# Is it possible to implement one 8-way 1-bit multiplexer?

I'm sorry, but shouldn't 8-way have 3 bits in order to traverse every address? Anyway, I have here a problem, and I am not sure if my solution is correct.

The problem: Given the boolean expression \begin{align*} C'(B+A) + A'B'C \end{align*} Using one 8-way 1-bit multiplexer, implement the boolean expression.

Solution. For our Addresses, we have: \begin{align*} D_7, D_6, D_5, D_4, D_3, D_2, D_1, D_0 \end{align*} For our selectors, \begin{align*} A, B, C \end{align*} Where A is the MSB and C is the LSB. I have 3 1-bit selectors so I can traverse every address. Then, we have our output Y.

Truth Table:

     A   B   C   Y
0   0   0   0
0   0   1   1
0   1   0   1
0   1   1   0
1   0   0   1
1   0   1   0
1   1   0   1
1   1   1   0



Hence, I have the boolean expression: \begin{align*} D_1A'B'C + D_2A'BC' + D_4AB'C' + D_6ABC' = Y \end{align*} What I did is I only included the rows with 1 in Y, and I negate the switch (A/B/C) if it is 0, and keep it as is otherwise.

• You should be able to solve this easy question. Assume (A, B, C) as the select lines (S2, S1, S0). Can you write down the values for D0 to D7 then? Hint: Truth table for the Boolean expression. Jun 11, 2021 at 8:53
• For eg: The meaning of D0 in the expression is, what's the value of Y if ABC = 111.... Jun 11, 2021 at 8:56
• I have the truth table posted. Does that mean my answer (the boolean expression) is correct? Jun 11, 2021 at 8:56
• Truth table looks fine. What you have written as answer is generic expression of an 8-bit mux. You have to incorporate (A,B,C) inputs, and values of D0-D7 in it to write the final expression. Then you can draw the block diagram/schematic straight away. Jun 11, 2021 at 9:00
• So, for example, since Y is 0 in D_0, that would mean I would negate S2, S1, S0? That would make it D_0S_2'S_1'S_0' for the first term? Jun 11, 2021 at 9:02