Page 35 - Computer science 868 Class 12
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= π(110,111,001,101) [Writing 1 in complemented literal and 0 in normal literal]
= π(6, 7, 1, 5)
Alternatively, using the truth table method
X Y Z (X'+Y') (Y+Z') (X'+Y').(Y+Z')
0 0 0 1 1 1
0 0 1 1 0 0
0 1 0 1 1 1
0 1 1 1 1 1
1 0 0 1 1 1
1 0 1 1 0 0
1 1 0 0 1 0
1 1 1 0 1 0
Taking the maxterms with 0’s in the final column, we get
(X+Y+Z').(X'+Y+Z').(X'+Y'+Z).(X'+Y'+Z') as the POS
OR
= π(1, 5, 6, 7)
Example 2: Convert the following SOP to canonical SOP
XY + Z'
Solution:
st
nd
= X.Y.1 + Z'.1.1 [Z is missing in 1 term and X and Y are missing in the 2 term]
= X.Y.(Z+Z') + Z'.(X+X').(Y+Y') [Replacing 1 with the missing literal]
= X.Y.Z + X.Y.Z' + (X.Z'+X'.Z').(Y+Y') [Distributive law]
= X.Y.Z + X.Y.Z' + X.Y.Z' + X'.Y.Z' + X.Y'.Z' + X'.Y'.Z'
= X.Y.Z + X.Y.Z' + X'.Y.Z' + X.Y'.Z' + X'.Y'.Z' [Eliminating duplicates]
= Σ(111, 110, 010, 100, 000) [Writing 0 for complemented and 1 for normal literal]
OR
= Σ(7, 6, 2, 4, 0)
Alternatively, using the truth table method
X Y Z X.Y Z' X.Y+Z'
0 0 0 0 1 1
0 0 1 0 0 0
0 1 0 0 1 1
0 1 1 0 0 0
1 0 0 0 1 1
1 0 1 0 0 0
1 1 0 1 1 1
1 1 1 1 0 1
33
Boolean Algebra 33

