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I haven't done EE for quite a long time. I need to implement some MUX, so I started with a two-way MUX. But I don't understand how to make the decision:

enter image description here

Wiki- Multiplexer

A truth table will show that

This truth table shows that when S=0 then Z=A but when S=1 then Z=B.

I don't get the logic here. The boolean expression is Z = (A and S') or (B and S)


Let S = 0, then each sub expression will become

                z1

A = 0,  S' = 1   => 0

A = 1,  S' = 1   =>  1

                z2

B = 0,  S = 0    => 0

B = 1,  S = 0    => 0

Can someone please explain to me how do we reach that conclusion? Thank you very much!

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  • 3
    \$\begingroup\$ Your exact question is not clear. The boolean expression is correct. BUT the last two lines MAY be incorrect depending on what you mean by them - eg Z1 and Z2 are undefined and what you mean by them is uncertain. You may find it useful to introduce "don't care" states. Essentially, while the boolean expression is correct you have really got a don't care state when the related mux controls are not activated. \$\endgroup\$ – Russell McMahon Aug 30 '11 at 3:37
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The best way to get a proper understanding is to write down the truth table completely. You have 3 inputs (A, B, S), so this will give you 2\$^3\$ = 8 combinations:

S  A  B | S' C  D | Z
--------+---------+--
0  0  0 | 1  0  0 | 0
0  0  1 | 1  0  0 | 0
0  1  0 | 1  1  0 | 1
0  1  1 | 1  1  0 | 1
1  0  0 | 0  0  0 | 0
1  0  1 | 0  0  1 | 1
1  1  0 | 0  0  0 | 0
1  1  1 | 0  0  1 | 1

It's often useful to add intermediate results to make things more clear. I added a term \$C = (A \land S')\$ and \$D = (B \land S)\$. Now it should be clear that \$Z = (C \lor D)\$.

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One point not yet mentioned is that in many cases, the proper expression is Z = (A and S') or (B and S) or (A and B). The last term is used if A and B are both set and S changes from zero to 1 or vice versa. In that scenario, without the last term, the output might go low briefly. Adding the last term will ensure that when A and B are both set, the output will remain set. This has at least two useful effects: (1) it will prevent problems if the Mux is used as part of the sequencing logic in a circuit. For example, a Mux whose output is tied to an input makes a nice latch, but only if the output doesn't glitch when S changes; (2) in CMOS logic, such a design may reduce energy consumption when S changes, since the momentary glitch resulting from a change in S might cause shoot-through currents.

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  • \$\begingroup\$ Standard HCMOS devices like the 74HC157 don't seem to have this third AND term, or at least it isn't shown in the logic diagram. I also wonder if an FPGA synth won't optimize it away. \$\endgroup\$ – stevenvh Aug 30 '11 at 14:25
  • \$\begingroup\$ @stevenvh: Logic compilers generally have options to indicate whether such terms must be kept (or even to indicate that they should be added when they don't exist). If the output of the circuit is going to be 'don't-care' any time S changes, the gate may be omitted. In the one circuit I designed with a mux as a latch, I used a small RC delay to cover the hazard condition. Actually, it was a cute memory-controller circuit. One chip generated two outputs: one of which was simply the inverse of A12, and the other of which was a latch which was strobed only when A12 was low and A11 high. \$\endgroup\$ – supercat Aug 30 '11 at 14:33
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Let S = 0, then each sub expression will become

                z1

A = 0,  S' = 1   => 0

A = 1,  S' = 1   =>  1

                z2

B = 0,  S = 0    => 0

B = 1,  S = 0    => 0

So, This means that When S=0, Irrespective of Input B, Output will depend only on Input A. If you make S=1, MUX will select Input B. Which means Output will reflect only B. It will not have any effect of A.

Let's take Main Boolean Expression:- Z = (A and S') or (B and S)

When S = 0;

Z = (A and 1) | (B and 0)

Z = A | 0

Hence Z = A

When S = 1;

Z = (A and 0) | (B and 1)

Z = 0 | B

Hence Z = B;

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In this simple case you can (almost?) design the circuit in English.

Depending on S, you want either A or B to appear on Z. So you want a 1 on Z when ( A is selected and A is one), or when ( B is selected and B is one ). In all other cases the selected input is 0, so you want a zero on the output.

Distilling the expression: ( A is one and A is selected ) or ( B is one and B is selected ) A is selected means: ( S == 0 ) B is selected means: ( S == 1 )

make it more formal: ( A ^ ~ S ) V ( B ^ S )

so you need an OR, tow ANDs and one inverter.

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A MUX is simply a logic switch of sorts. The S_0 signal (select signal) will pass the A signal if it is low (logic 0 or 0v) and pass the B signal if it is high (logic 1 or +5v or whatever voltage the system uses).

The same can be applied to a 4 to 1 MUX. The select signal acts as a binary number selecting which incoming signal's truth value is passed through. If you think of the MUX as vertical like this:

     ________
    |4to1MUX |
    |y0      |
    |y1      |
    |y2     F|
    |y3      |
    |        |
    |_s1__s0_|

The input signals are laid out from least significant to most significant and the select inputs are laid out most significant to least significant. When a "00" is applied to the select lines then y0 is selected. When "01" is applied, y1 is selected; "10",y2; "11",y3.

The other post are very helpful in explaining the Boolean algebra parts, which are found by examining the truth table as said elsewhere.

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