Digital Electronics - Boolean Expression and Its Simplification
The knowledge about boolean alegebra and evalation of boolean expression is already learnt in the discrete mathematics, so I will not write again, but I will still take notes about something that has forgotten.
Boolean Expression Simplification
Special Distributive Law
\[\begin{aligned} (A+B)(A+C) & = AA + AC + AB + BC \\ & = A + AC + AB + BC \\ & = A(1 + C) + AB + BC \\ & = A \cdot 1 + AB + BC \\ & = A(1 + B) + BC \\ & = A \cdot 1 + BC \\ & = A + BC \end{aligned}\]So, we have the following rules.
\[A + AB = A(1+B) = A \cdot 1 = A \\ A + A^\prime B = (A + A^\prime )(A + B) = 1 \cdot (A + B) = A + B\]Transforming into All NAND Expression through DeMorgan’s Theorems
Take twice complement of the original expression.
\[\begin{aligned} Y &= (B(A^\prime+C^\prime))^\prime \\ &= (B(AC)^\prime)^\prime \end{aligned}\]The Karnaugh Map
The main method is grouping the adjacent “1”s following the rules below.
- Maximize the size of groups and minimize the number of groups.
- Ensure the number of cells that every group contains is the power of 2.
- Must include all the “1”s.
- Cell already in a group can be used again, as long as the new group you intend to create includes a new cell.