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EKT 124 / 3 ELEKTRONIK DIGIT 1

EKT 124 / 3 ELEKTRONIK DIGIT 1. CHAPTER 1 INTRODUCTION TO DIGITAL LOGIC. De Morgan’s Theorem. Theorems of Boolean Algebra(1). 1) A + 0 = A 2) A + 1 = 1 3) A • 0 = 0 4) A • 1 = A 5) A + A = A 6) A + A = 1 7) A • A = A 8) A • A = 0. Theorems of Boolean Algebra(2). 9) A = A

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EKT 124 / 3 ELEKTRONIK DIGIT 1

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  1. EKT 124 / 3ELEKTRONIK DIGIT 1 CHAPTER 1 INTRODUCTION TO DIGITAL LOGIC

  2. De Morgan’s Theorem

  3. Theorems of Boolean Algebra(1) 1) A + 0 = A 2) A + 1 = 1 3) A • 0 = 0 4) A • 1 = A 5) A + A = A 6) A + A = 1 7) A • A = A 8) A • A = 0

  4. Theorems of Boolean Algebra(2) 9) A = A 10) A + AB = A 11) A + AB = A + B 12) (A + B)(A + C) = A + BC 13) Commutative : A + B = B + A AB = BA 14) Associative : A+(B+C) =(A+B) + C A(BC) = (AB)C 15) Distributive : A(B+C) = AB +AC (A+B)(C+D)=AC + AD + BC + BD

  5. De Morgan’s Theorems 16) (X+Y) = X . Y 17) (X.Y) = X + Y • Two most important theorems of Boolean Algebra were contributed by De Morgan. • Extremely useful in simplifying expression in which product or sum of variables is inverted. • The TWO theorems are :

  6. Implications of De Morgan’s Theorem • Input Output • X Y X+Y XY • 0 0 1 1 • 0 1 0 0 • 0 0 0 • 1 1 0 0 (a) (b) (c) (a) Equivalent circuit implied by theorem (16) (b) Negative- AND (c) Truth table that illustrates DeMorgan’s Theorem

  7. Implications of De Morgan’s Theorem (a) • Input Output • X Y XY X+Y • 0 0 1 1 • 0 1 1 1 • 0 1 1 • 1 1 0 0 (b) (c) (a) Equivalent circuit implied by theorem (17) (b) Negative-OR (c) Truth table that illustrates DeMorgan’s Theorem

  8. De Morgan’s Theorem Conversion Step 1: Change all ORs to ANDs and all ANDs to Ors Step 2: Complement each individual variable (short overbar) Step 3: Complement the entire function (long overbars) Step 4: Eliminate all groups of double overbars Example: A . B A .B. C = A + B = A + B + C = A + B = A + B + C = A + B = A + B + C = A + B

  9. De Morgan’s Theorem Conversion ABC + ABC (A + B +C)D = (A+B+C).(A+B+C) = (A.B.C)+D = (A+B+C).(A+B+C) = (A.B.C)+D = (A+B+C).(A+B+C) = (A.B.C)+D = (A+B+C).(A+B+C) = (A.B.C)+D

  10. Example: Analyze the circuit below Y 1. Y=??? 2. Simplify the Boolean expression found in 1

  11. Follow the steps list below (constructing truth table) • List all the input variable combinations of 1 and 0 in binary sequentially • Place the output logic for each combination of input • Base on the result found write out the boolean expression.

  12. Exercises: • Simplify the following Boolean expressions • (AB(C + BD) + AB)C • ABC + ABC + ABC + ABC + ABC • Write the Boolean expression of the following circuit.

  13. Standard Forms of Boolean Expressions • Sum of Products (SOP) • Products of Sum (POS) Notes: • SOP and POS expression cannot have more than one variable combined in a term with an inversion bar • There’s no parentheses in the expression

  14. Standard Forms of Boolean Expressions • Converting SOP to Truth Table • Examine each of the products to determine where the product is equal to a 1. • Set the remaining row outputs to 0.

  15. Standard Forms of Boolean Expressions • Converting POS to Truth Table • Opposite process from the SOP expressions. • Each sum term results in a 0. • Set the remaining row outputs to 1.

  16. Standard Forms of Boolean Expressions • The standard SOP Expression • All variables appear in each product term. • Each of the product term in the expression is called as minterm. • Example: • In compact form, f(A,B,C) may be written as

  17. Standard Forms of Boolean Expressions • The standard POS Expression • All variables appear in each product term. • Each of the product term in the expression is called as maxterm. • Example: • In compact form, f(A,B,C) may be written as

  18. Standard Forms of Boolean Expressions • Example: • Convert the following SOP expression to an equivalent POS expression: • Example: • Develop a truth table for the expression:

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