Lecture 5 - 5
pyo Hong, Dongguk University
, and Logic
A
1
0
F
Map
Diagramuit Represenation)
You
Logic Expression, Truth Table, K-MapDiagram
A
B
B
F = A B + AB 1
0
A
B
AB
Logic Expression
K-
Logic
A B F
0 0
00
1
1
01
11
10
Truth Table(= Digital Circ
Lecture 5 - 6
pyo Hong, Dongguk University
Function
A
1
1
F
⋅B = AB + A⋅B + A⋅B
You
Non-Unique Circuits for the Same
A
B
B
F = A + A⋅B
1
0
A
B
AB
Because F = A+A⋅B = A⋅(B + B)+A
AB
F AB
A
Lecture 5 - 7
pyo Hong, Dongguk University
. ex) X, Y, X
t of two or more lit-e represented by a
x) W+Y, XY+Z+A
ar. appears more
literals. ex) WXY,
You
Key Terms
• Literal - a variable or the complement of a variable
• Product Term (=Cube) : a single literal or a producerals. ex) Z, XY, WYZ, W⋅Y⋅Z. A product term can brectangle in a K-map and we will see why later.
• Sum-of-product (=SOP) : a sum of product terms. e
• Sum Term.
• Product-of-Sums (POS).
• Normal Term - a product or sum term in which no v
than once. ex) XYZ, Y+W
• n-variable minterm - a normal product term with n WXY (3-variable minterm)
Lecture 5 - 8
pyo Hong, Dongguk University
le.
naturally.
m naturally.s we will see.
You
True Table Again
• Let’s derive the logical expression from a truth tab
• A truth table provides a Sum-of-Product (SOP) form : each product is a minterm in this case.
• A K-map also provides a Sum-of-Product (SOP) for : each product does not have to be a minterm a
X Y Z F
00001010
00001111
00110011
01010101
F = X⋅Y⋅Z + X⋅Y⋅Z
(= XZ)
Lecture 5 - 9
pyo Hong, Dongguk University
Our Choice
You
K-Map Indexing Methods
Y
Z
XYZ 00 01 11 10
0
1
X
Lecture 5 - 10
pyo Hong, Dongguk University
uits “easily” from
ble in positive and
X⋅Y⋅Z
X⋅Y⋅Z
You
Property of K-Map
• One cell in K-map represents a minterm.
• We can get an AND-OR (Sum-of-Product) style circK-map.
• Two adjecent cells in K-map contain the same varianegative forms.
XYZ 00 01 11 10
0
1
1 10 0
0 0 0 0
Lecture 5 - 11
pyo Hong, Dongguk University
are a line..
11 10
1 1
1 1
1 0
1 1
You
Adjacency in K-Map (1)
A cell is adjacent to the cells if they sh
WXYZ
00 01
1 1
0 1
XYZ
00 01 11 10
0
1
1 11 1
0 1 0 00 0
0 0
000111
10
Lecture 5 - 12
pyo Hong, Dongguk University
to the cell on the !)
11 10
1 1
1 0
0 0
1 1
You
Adjacency in K-Map (2)
Note that the cell on the boundary are adjacentother side. (You need some imagination power!
WXYZ
00 01
1 1
0 0
XYZ
00 01 11 10
0
1
0 11 1
1 0 0 10 0
0 1
000111
10
Lecture 5 - 13
pyo Hong, Dongguk University
tion
he cells with 1
be combined into
ssion from K-map?
X⋅Y⋅Z
X⋅Y⋅Z
You
Motivation of K-Map Simplifica
• If we express K-map using logic expression, only tshow up in Sum-of-Product form.
• Two product terms that differ only in a variable canone.
Ex) X⋅Y⋅Z + X⋅Y⋅Z = XZ(Y+Y) = XZ
• We like simpler form. How can we get simler expre
XYZ 00 01 11 10
0
1
1 1
F = X⋅Y⋅Z + X⋅Y⋅Z
0 0 0 0
0 0
Lecture 5 - 14
pyo Hong, Dongguk University
ation
in that ractangle
r negative) is con-
X⋅Z
You
Introduction to K-Map Simplific
• We use a ractangle to specify that the product termcan be combined to one product.
• How to read the expression for a ractangle?
- Find out the variable whose polarity (positive osistent in the ractangle.
XYZ 00 01 11 10
0
1
1 1
0 0 0 0
0 0
Lecture 5 - 15
pyo Hong, Dongguk University
ding
You
Excercise on a Ractangle Rea
XYZ 00 01 11 10
0
1
1 1
0 0 1 0
0 0
WXYZ
00 01 11 10
1 11 0
0 1 1 0
0 1 1 0
0 0 1 1
000111
10
Lecture 5 - 16
pyo Hong, Dongguk University
d K-Map
6 4
7 5
, 7)
11 10
You
Minterm Numbers in Truth Table an
X Y Z minterm no.
01234567
00001111
00110011
01010101
0 2
1 3
F =X⋅Y⋅Z + X⋅Y⋅Z can be described by F = (0
XYZ 00 010
1
Lecture 5 - 17
pyo Hong, Dongguk University
les
0 0
0 0
+ ABC = A C + AC
0 0
0 0
11 10
11 10
You
K-Map Simplification Examp
A B C F
11110000
00001111
00110011
01010101
F = (0,1,2,3)1 1
1 1
F = A B C + A BC + ABC
1 1
1 1
F = A C + AC = A
ABC 00 01
ABC 00 01
0
1
0
1
Lecture 5 - 18
pyo Hong, Dongguk University
You1 1
0 0
1 1
0 0
1 0
1 0
0 1
0 1
XYZ 00 01 11 10
XYZ 00 01 11 10
0
1
0
1
Lecture 5 - 19
pyo Hong, Dongguk University
ible.
+ Y⋅Z (Better)
+ X⋅Y⋅Z
You
• Lesson
- Combine cells using as large ractangle as poss
1 1
0 0
1 0
0 0
F = X⋅Z
1 1
0 0
1 0
0 0
F = X⋅Z
XYZ 00 01 11 10
XYZ 00 01 11 10
0
1
0
1
Lecture 5 - 20
pyo Hong, Dongguk University
ant cover
ible cover
lls in a rectangle circled. Why?)
You
• Lesson
- Use as few ractangless possible.
1 1
1
1 1
1 1Redund
XYZ 00 01 11 10
XYZ 00 01 11 10
0
1
0
1
Imposs
(Only cecan be
Lecture 5 - 21
pyo Hong, Dongguk University
into one cover.
bles
You
Adjacent Covers
ABCD
00 01 11 10
1 11 0
1 1 1 1
1 1 1 1
0 0 1 1
000111
10
A ⋅ D
A ⋅ D
Then the two covers can be combined
if the two covers are adjecent.
The two covers differ in only one varia
Lecture 5 - 22
pyo Hong, Dongguk University
s
ned.
d.
e variables.
You
Partially Overlapped Cover
ABCD
00 01 11 10
0 00 0
1 1 1 0
1 1 1 0
0 0 0 0
000111
10
A ⋅ D
B ⋅ D
Then the two covers cannot be combi
The two covers are partially overlappe
The two covers differ in more than on
Lecture 5 - 23
pyo Hong, Dongguk University
Sizes
ned.
e variables.
You
Adjacent Covers with Different
ABCD
00 01 11 10
0 00 0
1 1 1 0
1 1 1 0
0 0 0 0
000111
10
A ⋅ D
A ⋅ B ⋅D
Then the two covers cannot be combi
The two covers have different size.
The two covers differ in more than on
Lecture 5 - 24
pyo Hong, Dongguk University
a segment along
11 10
1 1
1 1
1 1
1 1
You
Adjacency in K-Map (3)
A cube is adjacent to another cube if they shareone side and their sizes are the same.
WXYZ
00 01
1 1
1 1
XYZ
00 01 11 10
0
1
1 11 1
1 1 0 11 1
0 0
000111
10
Lecture 5 - 25
pyo Hong, Dongguk University
ined
...
You
The Covers That Can Be Comb
• They must be adjecent.
• Correct covers can have 2n cells in it.
2 + 2 = 4, 4 + 4 = 8, 8 + 8 = 16, so on..