Digital System�Lecture - 05

Canonical Form

• Canonical Form – In Boolean algebra, Boolean function can be expressed as Canonical Disjunctive Normal Form known as minterm and some are expressed as Canonical Conjunctive Normal Form known as maxterm .
• In Minterm, we look for the functions where the output results in “1” while in Maxterm we look for function where the output results in “0”.
• We perform Sum of minterm also known as Sum of products (SOP) .
• We perform Product of Maxterm also known as Product of sum (POS).
• Boolean functions expressed as a sum of minterms or product of maxterms are said to be in canonical form.

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Canonical Form

• Minterm means the term that is true for a minimum number of combination of inputs. That is true for only one combination of inputs.

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• Maxterm means the term or expression that is true for a maximum number of input combinations or that is false for only one combination of inputs.

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Canonical Form

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Product of Sum (POS)

• A canonical product of sum is a boolean expression that entirely consists of maxterms. The Boolean function F is defined on two variables X and Y. The X and Y are the inputs of the boolean function F whose output is true when only one of the inputs is set to true. The truth table for Boolean expression F is as follows:

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Product of Sum (POS)

• In our previous section, we learned about how we can form the minterm from the variable's value. Now, a column will be added for the minterm in the above table. The complement of the variables is taken whose value is 0, and the variables whose value is 1 will remain the same.

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Product of Sum (POS)

• Now, we will add all the minterms for which the output is true to find the desired canonical SOP(Sum of Product) expression.

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• F=X' Y+XY'+XY

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Converting Sum of Products (SOP) to shorthand notation

• The process of converting SOP form to shorthand notation is the same as the process of finding shorthand notation for minterms. There are the following steps to find the shorthand notation of the given SOP expression.

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• Write the given SOP expression.
• Find the shorthand notation of all the minterms.
• Replace the minterms with their shorthand notations in the given expression.

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Converting Sum of Products (SOP) to shorthand notation

• Example: F = X'Y+XY’+XY
• Firstly, we write the SOP expression:
• F = X'Y+XY’+XY
• Now, we find the shorthand notations of the minterms X'Y, XY', and XY.
• X'Y = (01)₂ = m₁
• XY' = (10)₂ = m₂
• XY = (11)₂ = m₃
• In the end, we replace all the minterms with their shorthand notations:
• F=m₁+m₂+m₃

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Converting shorthand notation to SOP expression

• Example: Let us assume that we have a boolean function F, which defined on two variables X and Y. The minterms for the function F are expressed as shorthand notation is as follows:
• F=∑(1,2,3)

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• Now, from this expression, we will find the SOP expression. The Boolean function F has two input variables X and y and the output of F=1 for m₁, m₂, and m₃, i.e., 1st, 2nd, and 3rd combinations. So,
• F=∑(1,2,3)
• F= m₁ + m₂ + m₃
• F= 01 + 10 + 11

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Converting shorthand notation to SOP expression

• Now, we replace zeros with either X' or Y' and ones with either X or Y. Simply, the complement variable is used when the variable value is 1 otherwise the non-complement variable is used.

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• F = ∑(1,2,3)
• F=01+10+11
• F= A'B + AB' + AB

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Sum of product(SOP)

• A canonical sum of products is a boolean expression that entirely consists of minterms. The Boolean function F is defined on two variables X and Y. The X and Y are the inputs of the boolean function F whose output is true when any one of the inputs is set to true. The truth table for Boolean expression F is as follows:

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Product of Sum (POS)

• In our minterm and maxterm section, we learned about how we can form the maxterm from the variable's value. A column will be added for the maxterm in the above table. The complement of the variables is taken whose value is 0, and the variables whose value is 1 will remain the same.

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Product of Sum (POS)

• Now, we will multiply all the minterms for which the output is false to find the desired canonical POS(Product of sum) expression.

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Converting Product of Sum (POS) to shorthand notation

• The process of converting POS form to shorthand notation is the same as the process of finding shorthand notation for maxterms. There are the following steps used to find the shorthand notation of the given POS expression.

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• Write the given POS expression.
• Find the shorthand notation of all the maxterms.
• Replace the minterms with their shorthand notations in the given expression.

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Converting shorthand notation to POS expression

• Example: F = (X'+Y').(X+Y)
• Firstly, we will write the POS expression:
• F = (X'+Y').(X+Y)
• Now, we will find the shorthand notations of the maxterms X'+Y' and X+Y.
• X'+Y' = (00)₂ = M₀
• X+Y = (11)₂ = M₃
• In the end, we will replace all the minterms with their shorthand notations:
• F=M₀.M₃

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Converting shorthand notation to POS expression

• Example: Let us assume that we have a boolean function F, defined on two variables X and Y. The maxterms for the function F are expressed as shorthand notation is as follows:
• F=∏(1,2,3)
• Now, from this expression, we find the POS expression. The Boolean function F has two input variables X and Y and the output of F=0 for M1, M2, and M3, i.e., 1st, 2nd, and 3rd combinations. So,
• F=∏(1,2,3)
• F= M₁.M₂.M₃
• F= 01.10.11

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Converting shorthand notation to POS expression

• Next, we replace zeros with either X or Y and ones with either X' or Y'. Simply, if the value of the variable is 1, then we take the complement of that variable, and if the value of the variable is 0, then we take the variable "as is".

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• F = ∑(1,2,3)
• F=01.10.11
• F=(A+B').( A'+B).( A'+B')

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Conversion of POS to SOP form

• For getting the SOP form from the POS form, we have to change the symbol ∏ to ∑. After that, we write the numeric indexes of missing variables of the given Boolean function.

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• There are the following steps to convert the POS function F = Π x, y, z (2, 3, 5) = x y' z' + x y' z + x y z' into SOP form:

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• In the first step, we change the operational sign to Σ.
• Next, we find the missing indexes of the terms, 000, 110, 001, 100, and 111.
• Finally, we write the product form of the noted terms.

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Conversion of POS to SOP form

Example: F = Π x, y, z (2, 3, 5) = x y' z' + x y' z + x y z' into SOP form:

000 = x' * y' * z'

001 = x' * y' * z

100 = x * y' * z'

110 = x * y* z'

111 = x * y * z

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So the SOP form is:

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F = Σ x, y, z (0, 1, 4, 6, 7) = (x' * y' * z') + (x' * y' * z) + (x * y' * z') + (x * y* z') + (x * y * z)

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Conversion of SOP form to POS form

• For getting the POS form of the given SOP form expression, we will change the symbol ∏ to ∑. After that, we will write the numeric indexes of the variables which are missing in the boolean function.

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• There are the following steps used to convert the SOP function F = ∑ x, y, z (0, 2, 3, 5, 7) = x' y' z' + z y' z' + x y' z + xyz' + xyz into POS:

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• In the first step, we change the operational sign to ∏.
• We find the missing indexes of the terms, 001, 110, and 100.
• We write the sum form of the noted terms.

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Conversion of SOP form to POS form

• F = ∑ x, y, z (0, 2, 3, 5, 7) = x' y' z' + z y' z' + x y' z + xyz' + xyz into POS:

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001 = (x + y + z)

100 = (x + y' + z')

110 = (x + y' + z')

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• So, the POS form is:

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F = Π x, y, z (1, 4, 6) = (x + y + z) * (x + y' + z') * (x + y' + z')

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