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Chapter 2 <1>
Digital Design and Computer Architecture, 2nd Edition
Part 3
David Money Harris and Sarah L. Harris
2
Boolean Algebra 1/2• A set of elements B
– There exist at least two elements x, y Î B s. t. x ¹ y • Binary operators: + and ·
closure w.r.t. both + and ·x, y Î B, (x+y) Î B , (x ·y) Î B
additive identity ?0 : x + 0 = 0 + x = x
multiplicative identity ?1 : x ·1 = 1 ·x = x
commutative w.r.t. both + and ·associative w.r.t. both + and ·
• Distributive law:is · distributive over + ?
yes : x ·(y+z) = (x ·y)+(x ·z)is + distributive over · ?
yes : x+(y ·z) = (x+y) ·(x+z)We do not have the second one in ordinary algebra
3
Boolean Algebra 2/2• Complement– " x Î B, there exist an element x’ Î B '
a. x + x’ = 1 (multiplicative identity) andb. x · x’ = 0 (additive identity)
– Not available in ordinary algebra • No inverses in Boolean Algebra!• Differences between ordinary and Boolean algebra– Ordinary algebra deals with real numbers (infinite)– Boolean algebra deals with elements of set B (finite)– Complement– Distributive law– Do not substitute laws from one to another where they
are not applicable
4
Two-Valued Boolean Algebra • Check the axioms– Two distinct elements, 0 ¹ 1– Closure, associative, commutative, identity elements– Complement
x + x ’ = 1 and x · x ’ = 0– Distributive law
x y z
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
y×z x+(y·z) x + y x + z (x + y) · (x + z)
5
Two-Valued Boolean Algebra • Two-valued Boolean algebra is actually equivalent to
the binary logic that we defined heuristically beforeOperations:
Ø · à ANDØ + à ORØ Complement à NOT
• Binary logic is the application of Boolean algebra to gate-type circuits– Two-valued Boolean algebra is developed in a formal
mathematical manner– This formalism is necessary to develop theorems and
properties of Boolean algebra
6
Duality Principle• An important principle– every algebraic expression deducible from the axioms of
Boolean algebra remains valid if the operators and identity elements are interchanged together
• Example:§ x + x = x§ x + x = (x+x) · 1 (identity element)
= (x+x)(x+x’) (complement)= x+(x · x’) (+ over ·)= x (complement)
§ duality principlex + x = x è x · x = x
7
Duality Principle & Theorems• Theorem a:
§ x + 1 = 1 hmmm?§ x + 1 = 1 · (x + 1)
= (x + x’)(x + 1)= x + x’ · 1= x + x’= 1
• Theorem b: (using duality) § x . 0 = 0
8
Absorption Theorem
x + xy = x hmmmmm?= x.1+xy= x(1+y)= x(1)= x
9
Involution & DeMorgan’s Theorems• Involution Theorem:
§ (x’)’ = x § x + x’ = 1 and x · x’ = 0§ Complement of x’ is x§ Complement is unique
• DeMorgan’s Theorem: a. (x + y)’ = x’ · y’b. From duality ? (x.y)’=x’+y’
10
Truth Tables for DeMorgan’s Theorem(x + y)’ = x’ · y ’
x’ y ’ x’ · y ’ x’ + y’1 11 00 10 0
x y x+y (x+y)’ x · y (x · y)’0 00 11 01 1
11
Operator Precedence(Boolean Operator Priority Order)
1. Parentheses2. NOT3. AND4. OR
PiNA cOlada?
12
Boolean Functions• Consists of – binary variables (normal or complement form)– the constants, 0 and 1– logic operation symbols “+” and “·”
• Example:§ F1(x, y, z) = x + y’ z§ F2(x, y, z) = x’ y’ z + x’ y z + xy’
1
1
1
1
0
0
1
0
0
1
1
1
0
1
00000
111
011
101
001
110
010
100
F2F1zyx
Chapter 2 <13>
• Every element is combinational• Every node is either an input or connects
to exactly one output• The circuit contains no cyclic paths• Example:
Rules of Combinational Composition
Chapter 2 <14>
• Functional specification of outputs in terms of inputs
• Example: S = F(A, B, Cin)Cout = F(A, B, Cin)
A S
S = A Å B Å CinCout = AB + ACin + BCin
BCin
CL Cout
Boolean Equations
Chapter 2 <15>
• Complement: variable with a bar over itA, B, C
• Literal: variable or its complementA, A, B, B, C, C
• Implicant: product of literalsABC, AC, BC
• Minterm: product that includes all input variablesABC, ABC, ABC
• Maxterm: sum that includes all input variables(A+B+C), (A+B+C), (A+B+C)
Some Definitions
Chapter 2 <16>
• Two-level logic: ANDs followed by ORs• Example: Y = ABC + ABC + ABC
BA C
Y
minterm: ABC
minterm: ABC
minterm: ABC
A B C
From Logic to Gates
Chapter 2 <17>
• Inputs on the left (or top)• Outputs on right (or bottom)• Gates flow from left to right• Straight wires are best
Circuit Schematics Rules
Chapter 2 <18>
• Wires always connect at a T junction• A dot where wires cross indicates a
connection between the wires• Wires crossing without a dot make no
connectionwires connectat a T junction
wires connectat a dot
wires crossingwithout a dot do
not connect
Circuit Schematic Rules (cont.)
Chapter 2 <19>
A1 A00 00 11 01 1
Y3 Y2 Y1 Y0A3 A20 00 00 00 0
0 00 10 11 01 10 0
0 10 10 11 0
0 11 01 01 10 00 1
1 01 01 11 1
1 01 11 11 1
A0
A1
PRIORITYCiIRCUIT
A2
A3
Y0
Y1
Y2
Y3
• Example: Priority CircuitOutput assertedcorresponding tomost significantTRUE input
Multiple-Output Circuits
Chapter 2 <20>
0
A1 A00 00 11 01 1
0
00
Y3 Y2 Y1 Y00000
0011
0100
A3 A20 00 00 00 0
0 0 0 1 0 00 10 11 01 10 0
0 10 10 11 0
0 11 01 01 10 00 1
1 01 01 11 1
1 01 11 11 1
0001
1110
0000
0000
1 0 0 01111
0000
0000
0000
1 0 0 01 0 0 0
A0
A1
PRIORITYCiIRCUIT
A2
A3
Y0
Y1
Y2
Y3
• Example: Priority CircuitOutput assertedcorresponding tomost significantTRUE input
Multiple-Output Circuits
Chapter 2 <21>
A1 A00 00 11 01 1
0000
Y3 Y2 Y1 Y00000
0011
0100
A3 A20 00 00 00 0
0 0 0 1 0 00 10 11 01 10 0
0 10 10 11 0
0 11 01 01 10 00 1
1 01 01 11 1
1 01 11 11 1
0001
1110
0000
0000
1 0 0 01111
0000
0000
0000
1 0 0 01 0 0 0
A3A2A1A0Y3Y2
Y1
Y0
Priority Circuit Hardware
Chapter 2 <22>
A1 A00 00 11 01 1
0000
Y3 Y2 Y1 Y00000
0011
0100
A3 A20 00 00 00 0
0 0 0 1 0 00 10 11 01 10 0
0 10 10 11 0
0 11 01 01 10 00 1
1 01 01 11 1
1 01 11 11 1
0001
1110
0000
0000
1 0 0 01111
0000
0000
0000
1 0 0 01 0 0 0
A1 A00 00 11 XX X
0000
Y3 Y2 Y1 Y00001
0010
0100
A3 A20 00 00 00 1
X X 1 0 0 01 X
Don’t Cares
Chapter 2 <23>
• Contention: circuit tries to drive output to 1 and 0– Actual value somewhere in between– Could be 0, 1, or in forbidden zone– Might change with voltage, temperature, time, noise– Often causes excessive power dissipation
• Warnings: – Contention usually indicates a bug.
– X is used for “don’t care” and contention - look at the context to tell them apart
A = 1Y = X
B = 0
Contention: X
Chapter 2 <24>
• Floating, high impedance, open, high Z• Floating output might be 0, 1, or
somewhere in between– A voltmeter won’t indicate whether a node is floating
Tristate Buffer
E A Y0 0 Z0 1 Z1 0 01 1 1
A
E
Y
Floating: Z
Chapter 2 <25>
• Floating nodes are used in tristatebusses– Many different drivers– Exactly one is active at
once
en1
to busfrom bus
en2
to busfrom bus
en3
to busfrom bus
en4
to busfrom bus
sha
red
bu
sprocessor
video
Ethernet
memory
Tristate Busses
26
Logic Circuit Diagram of F1F1(x, y, z) = x + y’ z
27
Logic Circuit Diagram of F1F1(x, y, z) = x + y’ z
x
y
zy’
y’z
x + y’ z
Gate Implementation of F1 = x + y’ z
28
Logic Circuit Diagram of F2F2 = x’ y’ z + x’ y z + xy’
29
Logic Circuit Diagram of F2F2 = x’ y’ z + x’ y z + xy’
x
y
z
Algebraic manipulationF2 = x’ y’ z + x’ y z + xy’
= x’z(y’+y) + xy’
F2
30
Alternative Implementations of F2F2 = x’ z + xy’
31
Alternative Implementations of F2F2 = x’ z + xy’x
y
zF2
32
Alternative Implementations of F2F2 = x’ z + xy’x
y
zF2
F2 = x’ y’ z + x’ y z + xy’
F2
x
y
z
Example
Example
Example
Example
37
OTHER LOGIC OPERATORS - 1• AND, OR, NOT are logic operators
– Boolean functions with two variables– These are three of the 16 possible two-variable Boolean functions
x y F0 F1 F2 F3 F4 F5 F6 F7
0 0 0 0 0 0 0 0 0 00 1 0 0 0 0 1 1 1 11 0 0 0 1 1 0 0 1 11 1 0 1 0 1 0 1 0 1
x y F8 F9 F10 F11 F12 F13 F14 F15
0 0 1 1 1 1 1 1 1 10 1 0 0 0 0 1 1 1 11 0 0 0 1 1 0 0 1 11 1 0 1 0 1 0 1 0 1
38
OTHER LOGIC OPERATORS - 2• Some of the Boolean functions with two variables
§ Constant functions: F0 = 0 and F15 = 1§ AND function: F1 = xy§ OR function: F7 = x + y § XOR function: • F6 = x’ y + xy’ = x Å y (x or y, but not both)
§ XNOR (Equivalence) function:• F9 = xy + x’ y’ = (x Å y)’ (x equals y)
§ NOR function:• F8 = (x + y)’ = (x ¯ y) (Not-OR)
§ NAND function:• F14 = (x y)’ = (x y) (Not-AND)
Chapter 2 <39>
More Two-Input Logic Gates
Chapter 2 <40>
More Two-Input Logic Gates
Chapter 2 <41>
• Y = AB + AB
Simplifying Boolean EquationsExample 1:
Chapter 2 <42>
• Y = AB + AB= B(A + A)= B(1)= B
Simplifying Boolean EquationsExample 1:
Chapter 2 <43>
• Y = A(AB + ABC)Example 2:
Simplifying Boolean Equations
Chapter 2 <44>
• Y = A(AB + ABC)= A(AB(1 + C))= A(AB(1))= A(AB)= (AA)B= AB
Example 2:
Simplifying Boolean Equations
Chapter 2 <45>
• Y = AB = A + B
• Y = A + B = A B
AB Y
AB Y
AB Y
AB Y
DeMorgan’s Theorem
Chapter 2 <46>
• Backward:– Body changes– Adds bubbles to inputs
• Forward:– Body changes– Adds bubble to output
AB Y A
B Y
AB YA
B Y
Bubble Pushing
Chapter 2 <47>
AB
YCD
• What is the Boolean expression for this circuit?
Bubble Pushing
Chapter 2 <48>
AB
YCD
• What is the Boolean expression for this circuit?
Y = AB + CD
Bubble Pushing
Chapter 2 <49>
AB
C
DY
• Begin at output, then work toward inputs• Push bubbles on final output back • Draw gates in a form so bubbles cancel
Bubble Pushing Rules
Chapter 2 <50>
AB
C YD
Bubble Pushing Example
Chapter 2 <51>
AB
C YD
no outputbubble
Bubble Pushing Example
Chapter 2 <52>
bubble oninput and outputA
B
C
DY
AB
C YD
no outputbubble
Bubble Pushing Example
Chapter 2 <53>
AB
C
DY
bubble oninput and outputA
B
C
DY
AB
C YD
Y = ABC + D
no outputbubble
no bubble oninput and output
Bubble Pushing Example
54
Complement of a Function• F’ is complement of F– We can obtain F’, by interchanging of 0’s and 1’s in
the truth table
0
0
1
1
0
1
0
0
F
1
0
1
0
1
0
1
0
z
11
11
01
01
10
10
00
00
F’yx
F =
F’ =
55
Generalizing DeMorgan’s Theorem
• We can also utilize DeMorgan’s Theorem§ (x + y)’ = x’ y’§ (A + B + C)’
= (A+B)’C’= A’B’C’
• We can generalize DeMorgan’s Theorem§ (x1 + x2 + … + xN )’ = x1’ · x2’ · … · xN’
§ (x1 · x2 · … · xN )’ = x1’ + x2’ + … + xN’
56
Example: Complement of a Function• Example:
§ F1 = x’yz’ + x’y’z§ F1’ = (x’yz’ + x’y’z)’
= (x’yz’)’(x’y’z)’= (x + y’ + z)(x + y + z’)
§ F2 = x(y’z’ + yz)§ F2’ = (x(y’z’ + yz))’
= x’+(y’z’ + yz)’= x’ + (y + z) (y’ + z’)
• Easy Way to Complement: take the dual of the function and complement each literal
57
Universal Gates• NAND and NOR gates are universal• We know any Boolean function can be written in
terms of three logic operations:– AND, OR, NOT
• In return, NAND gate can implement these three logic gates by itself– So can NOR gate
00
10
01
11
0
1
1
1
11
01
10
00
(x’ y’ )’y ’x’(xy)’yx
58
NAND Gatex
xy
x
y
59
NOR Gatex
xy
x
y
60
Designs with NAND gatesExample 1/2
• A function:F1 = x’ y + xy’
x
y
61
Example 2/2F2 = x’ y’ + xy’
x
y
62
Multiple Input Gates• AND and OR operations:– They are both commutative and associative – No problem with extending the number of inputs
• NAND and NOR operations:– they are both commutative but not associative– Extending the number of inputs is not obvious
• Example: NAND gates§ ((xy)’z)’ ¹ (x(yz)’)’
§ ((xy)’z)’ = xy + z’§ (x(yz)’)’ = x’ + yz
63
Nonassociativity of NOR operation
z
64
Multiple Input Universal Gates• To overcome this difficulty, we define multiple-
input NAND and NOR gates in slightly different manner
Three input NAND gate: (x y z)’xy
z(x y z)’
(x + y + z)’xy
zThree input NOR gate:(x + y + z)’
65
Multiple Input Universal Gates(x + y + z)’
xyz
3-input NOR gate
xyz
(xyz)’
3-input NAND gate
ABC
DE
F =
Cascaded NAND gates
Chapter 2 <66>
• Multi-input XOR: Odd parity
Multiple-Input Logic Gates
67
XOR and XNOR Gates• XOR and XNOR operations are both commutative
and associative.• No problem manufacturing multiple input XOR
and XNOR gates• However, they are more costly from hardware
point of view.• Therefore, we usually have 2-input XOR and XNOR
gatesxy
z
x Å y Å z
68
3-input XOR Gates
Chapter 2 <69>
Y = F(A, B) =
• All equations can be written in SOP form• Each row has a minterm• A minterm is a product (AND) of literals• Each minterm is TRUE for that row (and only that row)• Form function by ORing minterms where the output is TRUE • Thus, a sum (OR) of products (AND terms)
Sum-of-Products (SOP) Form
A B Y0 00 11 01 1
0101
minterm
A BA BA B
A B
mintermnamem0m1m2m3
Chapter 2 <70>
Y = F(A, B) =
Sum-of-Products (SOP) Form
• All equations can be written in SOP form• Each row has a minterm• A minterm is a product (AND) of literals• Each minterm is TRUE for that row (and only that row)• Form function by ORing minterms where the output is TRUE • Thus, a sum (OR) of products (AND terms)
A B Y0 00 11 01 1
0101
minterm
A BA BA B
A B
mintermnamem0m1m2m3
Chapter 2 <71>
Y = F(A, B) = AB + AB = Σ(1, 3)
Sum-of-Products (SOP) Form
• All equations can be written in SOP form• Each row has a minterm• A minterm is a product (AND) of literals• Each minterm is TRUE for that row (and only that row)• Form function by ORing minterms where the output is TRUE • Thus, a sum (OR) of products (AND terms)
A B Y0 00 11 01 1
0101
minterm
A BA BA B
A B
mintermnamem0m1m2m3
Chapter 2 <72>
A + BA B Y0 00 11 01 1
0101
maxterm
A + BA + BA + B
maxtermnameM0M1M2M3
Y = F(A, B) = (A + B)(A + B) = Π(0, 2)
• All Boolean equations can be written in POS form• Each row has a maxterm• A maxterm is a sum (OR) of literals• Each maxterm is FALSE for that row (and only that row)• Form function by ANDing the maxterms for which the
output is FALSE• Thus, a product (AND) of sums (OR terms)
Product-of-Sums (POS) Form
Chapter 2 <73>
• You are going to the cafeteria for lunch– You won’t eat lunch (E) – If it’s not open (O) or– If they only serve corndogs (C)
• Write a truth table for determining if you will eat lunch (E).
O C E0 00 11 01 1
Boolean Equations Example
Chapter 2 <74>
• You are going to the cafeteria for lunch– You won’t eat lunch (E) – If it’s not open (O) or– If they only serve corndogs (C)
• Write a truth table for determining if you will eat lunch (E).
O C E0 00 11 01 1
0010
Boolean Equations Example
Chapter 2 <75>
• SOP – sum-of-products
• POS – product-of-sums
O C E0 00 11 01 1
mintermO CO CO CO C
O + CO C Y0 00 11 01 1
maxterm
O + CO + CO + C
SOP & POS Form
Chapter 2 <76>
• SOP – sum-of-products
• POS – product-of-sums
O + CO C E0 00 11 01 1
0010
maxterm
O + CO + CO + C
O C E0 00 11 01 1
0010
minterm
O CO CO C
O C
Y = (O + C)(O + C)(O + C)= Π(0, 1, 3)
Y = OC= Σ(2)
SOP & POS Form
77
Min- & Maxterms with n = 3
x y zMinterms Maxterms
term designation term designation0 0 0 x’y’z’ m0 x + y + z M0
0 0 1 x’y’z m1 x + y + z’ M1
0 1 0 x’yz’ m2 x + y’ + z M2
0 1 1 x’yz m3 x + y’ + z’ M3
1 0 0 xy’z’ m4 x’ + y + z M4
1 0 1 xy’z m5 x’ + y + z’ M5
1 1 0 xyz’ m6 x’ + y’ + z M6
1 1 1 xyz m7 x’ + y’ + z’ M7
Formal Expression with Minterms
M7=x’+y’+z’
M6=x’+y’+z
M5=x’+y+z’
M4=x’+y+z
M3=x+y’+z’
M2=x+y’+z
M1=x+y+z’
F(1,1,1)
F(1,1,0)
F(1,0,1)
F(1,0,0)
F(0,1,1)
F(0,1,0)
F(0,0,1)
F(0,0,0)M0=x+y+z
m7=xyz111
m6=xyz’110
m5=xy’z101
m4=xy’z’100
m3=x’yz011
m2=x’yz’010
m1=x’y’z001
m0=x’y’z’000
FMimixyz
F(x, y, z) = F(0,0,0)m0 + F(0,0,1)m1 + F(0,1,0)m2 + F(0,1,1)m3 +F(1,0,0)m4 + F(1,0,1)m5 + F(1,1,0)m6 + F(1,1,1)m7
78
Formal Expression with Maxterms
M7=x’+y’+z’
M6=x’+y’+z
M5=x’+y+z’
M4=x’+y+z
M3=x+y’+z’
M2=x+y’+z
M1=x+y+z’
F(1,1,1)
F(1,1,0)
F(1,0,1)
F(1,0,0)
F(0,1,1)
F(0,1,0)
F(0,0,1)
F(0,0,0)M0=x+y+z
m7=xyz111
m6=xyz’110
m5=xy’z101
m4=xy’z’100
m3=x’yz011
m2=x’yz’010
m1=x’y’z001
m0=x’y’z’000
FMimixyz
F(x, y, z) = (F(0,0,0)+M0) (F(0,0,1)+M1) (F(0,1,0)+M2) (F(0,1,0)+M3)(F(1,0,0)+M4) (F(1,0,1)+M5) (F(1,1,0)+M6) (F(1,1,1)+M7)
79
80
Example
M7=x’+y’+z’
M6=x’+y’+z
M5=x’+y+z’
M4=x’+y+z
M3=x+y’+z’
M2=x+y’+z
M1=x+y+z’
0
1
0
0
0
1
1
0M0=x+y+z
m7=xyz111
m6=xyz’110
m5=xy’z101
m4=xy’z’100
m3=x’yz011
m2=x’yz’010
m1=x’y’z001
m0=x’y’z’000
FMimixyz
F(x, y, z) = ? in mintersF(x, y, z) = ? in maxterms
81
Example
M7=x’+y’+z’
M6=x’+y’+z
M5=x’+y+z’
M4=x’+y+z
M3=x+y’+z’
M2=x+y’+z
M1=x+y+z’
0
1
0
0
0
1
1
0M0=x+y+z
m7=xyz111
m6=xyz’110
m5=xy’z101
m4=xy’z’100
m3=x’yz011
m2=x’yz’010
m1=x’y’z001
m0=x’y’z’000
FMimixyz
F(x, y, z) = x’y’z + x’yz’ + xyz’ F(x, y, z) = (x+y+z)(x+y’+z’)(x’+y+z)(x’+y+z’)(x’+y’+z’)
82
Boolean Functions in Canonical Form
• F1(x, y, z) =• F2(x, y, z) =
x y z F1 F2
0 0 0 0 10 0 1 1 00 1 0 0 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1
83
Important Properties• Any Boolean function can be expressed
§ as a sum of minterms§ as a product of maxterms
• Example:§ F’ = S (0, 2, 3, 5, 6)
=§ How do we find the complement of F’?§ F =
84
Important Properties• Any Boolean function can be expressed
§ as a sum of minterms§ as a product of maxterms
• Example:§ F’ = S (0, 2, 3, 5, 6)
= x’y’z’ + x’yz’ + x’yz + xy’z + xyz’§ How do we find the complement of F’?§ F =
85
Important Properties• Any Boolean function can be expressed
§ as a sum of minterms§ as a product of maxterms
• Example:§ F’ = S (0, 2, 3, 5, 6)
= x’y’z’ + x’yz’ + x’yz + xy’z + xyz’§ How do we find the complement of F’?§ F = (x + y + z)(x + y’ + z)(x + y’ + z’)(x’ + y + z’)(x’ + y’ + z)
= =
86
Canonical Form• If a Boolean function is expressed as a sum of
minterms or product of maxterms, the function is said to be in canonical form.
• Example: F = x + y’z à canonical form?§ No § But we can put it in canonical form.
§ F = x + y’z = S (7, 6, 5, 4, 1)• Alternative way:– Obtain the truth table first and then the canonical term.
87
Example: Product of Maxterms• F = xy + x’z• Use the distributive law of + over ·• F = xy + x’z
= xy(z+z’) + x’z(y+y’)= xyz + xyz’ + x’yz + x’y’z= S (7,6,3,1)
88
Example: Product of Maxterms• F = xy + x’z• Use the distributive law of + over ·• F = xy + x’z
= xy(z+z’) + x’z(y+y’)= xyz + xyz’ + x’yz + x’y’z= S (7,6,3,1)
= P (4, 5, 0, 2)
89
Conversion Between Canonical Forms
• Fact: – The complement of a function (given in sum of
minterms) can be expressed as a sum of mintermsmissing from the original function
• Example:§ F(x, y, z) = S (1, 4, 5, 6, 7)§ F’(x, y, z) =§ Now take the complement of F’ and make use of
DeMorgan’s theorem§ (F’ )’ = =§ F = M0 · M2 · M3 = P (0, 2, 3)
90
General Rule for Conversion• Important relation:
§ mj’ = Mj
§ Mj’ = mj
• The rule:– Interchange symbols P and S, and– list those terms missing from the original form
• Example: F = xy + x’z
F = S(1, 3, 6, 7) è F = P(?, ?, ?, ?)
91
Standard Forms• Fact:– Canonical forms are very seldom the ones with the
least number of literals• Alternative representation:– Standard form• a term may contain any number of literals
– Two types1. the sum of products2. the product of sums
– Examples: • F1 = y’ + xy + x’yz’• F2 = x(y’ + z)(x’ + y + z’)
92
Example: Standard Forms• F1 = y’ + xy + x’yz’• F2 = x(y’ + z)(x’ + y + z’)
93
Example: Standard Forms• F1 = y’ + xy + x’yz’• F2 = x(y’ + z)(x’ + y + z’)
‘
94
Nonstandard Forms• Example:§ F3 = AB(C+D) + C(D + E)§ This hybrid form yields three-level implementation
– The standard form: F3 = ABC + ABD + CD + CE
D
A
F3
B
C
C
ED
ABC
F3
ABD
DC
CE
95
Complexity of Digital Circuits
• Directly related to the complexity of the algebraic expression we use to build the circuit.
• Truth table– may lead to different implementations– Question: which one to use?
Gate-Level Minimization
• Finding an optimal gate-level implementation of Boolean functions.– So far: ad hoc.• Difficult to perform manually
• Need a more systematic (algorithmic) way– Can use computer-based logic synthesis tool• Exp: espresso logic minimization software
– Karnough Map (K-map) can be used for manual design of digital circuits.
The Map Method• The truth table representation of a function is unique.• But, not the algebraic expression– Several versions of an algebraic expression exist.– Difficult to minimize algebraic functions manually.
• The map method is a simple proceure to minimize Boolean functions.– Pictorial form of a truth table.– Called Karnough Map or K-Map.• Boolean expressions can be minimized by combining terms• This way, K-maps minimize equations graphically
98
Two-Variable K-Map• Two variables: x and y
à 4 minterms:• m0 = x’y’ à 00• m1 = x’y à 01• m2 = xy’ à 10• m3 = xy à 11
• Four squares (cells) for four minterms
• Figure (b) shows the relationship between the squares and the variables xand y.
yx 0 1
0 m0 m1
1 m2 m3
yx 0 1
0 x’y’ x’y1 xy’ xy
99
Example: Two-Variable K-Map
ØF = m0 + m1 + m2 = x’y’ + x’y + xy’ØF = …ØF = …ØF = …
yx 0 1
0 1 11 1 0
Remember the Shortcuts
Øx + x = x ↔ x · x = x
Øx + 1 = 1 ↔ x · 0 = 0
Øx + xy = x ↔ x · (x+y) = x [Absorption]
Ø(x + y)’ = x’ · y’ ↔ (x.y)’=x’+y’ [DeMorgan]
101
Example: Two-Variable K-Map
ØF = m0 + m1 + m2 = x’y’ + x’y + xy’ØF = …ØF = …ØF = …ØF = x’ + y’
• We can do the same optimization by combining adjacent cells.
yx 0 1
0 1 11 1 0
102
Example: Two-Variable K-MapStrategy: 1. Cover as many adjacent 1 cells as possible by the orders of two in circles in each
direction– e.g., 1x1, 1x2, 2x1, 2x2, 1x4, 4x1, 2x4, 4x2, 4x4, ...– Every 1 must be circled at least once.– Each circle must be as large as possible– Adjacency is established either vertically or horizontally (not diagonally: cells
touching each other by corners only are not adjacent ) 2. Express each circle as a product of literals by only including the literals that are
the same (all true or all complement) in the circle3. Form the final expression by OR(+)ing the circle expressions
yx 0 1
0 1 11 1 0
Chapter 2 <103>
• Complement: variable with a bar over itA, B, C
• Literal: variable or its complementA, A, B, B, C, C
• Implicant: product of literalsABC, AC, BC
• Prime implicant: implicant corresponding to the largest circle in a K-map
K-Map Definitions
104
Example: Two-Variable K-MapStrategy: 1. Cover as many adjacent 1 cells as possible by the orders of two in circles in each
direction– e.g., 1x1, 1x2, 2x1, 2x2, 1x4, 4x1, 2x4, 4x2, 4x4, ...– Every 1 must be circled at least once.– Each circle must be as large as possible– Adjacency is established either vertically or horizontally (not diagonally: cells
touching each other by corners only are not adjacent ) 2. Express each circle as a product of literals by only including the literals that are
the same (all true or all complement) in the circle3. Form the final expression by OR(+)ing the circle expressions
yx 0 1
0 1 11 1 0
Two-Variable K-Map (Cont.)• May only be useful to represent 16 Boolean
functions.
Two-Variable K-Map (Cont.)• May only be useful to represent 16 Boolean
functions.– Exp: If m1=m2=m3=1 then
m1+m2+m3=x’y+xy’+xy=x+y (OR function)
Two-Variable K-Map (Cont.)• May only be useful to represent 16 Boolean
functions.– Exp: If only m3=1 then
m1=xy (AND function)
Two-Variable K-Map (Cont.)• May only be useful to represent 16 Boolean
functions.– Exp: If only m3=1 then
m1=xy (AND function)
Three-Variable Map• There are 8 minterms for 3 variables.• So, there are 8 squares.
Three-Variable Map• Minterms are arranged not in a binary sequence, but in a
sequence similar to the Gray code.• Adjacent squares: they differ by only one variable, which is
complement (primed) in one square and true (not primed) in the other§ m2 « m6 , m3 « m7 § m2 « m0 , m6 « m4
Three-Variable Map (Cont.)• For instance, the square for square m5
corresponds to row 1 and column 01: 101• Another look to m5 is m5=xy’z.• When variables are 0, they are primed (ex: x’).
Otherwise not primed (x).
Three-Variable Map (Cont.)• Two adjacent squares differ by one variable (one primed, the
other is not).– So, they can be minimized– Ex: m5+m7=xy’z+xyz=xz(y’+y)=xz
• try to cover as many adjacent squares as possible by the orders of two.– 1,2,4,8… squares that has the logical value1.
Chapter 2 <113>
C 00 01
0
1
Y
11 10AB
1
0
0
0
0
0
0
1
B C0 00 11 01 1
A0000
0 00 11 01 1
1111
11000000
Y
Example
Chapter 2 <114>
C 00 01
0
1
Y
11 10AB
1
0
0
0
0
0
0
1
B C0 00 11 01 1
A0000
0 00 11 01 1
1111
11000000
Y
Example
Chapter 2 <115>
C 00 01
0
1
Y
11 10AB
ABC
ABC
ABC
ABC
ABC
ABC
ABC
ABC
1 0
B C Y0 0 00 1 01 01 1 1
Truth Table
C 00 01
0
1
Y
11 10ABA
0000
0 0 00 1 01 0 01 1 1
1111
0
1
1
1
0
0
0
K-Map
Example
Chapter 2 <116>
C 00 01
0
1
Y
11 10AB
ABC
ABC
ABC
ABC
ABC
ABC
ABC
ABC
1 0
B C Y0 0 00 1 01 01 1 1
Truth Table
C 00 01
0
1
Y
11 10ABA
0000
0 0 00 1 01 0 01 1 1
1111
0
1
1
1
0
0
0
K-Map
Y = AB + BC
Example
Adjacent squares• Squares on opposite edges are adjacent. These adjacent squares
don’t touch each other.>> Circles may wrap around the edges– m0 is adjacent to m2 and m4 is adjacent to m6.– m0+m2=x’y’z’+x’yz’=x’z’(y’+y)=x’z’– m4+m6=xy’z’+xyz’=xz’(y’+y)=xz’
118
In 3-Variable Karnaugh Maps
• 1 alone square represents one minterm with 3 literals
• 2 adjacent squares represent a term with 2 literals
• 4 adjacent squares represent a term with 1 literal
• 8 adjacent squares produce a function that is always equal to 1.
119
Example: Three-Variable K-Map• F1(x, y, z) = S (2, 3, 4, 5)
1
0
10110100yz
x
• F1(x, y, z) =
• F2(x, y, z) = S (3, 4, 6, 7)
yzx 00 01 11 10
0
1
• F2(x, y, z) =
120
Example: Three-Variable K-Map• F1(x, y, z) = S (2, 3, 4, 5)
11
11
00
00
1
0
10110100yz
x
• F1(x, y, z) =
• F2(x, y, z) = S (3, 4, 6, 7)
yzx 00 01 11 10
0 0 0 1 01 1 0 1 1
• F2(x, y, z) =
121
• F1(x, y, z) = S (2, 3, 4, 5)
11
11
00
00
1
0
10110100yz
x
• F1(x, y, z) = xy’ + x’y• F2(x, y, z) = S (3, 4, 6, 7)
yzx 00 01 11 10
0 0 0 1 01 1 0 1 1
• F2(x, y, z) = xz’ + yz
Example: Three-Variable K-Map
122
Example• F1(x, y, z) = S (0, 2, 4, 5, 6)
z
1
0
10110100yz
x
y
x
F1(x, y, z) =
123
Example
111
1
0
001
• F1(x, y, z) = S (0, 2, 4, 5, 6)
z
1
0
10110100yz
x
y
x
F1(x, y, z) =
124
Example
111
1
0
001
• F1(x, y, z) = S (0, 2, 4, 5, 6)
z
1
0
10110100yz
x
y
x
F1(x, y, z) =
Example • F=A’C+A’B+AB’C+BC– Express F as a sum of minterms.• F(A,B,C)=Σ(1,2,3,5,7)
– Find the minimal sum-of-products expression• F=C+A’B
126
Finding Sum of Minterms• If a function is not expressed in sum of minterms
form, it is possible to get it using K-map– Example: F(x, y, z) = x’z + x’y + xy’z + yz
1
0
10110100yz
x
127
Finding Sum of Minterms• If a function is not expressed in sum of minterms
form, it is possible to get it using K-map– Example: F(x, y, z) = x’z + x’y + xy’z + yz
1
0
10110100yz
x
F(x, y, z) = x’y’z + x’yz + x’yz’ + xy’z + xyz
128
Four-Variable K-Map• Four variables: x, y, z, t
>> 8 literals>> 16 minterms
ztxy 00 01 11 1000 m0 m1 m3 m201 m4 m5 m7 m611 m12 m13 m15 m1410 m8 m9 m11 m10
z
t
xy
Four-Variable Map
130
Example: Four-Variable K-MapF(x,y,z,t) = S (0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14)
10
11
01
00
10110100zt
xy
F(x,y,z,t) =
131
Example: Four-Variable K-MapF(x,y,z,t) = S (0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14)
00
0
0
0
11
111
111
111
10
11
01
00
10110100zt
xy
F(x,y,z,t) =
132
Example: Four-Variable K-MapF(x,y,z,t) = S (0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14)
00
0
0
0
11
111
111
111
10
11
01
00
10110100zt
xy
F(x,y,z,t) =
133
Example: Four-Variable K-Map• F(x,y,z,t) = x’y’z’ + y’zt’ + x’yzt’ + xy’z’
10
11
01
00
10110100zt
xy
• F(x,y,z,t) =
134
Example: Four-Variable K-Map• F(x,y,z,t) = x’y’z’ + y’zt’ + x’yzt’ + xy’z’
10
11
01
1011
0000
1000
101100
10110100zt
xy
• F(x,y,z,t) =
135
Example: Four-Variable K-Map• F(x,y,z,t) = x’y’z’ + y’zt’ + x’yzt’ + xy’z’
10
11
01
1011
0000
1000
101100
10110100zt
xy
• F(x,y,z,t) =
Chapter 2 <136>
01 11
01
11
10
00
00
10AB
CD
Y
0
C D0 00 11 01 1
B0000
0 00 11 01 1
1111
1
110111
YA00000000
0 00 11 01 1
0000
0 00 11 01 1
1111
11111111
11
100000
Example
Chapter 2 <137>
01 11
1
0
0
1
0
0
1
101
1
1
1
1
0
0
0
1
11
10
00
00
10AB
CD
Y
0
C D0 00 11 01 1
B0000
0 00 11 01 1
1111
1
110111
YA00000000
0 00 11 01 1
0000
0 00 11 01 1
1111
11111111
11
100000
Example
Chapter 2 <138>
01 11
1
0
0
1
0
0
1
101
1
1
1
1
0
0
0
1
11
10
00
00
10AB
CD
Y
Y = AC + ABD + ABC + BD
0
C D0 00 11 01 1
B0000
0 00 11 01 1
1111
1
110111
YA00000000
0 00 11 01 1
0000
0 00 11 01 1
1111
11111111
11
100000
Example
139
Prime Implicants
• Prime Implicant: is a product term obtained by combining maximum possible number of adjacent cells in the map
• If a minterm can be covered by only one prime implicant, that prime implicant is said to be an Essential Prime Implicant.– A single 1 on the map represents a prime implicant if it is
not adjacent to any other 1s.– Two adjacent 1s form a prime implicant, provided that
they are not within a group of four adjacent 1s. – So on
140
Example: Prime Implicants• F(x,y,z,t) = S (0, 2, 3, 5, 7, 8, 9, 10, 11, 13, 15)
ztxy 00 01 11 1000
01
11
10
141
Example: Prime Implicants• F(x,y,z,t) = S (0, 2, 3, 5, 7, 8, 9, 10, 11, 13, 15)
ztxy 00 01 11 1000 1 0 1 101 0 1 1 011 0 1 1 010 1 1 1 1
142
Example: Prime Implicants• F(x,y,z,t) = S (0, 2, 3, 5, 7, 8, 9, 10, 11, 13, 15)
ztxy 00 01 11 1000 1 0 1 101 0 1 1 011 0 1 1 010 1 1 1 1
F(x,y,z,t) = y’t’ + yt + xy’ + zt• Which ones are the essential prime implicants?
143
Example: Prime Implicants• F(x,y,z,t) = S (0, 2, 3, 5, 7, 8, 9, 10, 11, 13, 15)
ztxy 00 01 11 1000 1 0 1 101 0 1 1 011 0 1 1 010 1 1 1 1
• Why are these the essential prime implicants?
144
Example: Prime Implicants• F(x,y,z,t) = S (0, 2, 3, 5, 7, 8, 9, 10, 11, 13, 15)
ztxy 00 01 11 1000 1 0 1 101 0 1 1 011 0 1 1 010 1 1 1 1
• y’t’ – essential since m0 is covered only in it• yt - essential since m5 is covered only in it• They together cover m0, m2, m8, m10, m5, m7, m13, m15
ztxy 00 01 11 1000 m0 m1 m3 m201 m4 m5 m7 m611 m12 m13 m15 m1410 m8 m9 m11 m10
145
Example: Prime Implicants
• m3, m9, m11 are not yet covered. • How do we cover them?• There is actually more than one way.
ztxy 00 01 11 1000 1 0 1 101 0 1 1 011 0 1 1 010 1 1 1 1
ztxy 00 01 11 1000 m0 m1 m3 m201 m4 m5 m7 m611 m12 m13 m15 m1410 m8 m9 m11 m10
146
Example: Prime Implicants
• Both y’z and zt covers m3 and m11.• m9 can be covered in two different prime implicants: xt or xy’
• m3, m11 à zt or y’z• m9 à xy’ or xt
ztxy 00 01 11 1000 1 0 1 101 0 1 1 011 0 1 1 010 1 1 1 1
12
3
4
147
Example: Prime Implicants• F(x, y, z, t) = yt + y’t’ + zt + xt or• F(x, y, z, t) = yt + y’t’ + zt + xy’ or• F(x, y, z, t) = yt + y’t’ + y’z + xt or• F(x, y, z, t) = yt + y’t’ + y’z + xy’• Therefore, what to do
1. Find out all the essential prime implicants2. Then find the other prime implicants that cover the
minterms that are not covered by the essential prime implicants. There are more than one way to choose those.
3. Simplified expression is the logical sum of the essential implicants plus the other implicants
148
Five-Variable Map• Downside:– Karnaugh maps with more than four variables are not
simple to use anymore.– 5 variables à 32 squares, 6 variables à 64 squares– Somewhat more practical way for F(x, y, z, t, w) :
twyz 00 01 11 1000 m0 m1 m3 m201 m4 m5 m7 m611 m12 m13 m15 m1410 m8 m9 m11 m10
twyz 00 01 11 1000 m16 m17 m19 m1801 m20 m21 m23 m2211 m28 m29 m31 m3010 m24 m25 m27 m26
x = 0 x = 1
149
Many-Variable Maps• Adjacency:– Each square in the x = 0 map is adjacent to the
corresponding square in the x = 1 map.– For example, m4 à m20 and m15 à m31
• 6-variables: Use four 4-variable maps to obtain 64 squares required for six variable optimization
• Alternatively: Can use computer programs– Quine-McCluskey method– Espresso method
Five-Variable Map
• Not simple, is not usually used. 5 variables, 32 squares.
151
Example: Five-Variable Map
F(x, y, z, t, w) = S (0, 2, 4, 6, 9, 13, 21, 23, 25, 29, 31)
twyz 00 01 11 1000
01
11
10
twyz 00 01 11 1000
01
11
10
x = 0 x = 1
• F(x,y,z,t,w) =
152
Example: Five-Variable Map
F(x, y, z, t, w) = S (0, 2, 4, 6, 9, 13, 21, 23, 25, 29, 31)
twyz 00 01 11 1000 1 101 1 111 110 1
twyz 00 01 11 1000
01 1 111 1 110 1
x = 0 x = 1
• F(x,y,z,t,w) =
153
Example: Five-Variable Map
F(x, y, z, t, w) = S (0, 2, 4, 6, 9, 13, 21, 23, 25, 29, 31)
twyz 00 01 11 1000 1 101 1 111 110 1
twyz 00 01 11 1000
01 1 111 1 110 1
x = 0 x = 1
• F(x,y,z,t,w) =
Example • An alternative look would be:
155
Product of Sums Simplification• So far simplified expressions from Karnaugh maps are in sum
of products form.
• Simplified product of sums can also be derived from Karnaugh maps.
• Method:– A square with 1 actually represents a “minterm”– Similarly an empty square (a square with 0) represents a
“maxterm”.– Treat the 0s in the same manner as we treat 1s – Combine them as we did for sum-of-products
• We obtain F’ (complement of the function)• Take the complement of F’ : ((F’)’) to obtain F
156
Example: Product of Sums• F(x, y, z, t) = S (0, 1, 2, 5, 8, 9, 10)– Simplify this function in
a. sum of productsb. product of sums
ztxy 00 01 11 1000 1 1 101 111
10 1 1 1
F(x, y, z, t) =
157
Example: Product of Sums• F(x, y, z, t) = S (0, 1, 2, 5, 8, 9, 10)– Simplify this function in
a. sum of productsb. product of sums
ztxy 00 01 11 1000 1 1 101 111
10 1 1 1
F(x, y, z, t) =
158
Example: Product of Sums• F(x, y, z, t) = S (0, 1, 2, 5, 8, 9, 10)– Simplify this function in
a. sum of productsb. product of sums
ztxy 00 01 11 1000 1 1 101 111
10 1 1 1
F(x,y,z,t) = y’t’ + y’z’ + x’z’t
159
Example: Product of Sums1. Find F’(x,y,z,t) in Sum of Products form by
grouping 0’s2. Apply DeMorgan’s theorem to F’ (use dual
theorem) to find F in Product of Sums formzt
xy 00 01 11 1000 1 1 0 101 0 1 0 011 0 0 0 010 1 1 0 1
160
Example: Product of Sums
ztxy 00 01 11 1000 1 1 0 101 0 1 0 011 0 0 0 010 1 1 0 1
1. Find F’(x,y,z,t) in Sum of Products form by grouping 0’s
2. Apply DeMorgan’s theorem to F’ (use dual theorem) to find F in Product of Sums form
161
Example: Product of Sums
ztxy 00 01 11 1000 1 1 0 101 0 1 0 011 0 0 0 010 1 1 0 1
F’ = yt’ + zt + xy
1. Find F’(x,y,z,t) in Sum of Products form by grouping 0’s
2. Apply DeMorgan’s theorem to F’ (use dual theorem) to find F in Product of Sums form
162
Example: Product of Sums
ztxy 00 01 11 1000 1 1 0 101 0 1 0 011 0 0 0 010 1 1 0 1
F(x,y,z,t) = (y’+t)(z’+t’)(x’+y’)
1. Find F’(x,y,z,t) in Sum of Products form by grouping 0’s
2. Apply DeMorgan’s theorem to F’ (use dual theorem) to find F in Product of Sums form
163
Example: Product of Sums
z’
y’t’y’z’x’
F
t
y’tx’y’z’
F
t’
F(x,y,z,t) = y’t’ + y’z’ + x’z’t: sum of products implementation
F = (y’ + t)(x’ + y’)(z’ + t’): product of sums implementation
164
Product of Maxterms• If the function is originally expressed in the
product of maxterms canonical form, the procedure is also valid
• Example: – F(x, y, z) = P (0, 2, 5, 7)
1
0
10110100yz
x
F(x, y, z) =
165
Product of Maxterms• If the function is originally expressed in the
product of maxterms canonical form, the procedure is also valid
• Example: – F(x, y, z) = P (0, 2, 5, 7)
1
0
10110100yz
x
F(x, y, z) =
0 0
0 0
166
Product of Maxterms• If the function is originally expressed in the
product of maxterms canonical form, the procedure is also valid
• Example: – F(x, y, z) = P (0, 2, 5, 7)
1
0
10110100yz
x
F(x, y, z) =
0 0
0 0
F(x, y, z) = x’z + xz’
167
Product of Sums• To enter a function F, expressed in product of sums,
in the map1. take its complement, F’2. Find the squares corresponding to the terms in F’,3. Fill these square with 0’s and others with 1’s.
• Example: – F(x, y, z, t) = (x’ + y’ + z’)(y + t)– F’(x, y, z, t) = zt
xy 00 01 11 1000
01
11
10
168
Product of Sums• To enter a function F, expressed in product of sums,
in the map1. take its complement, F’2. Find the squares corresponding to the terms in F’,3. Fill these square with 0’s and others with 1’s.
• Example: – F(x, y, z, t) = (x’ + y’ + z’)(y + t)– F’(x, y, z, t) = zt
xy 00 01 11 1000 0 001
11 0 010 0 0
169
Don’t Care Conditions
• Some functions are not defined for certain input combinations– Such function are referred as incompletely
specified functions– therefore, the corresponding output values do
not have to be defined– This may significantly reduce the circuit
complexity– Example: Four bit binary codes has six unused
combinations.
170
Unspecified Minterms• For unspecified minterms, we do not care what value
the function produces.• Unspecified minterms of a function are called don’t
care conditions.• We use “X” symbol to represent them in a Karnaugh
map.• Useful for further simplification• The Xs in the map can be taken as 0 or 1 to realize the
best simplification. >> A “don't care” (X) is circled (i.e., taken as 1) only if it helps to minimize the equation
171
Example: Don’t Care Conditions
• F(x, y, z, t) = S(1, 3, 7, 11, 15) – function• d(x, y, z, t) = S(0, 2, 5) – don’t care conditions
ztxy 00 01 11 1000 X 1 1 X01 0 X 1 011 0 0 1 010 0 0 1 0
F =
F1 = or
F2 =
172
• F(x, y, z, t) = S(1, 3, 7, 11, 15) – function• d(x, y, z, t) = S(0, 2, 5) – don’t care conditions
ztxy 00 01 11 1000 X 1 1 X01 0 X 1 011 0 0 1 010 0 0 1 0
F =
F1 = or
F2 =
Example: Don’t Care Conditions
173
Example: Don’t Care Conditions• F1 = zt + x’y’ = S(0, 1, 2, 3, 7, 11, 15) • F2 = zt + x’t = S(1, 3, 5, 7, 11, 15) • The two functions are algebraically unequal– But as far as the function F with d is concerned: both
functions are acceptable• Look at the simplified product of sums expression
for the same function F.zt
xy 00 01 11 1000 X 1 1 X01 0 X 1 011 0 0 1 010 0 0 1 0
F’ =F =
174
Example: Don’t Care Conditions• F1 = zt + x’y’ = S(0, 1, 2, 3, 7, 11, 15) • F2 = zt + x’t = S(1, 3, 5, 7, 11, 15) • The two functions are algebraically unequal– But as far as the function F with d is concerned: both
functions are acceptable• Look at the simplified product of sums expression
for the same function F.zt
xy 00 01 11 1000 X 1 1 X01 0 X 1 011 0 0 1 010 0 0 1 0
F’ =F =
Chapter 2 <175>
0
C D0 00 11 01 1
B0000
0 00 11 01 1
1111
1
110X11
YA00000000
0 00 11 01 1
0000
0 00 11 01 1
1111
11111111
11
XXXXXX
01 11
01
11
10
00
00
10AB
CD
Y
K-Maps with Don’t Cares
Chapter 2 <176>
0
C D0 00 11 01 1
B0000
0 00 11 01 1
1111
1
110X11
YA00000000
0 00 11 01 1
0000
0 00 11 01 1
1111
11111111
11
XXXXXX
01 11
1
0
0
X
X
X
1
101
1
1
1
1
X
X
X
X
11
10
00
00
10AB
CD
Y
K-Maps with Don’t Cares
Chapter 2 <177>
0
C D0 00 11 01 1
B0000
0 00 11 01 1
1111
1
110X11
YA00000000
0 00 11 01 1
0000
0 00 11 01 1
1111
11111111
11
XXXXXX
01 11
1
0
0
X
X
X
1
101
1
1
1
1
X
X
X
X
11
10
00
00
10AB
CD
Y
Y = A + BD + C
K-Maps with Don’t Cares
183
NAND and NOR Gates • NAND and NOR gates are easier to fabricate
VDD
A
B
C = (AB)’
CMOS 2-input AND gates requires 6 CMOS transistors
CMOS 3-input NAND gates requires 6 CMOS transistors
184
Design with NAND or NOR Gates• It is beneficial to derive conversion rules from
Boolean functions given in terms of AND, OR, an NOT gates into equivalent NAND or NOR implementations
x (x x)’ = x’ è NOT
[ (x y)’ ]’ = x y è ANDxy
(x’ y’ )’ = x + y è OR
x
y
185
New Notation
• Implementing a Boolean function with NAND gates is easy if it is in sum of products form.
• Example: F(x, y, z, t) = xy + zt
xyz
(xyz)’
AND-invert
x’ + y’ + z’xyz
Invert-OR
xy
zt
xy
zt
F(x, y, z, t) = xy + zt
F(x, y, z, t) = ((xy)’)’ + ((zt)’)’
=
ALWAYS USE THIS INTERMEDIATE
FORM WITH BUBBLES TO AVOID
CONFUSION
186
The Conversion Method
• Example: F(x, y, z) = S(1, 3, 4, 5, 7)
xy
zt
yzx 00 01 11 10
0 1 11 1 1 1
F = z + xy’
F = (z’)’ + ((xy’)’)’
y
zt
x
[ (xy)’ (zt)’ ] ’((xy)’)’ + ((zt)’)’ = xy + zt =
187
Example: Design with NAND Gates
• Summary1. Simplify the function 2. Draw a NAND gate for each product term3. Draw a NAND gate for the OR gate in the 2nd level, 4. A product term with single literal needs an inverter in
the first level. Assume single, complemented literals are available.
F = z + xy’
x
z’
y’
F
x
z’
y’
F
F = (z’)’ + ((xy’)’)’
188
Multi-Level NAND Gate Designs• The standard form results in two-level
implementations• Non-standard forms may raise a difficulty• Example: F = x(zt + y) + yz’
4-level implementation
zt
x
yF
yz’
Example: Multilevel NAND…
F
F = x(zt + y) + yz’
zt
F
y’
xyz’
zty’
xyz’
MUST COMPENSATE FOR EVERY NON-COUPLED BUBBLE
190
Design with Multi-Level NAND Gates
Rules1. Convert all AND gates to NAND gates2. Convert all OR gates to NAND gates3. Insert an inverter (one-input NAND gate) at
the output if the final operation is AND4. Check the bubbles in the diagram. For every
bubble along a path from input to output there must be another bubble. If not so, complement the input literal
191
Another (Harder) Example• Example: F = (xy’ + xy)(z + t’) – (three-level implementation)
x
z
y’
yx
F
t’
192
Example: Multi-Level NAND Gates
G = [ (xy’ + xy)(z’ + t) ]’
F = (xy’ + xy)(z + t’)
F = (xy’ + xy)(z + t’)
x
z’
y’
yx
t
F = (xy’ + xy)(z + t’)
x
z’
y’
yx
t
193
Design with NOR Gates• NOR is the dual operation of NAND.– All rules and procedure we used in the design with
NAND gates apply here in a similar way.– Function is implemented easily if it is in product of
sums form.(x + x)’ = x’ è NOT
[ (x+ y)’ ]’ = x + y è OR
(x’ + y’ )’ = x · y è AND
x
xy
x
y
=
195
Example: Design with NOR Gates• F = (x+y) (z+t) w
F = (x + y) (z + t) w
zt
xy
F
w
zt
xy
w’
Example: Design with NOR Gates• F = (xy’ + zt) (z + t’)
zt
xy’
F
zt’
F = [((x’ + y)’ + (z’ + t’)’)’ + (z + t’)’]’= ((x’ + y)’ + (z’ + t’)’)(z + t’)= (xy’ + zt) (z + t’)
z’t’
x’y
zt’
Harder Example• Example: F = x(zt + y) + yz’zt
x
yF
yz’
F
z’t’
y
x’
z
y’
198
Exclusive-OR Function• The symbol: Å
Øx Å y = xy’ + x’yØ(x Å y)’ = xy + x’y’
• Properties1. x Å 0 = x2. x Å 1 = x’3. x Å x = 04. x Å x’ = 15. x Å y’ = x’ Å y = (x Å y)’ : XNOR
• Commutative & AssociativeØx Å y = y Å x Ø(x Å y) Å z = x Å (y Å z)
199
Exclusive-OR Function• XOR gate is not universal– Only a limited number of Boolean functions can be
expressed in terms of XOR gates• XOR operation has very important applications in
arithmetic and error-detection circuits.• Odd Function
(x Å y) Å z = (xy’ + x’y) Å z = (xy’ + x’y) z’ + (xy’ + x’y)’ z= xy’z’ + x’yz’ + (xy + x’y’) z= xy’z’ + x’yz’ + xyz + x’y’z= S (4, 2, 7, 1) yz
x 00 01 11 100 0 1 0 11 1 0 1 0
XOR Implementations
201
Odd Function• If an odd number of variables are equal to 1, then
the function is equal to 1.• Therefore, multivariable XOR operation is
referred as the Odd function.
yzx 00 01 11 10
0 0 1 0 11 1 0 1 0
Odd function
yzx 00 01 11 10
0 1 0 1 01 0 1 0 1
Even function
Odd Function
• n variable XOR function is odd function defined as– The logical sum of the 2n/2 minterms whose
binary numerical values have an odd number of 1’s. If n=3, 4 minterms have odd number of 1’s.
203
Odd & Even Functions
• (x Å y Å z)’ = ((x Å y) Å z)’
xy
z
x Å y Å z Odd function
xy
z
(x Å y Å z)’
Parity Generation
• A parity bit is an extra bit included with a binary message– e.g., odd parity: If number of 1’s is odd, parity bit
is 1; else 0.
• The circuit that generates the parity bit in the transmitter is called parity generator.
• Parity bit can be generated using XOR function.
3-Bit Parity Generator
Parity Checker
• Bits are transmitted to the destination with parity.
• The circuit that checks the parity in the receiver is called a parity checker.
• Parity checker can be implmeneted with XOR gates.
3-Bit Parity Checker
208
Adder Circuit for Integers
• Addition of two 2-bit numbers§ Z = X + Y§ X = (x1 x0) and Y = (y1 y0)§ Z = (z2 z1 z0)
• Bitwise addition1. z0 = x0 Å y0 (sum)
c1 = x0 y0 (carry)2. z1 = x1 Å y1Å c1
c2 = x1 y1 + x1 c1 + y1 c1
3. z2 = c2
209
FA
Adder Circuit
c2= z2
x0y0
z0
c1
x1
z1
y1
z0 = x0 Å y0
c1 = x0 y0
z1 = x1 Å y1 Å c1
c2 = x1 y1 + x1 c1 + y1 c1z2 = c2
Comparator Circuit
• F( X>Y ) X = (x1 x0) and Y = (y1 y0)
210
y1 y0x1 x0 00 01 11 10
00011110
• F( X>Y ) X = (x1 x0) and Y = (y1 y0)
211
y1 y0x1 x0 00 01 11 10
00 0 0 0 001 1 0 0 011 1 1 0 110 1 1 0 0
F(x1, x0, y1, y0) = x1y1‘ + x1x0y0‘ + x0y0‘y1‘
Comparator Circuit
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