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:
Another procedure to implement the function using MUX
- Take one variable for input lines and rest of the term for selection lines.
- Then list the min terms with the variable selected in complimented form in 1st row and list the
- The min terms with variable selected in un-complimented form in 2nd row.
- Then encircle the min terms which are present in the function.
- If we have no circled variable in the column, then we put 0 on the corresponding line
- If we have both circled variables, then we put 1 on the line
- If bottom variable is circled and top is not circled, apply A to input line
- If bottom variable is not circled and top is circled, apply A’ to input line
e.g. To implement the function F(A, B, C)= Σ (1, 2, 5, 7) using MUX.
Let’s now take the variable A for input lines and B & C for selection lines.
So we list the min terms as follow:
So the circuit is
This is same as the circuit obtained using earlier method.