# How could a digital multiplexer transmit at a rate higher than its input?

From Wikipedia, "a multiplexer, also known as data selector, is a device that selects between several analog or digital inputs and forwards it to a single output line."

https://en.wikipedia.org/wiki/Multiplexer

Each input take turns in connecting to the output line. There is only one input channel transmitting at a time. When an input is selected, it is as if there is a continuous path from that input to the output line. So therefore, the line rate of the output channel must be the rate of which input it is connected to. For example, consider the figure above. If input B is selected, and its rate is 64 kbps, the output rate should also be 64 kbps.

But we know that in T-carrier system this is not the case. The rate of T1 is 1.544 Mbps because it is 24 channels of 64 kbps each plus 1 framing bit of 8000 samples per second.

Now that's the mathematics. But in terms of electronics, how could the multiplexer transmit at a rate higher than the input rate?

• Drive the switch selectors fast enough and you can sample all 5 channels within one 64kb/s bit time. If the line can handle 320kb/s, that is. – Brian Drummond Mar 23 at 15:07
• That was my first intuition. But that would only work with bit-interleaving. T-carrier is byte-interleaving. – Noob_Guy Mar 24 at 5:08
• Then it has to assemble entire bytes, select one in an 8-way version of your switch, feed that to a seriallzer, and reverse the process at the other end. More complex logic, longer latency, that's all.About one card of TTL for each end, in the days I met it (E1, 2048 kbits but same basic idea) – Brian Drummond Mar 24 at 11:24