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A 4:1 MUX can be built by using three 2:1 MUX. But I've not done it this way: I've used a 2:4 decoder to make a 4:1 MUX. Does it matter? enter image description here

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closed as not a real question by Wouter van Ooijen, Brian Carlton, Olin Lathrop, Dave Tweed, Leon Heller Mar 10 '13 at 3:37

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Matter for what or to whom? (in other words, this is not a real question.) – Wouter van Ooijen Mar 9 '13 at 20:48
For an ALU or to measure a time difference due to propagation(?) – Programmer 400 Mar 9 '13 at 20:53
I your are optimizing for delay time you must first decide whether it is an address change or a data change that is your starting point (or maybe the maximum of the two delays?). After that, I think you can count gates yourself? – Wouter van Ooijen Mar 10 '13 at 9:22

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It matters depending on what you optimise for. Your way is probably better once you factor in the extra cost a higher part count causes, and the fact that a single chip solution is likely to be faster than one involving 3. For most purposes, there isn't too much of a difference. Though, I will note that your schematic and actual application are important pieces of information required to pass specific judgement.

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