 BOOLEAN FUNCTION IMPLEMENTATION USING MUXes-PART I

Q-How to implement any Boolean function using MUX?

Ans: While implementing any function using MUX, if we have N variables in the function then we take (N-1) variables on the selection lines and 1 variable is used for inputs of MUX. As we have N-1 variables on selection lines we need to have 2 N-1 to 1 MUX. We just have to connect A, A’, 0 or 1 to different input lines.

e.g. To implement the function F(A, B, C)= Σ (1, 2, 5, 7) using (a)8 to 1 MUX (b)4 to 1 MUX

Ans: (a) We can implement it using all three variables at selection lines. We put 1 on the min term lines which are present in functions and 0 on the rest. (b)F= A’B’C + A’BC’ + AB’C + ABC

N=3 so we use 2 N-1 = 2 2 = 4 to 1 MUX.

Suppose we have B, C on the selection lines. So when we have BC=00, put B=0, C=0 in the function and we see output of the function should be 0 hence we connect 0 to 0th input line.

When BC=01, then output of the function should be A’+ A = 1. Hence we connect 1 to 1st line.

When BC=10, then output of the function should be A’. Hence we connect A’ to 2nd line.

When BC=11, then output of the function should be A. Hence we connect A to 3rd line.

Hence we have the circuit as: 