1 EECS150 - Digital Design Lecture 8

Transcription

Outline
• Canonical Forms (continued)
– They give us a method to go from truth tables (TT) to Boolean
Equations
•
•
•
•
•
EECS150 - Digital Design
Lecture 8 - Boolean Algebra II
Feburary 9, 2003
John Wawrzynek
Spring 2005
EECS150 – Lec8-Bool2
Page 1
Algebraic simplification example
K-map method of two-level logic simplication
Multi-level Logic
NAND/NOR networks
EXOR revisited
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Relationship Among Representations
Canonical Forms
* Theorem: Any Boolean function that can be expressed as a truth table can
be written as an expression in Boolean Algebra using AND, OR, NOT.
?
not
unique
Boolean
Expression
?
gate
representation
(schematic)
[convenient for
manipulation]
• Sum of Products (disjunctive normal form, minterm expansion).
Example:
not
unique
[close to
implementaton]
How do we convert from one to the other?
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• Standard form for a Boolean expression - unique algebraic
expression directly from a true table (TT) description.
• Two Types:
* Sum of Products (SOP)
* Product of Sums (POS)
unique
Truth Table
Page 2
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minterms
a’b’c’
a’b’c
a’bc’
a’bc
ab’c’
ab’c
abc’
abc
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abc
000
001
010
011
100
101
110
111
f f’
01
01
01
10
10
10
10
10
One product (and) term for each 1 in f:
f = a’bc + ab’c’ + ab’c +abc’ +abc
f’ = a’b’c’ + a’b’c + a’bc’
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1
Sum of Products (cont.)
Canonical Forms
Canonical Forms are usually not minimal:
Our Example:
f = a’bc + ab’c’ + ab’c + abc’ +abc
(xy’ + xy = x)
= a’bc + ab’ + ab
= a’bc + a
(x’y + x = y + x)
= a + bc
• Product of Sums (conjunctive normal form, maxterm expansion).
Example:
maxterms
a+b+c
a+b+c’
a+b’+c
a+b’+c’
a’+b+c
a’+b+c’
a’+b’+c
a’+b’+c’
f’ = a’b’c’ + a’b’c + a’bc’
= a’b’ + a’bc’
= a’ ( b’ + bc’ )
= a’ ( b’ + c’ )
= a’b’ + a’c’
abc
000
001
010
011
100
101
110
111
f f’
01
01
01
10
10
10
10
10
One sum (or) term for each 0 in f:
f = (a+b+c)(a+b+c’)(a+b’+c)
f’ = (a+b’+c’)(a’+b+c)(a’+b+c’)
(a’+b’+c)(a+b+c’)
Mapping from SOP to POS (or POS to SOP): Derive truth table then
proceed.
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Algebraic Simplification Example
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Algebraic Simplification
Ex: full adder (FA) carry out function (in canonical form):
Cout = a’bc + ab’c + abc’ + abc
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Ex: full adder (FA) carry out function
Cout = a’bc + ab’c + abc’ + abc
= a’bc + ab’c + abc’ + abc + abc
= a’bc + abc + ab’c + abc’ + abc
= (a’ + a)bc + ab’c + abc’ + abc
= (1)bc + ab’c + abc’ + abc
= bc + ab’c + abc’ + abc + abc
= bc + ab’c + abc + abc’ + abc
= bc + a(b’ +b)c + abc’ +abc
= bc + a(1)c + abc’ + abc
= bc + ac + ab(c’ + c)
= bc + ac + ab(1)
= bc + ac + ab
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2
General Two-level Logic Simplication
Boolean Cubes
Visual technique for identifying when the Uniting Theorem can be applied
Key tool: The Uniting Theorem
1-cube
xy’ + xy = x (y’ + y) = x (1) = x
ab
00
01
10
11
f
0
0
1
1
ab
00
01
10
11
g
1
0
1
0
0
3-cube
11
011
111
010
xy
110
y
001
000
x
10
100
101
z
• Sub-cubes of on-nodes can be used for simplification.
– On-set: filled in nodes, off-set: empty nodes
ab
00
01
10
11
g = a’b’+ab’ = (a’+a)b’ =b’
b values stay the same
a values changes
f
0
0
1
1
g
1
0
1
0
01
11
b
f
00
a
a asserted & unchanged
b varies
ab + ab'= a
10
b’ remains, a eliminated
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3-variable cube example
• K-map is an alternative method of representing the TT and
to help visual the adjacencies.
ab(c'
+c)
011
111
Note: “gray code” labeling.
010
b
000
a
100
a
a(b+b'
)c
110
001
b
101
c
cout = bc + ab + ac
What about larger sub-cubes?
011
111
010
110
b
001
000
a
100
101
c
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Karnaugh Map Method
(a'+ a)bc
FA carry out:
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2-cube
00
b values change within
the on-set rows
a values don’t change
b is eliminated, a remains
cout
0
0
0
1
0
1
1
1
1
x
f = ab’ + ab = a(b’+b) = a
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abc
000
001
010
011
100
101
110
111
01
c
ab’c’ + ab’c + abc’ + abc
ac’ + ac + ab = a + ab = a
0 1
0
1
ab
00 01 11 10
0
1
ab
cd
00 01 11 10
00
01
11
10
5 & 6 variable k-maps possible
• Both b & c change, a is
asserted & remains constant.
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3
Karnaugh Map Method
K-map Simplification
1. Draw K-map of the appropriate number of variables
(between 2 and 6)
2. Fill in map with function values from truth table.
3. Form groups of 1’s.
• Adjacent groups of 1’s represent product terms
a
b
a
0 1
0 0 1
1 0 1
b
f=a
0 1
0 1 1
1 0 0
Dimensions of groups must be even powers of two (1x1, 1x2, 1x4,
…, 2x2, 2x4, …)
Form as large as possible groups and as few groups as possible.
Groups can overlap (this helps make larger groups)
Remember K-map is periodical in all dimensions (groups can cross
over edges of map and continue on other side)
g = b'
ab
c
00 01 11 10
0 0 0 1 0
1 0 1 1 1
c
ab
00 01 11 10
0 0 0 1 1
1 0 0 1 1
cout = ab + bc + ac
4. For each group write a product term.
the term includes the “constant” variables (use the
uncomplemented variable for a constant 1 and complemented
variable for constant 0)
f=a
5. Form Boolean expression as sum-of-products.
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K-maps (cont.)
c
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Product-of-Sums Version
ab
00 01 11 10
0 1 0 0 1
1 0 0 1 1
1. Form groups of 0’s instead of 1’s.
2. For each group write a sum term.
the term includes the “constant” variables (use the
uncomplemented variable for a constant 0 and complemented
variable for constant 1)
f = b'
c'+ ac
3. Form Boolean expression as product-of-sums.
ab
cd
00
01
11
10
ab
00
1
0
1
1
01
0
1
1
1
11
0
0
1
1
10
1
0
1
1
cd
00
01
11
10
f = c + a'
bd + b'
d'
(bigger groups are better)
00
1
0
1
1
01
0
1
1
1
11
0
0
1
1
10
1
0
1
1
f = (b'+ c + d)(a'+ c + d'
)(b + c + d'
)
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4
BCD incrementer example
abcd
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
wxyz
0001
0010
0011
0100
0101
0110
0111
1000
1001
0000
- - - - - - - - - - - - - - - - - - -
cd
ab
00
01
11
10
cd
ab
00
01
11
10
w
cd
00
00 01
01 11
11 10
11
0
0
0
0
0
0
1
0
-
1
0
-
00
01
11
10
y
cd
00
00 01
01 11
11 10
11
0
1
0
1
0
1
0
1
-
0
0
-
ab
00
01
11
10
w=
BCD Incrementer Example
x
• Note one map for each output variable.
• Function includes “don’t cares” (shown as “-” in the table).
00
00 01
01 11
11 10
11
0
0
1
0
1
1
0
1
-
0
0
-
– These correspond to places in the function where we don’t care
about its value, because we don’t expect some particular input
patterns.
– We are free to assign either 0 or 1 to each don’t care in the
function, as a means to increase group sizes.
z
00
00 01
01 11
11 10
11
1
0
0
1
1
0
0
1
-
• In general, you might choose to write product-of-sums or
sum-of-products according to which one leads to a simpler
expression.
1
0
-
x=
y=
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ab
z=
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BCD incrementer example
cd
ab
00
01
11
10
cd
ab
00
01
11
10
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w
00 01
01 11
11 11
10
00
0
0
0
0
0
0
1
0
-
cd
1
0
-
00
01
11
10
y
00
00 01
01 11
11 10
11
0
1
0
1
0
1
0
1
-
0
0
-
ab
cd
ab
00
01
11
10
Higher Dimensional K-maps
x
00
01 11
11 11
10
00 01
0
0
1
0
1
1
0
1
-
0
0
-
z
a=1
w=
x=
a=0
y=
de
00
bc
00 01 11
10
ab = 10
10
11
10
00
10
11
10
00 01 11
10
ab = 11
ab = 01
00
00 01
01 11
11 10
11
1
0
0
1
1
0
0
1
-
1
0
-
Page 18
z=
ab = 00
Page 19
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ef
00
cd
00 01 11
10
00 01 11
10
00 01 11
10
00 01 11
10
10
11
10
00
10
11
10
00
10
11
10
00
10
11
10
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5
Multi-level Combinational Logic
Example: reduced sum-of-products form
x = adf + aef + bdf + bef + cdf + cef + g
Implementation in 2-levels with gates:
Multi-level Combinational Logic
Another Example: F = abc + abd +a’c’d’ + b’c’d’
a
let x = ab y = c+d
b
c
f = xy + x’y’
a
d
f
cost: 1 7-input OR, 6 3-input AND
=> 50 transistors
delay: 3-input OR gate delay + 7-input AND gate delay
Factored form:
x = (a + b +c)(d + e)f + g
cost: 1 3-input OR, 2 2-input OR, 1 3-input AND
=> 20 transistors
delay: 3-input OR + 3-input AND + 2-input OR
a
b
c
d
e
f
g
x
Which is faster?
In general: Using multiple levels (more than 2) will reduce the cost. Sometimes also
delay.
Sometime a tradeoff between cost and delay.
Page 21
NAND-NAND & NOR-NOR Networks
DeMorgan’s Law:
(a + b)’ = a’ b’
a + b = (a’ b’)’
a
b
x
f
c
d
y
No convenient hand methods exist for multi-level logic simplification:
Footnote: NAND would be used in
place of all ANDs and ORs.
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a
b
d
a
c
d
b
c
d
=
=
=
Are these optimizations still relevant for LUT implementations?
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Page 22
• Mapping from AND/OR to NAND/NAND
(a b)’ = a’ + b’
(a b) = (a’ + b’)’
=
a) CAD Tools
b) exploit some special structure, example adder
a)
b)
c)
d)
a
b
c
d
push bubbles or introduce in pairs or remove pairs.
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6
•
Mapping AND/OR to NOR/NOR
a)
a
b
c
d
Multi-level Networks
Mapping OR/AND to NOR/NOR
a)
b)
a
b
c
d
Convert to NANDs:
F = a(b + cd) + bc’
b)
c)
c)
a
b
c
d
•
•
pair
c
d
b
a
f
b
c'
OR/AND to NAND/NAND
(by symmetry with above)
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(note fanout)
a'
b'
c'
d'
a
b
c
d
Page 25
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Page 26
EXOR Function
Parity, addition mod 2
x'
x ⊕ y = x’y + xy’
x y xor xnor
00 0 1
x
01
10
1
1
0
0
y
11
0
1
y
x
y'
Another approach:
x
y
Spring 2005
0
1
if x=0 then y else y'
Page 27
7

1 EECS150 - Digital Design Lecture 8

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