Implement the following Boolean function using 4:1 multiplexer using A and B variables to the selection lines. F(A, B) = √ (2, 3, 7)

To implement the given Boolean function using a 4:1 multiplexer with A and B as the selection lines, we need to follow these steps:

Step 1: Understand the Boolean Function

The given Boolean function is represented in terms of a sum of minterms (using √ to denote the sum of minterms):

F(A, B) = √(2, 3, 7)

This means that the function F(A, B) is true (or '1') for the minterms 2, 3, and 7. In binary, these minterms correspond to the following combinations of A and B:

• Minterm 2: A = 0, B = 1 (binary 10)
• Minterm 3: A = 1, B = 1 (binary 11)
• Minterm 7: A = 1, B = 1 (binary 11)

Step 2: Create the Truth Table

To visualize the function, we can create a truth table for two variables, A and B.

A	B	F(A, B)
0	0	0
0	1	1
1	0	0
1	1	1

From the truth table, we can see that F(A, B) is '1' for the inputs (0,1) and (1,1), which correspond to the minterms 2, 3, and 7.

Step 3: Configure the 4:1 Multiplexer

A 4:1 multiplexer has 4 data inputs (D0, D1, D2, D3), 2 selection lines (S1, S0), and 1 output (Y). The selection lines S1 and S0 will be connected to the inputs A and B, respectively.

The data inputs of the multiplexer will be set based on the truth table. Since we want the output to be '1' for minterms 2, 3, and 7, we will set the corresponding data inputs to '1'. For all other minterms, the data inputs will be '0'.

Here is how we connect the inputs:

Data Input	Value	Minterm
D0	0	0
D1	1	2
D2	0	4
D3	1	6

Step 4: Implement the Multiplexer

The output Y of the multiplexer is determined by the selection lines and the data inputs. The output can be represented by the following formula:

Y = (¬S1 ∧ ¬S0 ∧ D0) ∨ (¬S1 ∧ S0 ∧ D1) ∨ (S1 ∧ ¬S0 ∧ D2) ∨ (S1 ∧ S0 ∧ D3)

Since we have set D0 and D2 to '0', and D1 and D3 to '1', the formula simplifies to:

Y = (¬S1 ∧ S0) ∨ (S1 ∧ S0)

Substituting S1 with A and S0 with B, we get:

Y = (¬A ∧ B) ∨ (A ∧ B)

Step 5: Verify the Implementation

To verify that our implementation is correct, we can check the output Y for all combinations of A and B:

A	B	Y (Output)
0	0	0
0	1	1
1	0	0
1	1	1

This matches the truth table we created in Step 2, confirming that the implementation of the Boolean function using a 4:1 multiplexer is correct.

Example Implementation

Here is a simple example of how the multiplexer would be connected:

    A (S1) ----+
               |
               |    +--- Y (Output)
    B (S0) --+ |    |
              | |    |
              | |    |
             _|_|____|_
            |           |
            |  4:1 MUX  |
            |           |
            |___|___|___|
               |   |
               |   |
              D1  D3
               |   |
               +---+

In this diagram, D0 and D2 are not shown as they are connected to '0' and do not contribute to the output.