Canonical Forms

A Canonical Form is a unique way of representing a Boolean function where each term includes all variables of the function, either in complemented or uncomplemented form.

1.1 Canonical Sum of Products (SOP)

Expressed as a sum (OR) of minterms.
Minterm = product (AND) of all variables in true or complemented form.
Each minterm corresponds to one row of the truth table where the function output = 1.

📊 Example:
For F(A, B, C), where F = 1 at inputs (0,1,1), (1,0,1), (1,1,0), (1,1,1):

A	B	C	F	Minterm
0	1	1	1	A'BC
1	0	1	1	AB'C
1	1	0	1	ABC’
1	1	1	1	ABC

So,

F(A,B,C)=A′BC+AB′C+ABC′+ABC=Σm(3,5,6,7)F(A,B,C) = A'BC + AB'C + ABC' + ABC = Σm(3, 5, 6, 7)F(A,B,C)=A′BC+AB′C+ABC′+ABC=Σm(3,5,6,7)

📷 Diagram (SOP logic circuit):

Inputs A, B, C → AND gates (each for minterms) → OR gate (sums them).

1.2 Canonical Product of Sums (POS)

Expressed as a product (AND) of maxterms.
Maxterm = sum (OR) of all variables in true or complemented form.
Each maxterm corresponds to one row of the truth table where the function output = 0.

📊 Example:
Using the same F(A, B, C), zeros occur at (0,0,0), (0,0,1), (0,1,0), (1,0,0).

A B C F Maxterm

0 0 0 0 (A + B + C)

0 0 1 0 (A + B + C')

0 1 0 0 (A + B' + C)

1 0 0 0 (A' + B + C)

So,

F(A,B,C)=(A+B+C)(A+B+C’)(A+B’+C)(A’+B+C)=ΠM(0,1,2,4)F(A,B,C) = (A + B + C)(A + B + C’)(A + B’ + C)(A’ + B + C) = ΠM(0,1,2,4)F(A,B,C)=(A+B+C)(A+B+C’)(A+B’+C)(A’+B+C)=ΠM(0,1,2,4)

📷 Diagram (POS logic circuit):

Inputs A, B, C → OR gates (each for maxterms) → AND gate (multiplies them).

2. Standard Forms

While Canonical forms are unique and complete, they are not minimal.
Standard Forms are simplified versions, still in SOP or POS structure, but with fewer terms and fewer literals per term.

2.1 Standard SOP

Not all variables need to appear in each term.
Simplified using Boolean algebra or K-maps.

🔹 Example:
From Canonical SOP:

F=A′BC+AB′C+ABC′+ABCF = A'BC + AB'C + ABC' + ABCF=A′BC+AB′C+ABC′+ABC

Simplified to:

F=AB+BC+ACF = AB + BC + ACF=AB+BC+AC

2.2 Standard POS

Not all variables appear in each sum term.

🔹 Example:
From Canonical POS:

F=(A+B+C)(A+B+C’)(A+B’+C)(A’+B+C)F = (A + B + C)(A + B + C’)(A + B’ + C)(A’ + B + C)F=(A+B+C)(A+B+C’)(A+B’+C)(A’+B+C)

Simplifies to:

F=(A+B)(B+C)(A+C)F = (A + B)(B + C)(A + C)F=(A+B)(B+C)(A+C)

Examples :

Example 1: Converting from Canonical Sum-of-Products (SOP) to Canonical Product-of-Sums (POS)

Problem: Convert the following Boolean function from its canonical Sum-of-Products (SOP) form to its canonical Product-of-Sums (POS) form: f(A,B,C)=∑m(0,1,2,5,7)

Solution: The given Boolean function, f(A,B,C)=∑m(0,1,2,5,7), indicates that the function outputs a '1' (True) for the minterms m0,m1,m2,m5, and m7.

For a three-variable function (A, B, C), there are 8 possible minterms, ranging from m0 to m7. The minterms that are not present in the canonical SOP form are those where the function's output is '0' (False). In this case, the missing minterms are m3,m4, and m6.
When converting from SOP to POS, these missing minterms directly correspond to the maxterms in the canonical POS form. Therefore, the canonical POS form of the given Boolean function is: f(A,B,C)=∏M(3,4,6)=M3⋅M4⋅M6

Example 2: Converting from Canonical Product-of-Sums (POS) to Canonical Sum-of-Products (SOP)

Problem: Convert the following Boolean function from its canonical Product-of-Sums (POS) form to its canonical Sum-of-Products (SOP) form: f(A,B,C)=∏M(0,2,4,7)

Solution: The provided Boolean function, f(A,B,C)=∏M(0,2,4,7), signifies that the function outputs a '0' (False) for the maxterms M0,M2,M4, and M7.

Similar to the previous example, for a three-variable function (A, B, C), there are 8 possible maxterms, from M0 to M7. The max terms that are not included in the canonical POS form are those where the function's output is '1' (True). Here, the missing maxterms are M1,M3,M5, and M6.
When converting from POS to SOP, these missing maxterms correspond directly to the minterms in the canonical SOP form. Consequently, the canonical SOP form of the given Boolean function is: f(A,B,C)=∑m(1,3,5,6)=m1+m3+m5+m6