Implementation of Boolean function f(a,b,c)=b ac using 2*1 Mux

Boolean function f(a,b,c)=b ac can be implemented using 2*1 Mux. In this implementation, we need to use minimum number of 2*1 Mux.

Truth Table

Let's first create the truth table for the given Boolean function.

| a | b | c | f(a,b,c) |
|---|---|---|----------|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |

K-Map

From the truth table, we can create the K-Map for the output function f(a,b,c).

| | 00 | 01 | 11 | 10 |
|---|----|----|----|----|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 |

From the K-Map, we can see that the output function f(a,b,c) is equal to b AND (NOT a) AND c.

Implementation using 2*1 Mux

We can implement the output function f(a,b,c) using 2*1 Mux as follows:

1. Connect a to the select input of the first Mux.
2. Connect NOT a to the select input of the second Mux.
3. Connect b to the data input of both Muxes.
4. Connect c to the data input of the first Mux.
5. Connect 0 (ground) to the data input of the second Mux.
6. Connect the output of the first Mux to the input of an AND gate.
7. Connect the output of the second Mux to the other input of the AND gate.
8. The output of the AND gate is the output function f(a,b,c).

Minimum number of 2*1 Mux

From the implementation using 2*1 Mux, we can see that we need two 2*1 Muxes to implement the output function f(a,b,c) with minimum number of Muxes.