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1 0 0 0 A.B'.C'.D' A'+B+C+D 8
1 0 0 1 A.B'.C'.D A'+B+C+D' 9
1 0 1 0 A.B'.C.D' A'+B+C'+D 10
1 0 1 1 A.B'.C.D A'+B+C'+D' 11
1 1 0 0 A.B.C'.D' A'+B'+C+D 12
1 1 0 1 A.B.C'.D A'+B'+C+D' 13
1 1 1 0 A.B.C.D' A'+B'+C'+D 14
1 1 1 1 A.B.C.D A'+B'+C'+D' 15
Any Boolean expression can either be the sum of its minterms or Sum of Products, abbreviated as SOP or the product
of its maxterms or Product of Sums in short POS.
1.10.1 Canonical SOP and POS Forms
An SOP can be in Canonical SOP form in which each product term contains all literals. So, these product terms are
actually the minterms. Hence, the canonical SOP form is also called as the sum of minterms form. It is represented by
the symbol Σ (sigma).
Similarly, a POS can be in Canonical POS form in which each sum term contains all variables present in the expression.
Thus they are actually the maxterms. Hence, canonical POS form is also called as the product of maxterms form.
Symbol π (pi) is used to represent canonical POS.
In ordinary SOP and POS forms, each term may not be a minterm or a maxterm. But a canonical SOP or POS expression
is represented using minterm or maxterm only.
Steps to convert SOP into canonical SOP or a POS into canonical POS
The following steps can be taken to convert an SOP into its canonical SOP form or a POS into its canonical POS form:
1. Identify the missing literal in each term.
2. To add for missing literal, multiply each term by 1 in case of SOP expression; and add 0 to each term in case of POS
expression.
3. If the missing literal is say X, then replace 1 with X+X' or 0 with X.X'.
4. Apply distributive law to get the maxterms or minterms.
Alternatively, the following steps can be taken:
1. Draw the truth table of the Boolean expression.
2. In the case of SOP, multiply the minterms resulting in 1.
3. In the case of POS, add the maxterms resulting in 0.
Let us some examples for better understanding.
Example 1: Given F(X, Y, Z) = (X'+Y').(Y+Z').
Write the function in canonical POS form. [ISC 2012]
Solution:
(X'+Y').(Y+Z')
st
nd
= (X'+Y'+0).(Y+Z'+0) [In 1 term, Z is missing and in the 2 term, X is missing]
st
= (X'+Y'+Z.Z').(Y+Z'+X.X') [Replacing 0 with Z.Z' and X.X' in 1 and 2 terms respectively]
nd
= (X'+Y'+Z).(X'+Y'+Z').(X+Y+Z').(X'+Y+Z') [Distributive law]
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