Question 3 of 6: Canonical and Minimized Forms from a POS Expression

Question 3: Canonical and Minimized Forms from a POS Expression (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $F(W,X,Y,Z)=(W+\bar X+\bar Y)(\bar W+\bar Z)(W+Y)$, an un-minimized 4-variable POS expression.

Find. (a) canonical SOP ("little $m$" form); (b) minimized SOP; (c) $\bar F$ minimized SOP; (d) $\bar F$ minimized POS.

Approach. Evaluate each of the three OR-factors' FALSE condition, union them to find where $F=0$, complement that set to get the $F=1$ minterms (part a), then K-map/algebraically minimize $F$ (part b) and, via De Morgan on the minimized SOP, obtain $\bar F$ in both minimized forms (parts c, d) without a second independent K-map.

1. Part (a) — canonical SOP by elimination. Each factor is false only for one specific input pattern: $(W+\bar X+\bar Y)$ is false only at $W{=}0,X{=}1,Y{=}1$; $(\bar W+\bar Z)$ is false only at $W{=}1,Z{=}1$; $(W+Y)$ is false only at $W{=}0,Y{=}0$. Checking all 16 combinations of $(W,X,Y,Z)$ against these three conditions (any one true $\Rightarrow F=0$) leaves exactly six minterms where none applies: $$F(W,X,Y,Z)=\boxed{\Sigma m(2,3,8,10,12,14)}=m_2+m_3+m_8+m_{10}+m_{12}+m_{14}$$ Written out: $m_2=\bar W\bar XY\bar Z,\ m_3=\bar W\bar XYZ,\ m_8=W\bar X\bar Y\bar Z,\ m_{10}=W\bar XY\bar Z,\ m_{12}=WX\bar Y\bar Z,\ m_{14}=WXY\bar Z$.
2. Part (b) — minimize by grouping. On the K-map, minterms $\{8,10,12,14\}$ all share $W{=}1,Z{=}0$ with $X,Y$ free — a quad giving $W\bar Z$. Minterms $\{2,3\}$ share $W{=}0,X{=}0,Y{=}1$ with $Z$ free — a pair giving $\bar W\bar XY$. These two prime implicants cover all six minterms and nothing extra: $$F=\boxed{W\bar Z+\bar W\bar XY}$$
3. Part (c) — complement via De Morgan on the minimized SOP. $\bar F=\overline{W\bar Z+\bar W\bar XY}=\overline{W\bar Z}\cdot\overline{\bar W\bar XY}=(\bar W+Z)(W+X+\bar Y)$. Multiplying out and re-grouping the resulting 6 product terms on a K-map for the ten minterms $\{0,1,4,5,6,7,9,11,13,15\}$ collapses to three prime implicants ($\bar W X$ covers $\{4,5,6,7\}$; $WZ$ covers $\{9,11,13,15\}$; $\bar W\bar X\bar Y$ covers $\{0,1\}$), each essential: $$\bar F=\boxed{\bar WX+WZ+\bar W\bar X\bar Y}$$
4. Part (d) — $\bar F$ in minimized POS. The fastest route to a minimized POS of $\bar F$ is to complement the minimized SOP of $F$ found in Step 2 (since $(\bar F)=(\overline{F})$, and $\overline{F_{\min SOP}}$ is automatically a minimized POS of $\bar F$ — De Morgan preserves minimality on this size of expression): $$\bar F=\overline{W\bar Z+\bar W\bar XY}=\overline{W\bar Z}\cdot\overline{\bar W\bar XY}=\boxed{(\bar W+Z)(W+X+\bar Y)}$$