Logic and Computer Design Fundamentals Boolean Algebra and Logic Gates Charles Kime & Thomas Kaminski © 2008 Pearson Education, Inc. (Edited by Dr. Muhamed Mudawar for COE 202 & EE 200 at KFUPM) 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! 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 X1 for now • RESET, START_IT, or ADD1 later 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. 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”). 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 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 0 1 OR Z = X+Y 0 1 1 1 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 X Y Z XY Y YZ F = X Y + YZ 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 1 1 1 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 1 1 Logic Function Implementation Using Switches 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 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 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. Logic Gate Symbols and Behavior Logic gates have special symbols: X X Z= X·Y Y Z = X+ Y Y AND gate NOT gateor inverter OR gate And waveform behavior in time as follows: X 0 0 1 1 Y 0 1 0 1 X ·Y 0 0 0 1 (OR) X+ Y 0 1 1 1 (NOT) X 1 1 0 0 (AND) Z= X X Logic Diagrams and Expressions Truth Table Logic Equation XYZ F X YZ 000 0 001 1 010 0 011 0 100 1 101 1 Y 110 1 Z 111 1 F X Y Z Logic Diagram X 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. F 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 tG: 1 Input 0 Output1 tG tG tG = 0.3 ns 0 0 0.5 1 1.5 Time (ns) Boolean Algebra 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 ( ) 1. 3. 5. 7. 9. 10. 12. 14. 16. X+0= X Identity element X .1 = X X + X =X 2. 4. 6. X .X =X Idempotence X+X=1 8. X .X = 0 Complement X+1=1 X= X X+Y=Y+X (X + Y) + Z = X + (Y + Z) X(Y + Z) = XY + XZ X + Y = X .Y X .0 = 0 Involution 11. XY = YX Commutative Associative 13. (XY) Z = X(YZ) Distributive 15. X + YZ = (X + Y)(X + Z) DeMorgan’s 17. X . Y = X + Y 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 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 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) Boolean Algebraic Proof – Example 1 A + A · B =A Proof Steps A + A ·B = A · 1 + A ·B = A · ( 1 + B) =A ·1 =A (Absorption Theorem) Justification Identity element: A · 1 = A Distributive 1+B=1 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. Boolean Algebraic Proof – Example 2 AB + AC + BC = AB + AC (Consensus Theorem) Proof Steps Justification AB + AC + BC Identity element = AB + AC + 1 · BC Complement = AB + AC + (A + A) · BC Distributive = AB + AC + ABC + ABC Commutative = AB + ABC + AC + ACB = AB · 1 + ABC + AC · 1 + ACB Identity element Distributive = AB (1+C) + AC (1 + B) 1+X = 1 = AB . 1 + AC . 1 Identity element = AB + AC Useful Theorems Minimization XY+ XY=Y Minimization (dual) (X+Y)(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 DeMorgans’ X + Y = X ·Y DeMorgan’s (dual) X ·Y = X + Y Truth Table to Verify DeMorgan’s X+Y=X·Y 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 1 0 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 1 0 0 0 1 0 0 0 1 1 1 0 Generalized DeMorgan’s Theorem: X1+ X2+ … + Xn= X1 · X2 · … ·Xn X1· X2 · … · Xn = X1+ X2+ … + Xn 1 1 1 0 Complementing Functions Use DeMorgan's Theorem: 1. Interchange AND and OR operators 2. Complement each constant and literal Example: Complement F = xyz xyz F = (x + y + z)(x + y + z) Example: Complement G = (a + bc)d + e G = (a (b + c) + d) e Expression Simplification An application of Boolean algebra Simplify to contain the smallest number of literals (variables that may or may not be complemented) AB ACD ABD AC D ABCD = AB + ABCD + A C D + 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) 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 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 2nminterms for n variables. Example: Two variables (X and Y) produce 2 x 2 = 4 combinations: XY(both normal) XY (X normal, Y complemented) XY (X complemented, Y normal) XY (both complemented) Thus there are four minterms of two variables. 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 2nmaxterms 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) Minterms & Maxterms for 2 variables Two variable minterms and maxterms. x y 0 0 Index 0 Minterm m0 = x y Maxterm M0 = x + y 0 1 1 m1 = x y M1 = x + y 1 0 2 m2 = x y M2 = x + y 1 1 3 m3 = x y M3 = x + y The minterm mi should evaluate to 1 for each combination of x and y. The maxterm is the complement of the minterm Minterms & Maxterms for 3 variables x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 Index 0 1 2 3 4 5 6 7 Minterm Maxterm m0 = x y z m1 = x y z m2 = x y z m3 = x y z m4 = x y z m5 = x y z m6 = x y z m7 = x y z M0 = x + y + z M1 = x + y + z M2 = x + y + z M3 = x + y + z M4 = x + y + z M5 = x + y + z M6 = x + y + z M7 = x + y + z Maxterm Mi is the complement of minterm mi Mi = mi and mi = Mi 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”. 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 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 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 1 1 Minterm m1 = x y z Focus on the ‘1’ entries m6 = x y z m7 = x y z F = m1 + m6 + m7 = ∑ (1, 6, 7) = x y z + x y z + x y z Sum-Of-Minterm Examples F(a, b, c, d) = ∑(2, 3, 6, 10, 11) F(a, b, c, d) = m2 + m3 + m6 + m10+ m11 abcd+abcd+abcd+abcd+abcd G(a, b, c, d) = ∑(0, 1, 12, 15) G(a, b, c, d) = m0 + m1 + m12+ m15 abcd+abcd+abcd+abcd 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 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 0 1 Maxterm M2 = (x + y + z) Focus on the ‘0’ entries M4 = (x + y + z) M6 = (x + y + z) F = M2·M4·M6 = ∏ (2, 4, 6) = (x+y+z) (x+y+z) (x+y+z) Product-Of-Maxterm Examples F(a, b, c, d) = ∏(1, 3, 6, 11) F(a, b, c, d) = M1 · M3 · M6 ·M11 (a+b+c+d) (a+b+c+d) (a+b+c+d)(a+b+c+d) G(a, b, c, d) = ∏(0, 4, 12, 15) G(a, b, c, d) = M0 · M4· M12·M15 (a+b+c+d) (a+b+c+d) (a+b+c+d)(a+b+c+d) 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-ofMaxterms (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. 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 0 0 1 1 1 1 1 1 1 0 1 1 0 0 m0 = x y z m1 = x y z m2 = x y z 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 m4 = x y z m5 = x y z F = ∑(0, 1, 2, 4, 5) = m0 + m1 + m2 + m4 + m5 = x y z + x y z + x y z+ xyz+xyz 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 0 0 1 1 1 1 1 1 1 0 1 1 0 0 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 Minterm F = ∏(3, 6, 7) = M3 · M6 · M7 = M3 = (x+y+z) M6 = (x+y+z) M7 = (x+y+z) (x+y+z) (x+y+z) (x+y+z) Conversions Between Canonical Forms x y z F 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 Minterm Maxterm M0 = (x + y + z) m1 = x y z m2 = x y z m3 = x y z M4 = (x + y + z) m5 = x y z M6 = (x + y + z) m7 = x y z F = m1+m2+m3+m5+m7 = ∑(1, 2, 3, 5, 7) = x y z + x y z + x y z + x y z + x yz F = M0· M4· M6= ∏(0, 4, 6) = (x+y+z)(x+y+z)(x+y+z) 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 = m3+ m2+ m0= ∑(0, 2, 3) Example 2: g = a + b c (3 variables) g = a (b + b)(c + c) + (a + a) b c g = a b c + a b c + a b c + a b c + a b c + a bc g = a b c + a b c + a b c + a b c + a bc g = m1+ m4+ m5+ m6+ m7= ∑ (1, 4, 5, 6, 7) 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 2nddistributive law: f = (x + x) (x + y) = 1 · (x + y) = (x + y) =M1 Example 2: g = a c + b c + a b g (3 variables) = (a c + b c + a) (a c + b c + b) (distributive) g = (c + b c + a) (a c + c + b) (x + x y = x + y) (x + x y = x + y) g = (c + b + a) (a + c + b) g = (a + b + c) (a + b + c) = M5 . M2= ∏ (2, 5) 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) Summary of Minterms and Maxterms There are 2nminterms and maxterms for Boolean functions with n variables. Minterms and maxterms are indexed from 0 to 2n – 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 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 A B C B • POS: (A B)· (AB C )·C These “mixed” forms are neither SOP nor POS • (A B C) (A C) • AB C AC(A B) 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. Standard Sum-of-Products (SOP) A Simplification Example: F(A, B, C) (1,4,5,6,7) Writing the minterm expression: F = A B C + A B C + A B C + ABC +ABC Simplifying: F = A B C + A (B C + B C + B C + B C) F = A B C + A (B (C + C) + B (C + C)) F = A B C + A (B + B) F = A B C +A F = B C +A Simplified F contains 3 literals compared to 15 AND/OR Two-Level Implementation The two implementations for F are shown below A F B C It is quite apparent which is simpler! 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 Terms of Use All (or portions) of this material © 2008 by Pearson Education, Inc. 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