logic design - chapter 3: boolean algebra

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CHAPTER 3 Boolean Algebra

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- Describing Logic Circuits Algebraically. - rules of evaluating a Boolean expression. - Boolean Theorems. - DeMorgan's Theorem. - Universality of NAND and NOR Gates. - Alternate Logic Gate Representations. - Minterms and Maxterms. - STANDARD FORMS.

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Page 1: Logic Design - Chapter 3: Boolean Algebra

CHAPTER 3

Boolean Algebra

Page 2: Logic Design - Chapter 3: Boolean Algebra

Contents

Describing Logic Circuits Algebraically rules of evaluating a Boolean expression Boolean Theorems DeMorgan's Theorem Universality of NAND and NOR Gates Alternate Logic Gate Representations Minterms and Maxterms STANDARD FORMS

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Page 3: Logic Design - Chapter 3: Boolean Algebra

Describing Logic Circuits Algebraically OR gate, AND gate, and NOT circuit are

the basic building blocks of digital systems

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Page 4: Logic Design - Chapter 3: Boolean Algebra

Circuits containing Inverters

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We can say that X = A’BC (A’D’).
Page 5: Logic Design - Chapter 3: Boolean Algebra

Evaluating Logic Circuit Outputs Let A=0, B=0, C=1, D=1, E=1

X = [D+ ((A+B)C)'] • E

= [1 + ((0+0)1 )'] • 1

= [1 + (0•1)'] • 1

= [1+ 0'] •1

= [1+ 1 ] • 1

= 1

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X = [D+ ((A+B)C)']•E = [D+((A+B)’+C’)]•E = [D+(A’B’+C’)]•E = [1+(1•1+0)]•1 = [1+(1+0)]•1 = [1+1] •1 = 1•1 = 1.
Page 6: Logic Design - Chapter 3: Boolean Algebra

Rules of evaluating a Boolean expression First, perform all inversions of single

terms; that is, 0 = 1 or 1 = 0. Then perform all operations within

parentheses. Perform an AND operation before an OR

operation unless parentheses indicate otherwise.

If an expression has a bar over it, perform the operations of the expression first and then invert the result.

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Page 7: Logic Design - Chapter 3: Boolean Algebra

Determining Output Level from a Diagram

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X in the first diagram can = A’BC (A’D’)
Page 8: Logic Design - Chapter 3: Boolean Algebra

Boolean Theorems

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Page 9: Logic Design - Chapter 3: Boolean Algebra

Multivariable Theorems(9) x + y = y + x (Commutative law)

(10) x • y = y • x (Commutative law)

(11) x+ (y+z) = (x+y) +z = x+y+z (Associative law)

(12) x (yz) = (xy) z = xyz (Associative law)

(13.a) x (y+z) = xy + xz (Distributive law)

(13.b) x + yz = (x + y) (x + z) (Distributive law)

(13.c) (w+x)(y+z) = wy + xy + wz + xz

(14) x + xy = x (Absorption) [proof]

(15) x + x'y = x + y

(16) (x +y)(x + z) = x +yz

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Commutative = Eltabdel - Associative = El2edmag - Distributive = Eltawze3 - Absorption = El2emtsas (mass)Proof:- x+xy = x.1+x.y = x(1+y) = x.1 = x
Page 10: Logic Design - Chapter 3: Boolean Algebra

Proof of (14, 15, 16)x + xy = x (1+y)

= x • 1 [using theorem (6)]

= x [using theorem (2)]

x + x’y = ( x + x’) (x + y) [theorem 13b]

= 1 (x +y)

= (x + y)

(x +y)(x + z) =xx + xz + yx + yz

= x + xz + yx + yz

= x (1+z+y) +yz

= x . 1 + yz

= x + yz

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Page 11: Logic Design - Chapter 3: Boolean Algebra

DeMorgan's Theorem (18) (x+y)' = x' • y' (19) (x•y)' = x' + y' Example

X = [(A'+C) • (B+D')]'

= (A'+C)' + (B+D')'

= (AC') + (B'D)

= AC' + B'D

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AND will be OR and nwza3 el bar 3lehom el 2, w el3ks sa7e7
Page 12: Logic Design - Chapter 3: Boolean Algebra

Three Variables DeMorgan's Theorem (20) (x+y+z)' = x' • y' • z' (21) (xyz)' = x' + y' + z‘

EXAMPLE: Apply DeMorgan’s theorems to each of the

following expressions: (a) (b) (c)

D)CBA( DEFABC

EFDCBA 12

Page 13: Logic Design - Chapter 3: Boolean Algebra

Universality of NAND Gates

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Universal because it do the 3 operation INVERTER, AND & OR
Page 14: Logic Design - Chapter 3: Boolean Algebra

Universality of NOR Gates

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Universal because it do the 3 operation INVERTER, OR & AND.
Page 15: Logic Design - Chapter 3: Boolean Algebra

Alternate Logic Gate Representations

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Page 16: Logic Design - Chapter 3: Boolean Algebra

Minterms and Maxtermsx

y

z

Minterms

Maxterms

Term

Designation

Term

Designation

0

0

0

x' y’ z'

m0

x+y+z

+y

-t- Z

M0

0

0

1

x' y' z

M1

x+y+z’

+y

+ 2

, t

M1

0

1

0

x' y z’ m2 x+y’+z

+ y'

+

z

M2

0

1

1

x' y z

m3 x+y’+z’

+y'

+

2'

M3

1

0

0

x y' z’

m4 x'+y+z

'+y

4-

2

M4

1

0

1

x y' z

m5 x

'+y+z’ M5

1

1

0

x y z’

m6 x

'+y’+z

M6

1

1

1

x y z

m7 x

'+y’+z’

M7

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Page 17: Logic Design - Chapter 3: Boolean Algebra

Canonical FORMS There are two types of canonical forms:

the sum of minterms The product of maxterms

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Page 18: Logic Design - Chapter 3: Boolean Algebra

Sum of minterms f1 = x'y'z + xy'z' + xyz = m1 + m4 +m7

f2 = x'yz + xy'z + xyz’ + xyz = m3 + m5+ m6 + m7

x y Z f1 f2 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1

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Page 19: Logic Design - Chapter 3: Boolean Algebra

Product of maxterms The complement of f1 is read by forming a

minterm for each combination that produces a 0 as:

f1’=x’y’z’ + x’yz’ + x’yz + xy’z + xyz’

f1 = (x + y + z)(x + y' + z)(x + y' + z' )(x’+ y + z)(x’ + y' + z)

= Mo M2 M3 M5 M6

Similarly f2 = ?

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Page 20: Logic Design - Chapter 3: Boolean Algebra

Example: Sum of Minterms Express the Boolean function F = A + B'C

in a sum of minterms. F=A+B'C = ABC + ABC' + AB'C + AB'C' + AB'C + A'B'C

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Page 21: Logic Design - Chapter 3: Boolean Algebra

Example: Product of Maxterms Express the Boolean function F =xy' + yz in

a product of maxterm form. F = xy' + yz = (xy' + y)(xy' + z) = (x + y)(y' + y)(x + z)(y' + z)

= (x + y)(x + z)(y' + z) = (x + y + zz')(x + yy' + z)(xx' + y' + z) = (x + y + z)(x + y + z')(x+y + z)(x+y’+ z)(x + y' + z)(x'+y'+z) = (x + y + z)(x + y + z') (x + y' + z) (x'+y'+z) = M0 M1 M2 M6

= Π (0,1,2,6)

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Page 22: Logic Design - Chapter 3: Boolean Algebra

STANDARD FORMS There are two types of standard forms:

the sum of products (SOP) The product of sums (POS).

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Page 23: Logic Design - Chapter 3: Boolean Algebra

Sum of Products The sum of products is a Boolean

expression containing AND terms, called product terms, of one or more literals each. The sum denotes the ORing of these terms.

F = xy + z +xy'z'. (SOP)

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Page 24: Logic Design - Chapter 3: Boolean Algebra

Product of Sums A product of sums is a Boolean expression

containing OR terms, called sum terms. Each term may have any number of literals. The product denotes the ANDing of these terms.

F = z(x+y)(x+y+z) (POS) F = x (xy' + zy) (nonstandard form)

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