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TEKNIK DIGITAL (B), Summaries of Algebra

National University of Ireland (NUI)Algebra

BOOLEAN ALGEBRA AND LOGIC. SIMPLICATION. Boolean Algebra ... (simplify) Boolean functions ... Boolean Algebra Properties (cont.) Let X: boolean variable, ...

Typology: Summaries

2021/2022

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TEKNIK TEKNIK DIGITAL DIGITAL (B)(B)
Materi Kuliah ke-5
BOOLEAN ALGEBRA AND LOGIC
SIMPLICATION
Boolean AlgebraBoolean Algebra
• VERY nice machinery used to manipulate
(simplify) Boolean functions
• George Boole (1815-1864): “An
investigation of the laws of thought”
• Terminology:
–Literal: A variable or its complement
–Product term:literals connected by •
–Sum term: literals connected by +
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TEKNIKTEKNIK DIGITALDIGITAL (B)(B)

Materi Kuliah ke- 5 BOOLEAN ALGEBRA AND LOGIC SIMPLICATION

Boolean AlgebraBoolean Algebra

  • VERY nice machinery used to manipulate (simplify) Boolean functions
  • George Boole (1815-1864): “ An investigation of the laws of thought ”
  • Terminology:
    • Literal: A variable or its complement
    • Product term: literals connected by •
    • Sum term: literals connected by +

Boolean Algebra PropertiesBoolean Algebra Properties

Let X: boolean variable, 0,1: constants

  1. X + 0 = X  Zero Axiom
  2. X • 1 = X  Unit Axiom
  3. X + 1 = 1  Unit Property
  4. X • 0 = 0  Zero Property

Boolean Algebra Properties (cont.)Boolean Algebra Properties (cont.)

Let X: boolean variable, 0,1: constants

  1. X + X = X  Idempotence
  2. X • X = X  Idempotence
  3. X + X’ = 1  Complement
  4. X • X’ = 0  Complement
  5. (X’)’ = X  Involution

More Boolean Algebra PropertiesMore Boolean Algebra Properties

  1. X + Y = Y + X 11. X • Y = Y • X Commutative

  2. X + (Y+Z) = (X+Y) + Z 13. X•(Y•Z) = (X•Y)•Z Associative

  3. X•(Y+Z) = X•Y + X•Z 15. X+(Y•Z) = (X+Y) • (X+Z) Distributive

  4. (X + Y)’ = X’ • Y’ 17. (X • Y)’ = X’ + Y’ DeMorgan’s In general, ( X 1 + X 2 + … + Xn )’ = X 1 ’•X 2 ’ • … • Xn’, and ( X 1 • X 2 • … • Xn )’ = X 1 ’ + X 2 ’ + … + Xn’ Let X,Y, and Z: boolean variables Absorption Property (Covering)Absorption Property (Covering)

  5. x + x•y = x

  6. x•(x+y) = x (dual)

  • Proof: x + x•y = x•1 + x•y = x•(1+y) = x• 1 = x QED ( 2 true by duality)

Consensus TheoremConsensus Theorem

  1. xy + x’z + yz = xy + x’z
  2. (x+y)•(x’+z)•(y+z) = (x+y)•(x’+z) -- (dual)
  • Proof: xy + x’z + yz = xy + x’z + (x+x’)yz = xy + x’z + xyz + x’yz = (xy + xyz) + (x’z + x’zy) = xy + x’z QED ( 2 true by duality).

Truth Tables (revisited)Truth Tables (revisited)

  • Enumerates all possible combinations of variable values and the corresponding function value
  • Truth tables for some arbitrary functions F 1 (x,y,z), F 2 (x,y,z) , and F 3 (x,y,z) are shown to the right. x y z F 1 F 2 F 3 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 0 0 1 0 1 1 0 1 1 1 0 0 0 1 0 1 0 1 0 1 0 1 1 0 0 0 0 1 1 1 1 0 1

Algebraic ManipulationAlgebraic Manipulation

  • Boolean algebra is a useful tool for

simplifying digital circuits.

  • Why do it? Simpler can mean cheaper,

smaller, faster.

  • Example: Simplify F = x’yz + x’yz’ + xz.
  • F = x’yz + x’yz’ + xz

= x’y(z+z’) + xz

= x’y• 1 + xz

= x’y + xz

Algebraic Manipulation (cont.)Algebraic Manipulation (cont.)

  • Example: Prove x’y’z’ + x’yz’ + xyz’ = x’z’ + yz’
  • Proof: x’y’z’+ x’yz’+ xyz’ = x’y’z’ + x’yz’ + x’yz’ + xyz’ = x’z’(y’+y) + yz’(x’+x) = x’z’•1 + yz’• 1 = x’z’ + yz’ QED.

Complement of a FunctionComplement of a Function

  • The complement of a function is

derived by interchanging (• and +),

and (1 and 0) , and complementing

each variable.

  • Otherwise, interchange 1s to 0s in the

truth table column showing F.

  • The complement of a function IS NOT

THE SAME as the dual of a function.

Complementation: ExampleComplementation: Example

  • Find G(x,y,z), the complement of F(x,y,z) = xy’z’ + x’yz
  • G = F’ = (xy’z’ + x’yz)’ = (xy’z’)’ • (x’yz)’ DeMorgan = (x’+y+z) • (x+y’+z’) DeMorgan again
  • Note: The complement of a function can also be derived by finding the function’s dual, and then complementing all of the literals

MintermMinterm

  • Represents exactly one combination in the truth table.
  • Denoted by mj , where j is the decimal equivalent of the minterm’s corresponding binary combination (bj).
  • A variable in mj is complemented if its value in bj is 0, otherwise is uncomplemented.
  • Example: Assume 3 variables (A,B,C), and j =3. Then, b j = 011 and its corresponding minterm is denoted by mj = A’BC

MaxtermMaxterm

  • Represents exactly one combination in the truth table.
  • Denoted by Mj , where j is the decimal equivalent of the maxterm’s corresponding binary combination (bj).
  • A variable in Mj is complemented if its value in bj is 1, otherwise is uncomplemented.
  • Example: Assume 3 variables (A,B,C), and j =3. Then, b j = 011 and its corresponding maxterm is denoted by M j = A+B’+C’

Truth Table notation forTruth Table notation for MintermsMinterms andand MaxtermsMaxterms

  • Minterms and Maxterms are easy to denote using a truth table.
  • Example: Assume 3 variables x,y,z (order is fixed) x y z Minterm Maxterm 0 0 0 x’y’z’ = m 0 x+y+z = M 0 0 0 1 x’y’z = m 1 x+y+z’ = M 1 0 1 0 x’yz’ = m 2 x+y’+z = M 2 0 1 1 x’yz = m 3 x+y’+z’= M 3 1 0 0 xy’z’ = m 4 x’+y+z = M 4 1 0 1 xy’z = m 5 x’+y+z’ = M 5 1 1 0 xyz’ = m 6 x’+y’+z = M 6 1 1 1 xyz = m 7 x’+y’+z’ = M 7

Canonical Forms (Unique)Canonical Forms (Unique)

  • Any Boolean function F( ) can be expressed as a unique sum of min terms and a unique product of max terms (under a fixed variable ordering).
  • In other words, every function F() has two canonical forms: - Canonical Sum-Of-Products (sum of minterms) - Canonical Product-Of-Sums (product of maxterms)

Shorthand:Shorthand: ∑∑ andand ∏∏

  • f 1 (a,b,c) = ∑ m(1,2,4,6), where ∑ indicates that this is a sum-of-products form, and m(1,2,4,6) indicates that the minterms to be included are m 1 , m 2 , m 4 , and m 6.
  • f 1 (a,b,c) = ∏ M(0,3,5,7), where ∏ indicates that this is a product-of-sums form, and M(0,3,5,7) indicates that the maxterms to be included are M 0 , M 3 , M 5 , and M 7.
  • Since mj = Mj’ for any j , ∑ m(1,2,4,6) = ∏ M(0,3,5,7) = f 1 (a,b,c) Conversion Between Canonical FormsConversion Between Canonical Forms
  • Replace ∑ with ∏ (or vice versa ) and replace those j’ s that appeared in the original form with those that do not.
  • Example: f 1 (a,b,c) = a’b’c + a’bc’ + ab’c’ + abc’ = m 1 + m 2 + m 4 + m 6 = ∑(1,2,4,6) = ∏(0,3,5,7) = (a+b+c)•(a+b’+c’)•(a’+b+c’)•(a’+b’+c’)

Standard Forms (NOT Unique)Standard Forms (NOT Unique)

  • Standard forms are “ like” canonical

forms, except that not all variables need

appear in the individual product (SOP) or

sum (POS) terms.

  • Example:

f 1 (a,b,c) = a’b’c + bc’ + ac’

is a standard sum-of-products form

  • f 1 (a,b,c) = (a+b+c)•(b’+c’)•(a’+c’)

is a standard product-of-sums form.

Conversion of SOP fromConversion of SOP from

standard to canonical formstandard to canonical form

  • Expand non-canonical terms by inserting equivalent of 1 in each missing variable x: (x + x’) = 1
  • Remove duplicate minterms
  • f 1 (a,b,c) = a’b’c + bc’ + ac’ = a’b’c + (a+a’)bc’ + a(b+b’)c’ = a’b’c + abc’ + a’bc’ + abc’ + ab’c’ = a’b’c + abc’ + a’bc + ab’c’