ECE U322 Digital Logic Design

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ECE U322Digital Logic Design
Sept. 14, 2005

• Lecture 4
• Binary numbers
• Boolean Algebra
• Boolean Logic
• Reading Markovitz Section 2-2

lect04.ppt
ECE U322 F05
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Announcements

• HW1 due Thursday, Sept 22
• Assignment will be posted on Blackboard.

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Signed Values

• MSB is sign bit.
• Three methods for interpreting the remaining
  (non-sign) bits
• sign magnitude
• ones complement
• twos complement

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Representing Signed Values
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To form the 2s complement

• To form the 2s complement of a number
• take the 1s complement (bit by bit complement)
• add 1 (using binary addition)
• ignore carry out of MSB

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Binary Addition

• 0 0 00 1
• 1 0 11 1 10
• The last line means that the sum is 0 and there
  is a carry out into the next bit position.

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Addition - Decimal

• Example add 2 decimal numbers together 57
  68
• Add 7 8 which is 5 carry 1
• Add 5 6 carry which is 2 carry 1
• The result is .

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Addition - Binary

• Binary addition works the same
• Add 1012 to 1112. Start from LSB. 1 1
  which is 0 carry 1,
• 1 0 carry 1 which is 0 carry
• 1 1 carry 1 which is
  
• Result is ( )2

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Form the 2s complement

• of 0011
• of 1101
• of 0100
• of 1100

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Property of 2s complement

• 2s comp(2s comp (A)) A
• 2s comp (0011) 1101
• 2s comp (1101) 0011
• 2s comp (0100) 1100
• 2s comp (1100) 0100

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Represent -12 in 5-bits

• Using sign-magnitude
• 2s complement

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Values of some 5-bit numbers

• What is the value of 01101
• if it is unsigned
• if it is sign magnitude
• if it is 2s complement

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Values of some 5-bit numbers

• What is the value of 11101
• if it is unsigned
• if it is sign magnitude
• if it is 2s complement

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Values of some 5-bit numbers

• What is the value of 10000
• if it is unsigned
• if it is sign magnitude
• if it is 2s complement

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Take twos comp

• What is 2s complement of 10000?
• What does that tell us ?

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Boolean Algebra

• Definition
• A Boolean algebra consists of a set k of 2
  elements and 2 operators and . For every a,
  b e k, ab e k, a b e k.
• Variables are designated by letters.
• Three basic logic operations AND, OR, and NOT.

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Boolean Function

• Consists of a binary variable denoting the
  function, an equals sign, and an algebraic
  expression formed by using binary variables,
  constants 0 and 1, and logic operands.
• Example
• F X YZ
• X and YZ are called ______.
• Any Boolean function can be represented by a
  truth table

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• Truth table for F X YZ
• There is only one way that a Boolean function can
  be represented in a truth table.

___ variables
Number of rows in a truth table is ____.
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• A Boolean function can be transformed into a
  circuit diagram X YZ
• Gates are interconnected by wires.
• Variables are combined by logical operations.

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• F X YZ
• X XYZ XYZ
• XZ XY
• Boolean functions can be expressed in many
  _________ algebraic forms.
• We can manipulate a Boolean expression using
  Boolean algebraic rules to obtain a simpler
  expression.
• Simpler expression results in _______ gates and
  _______ inputs to the gates.

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Boolean Algebra

• Operators
• AND ?
• OR v
• NOT A ?A
• Values
• 1 (true) 0 false

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Boolean Algebra rules

• Rules hold for any boolean value
• a,b, c ... stand for these values
• Identity element
• a 0 a
• a ? a
• and ? are commutative
• a b b a
• a ? b b

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Associative and Distributive

• Associative rules
• a (b c) (a b) c
• a ? (b ? c)
• Distributive rules
• a (b ? c) (a b) ? (a c)
• a ? (b c)
• Note these rules look like algebra !

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Complement (NOT)

• a
• a a 1
• a 0
• Additional rules
• a a a
• a ? a a

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Properties of 0 and 1

• a 0 a
• a ? 1
• 0
• 1

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Basic Identities of Boolean Algebra
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Venn Diagram

• You can think of Boolean equations as sets 1 is
  everything, and 0 is nothing.

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Absorption Theorem

• a ab a
• a ( a b) a
• Venn Diagram

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An expression is true

• A valid expression is true
• true true
• false false
• x x
• false true

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Absorption Theorem

• a ab a
• Proof
• a ab apply distributive law (14)
• a ab a (1b) apply 3 1 b 1
• a ab a ? 1 apply 2
• a ab

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Dual

• The dual of an algebraic expression is obtained
  by interchanging OR and AND operations, and
  replacing 1s by 0s and 0s by 1s.
• Replace
• 1
• 0

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Why do rules come in pairs ?

• Duality Principle
• The duality principle of Boolean algebra states
  that a Boolean equation remains valid (true) if
  we take the dual of the expression
• If an expression is true, its dual is true.

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Proof of second absorption Theorem

• a ( a b) a
• Proof by principle of duality
• a ab a is true we proved it
• Apply duality

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DeMorgans Laws

• (a ? b) a b
• (a b) a ? b
• Replace AND with OR and OR with AND.
• Remove complement from the entire expression and
  place over each variable instead.
• These laws are duals of one another.

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DeMorgans Laws in Pictures
A
A
B
B
A
A
B
B
A
A
B
B
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• Truth tables can be used to verify expressions.
• Example, verify DeMorgans Theorem

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Algebraic Manipulation

• Ex F XYZ XYZ XZ
• Boolean algebra is a useful tool for simplifying
  digital circuits.
• Literal single variable within a term that may
  or may not be complemented.

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• Simplify
• F XYZ XYZ XZ

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• Truth Table
• Truth table for both expressions are equivalent.
• By reducing the number of terms and number of
  literals, it is possible to obtain a simpler
  circuit.

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Consensus Theorem

• XY XZ YZ XY XZ
• Note Y and Z are associated with X and X, and
  appear together in the term that is eliminated.
• The dual of the consensus theorem is

redundant
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Proof

• XY XZ YZ

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Complement of a Function

• F can be obtained by interchanging 1s to 0s and
  0s to 1s for values of F in the truth table.
• Can apply DeMorgans theorem as many times as
  necessary to find F.
• F1 XYZ XYZ
• Obtain F1
• F1 XYZ XYZ

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Lab 2 Seven segment decoder

• Seven Segment display for calculator
• Input to display is 4 bit number
• Unsigned values that can be represented with 4
  bits_______________________
• Want to display each of these on the display

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Hex to seven segment display
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Hex to seven segment display
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Hex to Seven-Segment Display Truth Table
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Minimized Boolean Expressions