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Author Name: Arun Paul.

Canonical Form of Boolean Expression With Examples

Here we have listed Canonical Product of Sum or POS Boolean Expression and Canonical Sum of Product or SOP Boolean Expression

Canonical Product of Sum Expression

This is defined as the logical product of all the maxterms derived from the rows of truth table, for which the value of function is 0. It is also known as the maxterm canonical form. The canonical product of sum expression can be given in a compact form by listing the decimal codes corresponding to the maxterms containing a function value of 0.

For example, if the canonical product of sum form of a 3-variable logic function $Y$Y has four maxterms – (A+B+C),(A+B’+C),(A’+B+C) and (A’+B’+C’),
then it can be expressed as the product of decimal codes as given below:

• Y=Π(0,2,4,7)
• =M₀​⋅M₂​⋅M₄​⋅M_7​
• =(A+B+C)(A+B’+C)(A’+B+C)(A’+B’+C’)

The following procedure can be used to obtain the canonical product of the sum form of a logic function:

1. Examine each term in the given logic function. Retain it if it is a maxterm; continue to examine the next term in the same manner.
2. Check for variables that are missing in each sum, which is not a maxterm. Add $(X + \overline{X})$ to the sum term, for each variable $X$ that is missing.
3. Expand the expression using the distributive property and eliminate the redundant terms.

The above procedures can be explained with the following examples.

Canonical POS (Product of Sums) – Examples

Question 1:
Write the canonical POS for:

Y = Π_M(1, 3, 6)

Answer:
Y = (A + B + C′)(A + B′ + C′)(A′ + B′ + C)

Question 2:
Convert into canonical POS form.

Y = A + B′C

Answer:
Y = (A + B′ + C)(A + B + C′)(A + B + C)
Y = Π_M(0, 2, 4)

Question 3:
Write the canonical POS of:

Y = Π_M(0, 5, 7)

Answer:
Y = (A + B + C)(A′ + B + C′)(A′ + B′ + C′)

Canonical Sum of Product Expression

It is defined as the logical sum of all the minterms derived from the rows of a truth table, for which the value of the function is 1. It is also called a minterm canonical form. The canonical sum of product expression can be given in a compact form by listing the decimal codes in correspondence with the minterm containing a function value of 1.

For example, if the canonical sum of product form of a 3-variable logic function $Y$ has three minterms A’B’C’, AB’C,and ABC’, this can be expressed as the sum of the decimal codes corresponding to these minterms as stated below:

• Y=Σm(0,5,6)
• =m₀​+m₅​+m_6​
• =A’B’C’+AB’C+ABC’

where $\Sigma m (0,5,6)$ represents the summation of minterms corresponding to the decimal codes 0, 5 and 6.

Using the following procedure, the canonical sum of product form of a logic function can be obtained:

1. Examine each term in the given logic function. Retain it if it is a minterm; continue to examine the next term in the same manner.
2. Check for variables that are missing in each product which is not a minterm.
   Multiply the product by $(X + \overline{X})$(X+X), for each variable $X$X that is missing.
3. Multiply all the products and omit the redundant terms.

The above procedures can be explained with the following examples.

Canonical SOP (Sum of Products) – Examples

Question 1:
Convert into canonical SOP form.

Y = A′B + C

Answer:
Y = A′B(C + C′) + C(A + A′)(B + B′)
Y = A′BC + A′BC′ + ABC + AB′C
Y = Σm(1, 3, 6, 7)

Question 2:
Write the canonical SOP of:

Y = AB + AC

Answer:
Y = AB(C + C′) + AC(B + B′)
Y = ABC + ABC′ + AB′C
Y = Σm(5, 6, 7)

Question 3:
Express in canonical SOP form.

Y = A′ + BC

Answer:
Y = A′(B + B′)(C + C′) + BC(A + A′)
Y = A′BC + A′BC′ + A′B′C + A′B′C′ + ABC
Y = Σm(0, 1, 2, 3, 7)

Question 4:
Write the canonical SOP for:

Y = Σm(0, 2, 5, 7)

Answer:
Y = A′B′C′ + A′BC′ + AB′C + ABC