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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