Logic and Computer Design Fundamentals Boolean Algebra and

Binary Logic and Gates § Binary variables take on one of two values. § Logical operators operate on binary values and binary variables. § Basic logical operators are the logic functions AND, OR and NOT. § Logic gates implement logic functions. § Boolean Algebra: a useful mathematical system for specifying and transforming logic functions. § We study Boolean algebra as a foundation for designing and analyzing digital systems! 2

Binary Variables § Recall that the two binary values have different names: • • True/False On/Off Yes/No 1/0 § We use 1 and 0 to denote the two values. § Variable identifier examples: • A, B, y, z, or X 1 for now • RESET, START_IT, or ADD 1 later 3

Logical Operations § The three basic logical operations are: • AND • OR • NOT § AND is denoted by a dot (·). § OR is denoted by a plus (+). § NOT is denoted by an overbar ( ¯ ), a single quote mark (') after, or (~) before the variable. 4

Notation Examples § Examples: • Y = A × B is read “Y is equal to A AND B. ” • z = x + y is read “z is equal to x OR y. ” • X = A is read “X is equal to NOT A. ” § Note: The statement: 1 + 1 = 2 (read “one plus one equals two”) is not the same as 1 + 1 = 1 (read “ 1 or 1 equals 1”). 5

Operator Definitions § Operations are defined on the values "0" and "1" for each operator: AND 0· 0=0 0· 1=0 1· 0=0 1· 1=1 OR NOT 0+0=0 0+1=1 1+0=1 1+1=1 0=1 1=0 6

Truth Tables § Tabular listing of the values of a function for all possible combinations of values on its arguments § Example: Truth tables for the basic logic operations: X 0 0 1 1 AND Y Z = X·Y 0 0 1 0 0 0 1 1 X 0 0 1 1 Y 0 1 OR Z = X+Y 0 1 1 1 7 NOT X 0 1 Z=X 1 0

Truth Tables – Cont’d § Used to evaluate any logic function § Consider F(X, Y, Z) = X Y + Y Z XY Y YZ F=XY+YZ 0 0 1 1 0 1 0 1 0 0 0 1 1 0 0 0 1 1 1 8

Logic Function Implementation § Using Switches in parallel => OR • Inputs: § logic 1 is switch closed § logic 0 is switch open • Outputs: Switches in series => AND § logic 1 is light on § logic 0 is light off. • NOT input: Normally-closed switch => NOT C § logic 1 is switch open § logic 0 is switch closed 9

Logic Function Implementation – cont’d § Example: Logic Using Switches B C A D § Light is on (L = 1) for L(A, B, C, D) = A (B C + D) = A B C + A D and off (L = 0), otherwise. § Useful model for relay and CMOS gate circuits, the foundation of current digital logic circuits 10

Logic Gates § In the earliest computers, switches were opened and closed by magnetic fields produced by energizing coils in relays. The switches in turn opened and closed the current paths. § Later, vacuum tubes that open and close current paths electronically replaced relays. § Today, transistors are used as electronic switches that open and close current paths. 11

Logic Gate Symbols and Behavior § Logic gates have special symbols: X X Z = X ·Y Y Z= X+ Y Y X OR gate AND gate § And waveform behavior in time as follows: X 0 0 1 1 Y 0 1 X ·Y 0 0 0 1 (OR) X+ Y 0 1 1 1 (NOT) X 1 1 0 0 (AND) 12 Z= X NOT gate or inverter

Logic Diagrams and Expressions Truth Table XYZ 000 001 010 011 100 101 110 111 F = X + Y ×Z 0 1 0 X 0 1 Y 1 1 Z 1 Logic Equation F = X +Y Z Logic Diagram F § Boolean equations, truth tables and logic diagrams describe the same function! § Truth tables are unique, but expressions and logic diagrams are not. This gives flexibility in implementing functions. 13

Gate Delay § In actual physical gates, if an input changes that causes the output to change, the output change does not occur instantaneously. § The delay between an input change and the output change is the gate delay denoted by t. G: 1 Input 0 1 Output 0 0 t. G 0. 5 1 t. G = 0. 3 ns Time (ns) 1. 5 14

Boolean Algebra § § 1. 3. 5. 7. 9. Invented by George Boole in 1854 An algebraic structure defined by a set B = {0, 1}, together with two binary operators (+ and ·) and a unary operator ( ) X+0= X X+1 =1 X+X =X X+X =1 2. 4. 6. 8. Identity element X. 1 =X X. 0 =0 X. X = X Idempotence X. X = 0 Complement X=X 10. X + Y = Y + X 12. (X + Y) + Z = X + (Y + Z) 14. X(Y + Z) = XY + XZ 16. X + Y = X. Y Involution 11. XY = YX Commutative Associative 13. (XY) Z = X(YZ) 15. X + YZ = (X + Y) (X + Z) Distributive De. Morgan’s 17. X. Y = X + Y 15

Some Properties of Boolean Algebra § Boolean Algebra is defined in general by a set B that can have more than two values § A two-valued Boolean algebra is also know as Switching Algebra. The Boolean set B is restricted to 0 and 1. Switching circuits can be represented by this algebra. § The dual of an algebraic expression is obtained by interchanging + and · and interchanging 0’s and 1’s. § The identities appear in dual pairs. When there is only one identity on a line the identity is self-dual, i. e. , the dual expression = the original expression. § Sometimes, the dot symbol ‘ ’ (AND operator) is not written when the meaning is clear 16

Dual of a Boolean Expression § Example: F = (A + C) · B + 0 dual F = (A · C + B) · 1 = A · C + B § Example: G = X · Y + (W + Z) dual G = (X+Y) · (W · Z) = (X+Y) · (W+Z) § Example: H = A · B + A · C + B · C dual H = (A+B) · (A+C) · (B+C) § Unless it happens to be self-dual, the dual of an expression does not equal the expression itself § Are any of these functions self-dual? H is self-dual (A+B)(A+C)(B+C)=(A+BC)(B+C)=AB+AC+BC 17

Boolean Operator Precedence § The order of evaluation is: 1. Parentheses 2. NOT 3. AND 4. OR § Consequence: Parentheses appear around OR expressions § Example: F = A(B + C)(C + D) 18

Boolean Algebraic Proof – Example 1 § A+A·B=A (Absorption Theorem) Proof Steps Justification A+A·B =A· 1+A·B Identity element: A · 1 = A · ( 1 + B) Distributive =A· 1 1+B=1 = A Identity element § Our primary reason for doing proofs is to learn: • Careful and efficient use of the identities and theorems of Boolean algebra, and • How to choose the appropriate identity or theorem to apply to make forward progress, irrespective of the application. 19

Boolean Algebraic Proof – Example 2 § AB + AC + BC = AB + AC (Consensus Theorem) Proof Steps Justification = AB + AC + BC = AB + AC + 1 · BC Identity element = AB + AC + (A + A) · BC Complement = AB + AC + ABC Distributive = AB + ABC + ACB Commutative = AB · 1 + ABC + AC · 1 + ACB Identity element = AB (1+C) + AC (1 + B) Distributive = AB. 1 + AC. 1 1+X = 1 = AB + AC Identity element 20

Useful Theorems § Minimization XY+XY=Y § Minimization (dual) (X+Y) = Y § Absorption X+XY=X § Absorption (dual) X · (X + Y) = X § Simplification X+XY=X+Y § Simplification (dual) X · (X + Y) = X · Y § De. Morgan’s § X+Y=X·Y § De. Morgan’s (dual) § X·Y=X+Y 21

Truth Table to Verify De. Morgan’s X+Y=X·Y X·Y=X+Y X Y X·Y X+Y X · Y X·Y X+Y 0 0 1 1 0 1 0 0 0 1 1 1 0 0 1 0 1 0 0 0 1 1 1 0 § Generalized De. Morgan’s Theorem: X 1 + X 2 + … + Xn = X 1 · X 2 · … · Xn = X 1 + X 2 + … + Xn 22 1 1 1 0

Complementing Functions § Use De. Morgan's Theorem: 1. Interchange AND and OR operators 2. Complement each constant and literal § Example: Complement F = xy z + x y z F = (x + y + z) § Example: Complement G = (a + bc)d + e G = (a (b + c) + d) e 23

Expression Simplification § An application of Boolean algebra § Simplify to contain the smallest number of literals (variables that may or may not be complemented) A B + ACD + A BD + AC D + A BCD = AB + ABCD + A C D + A B D = AB + AB(CD) + A C (D + D) + A B D = AB + A C + A B D = B(A + AD) +AC = B (A + D) + A C (has only 5 literals) 24

Next … Canonical Forms § Minterms and Maxterms § Sum-of-Minterm (SOM) Canonical Form § Product-of-Maxterm (POM) Canonical Form § Representation of Complements of Functions § Conversions between Representations 25

Minterms § Minterms are AND terms with every variable present in either true or complemented form. § Given that each binary variable may appear normal (e. g. , x) or complemented (e. g. , x), there are 2 n minterms for n variables. § Example: Two variables (X and Y) produce 2 x 2 = 4 combinations: XY (both normal) X Y (X normal, Y complemented) XY (X complemented, Y normal) X Y (both complemented) § Thus there are four minterms of two variables. 26

Maxterms § Maxterms are OR terms with every variable in true or complemented form. § Given that each binary variable may appear normal (e. g. , x) or complemented (e. g. , x), there are 2 n maxterms for n variables. § Example: Two variables (X and Y) produce 2 x 2 = 4 combinations: X + Y (both normal) X + Y (x normal, y complemented) X + Y (x complemented, y normal) X + Y (both complemented) 27

Minterms & Maxterms for 2 variables § Two variable minterms and maxterms. x y Index Minterm Maxterm 0 0 0 m 0 = x y M 0 = x + y 0 1 1 m 1 = x y M 1 = x + y 1 0 2 m 2 = x y M 2 = x + y 1 1 3 m 3 = x y M 3 = x + y § The minterm mi should evaluate to 1 for each combination of x and y. § The maxterm is the complement of the minterm 28

Minterms & Maxterms for 3 variables x 0 0 1 1 y 0 0 1 1 z 0 1 0 1 Index 0 1 2 3 4 5 6 7 Minterm m 0 = x y z m 1 = x y z m 2 = x y z m 3 = x y z m 4 = x y z m 5 = x y z m 6 = x y z m 7 = x y z Maxterm M 0 = x + y + z M 1 = x + y + z M 2 = x + y + z M 3 = x + y + z M 4 = x + y + z M 5 = x + y + z M 6 = x + y + z M 7 = x + y + z Maxterm Mi is the complement of minterm mi Mi = mi and mi = Mi 29

Purpose of the Index § Minterms and Maxterms are designated with an index § The index number corresponds to a binary pattern § The index for the minterm or maxterm, expressed as a binary number, is used to determine whether the variable is shown in the true or complemented form § For Minterms: • ‘ 1’ means the variable is “Not Complemented” and • ‘ 0’ means the variable is “Complemented”. § For Maxterms: • ‘ 0’ means the variable is “Not Complemented” and • ‘ 1’ means the variable is “Complemented”. 30

Standard Order § All variables should be present in a minterm or maxterm and should be listed in the same order (usually alphabetically) § Example: For variables a, b, c: • Maxterms (a + b + c), (a + b + c) are in standard order • However, (b + a + c) is NOT in standard order (a + c) does NOT contain all variables • Minterms (a b c) and (a b c) are in standard order • However, (b a c) is not in standard order (a c) does not contain all variables 31

Sum-Of-Minterm (SOM) § Sum-Of-Minterm (SOM) canonical form: Sum of minterms of entries that evaluate to ‘ 1’ x y z F 0 0 1 1 0 1 0 1 0 1 0 0 1 1 Minterm m 1 = x y z Focus on the ‘ 1’ entries m 6 = x y z m 7 = x y z F = m 1 + m 6 + m 7 = ∑ (1, 6, 7) = x y z + x y z 32

Sum-Of-Minterm Examples § F(a, b, c, d) = ∑(2, 3, 6, 10, 11) § F(a, b, c, d) = m 2 + m 3 + m 6 + m 10 + m 11 abcd+abcd+abcd § G(a, b, c, d) = ∑(0, 1, 12, 15) § G(a, b, c, d) = m 0 + m 12 + m 15 abcd+abcd+abcd 33

Product-Of-Maxterm (POM) § Product-Of-Maxterm (POM) canonical form: Product of maxterms of entries that evaluate to ‘ 0’ x y z F 0 0 1 1 0 1 0 1 1 1 0 1 0 1 Maxterm M 2 = (x + y + z) Focus on the ‘ 0’ entries M 4 = (x + y + z) M 6 = (x + y + z) F = M 2·M 4·M 6 = ∏ (2, 4, 6) = (x+y+z) 34

Product-Of-Maxterm Examples § F(a, b, c, d) = ∏(1, 3, 6, 11) § F(a, b, c, d) = M 1 · M 3 · M 6 · M 11 (a+b+c+d) § G(a, b, c, d) = ∏(0, 4, 12, 15) § G(a, b, c, d) = M 0 · M 4 · M 12 · M 15 (a+b+c+d) 35

Observations § We can implement any function by "ORing" the minterms corresponding to the ‘ 1’ entries in the function table. A minterm evaluates to ‘ 1’ for its corresponding entry. § We can implement any function by "ANDing" the maxterms corresponding to ‘ 0’ entries in the function table. A maxterm evaluates to ‘ 0’ for its corresponding entry. § The same Boolean function can be expressed in two canonical ways: Sum-of-Minterms (SOM) and Product-of. Maxterms (POM). § If a Boolean function has fewer ‘ 1’ entries then the SOM canonical form will contain fewer literals than POM. However, if it has fewer ‘ 0’ entries then the POM form will have fewer literals than SOM. 36

Converting to Sum-of-Minterms Form § A function that is not in the Sum-of-Minterms form can be converted to that form by means of a truth table § Consider F = y + x z x y z F Minterm 0 0 1 1 1 1 0 0 m 0 = x y z m 1 = x y z m 2 = x y z 0 0 1 1 0 1 0 1 m 4 = x y z m 5 = x y z F = ∑(0, 1, 2, 4, 5) = m 0 + m 1 + m 2 + m 4 + m 5 = xyz+xyz+xyz 37

Converting to Product-of-Maxterms Form § A function that is not in the Product-of-Minterms form can be converted to that form by means of a truth table § Consider again: F = y + x z x y z F 0 0 1 1 1 1 0 0 1 1 0 1 0 1 Minterm F = ∏(3, 6, 7) = M 3 · M 6 · M 7 = M 3 = (x+y+z) M 6 = (x+y+z) M 7 = (x+y+z) 38

Conversions Between Canonical Forms x y z F 0 0 1 1 0 1 0 1 0 1 1 1 0 1 Minterm Maxterm M 0 = (x + y + z) m 1 = x y z m 2 = x y z m 3 = x y z M 4 = (x + y + z) m 5 = x y z M 6 = (x + y + z) m 7 = x y z F = m 1+m 2+m 3+m 5+m 7 = ∑(1, 2, 3, 5, 7) = xyz+xyz+xyz F = M 0 · M 4 · M 6 = ∏(0, 4, 6) = (x+y+z)(x+y+z) 39

Algebraic Conversion to Sum-of-Minterms § Expand all terms first to explicitly list all minterms § AND any term missing a variable v with (v + v) § Example 1: f = x + x y (2 variables) f = x (y + y) + x y f=xy+xy+xy f = m 3 + m 2 + m 0 = ∑(0, 2, 3) § Example 2: g = a + b c (3 variables) g = a (b + b)(c + c) + (a + a) b c g=abc+abc+abc+abc g=abc+abc+abc g = m 1 + m 4 + m 5 + m 6 + m 7 = ∑ (1, 4, 5, 6, 7) 40

Algebraic Conversion to Product-of-Maxterms § Expand all terms first to explicitly list all maxterms § OR any term missing a variable v with v · v § Example 1: f = x + x y (2 variables) Apply 2 nd distributive law: f = (x + x) (x + y) = 1 · (x + y) = M 1 § Example 2: g = a c + b c + a b (3 variables) g = (a c + b c + a) (a c + b) (distributive) g = (c + b c + a) (a c + b) (x + x y = x + y) g = (c + b + a) (a + c + b) (x + x y = x + y) g = (a + b + c) = M 5. M 2 = ∏ (2, 5) 41

Function Complements § The complement of a function expressed as a sum of minterms is constructed by selecting the minterms missing in the sum-of-minterms canonical form § Alternatively, the complement of a function expressed by a Sum of Minterms form is simply the Product of Maxterms with the same indices § Example: Given F(x, y, z) = ∑ (1, 3, 5, 7) F(x, y, z) = ∑ (0, 2, 4, 6) F(x, y, z) = ∏ (1, 3, 5, 7) 42

Summary of Minterms and Maxterms § There are 2 n minterms and maxterms for Boolean functions with n variables. § Minterms and maxterms are indexed from 0 to 2 n – 1 § Any Boolean function can be expressed as a logical sum of minterms and as a logical product of maxterms § The complement of a function contains those minterms not included in the original function § The complement of a sum-of-minterms is a product-ofmaxterms with the same indices 43

Standard Forms § Standard Sum-of-Products (SOP) form: equations are written as an OR of AND terms § Standard Product-of-Sums (POS) form: equations are written as an AND of OR terms § Examples: • SOP: A B C + B • POS: (A + B) · (A+ B + C )· C § These “mixed” forms are neither SOP nor POS • (A B + C) (A + C) • A B C + A C (A + B) 44

Standard Sum-of-Products (SOP) § A sum of minterms form for n variables can be written down directly from a truth table. • Implementation of this form is a two-level network of gates such that: • The first level consists of n-input AND gates • The second level is a single OR gate § This form often can be simplified so that the corresponding circuit is simpler. 45

Standard Sum-of-Products (SOP) § A Simplification Example: F( A, B, C) = S (1, 4, 5, 6, 7) § Writing the minterm expression: F = A B C + ABC § Simplifying: F = A B C + A (B C + B C) F = A B C + A (B (C + C) + B (C + C)) F = A B C + A (B + B) F=ABC+A F=BC+A § Simplified F contains 3 literals compared to 15 46

AND/OR Two-Level Implementation § The two implementations for F are shown below It is quite apparent which is simpler! 47

SOP and POS Observations § The previous examples show that: • Canonical Forms (Sum-of-minterms, Product-of-Maxterms), or other standard forms (SOP, POS) differ in complexity • Boolean algebra can be used to manipulate equations into simpler forms • Simpler equations lead to simpler implementations § Questions: • How can we attain a “simplest” expression? • Is there only one minimum cost circuit? • The next part will deal with these issues 48

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