Digital Logic & Microprocessor — Digital Logic, NEC licence examination syllabus (Nepal Engineering Council).
Product-of-Sums: the mirror image of SOP
Instead of listing where the output is 1, list where it's 0. Sometimes that's far shorter.
SOP builds a circuit from the rows where the output is 1. But suppose the output is 1 for fifteen of sixteen rows — then SOP needs fifteen product terms, while describing the single zero takes one term. POS is that complementary approach, and choosing the right one can halve your circuit.
Maxterms
A MAXTERM is a sum (OR) term containing every variable
exactly once, which equals 0 for exactly ONE input
combination.
For 3 variables:
M₀ = A+B+C (000) M₄ = A'+B+C (100)
M₁ = A+B+C' (001) M₅ = A'+B+C' (101)
M₂ = A+B'+C (010) M₆ = A'+B'+C (110)
M₃ = A+B'+C' (011) M₇ = A'+B'+C' (111)
The rule is the OPPOSITE of minterms:
variable = 0 → write it plain (A)
variable = 1 → write it barred (A')
Notation: F = ΠM(0,2,4) means the product of maxterms 0,2,4.
The complement rule is the trap. For a minterm, a 1 means write the variable plain. For a maxterm, a 0 means write it plain. They are exact opposites, and mixing them up produces an expression that is the complement of what you wanted. Write the rule at the top of your answer sheet before starting.
The relationship between the two forms
For any function of n variables:
F = Σm(list of 1-rows) = ΠM(list of 0-rows)
The two index lists are COMPLEMENTARY — together they cover
every number from 0 to 2ⁿ−1.
Example (3 variables, 8 rows):
F = Σm(0,1,4,5) → the 0-rows are 2,3,6,7
F = ΠM(2,3,6,7)
Also: m_i' = M_i (a minterm's complement is the
maxterm of the same index)
Worked numerical 1 — both forms of one function
For F = Σm(1,3,5,7), write the canonical POS form and simplify both.
A B C | F | form
────────┼────┼───────────────
0 0 0 | 0 | M₀ = A+B+C
0 0 1 | 1 | m₁
0 1 0 | 0 | M₂ = A+B'+C
0 1 1 | 1 | m₃
1 0 0 | 0 | M₄ = A'+B+C
1 0 1 | 1 | m₅
1 1 0 | 0 | M₆ = A'+B'+C
1 1 1 | 1 | m₇
SOP: F = Σm(1,3,5,7)
= A'B'C + A'BC + AB'C + ABC
Simplify: A'C(B'+B) + AC(B'+B) = A'C + AC = C(A'+A) = C
F = C ← remarkably simple!
POS: F = ΠM(0,2,4,6)
= (A+B+C)(A+B'+C)(A'+B+C)(A'+B'+C)
Simplify the first pair using (X+Y)(X+Y') = X:
(A+B+C)(A+B'+C) = A+C
And the second pair:
(A'+B+C)(A'+B'+C) = A'+C
Now: (A+C)(A'+C) = C + AA' = C + 0 = C
F = C ✔ same answer
Both forms collapse to F = C. Looking at the truth table
this is obvious in hindsight — the output equals C in every
row. Which is the lesson: always simplify, because canonical
forms hide simple functions behind four-term expressions.
Worked numerical 2 — when POS is the better choice
A function of 3 variables is 1 for every input except 000 and 111. Compare SOP and POS costs.
F is 0 only for rows 0 and 7, so:
F = Σm(1,2,3,4,5,6) ← six minterms
F = ΠM(0,7) ← two maxterms
SOP implementation:
6 three-input AND gates + 1 six-input OR = 7 gates
POS implementation:
F = (A+B+C)(A'+B'+C')
2 three-input OR gates + 1 two-input AND = 3 gates
POS uses less than half the hardware here.
RULE OF THUMB:
· few 1s in the truth table → use SOP
· few 0s in the truth table → use POS
Sanity check A=0,B=1,C=0:
(0+1+0)(1+0+1) = 1·1 = 1 ✔ (row 2 should be 1)
Check A=0,B=0,C=0:
(0+0+0)(1+1+1) = 0·1 = 0 ✔ (row 0 should be 0)
Worked numerical 3 — converting SOP to POS via complement
Given F = AB + A'C, find the POS form.
METHOD: find F', expand it, then complement back.
Step 1 — build the truth table:
A B C | AB | A'C | F
────────┼────┼─────┼───
0 0 0 | 0 | 0 | 0 ← M₀
0 0 1 | 0 | 1 | 1
0 1 0 | 0 | 0 | 0 ← M₂
0 1 1 | 0 | 1 | 1
1 0 0 | 0 | 0 | 0 ← M₄
1 0 1 | 0 | 0 | 0 ← M₅
1 1 0 | 1 | 0 | 1
1 1 1 | 1 | 0 | 1
F = Σm(1,3,6,7) → F = ΠM(0,2,4,5)
POS (canonical):
F = (A+B+C)(A+B'+C)(A'+B+C)(A'+B+C')
Simplify:
(A+B+C)(A+B'+C) = A+C [Y·Y' pair eliminated]
(A'+B+C)(A'+B+C') = A'+B
F = (A+C)(A'+B)
Verify by expanding:
(A+C)(A'+B) = AA' + AB + A'C + BC
= 0 + AB + A'C + BC
= AB + A'C + BC
By consensus, BC is redundant:
= AB + A'C ✔ matches the original
💡 Exam angle: the marks hinge on getting the maxterm rule right — 0 → plain, 1 → barred, opposite to minterms. Show the truth table with both m and M columns, give the Σm and ΠM notations, and state which form is cheaper for that particular function. The "count the 1s versus the 0s" rule of thumb is worth writing explicitly; it shows judgement rather than mechanical work.
Syllabus points
Maxterms; canonical & minimal POS
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