Q.Example 1: Simplify the Boolean expression $Y = \sum m(0,1,4,13,15) + \sum d(2,5,7)$ and then draw the logic diagram using only basic gates.

Plotting the minterms as 1s and the don't-cares as X, the map yields a quad $\overline{A}\,\overline{C}$ and a quad $BD$; ORing them gives the SOP form realis …

[!TLDR]

Map the minterms and don't-cares, group two quads ($\overline{A}\,\overline{C}$ and $BD$), OR them, and build the result from NOT + AND + OR gates: $Y = \overline{A}\,\overline{C} + BD$.

Plotting the map

Enter 1 for each minterm in $\sum m(0,1,4,13,15)$, X for each don't-care in $\sum d(2,5,7)$, and 0 elsewhere:

Grouping

• Quad 1: rows $\overline{A}\,\overline{B}$ and $\overline{A}B$, columns $\overline{C}\,\overline{D}$ and $\overline{C}D$ (minterms 0, 1, 4 and don't-care 5). $B$ and $D$ change; constants $A=0,\ C=0$ give $\overline{A}\,\overline{C}$.
• Quad 2: rows $\overline{A}B$ and $AB$, columns $\overline{C}D$ and $CD$ (minterms 13, 15 and don't-cares 5, 7). $A$ and $C$ change; constants $B=1,\ D=1$ give $BD$.

ORing the two groups gives the simplified SOP $Y = \overline{A}\,\overline{C} + BD$.

Realising with basic gates (Figure 10.4.1) …