Altifigence Academy

8 / 63 · Concept

From truth tables to equations: minterms, maxterms, SOP and POS

Derive the same function from its one rows and zero rows before minimizing it.

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Select exactly one truth-table row

Fix input order A, B, C with A most significant. Row 5 is 101, and m5=ABCm_5=A\overline B C is 1 only on that row. A minterm contains every variable once; the name does not mean minimum gate count.

If F is 1 on rows 1, 3, 5, 6 and 7:

F=Σm(1,3,5,6,7)=ABC+ABC+ABC+ABC+ABCF=\Sigma m(1,3,5,6,7)=\overline A\overline B C+\overline A B C+A\overline B C+AB\overline C+ABC

This canonical sum of products (SOP) ORs the one-row minterms. A general SOP need not include every variable in every term.

Select zero rows instead

The maxtermmaxterm An OR term containing every variable, false for exactly one input combination. AND together the maxterms of the false rows. M2=A+B+CM_2=A+\overline B+C is zero only on row 010. Pick each literal to be zero on that row. Since F is zero on rows 0, 2 and 4:

F=ΠM(0,2,4)=(A+B+C)(A+B+C)(A+B+C)F=\Pi M(0,2,4)=(A+B+C)(A+\overline B+C)(\overline A+B+C)

This is canonical product of sums (POS). Σm lists one rows; ΠM lists zero rows.

Two equations from the same table
  1. One rows: OR minterms for rows 1, 3, 5, 6, 7
  2. Same F: C is 1, or both A and B are 1
  3. Zero rows: AND maxterms for rows 0, 2, 4

Canonical forms are a starting point

The same function simplifies to F=C+ABF=C+AB: C covers four rows, and AB covers the remaining 110. This is SOP but not canonical SOP. The next Karnaugh-map lesson makes such grouping visible. Always verify the simplified expression against all eight rows.

Lab · Connect three representations of one function

Truth-table rows, canonical forms and simplified SOP

F = C + A · B

Three-input truth table · X is an unspecified output
ABCF
0000
0011
0100
0111
1000
1011
1101
1111

The A/B/C buttons change the input being observed. The cells change the function specification. The dashed outline marks the input row; blue borders mark the selected product term.

For the original function, set ABC=110. F remains 1 because AB is true even though C=0. Select the C and A·B groups and notice their shared m7 cell. Overlap does not change an OR result.

Click m6 twice: 1 → X → 0. The final function is just C. Row 6 leaves Σm and enters ΠM. The intermediate X means this row has no required output, not that a defined output may be ignored just to reduce cost.

Σm lists required ones, ΠM required zeros, and d unspecified rows. When d is nonempty, Σm or ΠM alone is not a complete truth-table specification. The lab minimizes product-term count first and literal count second, and shows one solution; this is not a claim about minimum physical gate area.

Essayez vous-même

If only m6 changes from required 1 to required 0 in F=C+AB, what expression remains? How does specifying X at m6 differ?

Lire l’explication

F=C. With m6=0 that row must produce 0. With m6=X either output is allowed on that row, but every other specified row must still be preserved.

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