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# application of Boolean algebra

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School
McMaster University
Department
Computer Engineering
Course
COMPENG 2DI4
Professor
Xudong Zhu
Semester
Fall

Description
APPLICATIONS OF BOOLEAN ALGEBRA MINTERM AND MAXTERM EXPANSIONS – Module 4 MULTIPLEXERS AND DECODERS – Module 9 A. Combinational Logic Design Using a Truth Table B. Minterm and Maxterm Expansions C. General Minterm and Maxterm Expansions D. Incompletely Specified Functions Click the mouse to move to the next page. Use the ESC key to exit this chapter. Conversion of English Sentences to Boolean Equations The three main steps in designing a single-output combinational switching circuit are 1. Find a switching function that specifies the desired behavior of the circuit. 2. Find a simplified algebraic expression for the function. 3. Realize the simplified function using available logic elements. Combinational Logic Design using a Truth Table Suppose we want the output of a circuit to be f = 1 if N ≥ 021 and f = 0 if N <2011 . Then the truth table is: Next, we will derive an algebraic expression for f from the truth table by using the combinations of values of A, B, and C for which f = 1. For example, the term A′BC is 1 only if A = 0, B = 1, and C = 1. Finding all terms such that f = 1 and ORing them together yields: f = A′BC + AB′C′ + AB′C + ABC′ + ABC (4-1) The equation can be simplified by first combining terms and then eliminating A′: f = A′BC + AB′ + AB = A′BC + A = A + BC (4-2) This equation leads directly to the following circuit: Instead of writing f in terms of the 1’s of the function, we may also write f in terms of the 0’s of the function. Observe that the term A + B + C is 0 only if A = B = C = 0. ANDing all of these ‘0’ terms together yields: f = (A + B + C)(A + B + C′)(A + B′ + C) (4-3) By combining terms and using the second distributive law, we can simplify the equation: f = (A + B + C)(A + B + C′)(A + B′ + C) (4-3) f = (A + B)(A + B′ + C) = A + B(B′ + C) = A + BC (4-4) Minterm and Maxterm Expansions f = A′BC + AB′C′ + AB′C + ABC′ + ABC(4-1) Each of the terms in Equation (4-1) is referred to as a minterm. In general, a minterm of n variables is a product of n literals in which each variable appears exactly once in either true or complemented form, but not both. (A literal is a variable or its complement) Section 4.3 (p. 93) Table 4-1 Minterms and Maxterms for Three Variables Minterm expansion for a function is unique. Equation (4-1) can be rewritten in terms of m-notation as: f = A′BC + AB′C′ + AB′C + ABC′ + ABC(4-1) f (A, B, C) = m 3 m + 4 + m5+ m 6 7 (4-5) This can be further abbreviated by listing only the decimal subscripts in the form: f (A, B, C) = Ʃ m(3, 4, 5, 6, 7) (4-5) Minterm Expansion Find the minterm expansion of f(a,b,c,d) = a'(b' + d) + acd'. Section 4.3 (p. 95) Design a comparator • Design a circuit returns 1 if number (a1 a0)2 >= (b1 b0)2. • We assume that the numbers are > 0 • Build the truth table of the comparator • Write the minterm expansion formula • Write the maxterm expansion formula Table 4-2. General Truth Table for Three Variables Table 4-2 represents a truth table for a general function of three variables. Each aiis a constant with a value of 0 or 1. General Minterm and Maxterm Expansions of three variables as follows:sion for a general function The maxterm expansion for a general function of three variables is: Section 4.4 (p. 97) Conversion of Forms QUIZ • Let’s assume that f( A, B, C) = ∑ m(0, 1, 2, 3); g( A, B, C) = ∏M(0, 1, 6, 7). What is the minimal form of h = f g A)A’B B)A’ + B’ C)AB + A’B’ D)B’C’ E)None of the above QUIZ • Let’s assume that f( A, B, C) = ∑ m(0, 3, 4,
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