Page 42 - Computer science 868 Class 12
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Example 2: Minimise a Boolean expression F(A, B, C) = π(0, 1, 4, 5)
Solution: First, draw a POS K-map with A at the left and B, C at the top.
Then plot 0 in the grids mentioned and 1 in the remaining boxes.
B+C B+C' B'+C' B'+C
A 0 0 0 1 1 3 1 2
A' 0 4 0 5 1 7 1 6
A quad is formed containing 4 0’s placed together horizontally or vertically.
A quad with squares M , M , M , M
1
0
5
4
is formed.
M = A+B+C
0
M = A+B+C'
1
M = A'+B+C
4
M = A'+B+C'
5
We find that literals A and C are complemented as well as uncomplemented. So A and C are cancelled and the minimised
POS expression is B.
Example 3: Minimise a Boolean expression F(A, B, C) = Σ(0, 1, 2, 3, 4, 5, 6, 7)
Solution: First draw the SOP K-map properly mentioning the literals and cardinal An octet with the blocks m , m , m ,
0
numbers and place 1 in the required grids. m , m , m , m , m is formed. 1 2
6
5
3
4
7
m = A'.B'.C'
We find that all the grids are filled with 1 in the K-map. 0
m = A'.B'.C
1
B'.C' B'.C B.C B.C' m = A'.B.C'
2
A' 1 0 1 1 1 3 1 2 m = A'.B.C
3
A 1 4 1 5 1 7 1 6 m = A. B'.C'
4
m = A.B'.C
Example 4: Minimise a Boolean expression F(A, B, C, D) = Σ(0, 1, 2, 3, 13, 15) 5
m = A.B.C'
6
Solution: First, draw the SOP K-map. Fill all the mentioned squares with 1 and m = A.B.C
7
the rest with 0. All literals in this expression have
C'.D' C'.D C.D C.D' changed their signs. Hence, the
A'.B' 1 0 1 1 1 3 1 2 reduced SOP expression is 1.
A'.B 0 4 0 5 0 7 0 6
A.B 0 12 1 13 1 15 0 14
A.B' 0 8 0 9 0 11 0 10
The following grouping can be made:
Quad 1 m , m , m , m 3 Pair 1 m , m 15
2
1
0
13
m = A'.B'.C'.D' m = A.B.C'.D
13
0
m = A'.B'.C'.D m = A.B.C.D
1
15
m = A'.B'.C.D Literal C in the expression changes
2
m = A'.B'.C.D' sign while A, B and D remain same.
3
Literals C and D in the expression are So, the reduced form = A.B.D
both in complemented and normal
form while A' and B' remain same.
So, the reduced form = A'.B'
Hence, the reduced expression is A'.B' + A.B.D
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