Page 36 - Computer science 868 Class 12
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Considering the minterms with 1 in their final column we get
X'.Y'.Z' + X'.Y.Z' + X.Y'.Z' + X.Y.Z' + X.Y.Z
OR
Σ(0, 2, 4, 6, 7)
Example 3: Convert the following SOP expression to canonical POS
X.Y + Y'.Z
Solution:
= (X.Y+Y').(X.Y+Z) [Distributive law]
= (X+Y').(Y+Y').(X+Z).(Y+Z) [Distributive law]
= (X+Y'+0).1.(X+Z+0).(Y+Z+0)
= (X+Y'+Z.Z').(X+Z+Y.Y').(Y+Z+X.X')
= (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)
= π(010, 011, 000, 100)
OR
= π(2, 3, 4, 0)
Alternatively, using the truth table method
X Y Z X.Y Y'.Z X.Y+Y'.Z
0 0 0 0 0 0
0 0 1 0 1 1
0 1 0 0 0 0
0 1 1 0 0 0
1 0 0 0 0 0
1 0 1 0 1 1
1 1 0 1 0 1
1 1 1 1 0 1
Considering the maxterms with 0 in their final column, we get
(X+Y+Z).(X+Y'+Z).(X+Y'+Z').(X'+Y+Z)
OR
π(0, 2, 3, 4)
Example 4: Convert the following POS expression to canonical SOP
(X+Y).Z'
Solution:
= X.Z'+Y.Z'
= X.Z'.1 + Y.Z'.1
= X.Z'.(Y+Y') + Y.Z'.(X+X')
= X.Y.Z' + X.Y'.Z' + X.Y.Z' + X'.Y.Z'
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