# How does a TDM multiplexer "catch" all bits?

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?

• 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.
– PkP
Mar 17, 2020 at 5:29
• 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? Mar 17, 2020 at 5:46