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Presented by: Ms. Maria Estrellita D. Hechanova, ECE.

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1 Presented by: Ms. Maria Estrellita D. Hechanova, ECE

2 Define Boolean algebra Identify axioms, theorems, corollaries, and laws pertaining to the manipulation of Boolean expressions Compute and manipulate or simplify given Boolean expressions Properly use Karnaugh Map (K-map) in simplification of Boolean expressions Relate this lesson to existing applications in computer Simplify digital circuits

3 Introduction to Boolean Algebra Basic Definitions and Axioms in Boolean Algebra Basic Theorems Product-of-sums and Sum-of-products Minimal Boolean Expressions and Prime Implicants Applications and other means of simplification: Logic gate and circuits Truth tables and Boolean functions Karnaugh map (K-map)

4 How can we communicate with our computers or laptops? How is it possible that my SMS from my mobile phone be sent hundreds of miles from my location? How does televisions be able to project images on a screen? Why does robots be able to do specific (and even complicated) tasks?

5 A statement is true if it agrees with reality, false if it doesnt. Two-state logic assumes that each statement is either true or false. The Greeks, especially Aristotle, worked out the theory of two-state logic in great detail. In 1854, George Boole came up with symbolic logic, better known as the Boolean Algebra.

6 Boolean algebra uses letters and symbols to represent statements and their logical connections. Each variable in Boolean algebra has either of two values: true or false. (this is why it is called a two-state or binary algebra) Boolean algebra was a far-out subject until 1938, when Claude Shannon used it to analyze and design telephone switching circuits. He let the variables represents closed and open relays.

7 Boolean algebra has become one of the major design tools of digital and computer electronics. LOGIC CIRCUITS!!!


9 The counting number we know of is called DECIMAL... Other number systems... BINARY OCTAL HEXADECIMAL

10 Let B be a nonempty set with two binary operations + (or) and * (and), a unary operation (not), and two distinct elements 0 and 1. Then B is called a Boolean algebra if the following axioms hold where a, b, c are any elements in B.

11 At least one (1) or more inputs of either logic 1 (true) or logic 0 (false) and a single desired output (either a 1 or a 0, depending on the inputs) Examples: 1. F = a+b 2. F = a*b 3. F = (a+b)*c 4. F = abc+(bd)+ab+acd Note that inputs a, b, c, and d should have a value either a logic 1 or logic 0 and the output F should acquire a value either 1 and 0.

12 a+b = b+a a*b = b*a a+(b*c) = (a+b)*(a+c) a*(b+c) = (a*b)+(a*c) a+0 = a a*1 = a a+a = 1 a*a = 0 Commutative Law Distributive Law Identity Law Complement Law

13 Let a be the values on the column side and b be the values on the row side. + Column Row

14 Let a be the values on the column side and b be the values on the row side. *

15 Simply change the input 0 to an output of 1 and the input 1 to an output 0!! 01 10

16 + * or and not

17 Subalgebras: Suppose C is a nonempty subset of a Boolean algebra B. We say C is a subalgebra of B if C itself is a Boolean algebra (with respect to the operations of B). We note that C is a subalgebra of B iff C is closed under the three operations of B, i.e., +,*, and.

18 Isomorphic Boolean Algebras: Two Boolean algebras B and B are said to be isomorphic if there is a one-to-one correspondence f:B B which preserves the three operations. f(a+b) = f(a) + f(b) f(a*b) = f(a) * f(b) f(a) = f(a) for any elements, a, b, in B

19 The dual of any statement in a Boolean algebra B is the statement obtained in interchanging the operations + and *, and interchanging their identity elements 0 and 1 in the original statement. Example: The dual of (1+a) * (b+0) = b is... Theorem 1. Principle of Duality (0*a) + (b*1) = b

20 Theorem 2: Idempotent Laws:Idempotent Laws a+a = aa*a = a Theorem 3: Boundedness Laws:Boundedness Laws a+1 = 1a*0 = 0 Theorem 4: Absorption Laws:Absorption Laws a+(a*b) = aa*(a+b) = a Theorem 5: Associative Laws:Associative Laws (a+b)+c = a+(b+c) (a*b)*c = a*(b*c)

21 Let a be any element of a Boolean algebra B. Theorem 6: Uniqueness of compliment:Uniqueness of compliment If a+x = 1 and a*x = 0, then x = a Theorem 7: Involution Law:Involution Law (a) = a Theorem 8: *Inversion Law: 0 = 11 = 0 Theorem 9: DeMorgans Laws:DeMorgans Laws (a+b) = a * b(a*b) = a + b


23 By the basic theorems and the commutative axiom, every Boolean algebra B satisfies the associative, commutative, and absorption laws and hence is a lattice where + and * are the join and meet operations, respectively. With respect to this, a+1 = 1 implies a 1 and a*0 = 0 implies 0 a, for any element a Є B. Thus B is a bounded lattice. Distributive and complement axioms show that B is also distributive and complemented.

24 So, we could say that, every bounded, distributive, and complemented lattice L satisfies the four (4) axioms. Alternative Definition: A Boolean algebra B is a bounded, distributive, and complemented lattice.

25 Since Boolean algebra is a lattice, it has a natural partial ordering. In the discussion of ordered sets and lattices, it was defined that ab if the equivalent conditions a+b = b and a*b = a are satisfied.

26 The following are equivalent in Boolean algebra: a+b = b a*b = a a+b = 1 a*b = 0 Therefore, in a Boolean algebra, we can define ab if all of the above conditions is known to be true.

27 Let B be a finite Boolean algebra. From the discussion of bounded lattices, an element a in B is an atom if a immediately succeeds 0, that is if 0<

28 Say, x = a 1 + a a r is such a representation. Consider the function f: B P(A) defined by f(x) = {a 1, a 2,..., a r } The mapping is well defined since the representation is unique.

29 Theorem 10: The above mapping f: B P(A) is an isomorphism. Thus we see the intimate relationship between the set theory and abstract Boolean algebra in the sense that every finite Boolean algebra is structurally the same as a Boolean algebra of sets. If a set A has n elements, then its power set P(A) has 2 n elements. Corollary 1: A finite Boolean algebra has 2 n elements for some positive integer n.

30 E will represent some area in the figure and hence will uniquely equal the union of one or more of the eight sets. Lets interpret union as a sum and an intersection as a product. Therefore, the eight sets are products, and the unique representation of E will be the sum (union) of products.

31 Boolean expression E is any variable or any expression built up from the variables using the Boolean operations. Naturally, the expression E must be well- formed, that is, where + and * are used as binary operations, and is used as a unary operation. Literal is a variable or complemented variable, such as x, x, y, y, and so on. Fundamental product is a literal or a product of two or more literals in which no two literals involve the same variable.

32 Note that, any product of literals can be reduced to either 0 or a fundamental product. A fundamental product P1 is said to be contained in (or included in) another fundamental product P2 if the literals of P1 are also literals of P2.

33 Definition: A Boolean expression E is called a sum-of-products expression if E is a fundamental product or the sum of two or more fundamental products none of which is contained in another. Definition: Let E be any Boolean expression. A sum-of-products form of E is an equivalent Boolean sum-of-products expression.

34 Consider the expressions E 1 = xz + yz + xyz and E 2 = xz + xyz + xyz!! E 1 = xz + yz + xyz E 1 = xz + xyz + yz by absorption law, a + ab = a, E 1 = xz +yz

35 Algorithm 1A: The input is a Boolean expression E. The output is a sum-of-products expression equivalent to E. 1. Use DeMorgans law and involution to move the complement operation into any parenthesis until finally the complement operation only applies to variables. Then E will consists only of sum of products of literals. 2. Use the distributive operation to next transform E into a sum of products.

36 3. Use the commutative, idempotent, and complement laws to transform each product in E into 0 or a fundamental product. 4. Use the absorption and identity laws to finally transform E into a sum-of-products expression.

37 E = ((xy)z)((x+z)(y+z)) Answer: E = xyz +xz

38 A Boolean expression E = E(x1, x2,..., xn) is said to be a complete sum-of-products expression if E is a sum-of- products expression where each product P involves all the n variables. Such a fundamental product P which involves all the variables is called a minterm, and there is a maximum of 2 n such products of n variables. Theorem 11: Every nonzero Boolean expression E=E(x1, x2,..., xn) is equivalent to a complete sum-of-products expression and such a representation is unique. This is called the complete sum-of-product form of E.

39 Algorithm 1B: The input is a Boolean sum-of-products expression E=E(x1, x2,..., xn). The output is a complete sum-of-products expression equivalent to E. 1. Find a product P in E which does not involve the variable xi, and then multiply P by xi+xi, deleting any repeated products. 2. Repeat step 1 until every product P in E is a minterm, i.e., every product P involves all the variables.

40 E(x,y,z) = x(yz) Answer: E = xyz + xyz + xyz

41 Sum-of-product form = Disjunctive normal form (DNF) Complete sum-of-product form = Full disjunctive normal form = Disjunctive canonical form = Minterm canonical form

42 Minimal Sum-of-Products Prime Implicants Consensus of Fundamental Products Consensus Method of Finding Prime Implicants Finding a Minimal Sum-of-Products Form

43 Consider a Boolean sum-of-products expression E. Let E L denote the number of literals in E (counted according to multiplicity), and let E S denote the number of summands in E. Example: E = xyz + xyt + xyzt +xyzt E L = = 14 E S = 4

44 Assuming that E and F are equivalent Boolean sum-of- products expression, E is simpler than F if: E L < F L and E S F L E L F L and E S < F L E is minimal if there is no equivalent sum-of-products expression which is simpler than E. Note that there can be more than one equivalent minimal sum-of- products expressions.

45 A fundamental product P is called prime implicant of a Boolean expression E if, P + E = E but no other fundamental product contained in P has this property. Example: E = xy + xyz + xyz Therefore, xz is a prime implicant of E.

46 Theorem 12: A minimal sum-of-products form of a Boolean expression E is a sum of prime implicants of E. (The following topics discuss on the methods of finding the prime implicants of E based on the notion of the consensus of fundamental products. This method can be used to find the minimal sum-of-products form for E. Geometric methods for finding prime implicants will also be discussed later...)

47 Let P 1 and P 2 be fundamental products such that exactly one variable, x k, appears uncomplemented in one of the fundamental products and complemented in the other. Then the consensus of P 1 and P 2 is the product (without repetitions) of the literals of P 1 and the literals of P 2 after x k and x k are deleted. Note that, it is not proper to define the consensus of P 1 = x and P 2 = x. Lemma 1: Suppose Q is the consensus of P 1 and P 2. Then P 1 + P 2 + Q = P 1 + P 2.

48 Example: Find the consensus Q of P 1 and P 2. P 1 = xyzs and P 2 = xyt Q = xzs * xt Q = xzst P 1 = xy and P 2 = y Q = x * 1 Q = x

49 Algorithm 2A (Consensus Method): The input is a Boolean expression E = P 1 + P P m where the P s are fundamental products. The output expresses E as a sum of its prime implicants (Theorem 13). 1. Delete any fundamental product Pi which includes any other fundamental product Pj. (Permissible by the absorption law) 2. Add the consensus of any Pi and Pj providing Q does not include any of the Ps. (Permissible by Lemma 1) 3. Repeat 1 and/or 2 until neither can be applied.

50 Theorem 13: The consensus method will eventually stop, and then E will be the sum of its prime implicants. Example:E = xyz + xz + xyz + xyz + xyz Answer:E = xz + xy + xy + yz Therefore E is a sum of prime implicants, that is, xz + xy + xy + yz

51 Algorithm 2B: The input is a Boolean expression E = P 1 + P P m where the P s are all the prime implicants of E. The output expresses E as a minimal sum-of- products. 1. Express each prime implicant P as a complete sum-of- products. 2. Delete one by one those prime implicants whose summands appear among the summands of the remaining prime implicants.

52 Example: E = xz + xy + xy + yz Answer: E = xy + xy + yz


54 Logic circuits (also known as logic networks) are structures which are built up from certain elementary circuits called logic gates. Each logic circuit may be viewed as a machine L which contains one or more input devices and exactly one output device. Each input device in L sends a signal, specifically a bit (binary digit), 0 or 1, to the circuit L, and L processes the set of bits to yield an output bit.

55 Logic Circuit L Input/sOutput

56 There are three basic logic gates: OR gate Denoted by a plus sign (addition) The output of an OR gate is logic high (1) if at least one of the inputs is logic high (1), else the output is logic low (0). AND gate Denoted by an asterisk sign (multiplication) The output of an AND gate is logic high (1) if all of the inputs are logic high (1), else the output is logic low (0). NOT gate Denoted by the prime (or a bar on top of the variable) The output is logic high if the input is logic low and the output is logic low if the input if logic high.

57 If at least one of the inputs has a value of logic high (1), the output is logic high (1). Else, the output is logic low (0)

58 If all of the inputs has a value of logic high (1), the output is logic high (1). Else, the output is logic low (0)

59 If the input is logic high (1), the output is logic low (0). And, vice versa.

60 Integration of AND, OR, and NOT gates.

61 Since that the truth tables for the OR, AND, and NOT gates are respectively equal to the truth tables for the propositions disjunction, conjunction, and negation, the logic circuit satisfy the same laws as do propositions and hence form a Boolean algebra. Theorem 14: Logic circuits form a Boolean algebra. Accordingly, all terms used with Boolean algebra may be used with out logic circuits.

62 This logic circuit corresponds to a Boolean sum-of-products. Some of the inputs or their complement are fed into each AND gate. The outputs of all the AND gates are fed into a single OR gate. The output of the OR gate is the output of the circuit.

63 NAND Gate Equivalent to an AND gate followed by a NOT gate. NOR Gate Equivalent to an OR gate followed by a NOT gate

64 2 n possible input combinations, where n is the number of inputs of the circuit. Example: Y = abc + abc + ab n = 3 2 n = 2 3 = 8 (there are 8 possible combinations for the input) Y will be having 8 outputs based on the combination of the inputs

65 Algorhm 3: The input is a Boolean sum-of-products expression Y = Y (A1, A2,..., An). 1. Write down the special sequences for the inputs A1, A2,..., An and their complements. 2. Find each product appearing in Y. 3. Find the sum Y of the products.


67 Let E be a Boolean expression with n variables x 1, x 2,..., x n. The entire discussion above can be applied to E where now the special sequences are assigned to the variables x 1, x 2,..., x n instead of the input devices A 1, A 2,..., A n. The truth table T = T(E) of E is defined in the same way as the truth table T = T (L) for logic circuit L. Example: E = xyz + xyz + xy T ( , , ) = T(E) =

68 Pictorial devices for finding prime implicants and minimal forms for Boolean expressions involving at most six (6) variables. Two fundamental products are said to be adjacent if both have the same variables and if they differ in exactly one literal. Therefore, there must be an uncomplemented variable in one product and complemented in the other. The sum of two such adjacent products will be a fundamental product with one less literal.

69 Case of Two Variables:

70 Case of Three Variables:

71 Case of Four Variables:

72 Examples: Find the prime implicants and a minimal sum- of-products form for each of the following complete sum-of-products Boolean expressions: 1. E 1 = xy + xy 2. E 2 = xy + xy + xy 3. E 3 = xy + xy 4. E 4 = xyz + xyz + xyz + xyz 5. E 5 = xyz + xyz + xyz + xyz 6. E 6 = xyz + xyz + xyz + xyz + xyz 7. E 7 = xy + xyz + xyz + xyzt

73 Problem: Design a three input minimal AND-OR circuit L with the following table: T = [A, B, C; L] = [ , , ; ]

74 But wait!! We have included problems for you to practice on with... Thank you for the time... Hope you learned something from our presentation!!

75 1. Schaums Outlines: Discrete Mathematics 2 nd Edition, by Lipschutz and Lipson, Digital Computer Electronics, by Malvino, Schaums Outline Series: Theory and Problems of Introduction to Computer Science, by Scheid, 1970.

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