CS 150 Computer Organization

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CS 150Computer Organization Architecture

• Lecture 3

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Overview

• Quiz 3
• Boolean Operations
• Truth tables
• Logic gates
• Alternate gate shapes
• Functionally complete gates
• Boolean expressions
• Boolean theorems
• Canonical forms (SOP POS)
• Boolean minimization
• The design process
• Karnaugh maps

3
A New Kind of Algebra
Boolean Algebra - a relatively simple
mathematical tool that allows us to describe the
relationship between a logic circuits output(s)
and its input(s) as an algebraic equation.
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Logical Not Operator

• Also known as complement or invert operation

Not Operation is usually denoted by X or X
This is a truth table. It gives input-output
mapping by simply enumerating all possible input
combinations and the associated output.
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NOT Gate

• Also known as complement or invert gate

X
X
This is a schematic. It is a graphical
representation of a circuit which implements the
operation.
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Logical And Operation

• Denotes AND condition Output is true iff all
  inputs are true

AND Operation is usually denoted by X A B X
AB
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AND Gate
A
X
B
Boolean Expressions
X A B X AB X A x B
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Logical Or Operation

• Denotes OR condition Output is true if any
  inputs are true

OR Operation is usually denoted by X A B
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OR Gate
A
X
B
C
Boolean Expression
X A B C
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NAND Gate
A
X
B
Denotes Inversion
Boolean Expressions
X (A B) X (AB) X
(A x B)
____ X A B __ X AB
____ X A x B
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NOR Gate
A
X
B
C
Denotes Inversion
Boolean Expression
X A B C X (A B C)
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XOR Gate
A
X
B
Boolean Expression
X AB AB
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Alternate Gate Shapes

• Sometimes the logic of a circuit is more easily
  understood if an alternative gate shape is used.
• Rules 1. Change the gate shape (AND gt OR OR
  gt AND).
• 2. Change the bubbles (add where missing
  remove if present).
• Examples

A
A
AB
AND
AB
B
B
(AB) AB
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Alternate Gate Shapes Cont
A
A
AB
AB
OR
B
B
(AB) AB
A
A
(AB)
(AB)
NAND
B
B
A B (AB)
A
A
(AB)
(AB)
NOR
B
B
AB (AB)
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One Last One
X
X
INV
X
X
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Functionally Complete Gates
Any Boolean logic equation can be implemented
with a combination of AND gates NOT
gates OR gates NOT gates Some gates are
functionally complete. These gates can implement
any combinational logic equation. These gates
are NAND gates (contain both AND NOT
functions) NOR gates (contain both OR NOT
functions)
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Functionally Complete Gates - NAND
A
A
NOT
A
AB
AND
B
A
AB
B
OR
What are the NOR Gate representation for each?
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Boolean Expressions

• Boolean expressions are made up of constants and
  variables combined by AND, OR and NOT operations
• Examples 1 A AB CD AB A(BC) ABC
• AB is the same as AB ( is omitted when obvious
  )
• Parentheses are used like in regular algebra for
  grouping
• A literal is each instance of a variable or
  constant
• This expression has 4 variables and 10 literals
• abd bcd ac ad

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Boolean Expressions

• Each Boolean expression can be specified by a
  truth table which lists all possible combinations
  of the values of all variables in the expression.

F A B C A B C A' B C F A' B
C 0 0 0 1 0 1 0 0 1
1 0 1 0 1 0 1
0 1 0 1 1 1 1
1 1 0 0 0 0 0 1 1
0 0 0 0 1 1 1 0
1 1
Which combination is missing?
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Boolean Expressions From Truth Tables

• Each 1 in the output of a truth table specifies
  one term in the corresponding boolean expression.
• The expression can be read off by inspection

• A B C F
• 0 0 0 0
• 0 0 1 0
• 0 1 0 1
• 0 1 1 0
• 0 0 0
• 1 0 1 0
• 1 1 0 1
• 1 1 1 0

F is true when A is false, B is true, and C
is false OR A is true, B is true, and C is
false F ABC ABC
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Another Example

• A B C F
• 0 0 0 0
• 0 0 1 1
• 0 1 0 1
• 0 1 1 0
• 0 0 1
• 1 0 1 0
• 1 1 0 1
• 1 1 1 0

• F ?

F ABC ABC ABC ABC
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Boolean Expressions

• Each Boolean expression has a corresponding
  realization with logic gates.
• F A B C

A
F
B
C
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Seat Belt Buzzer

• Three digital inputs
• Car on/off
• Seat belt attached/unattached
• Timer timed out/not timed out
• C 0 if car off, 1 if car on
• B 0 if unattached, 1 if attached
• T 0 if timed out, 1 if not timed out

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Seat Belt Buzzer

• Process
• Construct Truth Table
• Construct Boolean Equation

Result O CBT
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Seat Belt Buzzer Circuit
O CBT
T
O
C
B
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More Examples

• G AB ACDE ACF
• G A(B CDE CF)
• G A(B C(DE F))
• What are the circuit representations for these?

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More Examples

• G AB ACDE ACF

A
B
C
G
D
E
F
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More Examples

• G A(B CDE CF)

G
A
B
C
D
E
F
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More Examples

• G A(B C(DE F))

A
G
B
D
E
F
C
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Some Basic Boolean Theorems
From these
We can get these
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Some Basic Boolean Theorems
From these
We can get these
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Basic Boolean Algebra Theorems

• Here are the first Boolean Algebra theorems we
  will study and use.

X 0 X X 1
X X 1 1 X 0
0 X X X X X
X ( X' )' X X X' 1
X X' 0
Each can easily be proved by showing it valid for
X 0 and X 1 or by using a truth table
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Other Boolean Theorems

• Commutative Law
• X Y Y X
• X Y Y X
• Associative Law
• ( X Y ) Z X ( Y Z) X Y Z
• ( X Y ) Z X ( Y Z ) X Y Z
• Distributive Law
• X(Y Z) XY XZ
• (A B)(C D) AC AD BC BD
• X (YZ) (X Y)(X Z) ? NOT like regular
  Algebra
• AB CD (A C)(A D)(B C)(B D) ? NOT like
  regular Algebra

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DeMorgans Theorems

• It is sometimes difficult to understand a Boolean
  equation that is in the form of a NAND or NOR
  function.
• Using DeMorgans Theorem, it is possible to
  convert the equation into an AND or OR function,
  which is sometimes easier to deal with.
• ( X Y ) X Y ( X Y ) X Y

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General DeMorgans Conversion

• Conversion Steps
• 1. Change the logic symbol ( to to ).
• 2. Break the bar over the symbol that was
  changed.
• Examples
• Equation to convert X (A B C)
• step 1 (A B C) change logic symbol(s)
• step 2 A B C break bar over symbol
  change(s)
• X A B C
• Equation to convert X (A B C)
• step 1 (A B C) change logic symbol(s)
• step 2 A B C break bar over symbol
  change(s)
• X ABC

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Summary of Boolean Theorems

• Commutative Law X Y Y X
• X Y Y X
• Associative Law ( X Y ) Z X ( Y Z) X
  Y Z
• ( X Y ) Z X ( Y Z ) X Y Z
• Distributive Law X(Y Z) XY XZ
• X (YZ) (X Y)(X Z)
• (A B)(C D) AC AD BC BD
• AB CD (A C)(A D)(B C)(B D)
• Miscellaneous Laws X 0 X
  X 1 X
• X 1 1 X 0
  0
• X X X X X X
• ( X' )' X
• X X' 1 X X' 0
• Simplification Theorems X Y X Y X ( X Y )
  (X Y ) X
• X X Y X X ( X Y ) X

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Canonical Boolean Forms

• Sum-Of-Products form or SOP
• Product of Sums form or POS

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Sum of Products Form

• Sometimes called MINTERM form
• Indicates 1s in truth table
• All terms are products of single variables only
  (no parentheses)
• Terms are ANDed together
• X A B C D E F G H Yes
• X A B C D E Yes
• X A B C ( D E ) No

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Variations of SOP Form

• Standard SOP Form
• Every term contains every variable
• F ABC ABC ABC
• Numeric SOP or Minterm Expansion Form
• Uses numbers to represent rows in truth table
  where function is 1
• F(A,B,C) Sm(0, 3, 7)

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Minterm Expansion Form

• Minterm notation is a short hand way of writing
  Boolean expressions
• Each row can be numbered
• Call each rows AND term a minterm

What is the Boolean expression for this TT? F
ABC ABC ABC An alternate way of
expressing this is F m1 m3 m6 or F(A,B,C)
Sm(1, 3, 6)
m0 m1 m2 m3 m4 m5 m6 m7
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Product of Sums Form

• Sometimes called MAXTERM form
• Indicates 0s in truth table
• All terms are sums of single variables only
• Terms are ORed together
• X ( A B C ) ( D E ) Yes
• X ( A B )( C D )EF Yes
• X ( A B C ) ( D E ) H No
• X ( A BC ) ( D E ) No

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Variations of POS Form

• Standard POS Form
• Every term contains every variable
• F (A B C)(A B C)(A B C)
• Numeric POS or Maxterm Expansion Form
• Uses numbers to represent rows in truth table
  where function is 0
• F(A,B,C) pM(0, 3, 7)

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Maxterm Expansion Form

• Maxterm notation is a short hand way of writing
  Boolean expressions
• Each row can be numbered
• Call each rows OR term a maxterm

What is the boolean expression for this TT? F
(ABC)(ABC)(ABC) An alternate way of
expressing this is F (M0)(M2)(M4) or F(A,B,C)
pM(1, 2, 4)
M0 M1 M2 M3 M4 M5 M6 M7
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POS vs. SOP

• Any expression can be written either way
• Can convert from one to another using theorems
• Sometimes SOP looks simpler
• AB CD ( A C )( B C )( A D )( B D )
• Sometimes POS looks simpler
• (A B)( C D) BD AD BC AC
• SOP will be most commonly used in this class but
  learn both

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Algebraic Simplification

• Why simplify??
• Simplify equations -gt Less hardware to implement
  -gt less cost
• Example what is the Boolean expression for this
  TT?

F ABC ABC ABC ABC What hardware is
used for this? 4 3-input ANDs 1 4-input
OR What is simplified expression? F BC
AB What hardware is used for this? 2 2-input
ANDs 1 2-input OR
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Algebraic Simplification

• 3 Methods used to simplify Boolean expressions
• Combine terms (X Y X Y X)
• Eliminate terms and variables (X X Y X)
• Add redundant terms (X X X)

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Applying Methods of Algebraic Simplification

• F ABC ABC ABC
• Combine terms
• ABC ABC BC(AA) XX1, X1X
  BC
• F BC ABC
• Add redundant terms
• BC ABC ABC XXX
• Combine terms
• ABC ABC AC(BB) XX1, X1X
• AC
• F BC AC

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Applying Methods of Algebraic Simplification

• F (ABC)(ABC)(ABC)
• Combine terms
• (ABC)(ABC) (BC)(AA) XX0,
  X0X (BC)
• F (BC)(ABC)
• Add redundant terms
• (BC)(ABC)(ABC) XXX
• Combine terms
• (ABC)(ABC) (AC)(BB) XX0,
  X0X
• (AC)
• F (BC)(AC)

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Designing Combinational Logic Circuits

• Define the problem
• Set up the truth table
• Write the Boolean equation
• Simplify the output expression
• Draw the logic diagram

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Step 1 Define the Problem

• Converting English to Boolean
• The air conditioner should be turned on if and
  only if the temperature is greater than 75, the
  time is between 8am and 5pm, and it is not a
  holiday.

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Step 1 Define the Problem

• Identifying phrases
• The air conditioner should be turned on if and
  only if the temperature is greater than 75, the
  time is between 8am and 5pm, and it is not a
  holiday.
• F air conditioner should be turned on
• A temperature is greater than 75
• B time is between 8am and 5 pm
• C it is a holiday

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Step 1 Define the Problem

• Identifying connective words
• The air conditioner should be turned on if and
  only if the temperature is greater than 75, the
  time is between 8am and 5pm, and it is not a
  holiday.
• if and only if -gt
• , implied AND

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Step 1 Define the Problem

• Be careful Boolean algebra is precise, English
  is not.
• A B
• The roads will be very slippery if it snows or
  rains and there is oil on the road.
• C
• F A BC
• OR
• F (A B) C

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Step 2 - Set up the truth table

• The air conditioner should be turned on if and
  only if the temperature is greater than 75, the
  time is between 8am and 5pm, and it is not a
  holiday.
• F air conditioner should be turned on
• A temperature is greater than 75
• B time is between 8am and 5 pm
• C it is a holiday

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Step 3 Write Boolean Equation

• The air conditioner should be turned on if and
  only if the temperature is greater than 75, the
  time is between 8am and 5pm, and it is not a
  holiday.
• F air conditioner should be turned on
• A temperature is greater than 75
• B time is between 8am and 5 pm
• C it is a holiday

F A B C
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Step 4 Simplify Expression

• F A B C
• This one is already simplified.

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Step 5 Draw Logic Diagram

• F A B C

A
B
F
C
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New Form of Truth Table

• Alternate form of truth table

AB
CD
00
01
11
10
00
01
11
10
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K-Map Usefulness

• Easier to see all the data than truth tables.
  Smaller, more compact, etc
• Because of the way K-maps are constructed, a
  technique can be used to actually help minimize
  complex logic circuits

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Other Sized K-Maps
AB
AB
CD
00
01
11
10
C
00
01
11
10
00
0
01
1
3 Variable Karnaugh Map
11
10
A
B
0
1
4 Variable Karnaugh Map
0
1
2 Variable Karnaugh Map
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Reducing Logic Using K-Maps

• Lets look at the K-Map

AB
CD
00
01
11
10
00
01
We can write the reduced Boolean Equation
directly X ABC
11
10
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Step 1 Create K-Map

• Construct K-Map with variables along top and
  left-hand edges.
• Label rows and columns using Gray code.
• Place 1s in squares corresponding to 1s in the
  Truth Table. 0s elsewhere.

63
Step 2 For SOP Equation

• Examine map for adjacent 1s. Create loops as
  large as possible (size must be a power of 2 1,
  2, 4, 8, 16).
• Create fewest possible number of loops.
• Form the OR of all the terms generated by each
  loop

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Step 2 For POS Equation

• Examine map for adjacent 0s. Create loops as
  large as possible (size must be a power of 2 1,
  2, 4, 8, 16).
• Create fewest possible number of loops.
• Form the AND of all the terms generated by each
  loop

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Example 2

• Find the non-reduced Boolean expression
• Find the reduced equation

AB
CD
00
01
11
10
X ABCD ABCD ABCD ABCD
00
01
X ABC ABC
11
10
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Loop Size

• Here is a group of 8
• 3 variable disappear ( 23 8)

AB
CD
00
01
11
10
00
X A
01
11
10
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Edges are Adjacent

• The two outside columns are adjacent
• The top and bottom rows are also adjacent

AB
CD
00
01
11
10
00
X BD
01
11
10
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Incompletely Specified Functions

• Here is a 2 stage digital system
• We want F 1 for ABC
• ABC, and ABC, what does
• the TT look like?

Note N1 does not generate outputs A B C with
values of 101 or 111
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Incompletely Specified Functions

• Multiple ways to choose those Xs when
  minimizing. Cost of minimal solution will depend
  on which one chosen.

F1 ABC AC
F2 ABC AC AB
F3 BC AC
F4 A ABC A BC Let the K-Map decide!
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K-Maps with Dont Cares
AB
C
00
01
11
10
0
1
Minimum Boolean Equation
F A BC
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Summary

• Boolean Operations
• Truth tables
• Logic gates
• Alternate gate shapes
• Functionally complete gates
• Boolean expressions
• Boolean theorems
• Canonical forms (SOP POS)
• Boolean minimization
• The design process
• Karnaugh maps

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Homework

• HW 3 Chapt 3 Probs 1, 3, 12

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Next Time

• MSI Circuits
• Multiplexers
• Demultiplexers
• Encoders
• Decoders
• Comparators
• Adders

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