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CHAPTER 2 BOOLEAN ALGEBRA AND EXPRESSIONS
1. BOOLEAN ALGEBRA
1.1 Boolean operations
Variable, complement, and litteral are terms used in Boolean algebra. A variable is a symbol used to represent a logical quantity. A single variable can have a 1 or 0 value. The complement is the inverse of a variable and it is indicated by a bar over the variable (overbar). For example the complement of the variable A is 𝐴̅. If A = 1 then 𝐴̅ = 0, if A = 0 then 𝐴̅ = 1. A litteral is a variable or the complement of a variable.
1.1.1 Boolean addition Boolean addition is equivalent to the OR operation. In Boolean Algebra, a sum term is a sum of litterals. Some examples of sum terms are 𝐴 + 𝐵, 𝐴 + 𝐵̅, 𝐴 + 𝐵 + 𝐶̅, 𝐴̅ + 𝐵 + 𝐶 + 𝐷̅ A sum term is equal to 1 if one ore more of the litterals in the term are 1. A sum term is equal to 0 only when each of the litteral is 0. Example : determine the values of A, B, C, and D that make the sum term 𝐴̅ + 𝐵 + 𝐶 + 𝐷̅ = 0 Solution : 𝐴 = , 𝐵 = , 𝐶 = , 𝐷 =
1.1.2 Boolean multiplication Boolean multiplication is equivalent to the AND operation. In Boolean Algebra, a product term is a product of litterals. Some examples of product terms are 𝐴𝐵, 𝐴𝐵̅, 𝐴𝐵𝐶̅, 𝐴̅𝐵𝐶𝐷̅ A product term is equal to 1 only if each of the litteral in the term is 1. A product term is equal to 0 if one or more litterals in the product term are 0. Example : determine the values of A, B, C, and D that make the product term 𝐴𝐵̅𝐶𝐷̅ = 1 Solution : 𝐴 = , 𝐵 = , 𝐶 = , 𝐷 =
1.2 Laws and Rules of Boolean Algebra
1.2.1 Laws of Boolean Algebra
Laws Statements Commutative 𝐴 + 𝐵 = The order in which the variables are Ored makes no difference 𝐴𝐵 = 𝐵𝐴 The order in which the variables are ANDed makes no difference Associative 𝐴 + (𝐵 + 𝐶)^ = When Oring more than two variables, the result is the same regardless of the grouping of the variables
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𝐴(𝐵𝐶) = When ANDing more than two variables, the result is the same regardless of the grouping of the variables Distributive (^) 𝐴(𝐵 + 𝐶) (^) = (∙) is distributive over (+) 𝐴 + 𝐵𝐶 = (+) is distributive over (∙)
1.2.2 Rules of Boolean Algebra
A list of 12 basic rules that are usefull in manipulating and simplifying Boolean expressions.
1 𝐴 + 0 = 2 𝐴 ∙ 0 = 0 (^3) 𝐴 + 1 = 4 𝐴 ∙ 1 = 𝐴 5 𝐴 + 𝐴 = 6 𝐴 ∙ 𝐴 = 𝐴 7 𝐴 + 𝐴̅ = 8 𝐴 ∙ 𝐴̅ = 0 (^9) 𝐴̅̅ = 10 𝐴 + 𝐴̅ 𝐵 = 11 𝐴 + 𝐴𝐵 = 12 (𝐴 + 𝐵)(𝐴 + 𝐶) =
1.2.3 DeMorgan’s Theorems (^1) 𝐴̅̅ ̅+̅̅̅ ̅𝐵̅ = The complement of a sum term is equal to the product of the complements of each litteral in the sum term. (^2) 𝐴̅̅ ̅∙̅ ̅𝐵̅ = The complement of a product term is equal to the sum of the complements of each litteral in the product term.
Exercise 1 : Using Boolean Algebra techniques, simplify the following expressions.
a) 𝐴𝐵 + 𝐴(𝐵 + 𝐶) + 𝐵(𝐵 + 𝐶) b) 𝐴𝐵̅ + 𝐴(𝐵 + 𝐶̅̅̅̅̅̅̅̅) + 𝐵(𝐵 + 𝐶̅̅̅̅̅̅̅̅) c) [𝐴𝐵̅(𝐶 + 𝐵𝐷) + 𝐴̅𝐵̅]𝐶 d) 𝐴̅𝐵𝐶 + 𝐴𝐵̅𝐶̅ + 𝐴̅𝐵̅𝐶̅ + 𝐴𝐵̅𝐶 + 𝐴𝐵𝐶
Exercise 2 : Apply DeMorgan’s theorems to the following expressions.
a) 𝑋𝑌𝑍̅̅̅̅̅̅, 𝑋 + 𝑌 + 𝑍̅̅̅̅̅̅̅̅̅̅̅̅̅ b) 𝑊𝑋𝑌𝑍̅̅̅̅̅̅̅̅̅, 𝑊 + 𝑋 + 𝑌 + 𝑍̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ c) 𝐴 + 𝐵𝐶̅̅̅̅̅̅̅̅̅̅̅ + 𝐷(𝐸 + 𝐹̅)̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
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Exercise : 1) Identify each of the following expressions as SOP, standard SOP, POS, or standard POS. a) 𝐴𝐵 + 𝐴̅𝐵𝐷 + 𝐴̅𝐶𝐷̅ b) (𝐴 + 𝐵̅ + 𝐶)(𝐴 + 𝐵 + 𝐶̅) c) 𝐴̅𝐵𝐶 + 𝐴𝐵𝐶̅ d) 𝐴(𝐴 + 𝐶̅)(𝐴 + 𝐵) 2) Convert each SOP expression in question 1) to standard form. 3) Convert each POS expression in question 1) to standard form.
2.2 Canonical Forms of Boolean Expressions
Any Boolean function can be expressed as a minterm canonical formula , or a maxterm canonical formula.
A minterm is a standard product term including all the input variables and producing a logic ‘1’ for the corresponding input values.
A maxterm is a standard sum term including all the input variables and producing a logic ‘0’ for the corresponding input values.
Example : Minterms and Maxterms for a two-variable and three-variable functions
a) For a two-variable function 𝑓(𝑥, 𝑦)
𝒙 𝒚 Minterms Designation Maxterms Designation 0 0 𝑚 0 𝑀 0 (^0 1) 𝑚 1 𝑀 1 1 0 𝑚 2 𝑀 2 1 1 𝑚 3 𝑀 3
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b) For a three-variable function 𝑓(𝑥, 𝑦, 𝑧)
𝒙 𝒚 𝒛 Minterms^ Designation^ Maxterms^ Designation 0 0 0 𝑚 0 𝑀 0 0 0 1 𝑚 1 𝑀 1 (^0 1 0) 𝑚 2 𝑀 2 0 1 1 𝑚 3 𝑀 3 1 0 0 𝑚 4 𝑀 4 (^1 0 1) 𝑚 5 𝑀 5 1 1 0 𝑚 6 𝑀 6 1 1 1 𝑚 7 𝑀 7
Example 1 : Let’s consider the function F defined by the following truth table.
𝒙 𝒚 𝒛 F (^0 0 0) 0 0 0 1 1 0 1 0 0 (^0 1 1) 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1
This function can be expressed as a minterm canonical formula (standard SOP) or as a maxterm canonical formula (standard POS).
As a minterm caninical formula , add all the minterms resulting from the function values ‘1’
→ 𝐹(𝑥, 𝑦, 𝑧) = → algebraic form
= → m -notation
As a maxterms canonical formula , multiply all the maxterms resulting from the function values ‘0’
→ 𝐹(𝑥, 𝑦, 𝑧) = → algebraic form = = → m -notation
Example 2 : Express the Boolean function 𝐹(𝐴, 𝐵, 𝐶) = 𝐴 + 𝐵̅ 𝐶 as a minterms canonical formula and construct its truth table.
The Boolean function 𝐹(𝐴, 𝐵, 𝐶) = 𝐴 + 𝐵̅ 𝐶 is a nonstandard SOP.
i) Expanding the function as a minterms canonical formula using a numerical method
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2.3 Karnaugh-Map SOP Minimization
The Karnaugh-Map provides a systematic method for simplifying Boolean expressions and if properly used, will produce the simplest possible SOP or POS expressions known as the minimum expressions.
The number cells in a Karnaugh-Map is equal to the total number of input variables combinations, as is the number of rows in a truth table.
a) Format of a three-variable function → number of cells is 𝟐𝟑^ = 𝟖
b) Format of a four-variable function → number of cells is 𝟐𝟒^ = 𝟏𝟔
c) Cell adjacency The cells in a Karnaugh-Map are arranged so that there’s only one variable change between adjacent cells. Adjacency is defined by a single variable change.
2.3.1 Mapping a SOP expression on a Karnaugh-Map
Step 1: Determine the binary values for each product term in the SOP expression. These are the binary values that make each product term equal to 1. Step 2: As each product term is evaluated, place a 1 on the Karnaugh-Map in the corresponding cell.
Example 1 : Map the following Boolean expression on a Karnaugh-Map.
𝑓(𝐴, 𝐵, 𝐶, 𝐷) = 𝐴̅𝐵̅𝐶𝐷 + 𝐴̅𝐵𝐶̅𝐷̅ + 𝐴𝐵𝐶̅𝐷 + 𝐴𝐵𝐶𝐷 + 𝐴𝐵𝐶̅𝐷̅ + 𝐴̅𝐵̅𝐶̅𝐷 + 𝐴𝐵̅𝐶𝐷̅
BC
A
AB^00 01 11
CD
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Solution : This is a standard SOP expression, it can then be expressed as a canonical minterm formula by just finding the binary values that make each product term equal to 1.
𝑓(𝐴, 𝐵, 𝐶, 𝐷) = 𝐴̅𝐵̅𝐶𝐷 + 𝐴̅𝐵𝐶̅𝐷̅ + 𝐴𝐵𝐶̅𝐷 + 𝐴𝐵𝐶𝐷 + 𝐴𝐵𝐶̅𝐷̅ + 𝐴̅𝐵̅𝐶̅𝐷 + 𝐴𝐵̅𝐶𝐷̅
Example 2 : Map the following Boolean function on a Karnaugh-Map.
𝑓(𝐴, 𝐵, 𝐶, 𝐷) = 𝐵̅𝐶̅ + 𝐴𝐵̅ + 𝐴𝐵𝐶̅ + 𝐴𝐵̅𝐶𝐷̅ + 𝐴̅𝐵̅𝐶̅𝐷 + 𝐴𝐵̅𝐶𝐷
Solution : This a nonstandard SOP expression, it should then be expanded to obtain a minterms canonical formula. Using the numerical method leads to
𝑓(𝐴, 𝐵, 𝐶, 𝐷) = 𝐵̅𝐶̅ + 𝐴𝐵̅ + 𝐴𝐵𝐶̅ + 𝐴𝐵̅𝐶𝐷̅ + 𝐴̅𝐵̅𝐶̅𝐷 + 𝐴𝐵̅𝐶𝐷
Example 3 : Map the following Boolean function on a Karnaugh-Map.
𝑓(𝐴, 𝐵, 𝐶) = 𝐴̅ + 𝐴𝐵̅ + 𝐴𝐵𝐶̅
AB^00 01 11
CD
AB^00 01 11
CD
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Exercise : Use the Karnaugh-Map method to simplify the following Boolean expressions.
a) 𝐴 𝐵̅ 𝐶 + 𝐴̅ 𝐵 𝐶 + 𝐴̅ 𝐵̅ 𝐶 + 𝐴̅ 𝐵̅ 𝐶̅ + 𝐴 𝐵̅ 𝐶̅
b) 𝑋 𝑌̅ 𝑍 + 𝑋 𝑌 𝑍̅ + 𝑋̅ 𝑌 𝑍̅ + 𝑋 𝑌̅ 𝑍̅ + 𝑋 𝑌 𝑍
c) 𝐵̅ 𝐶̅ 𝐷̅ + 𝐴̅ 𝐵 𝐶̅ 𝐷̅ + 𝐴 𝐵 𝐶̅ 𝐷̅ + 𝐴̅ 𝐵̅ 𝐶 𝐷 + 𝐴 𝐵̅ 𝐶 𝐷 + 𝐴̅ 𝐵̅ 𝐶 𝐷̅ + 𝐴̅ 𝐵 𝐶 𝐷̅ + 𝐴 𝐵 𝐶 𝐷̅ + 𝐴 𝐵̅ 𝐶 𝐷̅
d) 𝑊̅ 𝑋̅ 𝑌̅ 𝑍̅ + 𝑊 𝑋̅ 𝑌 𝑍 + 𝑊 𝑋̅ 𝑌̅ 𝑍 + 𝑊̅ 𝑌 𝑍 + 𝑊 𝑋̅ 𝑌̅ 𝑍̅
A
BC
(c)
AB
CD
AB
CD
AB
CD
(d) (e)
(f)
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2.3.4 « Don’t Care » Combinaitions
“Don’t Care” Combinations are input combinations that are guaranteed never to occur during the normal operation in certain digital systems. The ‘don’t care’ combinations are so called with respect to their effect on the output function. That is, for these input combinations either a 1 or a 0 may be assigned to the output; it really does not matter since they will never occur. The ‘don’t care’ combinations are represented by X or dc. When an incompletely specified function, i.e., a function with ‘don’t care combinations is simplified to obtain minimum SOP expression, the value 1 can be assigned to the selected ‘don’t care’ combinations to make larger grouping. Similarly, selected ‘don’t care’ combinations may be assumed as 0s to form groups of 0s when finding the minimum POS expression.
Example : Obtain the minimum SOP for each of the following incompletely specified functions: a) 𝑓(𝐴, 𝐵, 𝐶, 𝐷) = ∑^ 𝑚(1,3,7,11,15) + 𝑑𝑐(0,2,5) b) 𝑓(𝐴, 𝐵, 𝐶, 𝐷) = ∑ (^) 𝑚(0,1,2,5,8,15) + 𝑑𝑐(6,7,10) c) 𝑓(𝑤, 𝑥, 𝑦, 𝑧) = ∑ (^) 𝑚(2,8,9,10,12,13) + 𝑑𝑐(7,11) d) 𝑓(𝑤, 𝑥, 𝑦, 𝑧) = ∑^ 𝑚(1,7,9,10,12,13,14,15) + 𝑑𝑐(4,5,8)
Solutions :
a) 𝐴̅ 𝐵̅ + 𝐶 𝐷 or 𝐴̅𝐷 + 𝐶𝐷 b) 𝐵̅ 𝐷̅ + 𝐴̅ 𝐶̅ 𝐷 + 𝐵 𝐶 𝐷 c) 𝑤 𝑦̅ + 𝑥̅ 𝑦 𝑧̅ d) 𝑦̅ 𝑧 + 𝑥 𝑧 + 𝑤 𝑧̅
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Solution: This expression is not a standard POS, the first term must be expanded (algebraically or numerically). The binary values making each term equal to 0 are found as:
(𝐵 + 𝐶 + 𝐷)(𝐴 + 𝐵 + 𝐶̅ + 𝐷)(𝐴̅ + 𝐵 + 𝐶 + 𝐷̅)(𝐴 + 𝐵̅ + 𝐶 + 𝐷)(𝐴̅ + 𝐵̅ + 𝐶 + 𝐷) 0 000 0010 1001 0100 1100 1 000
The equivalent SOP expression can be obtain by filling the empty cells with 1s and grouping as shown below:
EXERCISE :
Use Karnaugh Maps to minimize the following POS expressions; also obtain the equivalent simplest SOP.
a) 𝑓(𝑤, 𝑥, 𝑦, 𝑧) = (𝑤 + 𝑥̅ + 𝑧̅)(𝑤 + 𝑥̅ + 𝑦̅)(𝑥̅ + 𝑦̅ + 𝑧̅)(𝑤̅ + 𝑥̅ + 𝑧)(𝑤 + 𝑥 + 𝑦̅ + 𝑧̅) b) 𝑓(𝑤, 𝑥, 𝑦, 𝑧) = (𝑤 + 𝑦 + 𝑧̅)(𝑥̅ + 𝑦 + 𝑧̅)(𝑤̅ + 𝑥̅ + 𝑦)(𝑤 + 𝑥 + 𝑦 + 𝑧)(𝑤 + 𝑥̅ + 𝑦̅ + 𝑧̅)(𝑤̅ + 𝑥̅ + 𝑦̅ + 𝑧) c) 𝑓(𝑤, 𝑥, 𝑦, 𝑧) = (𝑤 + 𝑥̅)(𝑤 + 𝑦 + 𝑧)(𝑤̅ + 𝑥̅ + 𝑧̅)(𝑤 + 𝑦̅ + 𝑧) d) 𝑓(𝑤, 𝑥, 𝑦, 𝑧) = (𝑤 + 𝑦̅ + 𝑧̅)(𝑤̅ + 𝑥 + 𝑦̅)(𝑤 + 𝑥 + 𝑦̅ + 𝑧)(𝑤̅ + 𝑥̅ + 𝑦 + 𝑧)
AB
CD
For group 1(red), the resulting sum term is 𝐶 + 𝐷 For group 2(green), the resulting sum term is 𝐴̅ + 𝐵 + 𝐶 For group 3(blue), the resulting sum term is 𝐴 + 𝐵 + 𝐷 → the minimum POS expression: (𝐶 + 𝐷)(𝐴̅ + 𝐵 + 𝐶)(𝐴 + 𝐵 + 𝐷)
AB
CD
For group 1(red), the resulting product term is 𝐴̅ 𝐷 For group 2(green), the resulting product term is 𝐵𝐷 For group 3(blue), the resulting product term is 𝐵𝐶 For group 4(yellow), the resulting product term is 𝐴𝐶 → the minimum SOP expression: 𝐴̅ 𝐷 + 𝐵𝐷 + 𝐵𝐶 + 𝐴𝐶
