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CS 140 Lecture 3Combinational Logic
Professor CK Cheng
CSE Dept.
UC San Diego
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1. Specification2. Implementation3. K-maps
Part I.
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Literals xi or xi’
Product Term x2x1’x0
Sum Term x2 + x1’ + x0
Minterm of n variables: A product of n variablesin which every variable appears exactly once.
Definitions
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Implementation
Specification Schematic Diagram Net list,
Switching expressionObj min cost Search in solution space(max performance)
Cost: wires, gates Literals, product terms, sum terms
We want to minimize # of terms, # of literals
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Implementation (Optimization)
Karnaugh map – 2D truth table
ID A B f(A,B) minterm
0 0 0 0
1 0 1 1 A’B
2 1 0 1 AB’
3 1 1 1 AB
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Function can be represented by sum of minterms:f(A,B) = A’B+AB’+AB
This is not optimal however! We want to minimize the number of literals and terms.We factor out common terms –
A’B+AB’+AB= A’B+AB’+AB+AB=(A’+A)B+A(B’+B)=B+A
f(A,B) = A+B
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On the K-map however:
A = 0 A = 1
B = 0
B = 1
0 2
1 3
0 1
1 1
A’B
AB’
AB
f(A,B) = A + B
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ID A B f(A,B) minterm
0 0 0 0
1 0 1 1 A’B
2 1 0 0
3 1 1 1 AB
Another Example
f(A,B)=A’B+AB=(A’+A)B=B
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On the K-map:
A = 0 A= 1
B= 0
B = 1
0 2
1 3
0 0
1 1
A’B AB
f(A,B)=B
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ID A B f(A,B) Maxterm
0 0 0 0 A+B
1 0 1 1
2 1 0 0 A’+B
3 1 1 1
Using Maxterms
f(A,B)=(A+B)(A’+B)=(AA’)+B=0+B=B
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Two variable K-maps
Id a b f (a, b)
0 0 0 f (0, 0)
1 0 1 f (0, 1)
2 1 0 f (1, 0)
3 1 1 f (1, 1)
2 variables means we have 22 entries and thus we have 2 to the 22 possible functions for 2 bits, which is 16.
f(a,b)a
b
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Two-Input Logic Gates
AND
Y = AB
A B Y0 0 00 1 01 0 01 1 1
AB
Y
OR
Y = A + B
A B Y0 0 00 1 11 0 11 1 1
AB
Y
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More Two-Input Logic Gates
XNOR
Y = A + B
A B Y0 00 11 01 1
AB
Y
XOR NAND NOR
Y = A + B Y = AB Y = A + B
A B Y0 0 00 1 11 0 11 1 0
A B Y0 0 10 1 11 0 11 1 0
A B Y0 0 10 1 01 0 01 1 0
AB
Y AB
Y AB
Y
1001
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Three variables K-maps
Id a b c f (a,b,c)
0 0 0 0 1
1 0 0 1 02 0 1 0 13 0 1 1 04 1 0 0 15 1 0 1 06 1 1 0 17 1 1 1 0
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Corresponding K-map
0 2 6 4
1 3 7 5
b = 1
c = 1
a = 1
1 1 1 1
0 0 0 0
(0,0) (0,1) (1,1) (1,0)
c = 0
Gray code
f(a,b,c) = c’
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Karnaugh Maps (K-Maps)• Boolean expressions can be minimized by combining
terms• K-maps minimize equations graphically• PA + PA = P
C 00 01
0
1
Y
11 10AB
1
1
0
0
0
0
0
0
C 00 01
0
1
Y
11 10AB
ABC
ABC
ABC
ABC
ABC
ABC
ABC
ABC
B C0 00 11 01 1
A0000
0 00 11 01 1
1111
11000000
Y
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• Circle 1’s in adjacent squares• In the Boolean expression, include only the
literals whose true
K-map
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
y(A,B)=A’B’
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Another 3-Input example
Id a b c f (a,b,c)
0 0 0 0 0
1 0 0 1 02 0 1 0 13 0 1 1 04 1 0 0 15 1 0 1 16 1 1 0 -7 1 1 1 1
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Corresponding K-map
0 2 6 4
1 3 7 5
b = 1
c = 1
a = 1
0 1 - 1
0 0 1 1
(0,0) (0,1) (1,1) (1,0)
c = 0
f(a,b,c) = a + bc’
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Yet another example
Id a b c f (a,b,c,d)
0 0 0 0 1
1 0 0 1 12 0 1 0 -3 0 1 1 04 1 0 0 15 1 0 1 16 1 1 0 07 1 1 1 0
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Corresponding K-map
0 2 6 4
1 3 7 5
b = 1
c = 1
a = 1
1 - 0 1
1 0 0 1
(0,0) (0,1) (1,1) (1,0)
c = 0
f(a,b,c) = b’
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4-input K-map
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
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4-input K-map
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
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4-input K-map
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
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K-maps with Don’t Cares
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
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K-maps with Don’t Cares
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
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K-maps with Don’t Cares
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
