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ECE 2110: Introduction to Digital Systems. Further Examples on Simplifying Sum of Products using Karnaugh Maps. Previous…. Simplifying SOP: Draw K-map Find prime implicants Find distinguished 1-cell Determine essential prime implicants if available - PowerPoint PPT Presentation
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ECE 2110: Introduction to Digital Systems
Further Examples on
Simplifying Sum of Products
using Karnaugh Maps
2
Previous…
Simplifying SOP:Draw K-mapFind prime implicantsFind distinguished 1-cellDetermine essential prime implicants if availableSelect all essential prime implicants and the minimal set
of the remaining prime implicants that cover the remaining 1’s.
3
Example 2
- Combining (0,2) Product term : X’Z’- Combining (2,3) Product term : X’Y- Combining (3,7) Product term :YZ
X’Z’, X’Y, and YZ are prime implicants X’Z’, YZ are essential prime implicants X’Y is non-essential prime implicant (redundant) because all its minterms
are covered in the other essential prime implicants F= X’Z’+X’Y+YZ (complete sum) OR:
F = X’Z’+YZ ( the minimal sum of F )
0
1 3
2
XY
Z
X
Z
1 1
0 1
00 01
0
1 7
6
0
1
11
5
40
0
10
Y
4
Example 3
The essential prime implicants:- W’X’- W’Y’- WXY
Cell 7 is not covered by any of the essential prime implicants. Its covered by two non-essential prime implicant. We choosethe one with the less number of variables which is W’Z
F= W’X’+W’Y’+WXY+W’Z
0
1 5
4
WX
YZ
W
Z
1 1
1 1
00 01
00
13
12
0
0
11
9
80
0
10
X
3
2 6
71 1
1 014
151
110
110
0
01
11
10Y
1
1 1
The prime implicants :- Cells(0,1,2,3) : W’X’- Cells(0,1,4,5) : W’Y’- Cells(1,3,5,7) : W’Z- Cells(7,15) : XYZ- Cells(14,15) : WXY
5
Exercise
Row W X Y Z F 0 0 0 0 0 1 1 0 0 0 1 1 2 0 0 1 0 1 3 0 0 1 1 0 4 0 1 0 0 1 5 0 1 0 1 1 6 0 1 1 0 1 7 0 1 1 1 0 8 1 0 0 0 1 9 1 0 0 1 1 10 1 0 1 0 0 11 1 0 1 1 0 12 1 1 0 0 1 13 1 1 0 1 1 14 1 1 1 0 1 15 1 1 1 1 0
0
1 5
4
WX
YZ
W
Z
00 01
00
13
12
11
9
8
10
X
3
2 6
7
14
15
10
11
01
11
10Y
6
Exercise Solution:
Essential prime implicants:- cells (0,1,4,5,8,9,12,13) The product term : Y’- cells (2,6,0,4) The product term : W’Z’- Cells (4,6,12,14) The product term : XZ’
F=Y’+W’Z’+XZ’
0
1 5
4
WX
YZ
W
Z
1 1
1 1
00 01
00
13
12
1
1
11
9
81
1
10
X
3
2 6
70 0
1 114
150
110
110
0
01
11
10Y
7
Example 4 F=
The prime implicants: 1- (0,2) X’Z’ 2- (0,4) Y’Z’ 3- (2,3) X’Y 4- (3,7) YZ 5- (4,5) XY’ 6 -(5,7) XZ
No essential prime implicant!
Two possible minimal sums :1- Using the prime implicants 1,4,and 5 , F= X’Z’+YZ+XY’2- Using the prime implicants 2,3,and 6 , F= Y’Z’+ X’Y+XZ
0
1 3
2
XY
Z
X
Z
1 1
0 1
00 01
0
1 7
6
0
1
11
5
41
1
10
Y
zyx ,,)7,5,4,3,2,0(
8
Example 5 F=
- Cells (5,13,7,15) can be combined to form an essential prime implicant. W & Y change X & Z remain constant, X=1,Z=1- The product term : XZ
- Cells (0,8,2,10) can be combined to form an essential prime implicant. W & Y change Z & X remain constant, X=0, Z=0- The product term : X’Z’
F= XZ + X’Z’ Note that the corner cells (0,2),(0,8),(8,10),(2,10)
can be combined to form the implicants : W’X’Z’ , X’Y’Z’, WX’Z’, X’YZ’ but, they are not prime implicants.
0
1 5
4
WX
YZ
W
Z
1 0
0 1
00 01
00
13
12
0
1
11
9
81
0
10
X
3
2 6
70 1
1 014
151
010
110
1
01
11
10Y
zyx ,,)15,13,10,8,7,5,2,0(
9
Next…
Simplifying PoS
Read Chapter 4.4---4.7
