Logic Circuits and Computer Architecture

1
Logic Circuits and Computer Architecture

• Appendix A
• Digital Logic Circuits
• Part 1 Combinational Circuits
• and Minimization

2
Structured organization

• Problem-oriented language level
• Assembly language level
• Operating system machine level
• Instruction set architecture level
• Microarchitecture level
• Digital logic level

Abstraction lelvel
3
Digital Logic level

• Digital circuits
• Only two logical levels present (i.e., binary)
• low/high voltage
• Basic gates
• AND, OR, NOT
• Basic circuits
• Combinational (without memory, stateless)
• Sequential (with memory, state dependent
  behaviour)

4
Boolean Algebra

• Variables A, B,
• Domain of variables 2 values
• 1 or 0 Y or N true or false
• Fundamental Operations
• AND, OR, NOT
• Intended meaning (for humans - Laws of Thought)
• AND both inputs are true
• OR at least one input is true
• NOT negate the input
• Named from George Boole

5
George Boole (1815-1864)
An Investigation of the Laws of Thought, on
Which are founded the Mathematical Theories of
Logic and Probabilities (1854)
6
Claude Shannon (1916-2001)
A Symbolic Analysis of Relay and Switching
Circuits (1938)
ENIAC (Electronic Numerical Integrator And
Calculator) (1946)
7
Formal definition of functions (1)

• By means of truth tables
• Explicit representation of the output for all
  possible inputs

A B OR
0 0 0
0 1 1
1 0 1
1 1 1
A B AND
0 0 0
0 1 0
1 0 0
1 1 1
A NOT
0 1
1 0
8
truth table for a 3-variable function

• f(A,B,C) 1 if and only if at least 2 variables
  are equal to 1

A B C f(A,B,C)
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
9
Formal definition of functions (2)

• By means of boolean equation
• boolean equation consists of
• variables
• constants 0 and 1
• boolean operations (AND, OR, NOT)
• parentheses
• M OR( AND(NOT(A),NOT(B)), AND(A,B) )

10
Boolean functions

• Conventions
• NOT (negation) NOT(A) A A
• AND (conjunction) AND(A,B) AB A.B
• OR (disjunction) OR(A,B) AB

M OR( AND(NOT(A),NOT(B)), AND(A,B) ) M
(((A)(B)) (AB))
11
Boolean Operator Precedence

• The order of evaluation in boolean expression is
• Parentheses
• NOT
• AND
• OR
• Consequence parantheses appear around OR
  expressions
• Example FA(BC)(CD)

M (((A)(B)) (AB)) M AB AB
12
NOT gate - the simplest one

• NOT gate - inverts the signal
• If A is 0, X is 1
• If A is 1, X is 0
• A NOT gate is also called an inverter

13
AND gate

• Output is 1 if all inputs are 1
• In general, if the AND gate has N inputs, both
  input 1 AND input 2 AND AND input N must be 1
  for the output to be 1
• 2-input AND gate

14
OR gate

• Output is 1 if at least one input is 1
• In general, if the OR gate has N inputs, input 1
  OR input 2 OR OR input N must be 1 for the
  output to be 1
• 2-input OR gate

15
A more complex example

• 2-input equivalence circuit
• The output is 1 if the inputs are the same
• (i.e., both 0 or both 1)
• Truth table
• Boolean function
• M AB AB

A B M
0 0 1
0 1 0
1 0 0
1 1 1
16
Formal definition of functions (3)

• By means of logic circuits
• Combination of logic gates joined by wires

17
Conventions for logic circuits
18
Exercise (1)

• Write the truth table and the logic circuit for
• F X YZ

19
Truth table
X Y Z F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
20
Logic Circuit
21
Exercise (2)

• Write the boolean function and its truth table
  for the following logic circuit

22
Function and Truth Table

• F Y XYZ XY

X Y Z F
0 0 0 1
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
23
Exercise (3)

• Write the boolean function and its truth table
  for the following logic circuit

24
Function and Truth Table

• F XYZ XYZ XZ

25
Conversion between represent.

• Circuit -gt
• -gt Boolean formula (left-to-right inspection)
• -gt Truth table (explicit case-by-case
  computation)
• Boolean formula -gt
• -gt Circuit (bottom-up construction)
• -gt Truth table (explicit case-by-case evaluation)
• Truth table -gt
• -gt Circuit (through boolean formula)
• -gt Boolean formula (through canonical form see
  later)

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Boolean Identities
duality principle any algebraic equality remains
true when the operators OR and AND, and the
elements 0 and 1 are interchanged

• 1A A 0A A Identity
• 0A 0 1A 1 Null
• AA A AA A Idempotent
• AA 0 AA 1 Inverse
• AB BA AB BA Commutative
• (AB)C A(BC) (AB)C A(BC) Associative
• ABC (AB)(AC) A(BC) ABAC Distributive
• A(AB) A AAB A Absorption
• (AB) AB (AB) AB De Morgan

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Truth tables to verify De Morgans theorem
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Remark Each equality remains true if you
sobstitute any variable with any expression
Examples
(AB)(ACD) A BCD
(distributive)
((ABC)(DA)) (ABC) (DA)
(De Morgan) A
(BC) DA (De Morgan)
A(BC) DA
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algebraic manipulation
F XYZ XYZ XZ (distributive)
XY(Z Z) XZ (inverse) XY 1
XZ (identity) XY XZ
30
Boolean Algebra Vs Switching Algebra (1)
A Boolean Algebra is a structure A ltA, , ,
, 0, 1gt where

• A is a set
• and are binary operations
• is a unary operation
• 0, 1 ? A

satisfying the following axioms
(i) and are commutative
(ii) 0 and 1 satisfy a1a and a0a, ? a
?A
(iii) and distribute over each other
(iv) for each element a ?A, there exists an
element a ?A such that a
a 1 and aa0
31
Boolean Algebra Vs Switching algebra (2)
Switching Algebra is the following boolean algebra
A lt0,1, , , , 0, 1gt
32
Observation Axioms (i)-(iv) can be used to prove
all the other identities
An example Idempotent X X X
X X (X X)1 (ii)
(X X)(X X) (iv)
X (XX) (iii)
X 0
(iv) X
(ii)
33
De Morgan circuit equivalents

• AND/OR can be interchanged if you invert the
  inputs and outputs

bubble means inversion
34
NAND gate - the negation of AND

• The opposite of the AND gate is the NAND gate
  (output is 0 if all inputs are 1)
• Truth table
• Logic diagram

A B NAND
0 0 1
0 1 1
1 0 1
1 1 0
35
NOR gate - the negation of OR

• The opposite of the OR gate is the NOR gate
  (output is 0 if any input is 1)
• Truth table
• Logic diagram

A B NOR
0 0 1
0 1 0
1 0 0
1 1 0
36
Exercise

• Write the truth table for
• a 3 input NAND gate
• a 3 input NOR gate

37
XOR gate - the exclusive OR

• For a 2-input gate
• Output is 1 if exactly one of the inputs is 1
• Truth table
• Logic diagram
• For gt 2 inputs output is 1 if
• an odd number of inputs is 1

A B XOR
0 0 0
0 1 1
1 0 1
1 1 0
38
Universal Gates

• How many logical functions there are with n
  input?
• With n inputs there are 2(2n ) possible logical
  functions

A B 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
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Universal Gates (2)

• AND, OR, NOT can generate all possible boolean
  functions (see later)
• Is it possible to use fewer basic operations?
• Universal gate a gate type that can implement
  any Boolean function

40
Universal Gates (3)

• AND, NOT are enough !
• OR, NOT are enough !
• Even NAND alone or NOR alone are enough !

41
How NAND simulates AND, OR

• Simulation of NOT ???

42
Alternative NAND representations
43
How NOR simulates AND, OR

• Simulation of NOT ???

44
Alternative NOR representations
45
Gate equivalence

• Any AND, OR, NOT gate can be obtained using just
  NAND gates or just NOR gates
• Consequence any circuit can be constructed using
  just NAND gates or just NOR gates (easier to
  build)

46
Equivalence modifications (1)
47
Equivalence modifications (2)

• Substitute equivalent gates

48
Transforming OR, AND to NAND

• Transform the following circuit

49
Solution
50
Exercise

• Write a NAND only logic circuit for
• F XY XY Z

51
Solution
52
Exercise

• Write a NAND only logic circuit for the exclusive
  OR function (XOR)
• XOR(A,B) AB AB

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Solution (1)

• Truth table initial circuit

A B XOR
0 0 0
0 1 1
1 0 1
1 1 0
54
Solution (2) equivalence transform.
A B A B

• A
• B
• A
• B

A B A B
55
Exercise

• Write a NOR only logic circuit for
• F (AD)(CD)E

56
Solution

• Direct realization

57
SOP and POS rapresentations

• It is useful to specify boolean function in a
  particular form
• sum of products (SOP) rapresentation
• it is an OR of AND combinations of its inputs
• F1AB BC it is a SOP rapresentation
• F2 AB (BC) it is not a SOP rapresentation
• product of sums (POS) rapresentation
• it is an AND of OR combiations of its inputs
• F1A(B C)(BA) it is a POS rapresentation
• F2AB BC it is not a POS rapresentation

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Exercise

• Express Z(A(BC(AB))) as sum of products

59
Solution

• Z(A(BC(AB)))
• A (BC(AB))
• A B(C(AB))
• A B(C(AB))
• A B(CAB)
• A BC ABB
• A BC A0
• A BC 0 A BC

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Canonical Form for boolean functions

• It is a standard way of expressing SOP or POS
• It is
• a sum of minterms, for SOP
• a product of maxterms for POS
• A minterm is a product containing all variables,
  either in the positive form or in the negative
  form
• A maxterm is a sum containing all variables,
  either in the positive or in the negative form.
• Examples
• F (ABC) (BC) is not in a POS canonical
  form
• M AB ABC is not in a SOP canonical form

61
Minterms

• Given that each variable may appear normal (e.g.
  X) or complemented (e.g. X), there are 2n
  minterms for n variables
• Example Two variables (X and Y) produce 4
  combinations
• XY
• XY
• XY
• XY
• Thus there are 4 minterms of 2 variables

62
Maxterms

• Given that each variable may appear normal (e.g.
  X) or complemented (e.g. X), there are 2n
  maxterms for n variables
• Example Two variables (X and Y) produce 4
  combinations
• XY
• XY
• XY
• XY
• Thus there are 4 maxterms of 2 variables

63
Maxterms and Minterms

• Examples 2 variable minterms and maxterms
• The index above is important for describing which
  variables in the terms are true and which are
  complemented

Index Minterm Maxterm
0 XY XY
1 XY XY
2 XY XY
3 XY XY
64
Number Systems Representation

• Positive radix, positional number systems
• A number with radix r is represented by a string
  of digits An - 1An - 2 A1A0 . A- 1 A- 2
  A- m 1 A- m in which 0 Ai lt r and . is the
  radix point.
• The string of digits represents the power series

(
)
(
)
65
Number Systems Examples
General Decimal Binary
Radix (Base) r 10 2
Digits 0 gt r - 1 0 gt 9 0 gt 1
0 1 2 3 4 5 -1 -2 -3 -4 -5 r0 r1 r2 r3 r4 r5 r -1 r -2 r -3 r -4 r -5 1 10 100 1000 10,000 100,000 0.1 0.01 0.001 0.0001 0.00001 1 2 4 8 16 32 0.5 0.25 0.125 0.0625 0.03125
powers of radix
66
Standard Order

• Minterms and Maxterms are designated with
  subscript
• The subscript is a numer, corresponding to a
  binary pattern
• The bits in the pattern represent the
  complemented or normal state of each variable
  listed in a standard order
• All the variables will be present in a minterm or
  maxterm and will be listed in the same order
  (usually alphabetically)
• Example For variables A, B, C
• Maxterms (A B C), (A B C)
• Terms (B A C), ACB, and (C B A) are
  NOT in standard order
• Minterms ABC, ABC, ABC
• Terms (AC), BC, and (AB) do not contain
  all variables

67
Purpose of the Index

• The index for the minterm or maxterm, expressed
  as a binary number, is used to determine whether
  the variable is shown in the true form or
  complemented form.
• For Minterms
• 1 means the variable is Not Complemented and
• 0 means the variable is Complemented
• For Maxterms
• 0 means the variable is Not Complemented and
  
• 1 means the variable is Complemented

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X Y Z Minterm Symbol m0 m1 m2 m3 m4 m5 m6 m7
0 0 0 XYZ m0 1 0 0 0 0 0 0 0
0 0 1 XYZ m1 0 1 0 0 0 0 0 0
0 1 0 XYZ m2 0 0 1 0 0 0 0 0
0 1 1 XYZ m3 0 0 0 1 0 0 0 0
1 0 0 XYZ m4 0 0 0 0 1 0 0 0
1 0 1 XYZ m5 0 0 0 0 0 1 0 0
1 1 0 XYZ m6 0 0 0 0 0 0 1 0
1 1 1 XYZ m7 0 0 0 0 0 0 0 1
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X Y Z Maxterm Symbol M0 M1 M2 M3 M4 M5 M6 M7
0 0 0 XYZ M0 0 1 1 1 1 1 1 1
0 0 1 XYZ M1 1 0 1 1 1 1 1 1
0 1 0 XYZ M2 1 1 0 1 1 1 1 1
0 1 1 XYZ M3 1 1 1 0 1 1 1 1
1 0 0 XYZ M4 1 1 1 1 0 1 1 1
1 0 1 XYZ M5 1 1 1 1 1 0 1 1
1 1 0 XYZ M6 1 1 1 1 1 1 0 1
1 1 1 XYZ M7 1 1 1 1 1 1 1 0
70
Minterm and Maxterm Relationship





y
x
y
x
y
x

y


x

• DeMorgan's Theorem
• Two-variable example
• Thus M2 is the complement of m2 and vice-versa.
• Since DeMorgan's Theorem holds for n variables,
  the above holds for terms of n variables
• giving
• Thus Mi is the complement of mi.

y
x


m

y

x


M
2
2
71
Boolean function implementation

• Any function can be implemented as the OR of the
  AND combinations of its inputs
• canonical SOP
• Start from the truth table
• For each 1 in the output
• Write its inputs in AND
• Write these in OR
• MABCABCABCABC
• m3m5m6m7

A B C M
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
72
Equivalent Logic Circuit
73
Boolean function implem. (2)

• Any function can be implemented as the AND of the
  OR combinations of its inputs
• canonical POS
• Start from the truth table
• For each 0 in the output
• Write its inputs in AND
• Write these negated in AND
• Obtain F Z0 . Z1 . Z2 ...
• Finally, apply De Morgan to F
• It is an AND combination of maxterms

A B C F
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 0
74
Boolean function implem. (3)

• De Morgan (general) (ABC)ABC
• F(ABC).(ABC).(ABC).(ABC).(ABC)
• (ABC).(ABC).(ABC).
• .(ABC).(ABC)
• (ABC).(ABC).(ABC).
• .(ABC).(ABC)
• M0M1M4M5M7

75
Boolean function implem. (4)

• Alternative procedure for POS form (use only if
  you know what you are doing!)
• Complement the table by substituting everywhere a
  0 with a 1 and a 1 with a 0
• Write a SOP form for the complemented table
• Complement the formula by substituting everywhere
  and AND with an OR and an OR with an AND
• Why does it work ???

76
Example canonical SOP
X Y Z F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
m1 m3 m6 m7
0 0 0 0
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 0
0 0 1 0
0 0 0 1
canonical SOP
F m1 m3 m6 m7 XYZ XYZ XYZ
XYZ
77
Example canonical POS
X Y Z F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
M0 M2 M4 M5
0 1 1 1
1 1 1 1
1 0 1 1
1 1 1 1
1 1 0 1
1 1 1 0
1 1 1 1
1 1 1 1
canonical POS
F M0 M2 M4 M5 (XYZ) (XYZ)
(XYZ) (XYZ)
78
algebraic manipulation
F XYZ XYZ XZ XY(Z Z) XZ
XY 1 XZ XY XZ
a simpler SOP representation leads to a
simpler circuit
79
Circuit Optimization

• Goal To obtain the simplest implementation for a
  given function
• Optimization is a more formal approach to
  simplification that is performed using a specific
  procedure or algorithm
• Optimization requires a cost criterion to measure
  the simplicity of a circuit
• Optimization for two-level (SOP and POS)
  circuits
• minimal SOP
• minimum number of pruduct terms
• minimum number of literals for each product term
• similarly for POS

80
Minimization procedures

• Karnaughs maps (by hand)
• Used to minimize boolean functions of up to 4
  input variables
• For more variables use the method of
  Quine-McKluskey (programmable)

81
Karnaughs Maps (KM)

• Grid-like representation for boolean functions
• Each cell represents a minterm
• The K-map can be viewed as a reorganized version
  of the truth table
• Minterms with just one variable different
  occupies adjacent cells
• Alternative algebraic expressions for the same
  function are derived by recognizing patterns of
  squares
• Consider only 1s in the representation (focusing
  on a SOP representation)
• IDEA if 2 adjacent cells have a 1 the function
  can be simplified

82
A KM for 2-variable functions

• The generic KM
• Function F XY Function F X Y

83
A KM for 3-variable functions
84
example
X Y Z FZ
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 1
1
1
1
1
85
X Y Z FZ
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 0
1
1
1
1
86
X Y Z FYZ
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1
1
1
87
An example

• F XYZ XYZ XYZ XYZ

XY(ZZ) XY(ZZ) XY XY
X Y Z F
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 0

• Idea
• we want to cover all 1s by using rectangles of
  adjacent cells
• each rectangle of 2k adjacent cells (for some k)
  represents a literal
• product term
• bigger rectangles correspond to simpler product
  terms

88
Combining Squares

• By combining squares, we reduce number of
  literals in a product term
• On a 3-variable K-Map
• One square represents a minterm with three
  variables
• Two adjacent squares represent a product term
  with two variables
• Four adjacent terms represent a product term
  with one variable
• Eight adjacent terms is the function of all
  ones (no variables) 1.

89
Circular adjacencies for 3 variables
labels are useful to get the expression correspond
ing to a given rectangle
90
Three-Variable Maps

• Example Shapes of 2-cell Rectangles

y
XY
XZ
YZ
91
Three-Variable Maps

• Example Shapes of 4-cell Rectangles

y
92
Four 1 adjacents
F XYZ XYZ XYZ XYZ YZ(XX)
YZ (XX) YZ YZ Z(YY) Z
93
Exercise (1)

• Which is the minimal SOP expression for function
  F1m3m4m6m7?

94
Solution

• F1 YZ XZ

95
Exercise (2)

• Which is the minimal SOP expression for function
  F2m0m2m4m5m6?

96
Solution

• F2 Z XY

97
k-cube of 1s

• Is a set of 2k adjacent cells
• 0-cube, 1 cell, a minterm
• 1-cube, 2 adjacent cells
• 2-cube, 4 adjacent cells
• 3-cube, 8 adjacent cells
• .

98
Prime implicants

• F P1 P2 P3 ...
• The term corresponding to a k-cube is called an
  implicant
• Infact, it is a term Pn which implies the
  function F,
• i.e. if Pn is true then F is true
• An implicant is said to be a prime implicant for
  F if it does not imply any other implicant of F
• A prime implicant can be chosen by selecting a
  maximal k-cube, i.e. a k-cube in the KM which is
  not contained in any larger h-cube (hgtk)

99
Minimal representation

• F P1 P2 P3 ...
• has a minimal SOP representation if
• 1. Each Pi is a prime implicant
• 2. There is a minimum number of them

100
Minimality procedure

1. Find maximal k-cubes (prime implicants)
2. If a 1 is covered by only one maximal k-cube this
   has to be chosen (essential prime implicants)
3. Select a minimum number of the remaining k-cube
   so to cover all 1s not covered by essential prime
   implicants

101
Exercise (3)

• Find the minimal SOP expression for
• Fm1m3m4m5m6

1
1
1
1
1
FXZXZXY
102
Exercise (3)

• Find the minimal SOP expression for
• Fm1m2m3m5m7

1
1
1
1
1
FZXY
103
A KM for 4-variable functions
104
Circular adjacenciesfor 4 variables
105
Four Variable Terms

• Four variable maps can have rectangles
  corresponding to
• A single 1 4 variables, (i.e. Minterm)
• Two 1s 3 variables,
• Four 1s 2 variables
• Eight 1s 1 variable,
• Sixteen 1s zero variables (i.e. Constant "1")

106
Four-Variable Maps

• Example Shapes of Rectangles

XZ
YW
XZ
107
Four-Variable Maps

• Example Shapes of Rectangles

Y
X
W
Z
108
Example

• Simplify F(A, B, C, D) given on the K-map.

minimal SOP ABACDACDBCD
109
One more example
minimal SOP BDBDCDAB


ESSENTIAL Prime Implicants
C


110
Five Variable or More K-Maps

• For five variable problems, we use two adjacent
  K-maps. It becomes harder to visualize adjacent
  minterms for selecting k-cubes.
• You can extend the problem to six variables by
  using four K-Maps.

111
The KM method for POS

• Which is the POS expression of function F
  represented by this KM?
• Use the same method used for build POS canonical
  form from truth tables
• Find the minimal SOP for F
• apply DeMorgan

112
Example
1
0
1
1
1
0
0
0
0
0
0
0
1
1
0
1

• F (CDBDAB) (CD) . (BD) . (AB)
• (CD) . (BD) . (A B)
• (CD) . (BD) . (A B)

113
Don't Cares in K-Maps

• Sometimes a function table or map contains
  entries for which it is known
• the input values for the minterm will never
  occur, or
• the output value for the minterm is not used
• In these cases, the output value need not be
  defined
• Instead, the output value is defined as a don't
  care
• By placing don't cares ( an x entry) in the
  function table or map, the cost of the logic
  circuit may be lowered.

114
Example 1
A B C D f(A,B,C,D)
0 0 0 0 f(0,0,0,0)
0 0 0 1 f(0,0,0,1)
0 0 1 0 f(0,0,1,0)
0 0 1 1 f(0,0,1,1)
0 1 0 0 f(0,1,0,0)
0 1 0 1 f(0,1,0,1)
0 1 1 0 f(0,1,1,0)
0 1 1 1 f(0,1,1,1)
1 0 0 0 f(1,0,0,0)
1 0 0 1 f(1,0,0,1)
1 0 1 0 x
1 0 1 1 x
1 1 0 0 x
1 1 0 1 x
1 1 1 0 x
1 1 1 1 x

• A logic function having the binary codes for the
  BCD digits as its inputs.
• Only the codes for 0 through 9 are used.
• The six codes, 1010 through 1111 never occur, so
  the output values for these codes are x to
  represent dont cares.

115
Example 2

• Consider the following function f(A,B)
• f(A,B)1 if AB0
• f(A,B)0 if A ? B

Truth table on the left may be substitued by
anyone on the right
A B f
0 0 1
0 1 0
1 0 0
1 1 x
A B f
0 0 1
0 1 0
1 0 0
1 1 0
A B f
0 0 1
0 1 0
1 0 0
1 1 1
116
Example 3

• Consider the following function f(A,B)
• f(A,B)1 if AB0
• f(A,B)0 if A ? B
• f(A,B) is used just as input for another function
  g(f,A,B)(AB) f(A,B)

A B f
0 0 1
0 1 0
1 0 x
1 1 x
Notice that for A1 and B0 g(f,A,B) (10)
f(A,B) 0
117
Example
f(W,X,Y,Z) YZ XW
X
X
1
1
Simplify the choice, since each X can be
considered a 0 or a 1
X
1
1
1
118
Example
f(W,X,Y,Z) YZ ZW
X
X
1
1
a different choise is possible
X
1
1
1