Chapter 2:

Basic Definitions BB inary Operators ●A●AND z = x y = x yz=1 if x=1 AND y=1 ●O●OR z = x + yz=1 if x=1 OR y=1 ●N●NOT z = x = x’ z=1 if x=0 BB oolean Algebra ●B●Binary Variables: only ‘0’ and ‘1’ values ●A●Algebraic Manipulation

Boolean Algebra Postulates CC ommutative Law x y = y xx + y = y + x II dentity Element x 1 = x x + 0 = x CC omplement x x’ = 0 x + x’ = 1

Boolean Algebra Theorems DD uality ●T●The dual of a Boolean algebraic expression is obtained by interchanging the AND and the OR operators and replacing the 1’s by 0’s and the 0’s by 1’s. ● x ( y + z ) = ( x y ) + ( x z ) + ( y z ) = ( x + y ) ( x + z ) TT heorem 1 x = xx + x = x TT heorem 2 0 = 0x + 1 = 1 Applied to a valid equation produces a valid equation

Boolean Algebra Theorems  Theorem 3: Involution ● ( x’ )’ = x ( x ) = x  Theorem 4: Associative & Distributive ● ( x y ) z = x ( y z )( x + y ) + z = x + ( y + z ) ● x ( y + z ) = ( x y ) + ( x z ) x + ( y z ) = ( x + y ) ( x + z )  Theorem 5: DeMorgan ● ( x y )’ = x’ + y’ ( x + y )’ = x’ y’ ● ( x y ) = x + y ( x + y ) = x y  Theorem 6: Absorption ● x ( x + y ) = x x + ( x y ) = x

Operator Precedence  Parentheses (... ) (...)  NOT x’ + y  AND x + x y  OR

DeMorgan’s Theorem

Boolean Functions BB oolean Expression Example:F = x + y’ z TT ruth Table All possible combinations of input variables LL ogic Circuit xyzF

Algebraic Manipulation LL iteral: A single variable within a term that may be complemented or not. UU se Boolean Algebra to simplify Boolean functions to produce simpler circuits Example: Simplify to a minimum number of literals F = x + x’ y( 3 Literals) = x + ( x’ y ) = ( x + x’ ) ( x + y ) = ( 1 ) ( x + y ) = x + y( 2 Literals) Distributive law (+ over )

Complement of a Function  DeMorgan’s Theorm  Duality & Literal Complement

Standard Forms A Boolean function can be written algebraically in a variety of ways Standard form: is a standard algebraic expression of the function: Help simplification procedures and frequently results in more desirable logic circuits (e.g. less number of gates) Standard form: contains product terms and sum terms Product term: X’Y’Z (logical product with AND) Sum term: X + Y + Z’ (logical sum with OR)

Minterms and Maxterms A minterm: a product term in which all variables (or literals) of the function appear exactly once (complemented or not complemented) A maxterm: a sum term in which all the variables (or literals) of the function appear exactly once (complemented or not complemented) A function of n variables – have 2 n possible minterms and 2 n possible maxterms Example: for the function F(X,Y,Z), the term X’Y is not a minterm, but XYZ’ is a minterm The term X’+Z is not a maxterm, but X+Y’+Z’ is maxterm

Minterms Minterms for two variables F(X,Y) XY Product Terms Symbolm0m0 m1m1 m2m2 m3m3 00X’Y’m0m X’Ym1m XY’m2m XYm3m Variable complemented if 0 Variable uncomplemented if 1 m i indicated the i th minterm For each binary combination of X and Y there is a minterm The index of the minterm is specified by the binary combination m i is equal to 1 for ONLY THAT combination

Maxterms Maxterms for two variables F(X,Y) XYSum TermsSymbolM0M0 M1M1 M2M2 M3M3 00X+YM0M X+Y’M1M X’+YM2M X’+Y’M3M Variable complemented if 1 Variable not complemented if 0 M i indicated the i th maxterm For each binary combination of X and Y there is a maxterm The index of the maxterm is specified by the binary combination M i is equal to 0 for ONLY THAT combination

Example A Boolean function can be expressed algebraically from a give truth table by forming the logical sum of ALL the minterms that produce 1 in the function Example: XYZmF 000m0m m1m m2m m3m m4m m5m m6m m7m7 1 Consider the function defined by the truth table F(X,Y,Z)  3 variables  8 minterms F can be written as F = X’Y’Z’ + X’YZ’ + XY’Z + XYZ, or = m 0 + m 2 + m 5 + m 7 =  m(0,2,5,7)

Minterms and Maxterms In general, a function of n variables has 2 n minterms: m 0, m 1, …, m 2 n -1 2 n maxterms: M 0, M 1, …, M 2 n -1 m i ’ = M i or M i ’ = m i  Example: for F(X,Y):  m 2 = XY’  m 2 ’ = X’+Y = M 2

Example (Cont.) A Boolean function can be expressed algebraically from a give truth table by forming the logical product of ALL the maxterms that produce 0 in the function XYZMFF’ 000M0M M1M M2M M3M M M5M M6M M7M7 10 Example: Consider the function defined by the truth table F(X,Y,Z)  in a manner similar to the previous example, F’ can be written as F’ = m 1 + m 3 + m 4 + m 6 =  m(1,3,4,6) Now apply DeMorgan’s rule F = F’’ = [m 1 + m 3 + m 4 + m 6 ]’ = m 1 ’.m 3 ’.m 4 ’.m 6 ’ = M 1.M 3.M 4.M 6 =  M(1,3,4,6) Note the indices in this list are those that are missing from the previous list in  m(0,2,5,7)

Expressing Functions with Minterms and Maxterms F = X. ( Y’ + Z) XYZ F = XY’Z’ + XY’Z + XYZ = m 4 + m 5 + m 7 =  m(4,5,7) F=  M(0,1,2,3,6)

Summary A Boolean function can be expressed algebraically as: The logical sum of minterms The logical product of maxterms Given the truth table, writing F as  m i – for all minterms that produce 1 in the table, or  M i – for all maxterms that produce 0 in the table Another way to obtain the  m i or  M i is to use ALGEBRA – see next example

Example Write E = Y’ + X’Z’ in the form of  m i and  M i ? Solution: Method1 First construct the Truth Table as shown Second: E =  m(0,1,2,4,5), and E =  M(3,6,7) XYZmME 000m0m0 M0M m1m1 M1M m2m2 M2M m3m3 M3M m4m4 M41 101m5m5 M5M m6m6 M6M m7m7 M7M7 0

Example (Cont.) Solution: Method2_a E = Y’ + X’Z’ = Y’(X+X’)(Z+Z’) + X’Z’(Y+Y’) = (XY’+X’Y’)(Z+Z’) + X’YZ’+X’Z’Y’ = XY’Z+X’Y’Z+XY’Z’+X’Y’Z’+ X’YZ’+X’Z’Y’ = m 5 + m 1 + m 4 + m 0 + m 2 + m 0 = m 0 + m 1 + m 2 + m 4 + m 5 =  m(0,1,2,4,5) To find the form  Mi, consider the remaining indices E =  M(3,6,7) Solution: Method2_b E = Y’ + X’Z’ E’ = Y(X+Z) = YX + YZ = YX(Z+Z’) + YZ(X+X’) = XYZ+XYZ’+X’YZ E= (X’+Y’+Z’)(X’+Y’+Z)(X+Y’+Z’ ) = M 7. M 6. M 3 =  M(3,6,7) To find the form  m i, consider the remaining indices E =  m(0,1,2,4,5)

SOP vs POS Boolean function can be represented as Sum of Products (SOP) or as Product of Sums (POS) of their input variables AB+CD = (A+C)(B+C)(A+D)(B+D) The sum of minterms is a special case of the SOP form, where all product terms are minterms The product of maxterms is a special case of the POS form, where all sum terms are maxterms SOP and POS are referred to as the standard forms for representing Boolean functions

SOP vs POS  SOP  POS  F = AB + CD  = (AB+C)(AB+D)  = (A+C)(B+C)(AB+D)  = (A+C)(B+C)(A+D)(B+D)  Hint 1: Used X+YZ=(X+Y)(X+Z)  Hint 2: Factor POS  SOP F = (A’+B)(A’+C)(C+D) = (A’+BC)(C+D) = A’C+A’D+BCC+BCD = A’C+A’D+BC+BCD = A’C+A’D+BC Hint 1: Used (X+Y)(X+Z)=X+YZ Hint 2: Multiply

Implementation of SOP  Any SOP expression can be implemented using 2-levels of gates  The 1 st level consists of AND gates, and the 2 nd level consists of a single OR gate  Also called 2-level Circuit

Implementation of POS  Any POS expression can be implemented using 2-levels of gates  The 1 st level consists of OR gates, and the 2 nd level consists of a single AND gate  Also called 2-level Circuit

Implementation of SOP Consider F = AB + C(D+E) This expression is NOT in the sum-of-products form Use the identities/algebraic manipulation to convert to a standard form (sum of products), as in F = AB + CD + CE Logic Diagrams: F E C D C B A F C B A E D 3-level circuit2-level circuit

Logic Operators  AND  NAND (Not AND) xyNAND xyAND

Logic Operators  OR  NOR (Not OR) xyOR xyNOR

Logic Operators  XOR (Exclusive-OR)  XNOR (Exclusive-NOR) (Equivalence) xyXOR xyXNOR

Logic Operators  NOT (Inverter)  Buffer xNOT xBuffer 00 11

DeMorgan’s Theorem on Gates  AND Gate ●F = x yF = (x y) F = x + y  OR Gate ●F = x + yF = (x + y) F = x y  Change the “Shape” and “bubble” all lines

Assignments  Mano ●Chapter 2 ♦ 2-4 ♦ 2-5 ♦ 2-6 ♦ 2-8 ♦ 2-9 ♦ 2-10 ♦ 2-12 ♦ 2-15 ♦ 2-18 ♦ 2-19

Assignments  Mano 2-4Reduce the following Boolean expressions to the indicated number of literals: (a) A’C’ + ABC + AC’to three literals (b) (x’y’ + z)’ + z + xy + wzto three literals (c) A’B (D’ + C’D) + B (A + A’CD)to one literal (d) (A’ + C) (A’ + C’) (A + B + C’D)to four literals 2-5Find the complement of F = x + yz; then show that FF’ = 0 and F + F’ = 1

Assignments 2-6Find the complement of the following expressions: (a) xy’ + x’y(b) (AB’ + C)D’ + E (c) (x + y’ + z) (x’ + z’) (x + y) 2-8List the truth table of the function: F = xy + xy’ + y’z 2-9Logical operations can be performed on strings of bits by considering each pair of corresponding bits separately (this is called bitwise operation). Given two 8-bit strings A = and B = , evaluate the 8-bit result after the following logical operations: (a) AND, (b) OR, (c) XOR, (d) NOT A, (e) NOT B.

Assignments 2-10Draw the logic diagrams for the following Boolean expressions: (a) Y = A’B’ + B (A + C)(b) Y = BC + AC’ (c) Y = A + CD(d) Y = (A + B) (C’ + D) 2-12Simplify the Boolean function T 1 and T 2 to a minimum number of literals. ABCT1T1 T2T

Assignments 2-15Given the Boolean function F = xy’z + x’y’z + w’xy + wx’y + wxy (a) Obtain the truth table of the function. (b) Draw the logic diagram using the original Boolean expression. (c) Simplify the function to a minimum number of literals using Boolean algebra. (d) Obtain the truth table of the function from the simplified expression and show that it is the same as the one in part (a) (e) Draw the logic diagram from the simplified expression and compare the total number of gates with the diagram of part (b).

Assignments 2-18Convert the following to the other canonical form: (a) F (x, y, z) = ∑ (1, 3, 7) (b) F (A, B, C, D) = ∏ (0, 1, 2, 3, 4, 6, 12) 2-19Convert the following expressions into sum of products and product of sums: (a) (AB + C) (B + C’D) (b) x’ + x (x + y’) (y + z’)