--- outputs logical functions of inputs --- new outputs appear shortly after changed inputs (propagation delay) --- no feedback loops --- no clock Sequential

--- outputs logical functions of inputs--- new outputs appear shortly after changed inputs

(propagation delay)--- no feedback loops--- no clock

Sequential logic--- outputs logical functions of inputs and previous

history of circuit (memory)--- after changed inputs, new outputs appear in

the next clock cycle--- frequent feedback loops

Combinational logic

Fundamentals of Boolean algebra

Named after George Boole He presented an algebraic formulation of the

process of “logical thought and reason”

This formulation come to be known as Boolean Algebra

Postulates of Boolean algebra

1) Definition A Boolean algebra is a closed algebraic

system containing a set K of two or more elements and the two operators ‘•’ or ‘/\’ or ‘’, called AND, and ‘+’ or ‘\/’ or ‘’, called OR;

Closed system: for every a and b in set K, a•b belongs to K and a+b belongs to K.


2) Existence of 1 and 0 There exist unique elements 1 (one) and 0

(zero) in set A" such that for every a in Ka) a + 0 = a,

b) a • 1 = a, where 0 is the identity element for the +

operation and 1 is the identity element for the • operation.


3) Commutativity of the + and • operations For every a and b in K

a) a + b = b + a.

b) a • b = b • a

4) Associativity of the + and operations For every a, b, and c in K

a) a + (b + c) = (a + b) + c.

b) a • (b • c) = (a • b) • c.


5) Distributivity of + over • and • over + For every a, b, and c in K

a) a + (b • c) = (a + b) • (a + c),

b) a • (b + c) = (a • b) + (a • c).

6) Existence of the complement For every a in K there exists a unique

element called ā (complement of a) in K such that

a) a + ā = 1.

b) a • ā = 0.

Venn diagrams for the postulates• Operations on sets

Sets closed regionsSets correspond to elementsIntersection corresponds to •Union corresponds to +

Venn diagrams for the postulates

Venn diagramsExamples of Venn diagrams

Venn diagramsa + b • c = (a + b) • (a + c)

Venn diagrams

a + b • c = (a + b) • (a + c)

Boolean algebra

• Duality– If an expression

f(x1, x2, … xn, +, •, 0, 1)

is valid, thenf(x1, x2, … xn, •, +, 1, 0)

obtained by interchanging + and •, 0 and 1 is also valid

a • (b + c) = (a • b) + (a • c)

a + (b • c) = (a + b) • (a + c)

Postulates 2 – 6 are stated in dual form

Fundamental theorems of Boolean algebra

– Prove part (b) by exchanging + with •, and use the dual form of the postulates



a • ā = 0 [P6(b)]

a + ā = 1 [P6(a)]

Therefore, ā is the complement of a, and also a is the complement of ā. Because the complement of ā is unique, it must be equal to a.


• Why a + ab = a


a ab b






• Example using DeMorgan’s theorem



Boolean algebra postulates and theorems

Theorems

• Proofs by perfect inductionProofs by exhaustion:

Let variables assume all possible values and showvalidity of result in all cases

Example: Show X + 0 = X

00110

111

000

1'0

01

XX

XX

10110

000

111

0'1

10

XX

XX (a) Keep axioms handy

(b) Elaborate cases:if X = 0, have

X + 0 = 0 + 0 = 0 = Xif X = 1, have

X + 0 = 1 + 0 = 1 = X

More Theorems

Can prove by exhaustion....but have more casesFor distributive laws,

T8 looks like ordinary algebraT8’ also true (swap operators, factor, swap back)

T9, T10 for logic minimization - drop irrelevant terms

T9, T10, T11 for logic minimization - drop superfluous terms

Proof: X + XY = X1 + XY = X(1+Y) = X1 = X X(X+Y) = (X+0)(X+Y) = X+(0Y) = X+0 = X

T10 (Combining): XY + XY’ = X and (X + Y) (X + Y’) = XProof: XY + XY’ = X(Y + Y’) = X1 = X (X + Y)(X + Y’) = X + (YY’) = X + 0 = X

T11 (Consensus): XY+X’Z+YZ = XY+X’Z and (X+Y)(X’+Z)(Y+Z)= (X+Y)(X’+Z)Proof: If YZ = 0

XY+X’Z+YZ = XY+X’Z+ 0 = XY+X’Z else

Y = Z = 1left side: XY+X’Z+YZ = something + YZ = something + 1 =1right side: XY+X’Z = X + X’ = 1

So, in either case, XY+X’Z+YZ = XY+X’Z

If Y+Z = 1 (X+Y)(X’+Z)(Y+Z)= (X+Y)(X’+Z)1= (X+Y)(X’+Z)

elseY = Z = 0left side: (X+Y)(X’+Z)(Y+Z)= something (Y + Z) = something 0 = 0right side: (X+Y)(X’+Z) = (X+0)(X’+0) = XX’ = 0

So, in either case, (X+Y)(X’+Z)(Y+Z)= (X+Y)(X’+Z)

T9 (Covering): X + XY = X and X(X+Y)=X

Duality• De Morgan’s Theorems:

(X + Y)’ = X’ Y’(X Y)’ = X’ + Y’

• Dual: Swap 0 & 1, AND & OR, but leave variables unchanged– Result: Theorems still true

• Why?– f(X, Y) = g(X, Y)– complement[f(X, Y)] = complement[g(X, Y)]– dual[f(X’, Y’)] = dual[g(X’, Y’)]– but X’, Y’ just dummy variables, replace with originals

• Counterexample?X + X Y = X (T9)X X + Y = X (dual)X + Y = X (T3)!! error ?

X + (X Y) = X (T9)X (X + Y) = X (dual)(X X) + (X Y) = X (T8)X + (X Y) = X (T3)parentheses,operator precedence

N-variable Theorems

• Prove via induction• Most important: DeMorgan theorems

DeMorgan Symbol Equivalence

Bubble-pushing...

Likewise for OR

Documents

--- outputs logical functions of inputs --- new outputs appear shortly after changed inputs (propagation delay) --- no feedback loops --- no clock Sequential