Gate & Circuit Design

Overview Part 1 – Gate Circuits and Boolean Equations Binary Logic and Gates Boolean Algebra Standard Forms Part 2 – Circuit Optimization Two-Level Optimization Map Manipulation Multi-Level Circuit Optimization Part 3 – Additional Gates and Circuits Other Gate Types Exclusive-OR Operator and Gates High-Impedance Outputs

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 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 0 · 0 = 0 0 + 0 = 0 = 1 0 · 1 = 0 0 + 1 = 1 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 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 1 NOT 1 =

Truth Tables Truth table - a 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: 1 Z = X·Y Y X AND OR X Y Z = X+Y 1 1 X NOT Z =

Logic Function Implementation Using Switches For inputs: logic 1 is switch closed logic 0 is switch open For outputs: logic 1 is light on logic 0 is light off. NOT uses a switch such that: logic 1 is switch open logic 0 is switch closed Switches in parallel  OR Switches in series  AND C Normally-closed switch  NOT

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

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.

Transistors Today's computers use circuitry made with complementary metal oxide semiconductor (CMOS) technology. B = input (usually 1 or 0 represented by some voltage) C = voltage base E = output The transistor without the “bubble” on its input is an N-type field effect transistor. It acts like a closed switch between its top and bottom terminals with an H (1) applied to its input on its left. It acts like an open switch with an L (0) applied to its input. The transistor with the “bubble” on its input is a P-type field effect transistor. The +V at the top provides an H (1) and the Ground symbol at the bottom provides an L (0). By modeling the two types of field effect transistors as switches, one can see how the series and parallel interconnections can produce 1’s and 0’s on the outputs on the right in response to applied 1’s and 0’s on the inputs on the left. NOR and NAND are OR and AND, each followed by a NOT respectively.

Transistors Implementation of logic gates with transistors The transistor without the “bubble” on its input is an N-type field effect transistor. It acts like a closed switch between its top and bottom terminals with an H (1) applied to its input on its left. It acts like an open switch with an L (0) applied to its input. The transistor with the “bubble” on its input is a P-type field effect transistor. The +V at the top provides an H (1) and the Ground symbol at the bottom provides an L (0). By modeling the two types of field effect transistors as switches, one can see how the series and parallel interconnections can produce 1’s and 0’s on the outputs on the right in response to applied 1’s and 0’s on the inputs on the left. NOR and NAND are OR and AND, each followed by a NOT respectively. Transistor or tube implementations of logic functions are called logic gates or just gates Transistor gate circuits can be modeled by switch circuits

NAND NAND = not both X and Y = (X.Y) = X + Y = X or Y The transistor without the “bubble” on its input is an N-type field effect transistor. It acts like a closed switch between its top and bottom terminals with an H (1) applied to its input on its left. It acts like an open switch with an L (0) applied to its input. The transistor with the “bubble” on its input is a P-type field effect transistor. The +V at the top provides an H (1) and the Ground symbol at the bottom provides an L (0). By modeling the two types of field effect transistors as switches, one can see how the series and parallel interconnections can produce 1’s and 0’s on the outputs on the right in response to applied 1’s and 0’s on the inputs on the left. NOR and NAND are OR and AND, each followed by a NOT respectively. Switches in parallel  OR Switches in series  AND

NAND – Universal Gate NAND NOT AND OR NOT(NOT(A AND B)) The transistor without the “bubble” on its input is an N-type field effect transistor. It acts like a closed switch between its top and bottom terminals with an H (1) applied to its input on its left. It acts like an open switch with an L (0) applied to its input. The transistor with the “bubble” on its input is a P-type field effect transistor. The +V at the top provides an H (1) and the Ground symbol at the bottom provides an L (0). By modeling the two types of field effect transistors as switches, one can see how the series and parallel interconnections can produce 1’s and 0’s on the outputs on the right in response to applied 1’s and 0’s on the inputs on the left. NOR and NAND are OR and AND, each followed by a NOT respectively. NOT(NOT(A ) AND NOT(B))

Implementation of AND Connect output of NAND to an inverter since: The transistor without the “bubble” on its input is an N-type field effect transistor. It acts like a closed switch between its top and bottom terminals with an H (1) applied to its input on its left. It acts like an open switch with an L (0) applied to its input. The transistor with the “bubble” on its input is a P-type field effect transistor. The +V at the top provides an H (1) and the Ground symbol at the bottom provides an L (0). By modeling the two types of field effect transistors as switches, one can see how the series and parallel interconnections can produce 1’s and 0’s on the outputs on the right in response to applied 1’s and 0’s on the inputs on the left. NOR and NAND are OR and AND, each followed by a NOT respectively. Connect output of NAND to an inverter since: AND = NOT(NAND)

NAND – Universal Gate NOR XOR Can also use NOR gates to represent other gates (do as exercise). NOT(NOT(NOT(A) AND NOT(B))) The transistor without the “bubble” on its input is an N-type field effect transistor. It acts like a closed switch between its top and bottom terminals with an H (1) applied to its input on its left. It acts like an open switch with an L (0) applied to its input. The transistor with the “bubble” on its input is a P-type field effect transistor. The +V at the top provides an H (1) and the Ground symbol at the bottom provides an L (0). By modeling the two types of field effect transistors as switches, one can see how the series and parallel interconnections can produce 1’s and 0’s on the outputs on the right in response to applied 1’s and 0’s on the inputs on the left. NOR and NAND are OR and AND, each followed by a NOT respectively.

Logic Gate Symbols and Behavior Logic gates have special symbols: And waveform behavior in time as follows: +

Logic Diagrams and Expressions Truth Table 1 1 1 1 1 1 0 1 0 1 1 0 0 0 1 1 0 1 0 0 0 1 0 0 0 X Y Z Z Y X F × + = Equation Z Y X F + = X Y F Z Logic Diagram Boolean equations, truth tables and logic diagrams describe the same function! Truth tables are unique; expressions and logic diagrams are not. This gives flexibility in implementing functions.

Logic Diagrams and Expressions

Boolean Algebra An algebraic structure defined on a set of at least two elements, B, together with three binary operators (denoted +, · and ) that satisfies the following basic identities: 1. 3. 5. 7. 9. X + 0 = + 1 X + X X = X 2. 4. 6. 8. X . 1 = . 0 X . X Existence of 0 and 1 Idempotence Existence of Complement Involution 10. 12. 14. 16. X + Y Y + X = (X + Y) Z + X + (Y Z) X(Y + XY XZ X . Y 11. 13. 15. 17. XY YX = (XY) Z X(Y Z) X + YZ (X + Y) (X + Z) X . Y X + Y Commutative Associative Distributive DeMorgan ’ s

Some Properties of Identities & the Algebra If the meaning is unambiguous, we leave out the symbol “·” The dual of an algebraic expression is obtained by interchanging + and · and interchanging 0’s and 1’s. e.g., The dual of (1, 3, 5, 7)’s dual is (2, 4, 6, 8) 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. e.g., #9

Truth Tables for DeMorgan’s Law

Gate Implementation of Boolean Functions 011 010 ? Optimized to one gate? Karnaugh maps.

Truth Tables for DeMorgan’s Law

Truth Tables for DeMorgan’s Law

Some Properties of Identities & the Algebra Unless it happens to be self-dual, the dual of an expression does not equal the expression itself. Example: F = (A + C) · B + 0 dual F = (A · C + B) · 1 = A · C + B Example: G = X · Y + (W + Z) dual G = ? Example: H = A · B + A · C + B · C dual H = ? Are any of these functions self-dual? Dual G = ((X+Y) · (W · Z)') = ((X+Y) ·(W' + Z') Dual H = (A + B)(A + C)(B + C). Using the Boolean identities, = (A +BC) (B+C) = AB + AC + BC. So H is self-dual.

Some Properties of Identities & the Algebra (Continued) There can be more that 2 elements in B, i. e., elements other than 1 and 0. What are some common useful Boolean algebras with more than 2 elements? 1. 2. If B contains only 1 and 0, then B is called the switching algebra which is the algebra we use most often. Algebra of Sets Algebra of n-bit binary vectors Algebra of Sets correspondence to switching algebra: Set - variable, Union - OR, Intersect AND ., Universe - 1, Empty set - 0, Complement of Set - NOT, Subset – AND of variables and variable complements. Algebra of n-bit binary vectors correspondence to switching algebra: {n-bit binary vectors} – {0,1}, bitwise OR - OR, bitwise AND - AND, bitwise NOT - NOT, Vector of all 1s – 1, Vector of all 0s – 0.

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

Example 1: Boolean Algebraic Proof A + A·B = A (Absorption Theorem) Proof: Justification (identity or theorem) A + A·B = A · 1 + A · B X = X · 1 = A · ( 1 + B) X · Y + X · Z = X ·(Y + Z)(Distributive Law) = A · 1 1 + X = 1 = A X · 1 = X 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.

Example 2: Boolean Algebraic Proofs XY + XZ + YZ = XY + XZ (Consensus Theorem) Proof: Justification (identity or theorem) Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X

Example 2: Boolean Algebraic Proofs XY + XZ + YZ = XY + XZ (Consensus Theorem) Proof: Justification (identity or theorem) XY + XZ + YZ (LHS) Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X

Example 2: Boolean Algebraic Proofs XY + XZ + YZ = XY + XZ (Consensus Theorem) Proof: Justification (identity or theorem) XY + XZ + YZ (LHS) = XY + XZ + (YZ .1) Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X

Example 2: Boolean Algebraic Proofs XY + XZ + YZ = XY + XZ (Consensus Theorem) Proof: Justification (identity or theorem) XY + XZ + YZ (LHS) = XY + XZ + (YZ .1) = XY + XZ + (YZ .(X + X)) Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X

Example 2: Boolean Algebraic Proofs XY + XZ + YZ = XY + XZ (Consensus Theorem) Proof: Justification (identity or theorem) XY + XZ + YZ (LHS) = XY + XZ + (YZ .1) = XY + XZ + (YZ .(X + X)) = XY + XZ + XYZ + XYZ Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X

Example 2: Boolean Algebraic Proofs XY + XZ + YZ = XY + XZ (Consensus Theorem) Proof: Justification (identity or theorem) XY + XZ + YZ (LHS) = XY + XZ + (YZ .1) = XY + XZ + (YZ .(X + X)) = XY + XZ + XYZ + XYZ = XY + XYZ + XZ + XYZ Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X

Example 2: Boolean Algebraic Proofs XY + XZ + YZ = XY + XZ (Consensus Theorem) Proof: Justification (identity or theorem) XY + XZ + YZ (LHS) = XY + XZ + (YZ .1) = XY + XZ + (YZ .(X + X)) = XY + XZ + XYZ + XYZ = XY + XYZ + XZ + XYZ = XY(1+Z) + XZ(1+Y) Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X

Example 2: Boolean Algebraic Proofs XY + XZ + YZ = XY + XZ (Consensus Theorem) Proof: Justification (identity or theorem) XY + XZ + YZ (LHS) = XY + XZ + (YZ .1) = XY + XZ + (YZ .(X + X)) = XY + XZ + XYZ + XYZ = XY + XYZ + XZ + XYZ = XY(1+Z) + XZ(1+Y) = XY + XZ (RHS) Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X

Example 2: Boolean Algebraic Proofs ACD + ABD + BCD + ABC + ACD underlined term can be eliminated by consensus thm. Can we do better? ACD + ABD + BCD + ABC + ACD start over -- this time eliminate two other terms Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X

Example 2: Boolean Algebraic Proofs Now Consider: F = ABCD + BCDE + AB + BCE cannot reduce by consensus thm. F = ABCD + BCDE + AB + BCE + ACDE add the consensus term ACDE first F = BCE + ABCD + ACDE + AB +BCDE + ACDE Then the two underlined terms become redundant by consensus thm. Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X

( ) ( ) Useful Theorems x + x.y = x x.(x+y) = x Absorption x . y + x . Minimization x + x . y = x + y x . ( x + y = ) x . y Simplification x y + x . z + y . z = x . y + . x . z Consensus x + y . ( x + z ) . ( y + z ) = + ( ) ( x y ) . ( x + z ) x + y = x . y x . y = x + y DeMorgan‘s Law Minimization – useful for Karnaugh Maps!!

Boolean Function Evaluation = xy z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Boolean Function Evaluation = xy z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Boolean Function Evaluation = xy z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Boolean Function Evaluation = xy z F2 = x + y z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Boolean Function Evaluation = xy z F2 = x + y z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Boolean Function Evaluation = xy z F2 = x + y z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Boolean Function Evaluation = xy z F2 = x + y z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Boolean Function Evaluation = xy z F2 = x + y z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Boolean Function Evaluation = xy z F2 = x + y z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Boolean Function Evaluation = xy z F2 = x + y z F3 = x y z + x y z + x y F4 = x y + x z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Boolean Function Evaluation = xy z F2 = x + y z F3 = x y z + x y z + x y F4 = x y + x z F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0

Overview – Canonical Forms What are Canonical Forms? Minterms and Maxterms Index Representation of Minterms and Maxterms Sum-of-Minterm (SOM) Representations Product-of-Maxterm (POM) Representations Representation of Complements of Functions Conversions between Representations

Canonical Forms It is useful to specify Boolean functions in a form that: Allows comparison for equality. Has a correspondence to the truth tables Canonical Forms in common usage: Sum of Minterms (SOM) Product of Maxterms (POM)

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., ), there are 2n minterms for n variables. Example: Two variables (X and Y)produce 2 x 2 = 4 combinations: (both normal) (X normal, Y complemented) (X complemented, Y normal) (both complemented) Thus there are four minterms of two variables. x Y X X Y X Y

Maxterms (x normal, y complemented) (x complemented, y normal) Maxterms are OR terms with every variable in true or complemented form, or maxterm = !minterm Given that each binary variable may appear normal (e.g., x) or complemented (e.g., x), there are 2n maxterms for n variables. Example: Two variables (X and Y) produce 22 = 4 combinations: (both normal) (x normal, y complemented) (x complemented, y normal) (both complemented) Y X +

Maxterms and Minterms Index Minterm Maxterm 0 (00) x y x + y 1 (01) Examples: Two variable minterms and maxterms. The index above is important for describing which variables in the terms are true and which are complemented. Index Minterm Maxterm 0 (00) x y x + y 1 (01) 2 (10) 3 (11)

Standard Order Minterms and maxterms are designated with a subscript The subscript is a number, corresponding to a binary pattern The bits in the pattern represent the complemented or normal state of each variable listed in a standard order. All variables will be present in a minterm or maxterm and will be listed in the same order (usually alphabetically) Example: For variables a, b, c: Maxterms: (a + b + c), (a + b + c) Terms: (b + a + c), a c b, and (c + b + a) are NOT in standard order. Minterms: a b c, a b c, a b c Terms: (a + c), b c, and (a + b) do not contain all variables

Purpose of the Index The index for the minterm or maxterm, expressed as a binary number, is used to determine whether the variable is shown in the true form 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”.

Index Example in Three Variables Example: (for three variables) Assume the variables are called X, Y, and Z. The standard order is X, then Y, then Z. The index 0 (base 10) = 000 (base 2) for three variables). All three variables are complemented for minterm 0 ( ) and no variables are complemented for Maxterm 0 (X,Y,Z). Minterm 0, called m0 is . Maxterm 0, called M0 is (X + Y + Z). Minterm 6 ? Maxterm 6 ? Z , Y X Z Y X m6 = X Y Z’ M6 = (X’ + Y’ + Z)

Index Examples – Four Variables Index Binary Minterm Maxterm i Pattern mi Mi 0 0000 1 0001 3 0011 5 0101 7 0111 10 1010 13 1101 15 1111 a b c d a + b + c + d a b c d ? ? a + b + c + d a b c d a + b + c + d ? a + b + c + d M1 = a + b + c + d’ m3 = a’ b’ c d m7 = a’ b c d M 13 = a’ + b’ + c + d’ a b c d a + b + c + d a b c d ? a b c d a + b + c + d

Minterm and Maxterm Relationship Review: DeMorgan's Theorem and Two-variable example: Thus M2 is the complement of m2 and vice-versa. Since DeMorgan's Theorem holds for n variables, the above holds for terms of n variables giving: Thus Mi is the complement of mi. x · y = x + y x + y = x . y M = x + & y m = x· y 2 2 i m M = &

Function Tables for Both Minterms of Maxterms of 2 variables 2 variables Each column in the maxterm function table is the complement of the column in the minterm function table since Mi is the complement of mi. x y m 1 2 3 0 0 0 1 1 0 1 1 x y M 1 2 3 0 0 0 1 1 0 1 1

Observations In the function tables: Each minterm has one and only one 1 present in the 2n terms (a minimum of 1s). All other entries are 0. Each maxterm has one and only one 0 present in the 2n terms All other entries are 1 (a maximum of 1s). We can implement any function by "ORing" the minterms corresponding to "1" entries in the function table. These are called the minterms of the function. We can implement any function by "ANDing" the maxterms corresponding to "0" entries in the function table. These are called the maxterms of the function. This gives us two canonical forms: Sum of Minterms (SOM) Product of Maxterms (POM) for stating any Boolean function.

Minterm Function Example Example: Find F1 = m1 + m4 + m7 F1 = x y z + x y z + x y z x y z index m1 + m4 m7 = F1 0 0 0 = 0 0 0 1 1 = 1 0 1 0 2 0 1 1 3 1 0 0 4 1 0 1 5 1 1 0 6 1 1 1 7

Maxterm Function Example Example: Implement F1 in maxterms: F1 = M0 · M2 · M3 · M5 · M6 ) z y z)·(x ·(x z) (x F 1 + = x )·( ·( x y z i M0  M2 M3 M5 M6 = F1 0 0 0 1 = 0 0 0 1 = 1 0 1 0 2 0 1 1 3 1 0 0 4 1 0 1 5 1 1 0 6 1 1 1 7

Canonical Sum of Minterms Any Boolean function can be expressed as a Sum of Minterms. For the function table, the minterms used are the terms corresponding to the 1's For expressions, expand all terms first to explicitly list all minterms. Do this by “ANDing” any term missing a variable v with a term ( ). Example: Implement as a sum of minterms (prod terms). First expand terms: Then distribute terms: Express as sum of minterms: f = m3 + m2 + m0 v + y x f + = f = x ( y + y ) + x y f = xy + x y + x y

Another SOM Example C B A F + = Example: There are three variables, A, B, and C which we take to be the standard order. Expanding the terms with missing variables: A B C 1 x x (where x is a don’t care condition) x 0 1 1xx  100, 101, 110, 111  (4, 5, 6, 7) x01  001, 101  (1, 5) F = A(B + B’)(C + C’) + (A + A’) B’ C = ABC + ABC’ + AB’C + AB’C’ + AB’C + A’B’C = ABC + ABC’ + AB’C + AB’C’ + A’B’C = m7 + m6 + m5 + m4 + m1 = m1 + m4 + m5 + m6 + m7

Shorthand SOM Form Collect terms (removing all but one of duplicate terms):  (1, 4, 5, 6, 7) F = m1+m4+m5+m6+m7 This can be denoted in the formal shorthand: Note that we explicitly show the standard variables in order and drop the “m” designators.

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 forms. 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 ) 7 , 5 3 1 ( z y x F m S = ( x , y , z ) = S F ( , 2 , 4 , 6 ) m P F ( x , y , z ) = ( 1 , 3 , 5 , 7 ) M

Conversion Between Forms To convert between sum-of-minterms and product-of- maxterms form (or vice-versa) we follow these steps: Find the function complement by swapping terms in the list with terms not in the list. Change from products to sums, or vice versa. Example:Given F as before: Form the Complement: Then use the other form with the same indices – this forms the complement again, giving the other form of the original function: ) 6 , 4 2 ( z y x F m S =

Standard Forms A B C + A B C + B 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: POS: These “mixed” forms are neither SOP nor POS A B C + A B C + B (A + B) · (A + B + C ) · C (A · B + C) (A + C) A B C + A C (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, and The second level is a single OR gate (with fewer than 2n inputs). This form often can be simplified so that the corresponding circuit is simpler.

Standard Sum-of-Products (SOP) A Simplification Example: Writing the minterm expression: F = A B C + A B C + A B C + ABC + ABC Simplifying: F = A + B C Simplified F contains 3 literals compared to 15 in minterm F F = A’ B’ C + A (B’ C’ + B C’ + B’ C + B C) = A’ B’ C + A (B’ + B) (C’ + C) = A’ B’ C + A.1.1 = A’ B’ C + A = B’C + A

AND/OR Two-level Implementation of SOP Expression The two implementations for F are shown below – 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 two-level implementations Questions: How can we attain a “simplest” expression? Is there only one minimum cost circuit? The next part will deal with these issues.

Karnaugh Maps (K-map) A K-map is a collection of squares Each square represents a minterm The collection of squares is a graphical representation of a Boolean function Adjacent squares differ in the value of one variable Alternative algebraic expressions for the same function are derived by recognizing patterns of squares The K-map can be viewed as A reorganized version of the truth table A topologically-warped Venn diagram as used to visualize sets in algebra of sets

Some Uses of K-Maps Provide a means for: Finding optimum or near optimum SOP and POS standard forms, and two-level AND/OR and OR/AND circuit implementations for functions with small numbers of variables Visualizing concepts related to manipulating Boolean expressions, and Demonstrating concepts used by computer-aided design programs to simplify large circuits

y x a b c d K-Map K-Map and Truth Tables Input Values (x,y) Function The K-Map is just a different form of the truth table. Example – Two variable function: We choose a,b,c and d from the set {0,1} to implement a particular function, F(x,y). Function Table K-Map Input Values (x,y) Function Value F(x,y) 0 0 a 0 1 b 1 0 c 1 1 d y x a b c d

Two-variable Maps A 2-variable Karnaugh Map: Note that minterm m0 and minterm m1 are “adjacent” and differ in the value of the variable y Similarly, minterm m0 and minterm m2 differ in the x variable. Also, m1 and m3 differ in the x variable as well. Finally, m2 and m3 differ in the value of the variable y

Two-Variable Function F(A, B) 1 1 1

K-Map Function Representation Example: F(x,y) = x For function F(x,y), the two adjacent cells containing 1’s can be combined using the Minimization Theorem: 1 0 1 1 On a 2-variable K-Map: One square represents a minterm with two variables Two adjacent squares represent a product term with one variable. y x 1 x x y ) , ( F = +

K-Map Function Representation Example: G(x,y) = x + y For G(x,y), two pairs of adjacent cells containing 1’s can be combined using the Minimization Theorem: 1 y y x x ( ) ( ) y x xy ) , ( G + = Duplicate x y

Three-variable & Four-variable Maps

Three Variable Maps A three-variable K-map: Where each minterm corresponds to the product terms: Note that if the binary value for an index differs in one bit position, the minterms are adjacent on the K-Map yz=00 yz=01 yz=11 yz=10 x=0 m0 m1 m3 m2 x=1 m4 m5 m7 m6 yz=00 yz=01 yz=11 yz=10 x=0 x=1 z y x

Alternative Map Labeling Map use largely involves: Entering values into the map, and Reading off product terms from the map. Alternate labelings are useful: y z y y y x z x 1 2 4 3 5 6 7 y 00 01 11 10 x 1 3 2 x 1 4 5 7 6 z z z

Alternative Map Labeling F = x F = x y z x 1 2 4 3 5 6 7 y z x 1 2 4 3 5 6 7 y z x 1 y z x 1 y

Alternative Map Labeling y y z x 1 y z x 1 y y y z x 1 y z x 1 y

Example Functions y x z y x z By convention, we represent the minterms of F by a "1" in the map and leave the minterms blank Example: Learn the locations of the 8 indices based on the variable order shown (x, most significant and z, least significant) on the map boundaries y x 1 z 00 01 11 10 1 x y 1 z 00 01 11 10 1

Combining Squares By combining squares, we reduce number of literals in a product term, reducing the literal cost, thereby reducing the other two cost criteria On a 3-variable K-Map: One square represents a minterm with three variables Two adjacent squares represent a product term with two variables Four “adjacent” terms represent a product term with one variable Eight “adjacent” terms is the function of all ones (no variables) = 1.

Example: Combining Squares y 1 2 4 3 5 6 7 z Example: Let Applying the Minimization Theorem three times: Thus the four terms that form a 2 × 2 square correspond to the term "y". z y x ) , ( F + = z y yz + = y =

Three-Variable Maps Reduced literal product terms for SOP standard forms correspond to rectangles on K-maps containing cell counts that are powers of 2. Rectangles of 2 cells represent 2 adjacent minterms and will have two variables; Rectangles of 4 cells represent 4 minterms and will have one variable. Rectangles of 1 cell has all three variables.

Minterm vs. Variable Adjacent cells in a rectangle dictate the final reduced number of variable(s) in a minterm. total # vars # cells in rectangle final reduced # var 2 2 1 (e.g., F = y) (e.g., F = xy) 1 2 (e.g., F = xy) 3 4 1 (e.g., F = xyz) 2 2 1 3 (e.g., F = xyz) 4 8 1 (e.g., F = wxyz) 4 2 2 3 1 4

Three Variable Maps F(x, y, z) = z + xy y 1 x z K-Maps can be used to simplify Boolean functions by systematic methods. Terms are selected to cover the “1s”in the map, e.g., to simplify y 1 x z 00 01 11 10 1 F(x, y, z) = z + xy

Three-Variable Maps y x z Example Shapes of 2-cell Rectangles: Read off the product terms for the rectangles shown. Answer: yz + xz + xy y 00 01 11 10 1 x z 1 1 1 1 1

y x z Three-Variable Maps Example Shapes of 2-cell Rectangles: Read off the product terms for the rectangles shown. Answer: yz + xz + xy y 1 x z 00 01 11 10 1 1 1 1 1

y y z x x z z Three-Variable Maps Example Shapes of 4-cell Rectangles: Read off the product terms for the rectangles shown y y 1 3 2 5 6 4 7 z x x z z

y y 1 z x 1 1 1 x 1 1 z z Three-Variable Maps Break into groups of two and minimize: y y 1 z x 1 1 1 x 1 1 z z

Three-Variable Maps y y 1 z x 1 1 1 x 1 1 z z

Three-Variable Maps y y 1 z x 1 1 1 x 1 1 z z

Four Variable Terms Four variable maps can have rectangles corresponding to: A single 1 = 4 variables, (i.e. Minterm) Two 1s = 3 variables, Four 1s = 2 variables Eight 1s = 1 variable, Sixteen 1s = zero variables (i.e. Constant "1")

F = xz Four-Variable Maps 8 9 10 11 12 13 14 15 1 3 2 5 6 4 7 X Y Z W A single 1 = 4 variables, (i.e. Minterm) Two 1s = 3 variables, Four 1s = 2 variables Eight 1s = 1 variable F = xz 8 9 10 11 12 13 14 15 1 3 2 5 6 4 7 X Y Z W

F = xz Four-Variable Maps Example Shapes of Rectangles: 8 9 10 11 12 13 14 15 1 3 2 5 6 4 7 X Y Z W F = xz

Four-Variable Maps Example Shapes of Rectangles: Y 8 9 10 11 12 13 14 15 1 3 2 5 6 4 7 X W Z

Four-Variable Map Example f = wxyz + wxyz + wxyz + wxyz X Y Z W 00 01 11 10 00 01 11 10 1 1 1 1 Four 1s = 2 variables f = x y

Four-Variable Maps Example

Four-Variable Map Simplification Break into groups of two and minimize: F(W, X, Y, Z) = ∑ (3,4,5,7,9,1 3,14,15 ) m F = XZ + X'Z'

Original and Simplified Circuits Equivalent circuit after K-Map reduction Logic circuit from truth table

Case Study: Seven Segment Display

Case Study (cont.) L1 L 6 L2 L3 L 7 L 4 L 5

Case Study (cont.) Some gate level implementation Implement L4: Some gate level implementation of the Boolean function for L4

Systematic Simplification A Prime Implicant is a product term (minterm) obtained by combining the maximum possible number of adjacent squares in the map into a rectangle with the number of squares a power of 2. A prime implicant is called an Essential Prime Implicant if it is the only prime implicant that covers (includes) one or more minterms. Prime Implicants and Essential Prime Implicants can be determined by inspection of a K-Map. A set of prime implicants "covers all minterms" if, for each minterm of the function, at least one prime implicant in the set of prime implicants includes the minterm.

Example of Prime Implicants Find ALL Prime Implicants CD Minterms covered by single prime implicant C 1 B D A D B C B BD AD B A

Example of Prime Implicants ESSENTIAL Prime Implicants (EPI) ALL Prime Implicants PI that includes minterms not belonging to more than one PI CD C 1 B D A D B D B 1 B C D A C B BD BD AD B A minterms that do not belong to more than one PI

Prime Implicants (PI) PI: 7 Essential PI: 2

Prime Implicants (PI) PI: 6 Essential PI: 2

Prime Implicants (PI) PI: 5 Essential PI: 3

Prime Implicants (PI) PI: 4 Essential PI: 4

Prime Implicants (PI) PI: 8 Essential PI: 0

Homework Simplify: Find all prime implicants of: Determine all essential prime implicants too. Hint: There are seven prime implicants! F(W, X, Y, Z) = ∑ (0, 2,4,5,6,7, 8,10,13,15 ) m Prime implicants are: A, B'C, and B'D'

Don't Cares in K-Maps Sometimes a function table or map contains entries for which it is known: the input values for the minterm will never occur, or The output value for the minterm is not used In these cases, the output value need not be defined Instead, the output value is defined as a “don't care” By placing “don't cares” ( an “x” entry) in the function table or map, the cost of the logic circuit may be lowered. Example 1: A logic function having the binary codes for the BCD digits as its inputs. Only the codes for 0 through 9 are used. The six codes, 1010 through 1111 never occur, so the output values for these codes are “x” to represent “don’t cares.”