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Eng. Mohammed Timraz Electronics & Communication Engineer

University of Palestine Faculty of Engineering and Urban planning Software Engineering Department. Digital Logic Design ESGD2201. Lecture 8. Boolean Algebra and Logic Simplification. Eng. Mohammed Timraz Electronics & Communication Engineer. Saturday, 28 th June 2008. Agenda.

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Eng. Mohammed Timraz Electronics & Communication Engineer

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  1. University of Palestine Faculty of Engineering and Urban planning Software Engineering Department Digital Logic Design ESGD2201 Lecture 8 Boolean Algebra and Logic Simplification Eng. Mohammed Timraz Electronics & Communication Engineer Saturday, 28th June 2008

  2. Agenda Boolean Algebra and Logic Simplification 1. Boolean Operations and Expressions. 2. Laws and Rules of Boolean Algebra. 3. DeMorgan’s Theorems. 4. Boolean Analysis of Logic Circuits. 5. Simplification Using Boolean Algebra. 6. Standard Forms of Boolean Expressions. 7. Boolean Expressions and Truth Tables.

  3. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. The Sum of Products (SOP) Form: When two or more product terms are summed by Boolean addition, the resulting expression is a Sum of Products. Some Examples:

  4. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. The Sum of Products (SOP) Form: Examples 1: Make an Implementation of the SOP expression

  5. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. The Sum of Products (SOP) Form: Examples 1:

  6. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. The Sum of Products (SOP) Form: Examples 1: A AB B AB+BCD+AC B BCD C D A C AC

  7. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Conversion of a general Expression to SOP form: Any logic expression can be changed into SOP form by applying Boolean algebra techniques. For Example: The expression A(B+CD) can be converted to SOP form applying the distributive law: A(B+CD) = AB+ACD

  8. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Conversion of a general Expression to SOP form: Example 1: Convert the following Boolean expression into Standard SOP form. Solution: The domain of this SOP expression is A,B,C and D. Take one term at time.

  9. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Conversion of a general Expression to SOP form: Solution: The first term, , is missing variable D or , so multiply the first term by as follows: The second term, , is missing variable C or , so multiply the second term by as follows:

  10. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Conversion of a general Expression to SOP form: The two resulting terms are missing variable D or ,so multiply both terms by as follows: The third term, , is already in standard form.

  11. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Conversion of a general Expression to SOP form: The third term, , is already in standard form. The standard SOP form of the original expression is as follows:

  12. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Binary Representation of a Standard Product Term: A standard product term is equal to “ 1 “ for only one combination of variable values. For example , the product term is “ 1 “ when A=1, B=0, C=1, and D=0. And is “ 0 “ for all other combinations of values for the variables.

  13. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Binary Representation of a Standard Product Term: Example: Determine the binary values for which the following standard SOP expression is equal to “ 1 “. Solution: The term is equal to “ 1 “ when A=1, B=1, C=1, and D=1.

  14. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Binary Representation of a Standard Product Term: The term is equal to “ 1 “ when A=1, B=0, C=0, and D=1. The term is equal to “ 1 “ when A=0, B=0, C=0, and D=0. The SOP expression result is as follows: The SOP expression equals “ 1 “ when any or all of the three product term is “ 1 “.

  15. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. The Products of Sum (POS) Form: When two or more sum terms are multiplied, the resulting expression is a Product of Sums. Some Examples:

  16. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. The Products of Sum (POS) Form: Examples 1: Make an Implementation of the POS expression

  17. A A+B B (A+B).(B+C+D).(A+C) B B+C+D C D A C Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. The Product of Sums (POS) Form: Examples 1: A+C

  18. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. The Product of Sums (POS) Form: Conversion of a general Expression to POS form: Each sum term in an POS expression that does not contain all the variables in the domain can be expanded to standard form to include all variables in the domain and their complements.

  19. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. The Product of Sums (POS) Form: Conversion of a general Expression to POS form: Example 1: Convert the following Boolean expression into standard POS form.

  20. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. The Product of Sums (POS) Form: Conversion of a general Expression to POS form: Example 1: The domain of this POS expression is A,B,C,D. Take one term at a time.

  21. Boolean Algebra and Logic Simplification Conversion of a general Expression to POS form: Example 1: Solution: The first term , is missing variable D or , so add .

  22. Boolean Algebra and Logic Simplification Conversion of a general Expression to POS form: Example 1: Solution: The second term , is missing variable A or , so add .

  23. Boolean Algebra and Logic Simplification Conversion of a general Expression to POS form: Example 1: Solution: The third term , is already in standard form. The standard POS form of the original expression is as follows:

  24. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Example 1: Convert the following expression to an equivalent POS expression Solution: The binary evaluation for the above expression is as follows: 0 0 0 + 0 1 0 + 0 1 1 + 1 0 1 + 1 1 1 Because, there are three variables in the domain of this expression, there are a total eight (23) possible combination.

  25. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Example 1: Convert the following expression to an equivalent POS expression Solution: The SOP expression contains five of these combinations, so the POS must contain the other three which are 001, 100 and 110. Remember , these are the binary values that make the sum term 0. the equivalent POS expression is

  26. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Binary Representation of a Standard Sum Term: A standard sum term is equal to “ 0 “ for only one combination of variable values. For example , the sum term is “ 0 “ when A=0, B=1, C=0, and D=1. And is “ 1 “ for all other combinations of values for the variables.

  27. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Binary Representation of a Standard Sum Term: Example: Determine the binary values for which the following standard POS expression is equal to “ 0 “. Solution: The term is equal to “ 0 “ when A=0, B=0, C=0, and D=0.

  28. Boolean Algebra and Logic Simplification 6. Standard Forms of Boolean Expressions. Binary Representation of a Standard Product Term: The term is equal to “ 0 “ when A=0, B=1, C=1, and D=0. The term is equal to “ 0 “ when A=1, B=1, C=1, and D=1. The POS expression result is as follows: The POS expression equals “ 0 “ when any or all of the three product term is “ 0 “.

  29. Boolean Algebra and Logic Simplification 7. Boolean Expressions and Truth Tables. Example 1: Develop the truth table for the following standard SOP expression: Solution: There are three variable in the domain, so there are eight possible combinations of binary values of the variables. The binary values that make the product terms in the expression equal to 1 are : =001 =100 =111

  30. Boolean Algebra and Logic Simplification 7. Boolean Expressions and Truth Tables. The truth table for the standard SOP

  31. Boolean Algebra and Logic Simplification 7. Boolean Expressions and Truth Tables. Example 2: From the following truth table, determine the standard SOP expression and the equivalent standard POS expression.

  32. Boolean Algebra and Logic Simplification 7. Boolean Expressions and Truth Tables. Example 2: Solution: There are four “ 1s “ in the output column and the corresponding binary values are: 011 100 110 111 The resulting standard SOP expression for the output X is.

  33. Boolean Algebra and Logic Simplification 7. Boolean Expressions and Truth Tables. Example 2: Solution: For the POS expression, the output is “ 0 “ for binary values 000,001,010, and 101. These binary values are converted to sum terms as follows: 000 001 010 101 The resulting standard POS expression for the output X is.

  34. Boolean Algebra and Logic Simplification Exercises: Exercise 1: Realize this function using NAND gates only. Solution:

  35. Boolean Algebra and Logic Simplification Exercises: Exercise 1: Realize this function using NAND gates only. Solution:

  36. Boolean Algebra and Logic Simplification Exercises: Exercise 1: The implementation for result expression by using NAND gate Only.

  37. Boolean Algebra and Logic Simplification Exercises: Exercise 2: Realize this function using NAND gates only. Solution:

  38. Boolean Algebra and Logic Simplification Exercises: Exercise 2: Realize this function using NAND gates only. Solution:

  39. Boolean Algebra and Logic Simplification Exercises: Exercise 2: The implementation for result expression by using NAND gate Only.

  40. Boolean Algebra and Logic Simplification Exercises: Exercise 3: Realize this function using NOR gates only. Solution:

  41. Boolean Algebra and Logic Simplification Exercises: Exercise 3: Realize this function using NOR gates only. Solution:

  42. Boolean Algebra and Logic Simplification Exercises: Exercise 3: The implementation for result expression by using NOR gate Only.

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