup\$
    – PkP
    Mar 17, 2020 at 5:29
  • \$\begingroup\$ Does it mean there is a central controller that allocates the time to each participant? What if for example I connect my laptop to a multiplexer. How does it know when to transmit? \$\endgroup\$
    – Noob_Guy
    Mar 17, 2020 at 5:46

1 Answer 1


In my opinion, you are missing the simple input buffers. Let’s consider the traditional PDH system. A 4-1 multiplexer tries to construct an E4 signal from the 4 input E1 signals. All inputs must have buffers long enough to be able to tolerate the problem you mentioned. Apart from that, there is no guarantee that all inputs will have the same exact clock. What happens if input B has 999 bits when is was supposed to have 1000 bits in the buffer? This is where bit stuffing and rubbing help. These problems are solved using control bits in the next generation SDH systems. So basically, 2 standard TDM transmission systems solve the problem you mentioned by buffering + control bits (pointers).


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