chapter 2: basic definitions bb inary operators ●a●and z = x y = x yz=1 if x=1 and y=1 ●o●or...
TRANSCRIPT
Chapter 2:Chapter 2:
Basic DefinitionsBasic Definitions
Binary Operators
● AND
z = x • y = x y z=1 if x=1 AND y=1
● OR
z = x + y z=1 if x=1 OR y=1
● NOT
z = x = x’ z=1 if x=0
Boolean Algebra
● Binary Variables: only ‘0’ and ‘1’ values
● Algebraic Manipulation
Boolean Algebra PostulatesBoolean Algebra Postulates
Commutative Law
x • y = y • x x + y = y + x
Identity Element
x • 1 = x x + 0 = x
Complement
x • x’ = 0 x + x’ = 1
Boolean Algebra TheoremsBoolean Algebra Theorems
Duality
● 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 )
● x + ( y • z ) = ( x + y ) • ( x + z )
Theorem 1
● x • x = x x + x = x
Theorem 2
● x • 0 = 0 x + 1 = 1
Applied to a valid equation produces a valid equation
Boolean Algebra TheoremsBoolean 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 PrecedenceOperator Precedence
Parentheses
( . . . ) • ( . . .)
NOT
x’ + y
AND
x + x • y
OR
])([ xwzyx
])([
)(
)(
)(
)(
xwzyx
xwzy
xwz
xw
xw
DeMorgan’s TheoremDeMorgan’s Theorem
)]([ edcba
)]([ edcba
))(( edcba
))(( edcba
))(( edcba
)( edcba
Boolean FunctionsBoolean Functions
Boolean Expression
Example: F = x + y’ z
Truth Table
All possible combinationsof input variables
Logic Circuit
x y z F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
xyz
F
Algebraic ManipulationAlgebraic Manipulation
Literal:
A single variable within a term that may be complemented or not.
Use 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 FunctionComplement of a Function
DeMorgan’s Theorm
Duality & Literal Complement
CBAF
CBAF
CBAF
CBAF CBAF
CBAF
Standard FormsStandard 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 MaxtermsMinterms 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 2n possible minterms and 2n 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
MintermsMinterms
Minterms for two variables F(X,Y)
X Y Product Terms
Symbol m0 m1 m2 m3
0 0 X’Y’ m0 1 0 0 0
0 1 X’Y m1 0 1 0 0
1 0 XY’ m2 0 0 1 0
1 1 XY m3 0 0 0 1
Variable complemented if 0
Variable uncomplemented if 1
mi indicated the ith minterm
For each binary combination of X and Y there is a minterm
The index of the minterm is specified by the binary combination
mi is equal to 1 for ONLY THAT combination
MaxtermsMaxterms
Maxterms for two variables F(X,Y)
X Y Sum
TermsSymbol M0 M1 M2 M3
0 0 X+Y M0 0 1 1 1
0 1 X+Y’ M1 1 0 1 1
1 0 X’+Y M2 1 1 0 1
1 1 X’+Y’ M3 1 1 1 0
Variable complemented if 1
Variable not complemented if 0
Mi indicated the ith maxterm
For each binary combination of X and Y there is a maxterm
The index of the maxterm is specified by the binary combination
Mi is equal to 0 for ONLY THAT combination
ExampleExample
• 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:X Y Z m F0 0 0 m0 1
0 0 1 m1 0
0 1 0 m2 1
0 1 1 m3 0
1 0 0 m4 0
1 0 1 m5 1
1 1 0 m6 0
1 1 1 m7 1
Consider the function defined by the truth table
F(X,Y,Z) 3 variables 8 mintermsF can be written asF = X’Y’Z’ + X’YZ’ + XY’Z + XYZ, or = m0 + m2 + m5 + m7
= m(0,2,5,7)
Minterms and Maxterms Minterms and Maxterms
• In general, a function of n variables has• 2n minterms: m0, m1, …, m2
n-1
• 2n maxterms: M0, M1, …, M2n
-1
• mi’ = Mi or Mi’ = mi
Example: for F(X,Y): m2 = XY’ m2’ = X’+Y = M2
Example (Cont.)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
X Y Z M F F’0 0 0 M0 1 0
0 0 1 M1 0 1
0 1 0 M2 1 0
0 1 1 M3 0 11 0 0 M4 0 11 0 1 M5 1 0
1 1 0 M6 0 1
1 1 1 M7 1 0
• Example:Consider the function defined by the
truth tableF(X,Y,Z) in a manner similar to
the previous example, F’ can be written as
F’ = m1 + m3 + m4 + m6
= m(1,3,4,6)Now apply DeMorgan’s rule F = F’’ = [m1 + m3 + m4 + m6]’ = m1’.m3’.m4’.m6’ = M1.M3.M4.M6
= 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 Expressing Functions with Minterms and Maxtermsand Maxterms
F = X . ( Y’ + Z)
X Y Z F=X. (Y’+Z)
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 1
F = XY’Z’ + XY’Z + XYZ
= m4 + m5 + m7
= m(4,5,7)
F = M(0,1,2,3,6)
SummarySummary
• 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• mi – for all minterms that produce 1 in the table,
or Mi – for all maxterms that produce 0 in the table
• Another way to obtain the mi or Mi is to use ALGEBRA – see next example
ExampleExample
• Write E = Y’ + X’Z’ in the form of mi and Mi?
• Solution: Method1 First construct the
Truth Table as shown
Second:E = m(0,1,2,4,5), andE = M(3,6,7)
X Y Z m M E0 0 0 m0 M0 1
0 0 1 m1 M1 1
0 1 0 m2 M2 1
0 1 1 m3 M3 0
1 0 0 m4 M4 1
1 0 1 m5 M5 1
1 1 0 m6 M6 0
1 1 1 m7 M7 0
Example (Cont.)Example (Cont.)
Solution: Method2_aE = 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’= m5 + m1 + m4 + m0 + m2 + m0
= m0 + m1 + m2 + m4 + m5
= m(0,1,2,4,5)
To find the form Mi, consider the remaining indices
E = M(3,6,7)
Solution: Method2_bE = 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’)= M7 . M6 . M3 = M(3,6,7)
To find the form mi, consider the remaining indices
E = m(0,1,2,4,5)
SOP vs POSSOP 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 POSSOP 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+BCHint 1: Used (X+Y)(X+Z)=X+YZ
Hint 2: Multiply
Implementation of SOPImplementation of SOP
Any SOP expression can be implemented using 2-levels of gates
The 1st level consists of AND gates, and the 2nd level consists of a single OR gate
Also called 2-level Circuit
Implementation of POSImplementation of POS
Any POS expression can be implemented using 2-levels of gates
The 1st level consists of OR gates, and the 2nd level consists of a single AND gate
Also called 2-level Circuit
Implementation of SOPImplementation 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
FC
B
A
E
D
3-level circuit 2-level circuit
Logic OperatorsLogic Operators
AND
NAND (Not AND) x y NAND0 0 10 1 11 0 11 1 0
x y AND0 0 00 1 01 0 01 1 1
xy
x • y
xy
x • y
Logic OperatorsLogic Operators
OR
NOR (Not OR)
x y OR0 0 00 1 11 0 11 1 1
x y NOR0 0 10 1 01 0 01 1 0
xy
x + y
xy
x + y
Logic OperatorsLogic Operators
XOR (Exclusive-OR)
XNOR (Exclusive-NOR) (Equivalence)
x y XOR0 0 00 1 11 0 11 1 0
x y XNOR0 0 10 1 01 0 01 1 1
xy
x Å yx � y
x y + x y
xy
x Å yx y + x y
Logic OperatorsLogic Operators
NOT (Inverter)
Buffer
x NOT
0 1
1 0
x Buffer
0 0
1 1
x x
x x
DeMorgan’s Theorem on GatesDeMorgan’s Theorem on Gates
AND Gate
● F = x • y F = (x • y) F = x + y
OR Gate
● F = x + y F = (x + y) F = x • y
Change the “Shape” and “bubble” all lines
AssignmentsAssignments
Mano
● Chapter 2
♦ 2-4
♦ 2-5
♦ 2-6
♦ 2-8
♦ 2-9
♦ 2-10
♦ 2-12
♦ 2-15
♦ 2-18
♦ 2-19
AssignmentsAssignments
Mano2-4 Reduce the following Boolean expressions to the indicated
number of literals:
(a) A’C’ + ABC + AC’ to three literals
(b) (x’y’ + z)’ + z + xy + wz to 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
2-5 Find the complement of F = x + yz; then show thatFF’ = 0 and F + F’ = 1
AssignmentsAssignments
2-6 Find the complement of the following expressions:
(a) xy’ + x’y (b) (AB’ + C)D’ + E
(c) (x + y’ + z) (x’ + z’) (x + y)
2-8 List the truth table of the function:F = xy + xy’ + y’z
2-9 Logical 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 stringsA = 10101101 and B = 10001110, evaluate the 8-bit result after the following logical operations: (a) AND, (b) OR, (c) XOR, (d) NOT A, (e) NOT B.
AssignmentsAssignments
2-10 Draw 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)
2-12 Simplify the Boolean function T1 and T2 to a minimum number of literals.
A B C T1 T2
0 0 0 1 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 0 11 0 1 0 11 1 0 0 11 1 1 0 1
AssignmentsAssignments
2-15 Given the Boolean functionF = 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).
AssignmentsAssignments
2-18 Convert 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)
2-19 Convert the following expressions into sum of products and product of sums:
(a) (AB + C) (B + C’D)
(b) x’ + x (x + y’) (y + z’)