IKI10201 03aBoolean Algebra - PowerPoint PPT Presentation

Title: IKI10201 03aBoolean Algebra


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IKI1020103a-Boolean Algebra

• Bobby Nazief
• Semester-I 2005 - 2006

The materials on these slides are adopted from
those in CS231s Lecture Notes at UIUC, which is
derived from Howard Huangs work and developed by
Jeff Carlyle.
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Road Map
Logic Gates Flip-flops
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Boolean Algebra
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6
Finite-StateMachines
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Sequential DesignTechniques
Logic DesignTechniques
CombinatorialComponents
StorageComponents
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Binary Systems Data Represent.
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5
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Register-TransferDesign
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Generalized FSM
ProcessorComponents
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3
Boolean Algebra
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Boolean values

• Earlier, we used electrical voltages to
  representtwo discrete values 1 and 0, from which
  binary numberscan be formed.
• Its also possible to think of voltages as
  representingtwo logical values, true and false.
• For simplicity, we often still write digits
  instead
• 1 (high) is true
• 0 (low) is false
• We will use these values 1 and 0 as the elements
  of our Boolean System.

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Basic boolean operations

• There are two basic operations for logical values.

AND (product) of two inputs
OR (sum) of two inputs
Operation
Expression
xy, or x?y
x y
Truth table
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Axiomatic definition of Boolean Algebra

• A Boolean algebra requires
• A set of elements B, consisting of two elements
  (0 and 1)
• Two binary operations OR and AND
• The axioms below must always be true
• Based on axiom 5, we can develop a unary
  (one-argument) operation NOT

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Basic theorems of Boolean Algebra

• In addition to the axioms, additional laws can be
  derived they are called theorems of Boolean
  Algebra
• These theorems are useful in performing algebraic
  manipulations of Boolean expressions

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Proofing the theorems using axioms

• Idempotency x x x
• Proof x x (x x) ? 1 by
  identity (x x) ? (x x) by complement
  x x ? x by distributivity x 0 by
  complement x by identity
• Idempotency x ? x x
• Proof x ? x (x ? x) 0 by
  identity (x ? x) (x ? x) by complement
  x ? (x x) by distributivity x ? 1 by
  complement x by identity

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Duality principle

• The left and right columns of axioms are duals
• exchange all ANDs with ORs, and 0s with 1s
• So are the theorems

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Boolean Functions

• Computers take inputs and produce outputs, just
  like functions in math!
• Mathematical functions can be expressed in two
  ways
• We can represent logical functions in two
  analogous ways too
• A finite, but non-unique Boolean expression.
• A truth table, which will turn out to be unique
  and finite.

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Boolean expressions

• We can use these basic operations to form more
  complex expressions
• f(x,y,z) (x y)z x
• Some terminology and notation
• f is the name of the function.
• (x,y,z) are the input variables, each
  representing 1 or 0. Listing the inputs is
  optional, but sometimes helpful.
• A literal is any occurrence of an input variable
  or its complement. The function above has four
  literals x, y, z, and x.
• Precedence is important, but not too difficult.
• NOT has the highest precedence, followed by AND,
  and then OR.
• Fully parenthesized, the function above would be
  kind of messy
• f(x,y,z) (((x (y))z) x)

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Truth tables

• A truth table shows all possible inputs and
  outputs of a function.
• Remember that each input variable represents
  either 1 or 0.
• Because there are only a finite number of values
  (1 and 0), truth tables themselves are finite.
• A function with n variables has 2n possible
  combinations of inputs.
• Inputs are listed in binary orderin this
  example, from 000 to 111.

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Complement of a function

• The complement of a function always outputs 0
  where the original function outputted 1, and 1
  where the original produced 0.
• In a truth table, we can just exchange 0s and 1s
  in the output column(s)

f(x,y,z) x(yz yz)
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Complementing a function algebraically

• You can use DeMorgans law to keep pushing the
  complements inwards
• You can also take the dual of the function, and
  then complement each literal
• If f(x,y,z) x(yz yz)
• the dual of f is x (y z)(y z)
• then complementing each literal gives x (y
  z)(y z)
• so f(x,y,z) x (y z)(y z)

f(x,y,z) x(yz yz) f(x,y,z) ( x(yz
yz) ) complement both sides x (yz
yz) because (xy) x y x (yz)
(yz) because (x y) x y x (y
z)(y z) because (xy) x y, twice
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Algebraic manipulation

• We can now start doing some simplifications

xy xyz xy x(y y) xyz
Distributive xy xy x(y y) x?1
xyz Axiom 5 y y 1 x xyz Axiom
2 x?1 x (x x)(x yz)
Distributive 1 ? (x yz) Axiom 5 x x
1 x yz Axiom 2 x?1 x
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Canonical Forms

• Any boolean function that is expressed as a sum
  of minterms or as a product of maxterms is said
  to be in its canonical form.

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Minterms

• A minterm is a special product of literals, in
  which each input variable appears exactly once.
• A function with n variables has 2n minterms
  (since each variable can appear complemented or
  not)
• A three-variable function, such as f(x,y,z), has
  23 8 minterms
• Each minterm is true for exactly one combination
  of inputs

xyz xyz xyz xyz xyz xyz xyz xyz
Minterm Is true when Shorthand xyz x0, y0,
z0 m0 xyz x0, y0, z1 m1 xyz x0, y1,
z0 m2 xyz x0, y1, z1 m3 xyz x1, y0,
z0 m4 xyz x1, y0, z1 m5 xyz x1, y1,
z0 m6 xyz x1, y1, z1 m7
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Sum of minterms form

• Every function can be written as a sum of
  minterms, which is a special kind of sum of
  products form
• The sum of minterms form for any function is
  unique
• If you have a truth table for a function, you can
  write a sum of minterms expression just by
  picking out the rows of the table where the
  function output is 1 (1-minterm).

f xyz xyz xyz xyz xyz m0
m1 m2 m3 m6 ?(0,1,2,3,6)
f xyz xyz xyz m4 m5 m7
?(4,5,7)
f contains all the minterms not in f
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Sum of minterms practise

• F x yz, how to express this in the sum of
  minterms?
• x(y y)(z z) (x x)yz xyz
  xyz xyz xyz xyz xyz xyz
  xyz xyz xyz xyz ?(3,4,5,6,7)
• or, convert the expression into truth-table and
  then read the minterms from the table

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Maxterms

• A maxterm is a sum of literals, in which each
  input variable appears exactly once.
• A function with n variables has 2n maxterms
• The maxterms for a three-variable function
  f(x,y,z)
• Each maxterm is false for exactly one combination
  of inputs

x y z x y z x y z x y z x
y z x y z x y z x y z
Maxterm Is false when Shorthand x y z x0,
y0, z0 M0 x y z x0, y0, z1 M1 x y
z x0, y1, z0 M2 x y z x0, y1,
z1 M3 x y z x1, y0, z0 M4 x y
z x1, y0, z1 M5 x y z x1, y1,
z0 M6 x y z x1, y1, z1 M7
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Product of maxterms form

• Every function can be written as a unique product
  of maxterms
• If you have a truth table for a function, you can
  write a product of maxterms expression by picking
  out the rows of the table where the function
  output is 0 (0-maxterm).

f (x y z)(x y z)(x y z) M4
M5 M7 ?(4,5,7)
f (x y z)(x y z)(x y z) (x
y z)(x y z) M0 M1 M2 M3 M6
?(0,1,2,3,6)
f contains all the maxterms not in f
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Product of maxterms practise

• F xy xz, how to express this in the product
  of maxterms?
• (xy x)(xy z) (x x)(y
  x)(x z)(y z) (x y)(x z)(y z)
  (x y zz)(x z yy)(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)
  ?(2,3,4,6)
• or, convert the expression into truth-table and
  then read the minterms from the table

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Minterms and maxterms are related

• Any minterm mi is the complement of the
  corresponding maxterm Mi
• For example, m4 M4 because (xyz) x y
  z

Maxterm Shorthand x y z M0 x y z M1 x
y z M2 x y z M3 x y z M4 x
y z M5 x y z M6 x y z M7
Minterm Shorthand xyz m0 xyz m1 xyz m
2 xyz m3 xyz m4 xyz m5 xyz m6 xyz m
7
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Converting between canonical forms

• We can convert a sum of minterms to a product of
  maxterms
• In general, just replace the minterms with
  maxterms, using maxterm numbers that dont appear
  in the sum of minterms
• The same thing works for converting from a
  product of maxterms to a sum of minterms

From before f ?(0,1,2,3,6) and f
?(4,5,7) m4 m5 m7 complementing (f)
(m4 m5 m7) so f m4 m5 m7 DeMorgans
law M4 M5 M7 By the previous page
?(4,5,7)
f ?(0,1,2,3,6) ?(4,5,7)
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Standard Forms

• Any boolean function that is expressed as a sum
  of products (SOP) or as a product of sums (POS),
  where each product-term or sum-term may require
  fewer than (n-1) operations, is said to be in its
  standard form.
• Standard forms are not unique, there can be
  several different SOPs and POSs for a given
  function.
• A SOP expression contains
• Only OR (sum) operations at the outermost level
• Each term (implicant) must be a product of
  literals
• A POS expression contains
• Only AND (product) operations at the outermost
  level
• Each term (implicate) must be a sum of literals

f(x,y,z) xy xyz xyz
f(x,y,z) (x y)(x y z)(x y z)
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Strategy for operator (literal) reduction
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Other Logic Operations

• Weve already seen all the basic Boolean
  operations and the associated primitive logic
  gates.
• There is a few additional gates that are often
  used in logic design.
• They are all equivalent to some combination of
  primitive gates.
• But they have some interesting properties in
  their own right.

NAND (NOT-AND)
NOR (NOT-OR)
XOR (eXclusive OR)
Operation
Expressions
(xy) x y
(x y) x y
x ? y xy xy
Truth table