2
\$\begingroup\$

If a multiplexer has N inputs and 1 output, that's called an N:1 multiplexer.

But suppose a multiplexer has N inputs, each of which is an M-bit bus, and the output is also an M-bit bus. What is the correct terminology to describe that?

To take an example, say you have 4 busses of 8 bits each. So a total of 4×8 = 32 bits go in, and 8 bits come out. So is that a 32:8 mux? Or do you just describe it as 8 separate 4:1 mux? (I mean, that's how you implement it.) Or is there some other terminology here?

\$\endgroup\$
2
  • \$\begingroup\$ I'd call it an \$\sum^M_1\left(N:1\right)\$ but, you can call it what you want surely? \$\endgroup\$
    – Andy aka
    Commented Oct 8, 2022 at 9:53
  • \$\begingroup\$ Isn't that just the same as (M) N:1 muxes? \$\endgroup\$
    – jonk
    Commented Oct 8, 2022 at 10:24

1 Answer 1

1
\$\begingroup\$

There's a more important distinction. That's how many independent connections it can support. These are often described as 'blocking' and 'non-blocking', but I think it's more useful simply to say how many connections it has.

For instance, the cheapest way to make a 4:8 multiplexer is to have a 4:1 mux down to a single channel, then a 1:8 demux up to 8 outputs.

This is known as a 'blocking' mux/demux, as it obviously can support only one communication path at a time. It only requires 12 switches.

The most expensive way to make a 4:8 multiplexer is a 'crossbar' design, where any output can be routed with any input. This needs 32 switches.

You can have any intermediate option as well.

With this few inputs and outputs, it may be not obvious why there's much fuss about how it's implemented. When you get to the scale of trans-ocean telephony, with millions of inputs and outputs, yet possibly only thousands of connections, the distinction is quite important.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.