Page 39 - Computer science 868 Class 12
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The inputs are:

                                                    A              Has an excellent academic record
                                                    F              Financially sound
                                                    C              Belongs to a backward class
                                                     I             Is physically impaired
                 (In all the above cases 1 indicates Yes and 0 indicates No).

                 Output: X                            [1 indicates Yes, 0 indicates No for all cases]
                 Draw the truth table for the inputs and outputs given above and write the SOP expression for X(A, F, C, I).  [ISC 2018]

                 Solution:

                 In the first statement, we get A=1 and F=0.
                 In the second statement, we get A=0 and C=1.

                 And the third statement has A=0 and I=1.
                 Let us draw the truth table and put 1 (true) in the final column X having the above combinations.

                                    A           F           C           I          X       Cardinal Number

                                    0           0           0           0          0              0
                                    0           0           0           1          1              1
                                    0           0           1           0          1              2

                                    0           0           1           1          1              3
                                    0           1           0           0          0              4
                                    0           1           0           1          1              5

                                    0           1           1           0          1              6
                                    0           1           1           1          1              7

                                    1           0           0           0          1              8
                                    1           0           0           1          1              9
                                    1           0           1           0          1             10

                                    1           0           1           1          1             11
                                    1           1           0           0          0             12
                                    1           1           0           1          0             13

                                    1           1           1           0          0             14
                                    1           1           1           1          0             15

                 For deriving SOP expression, we add those minterms which have 1 in their output column. The final SOP expression in
                 the canonical form is:
                          A'.F'.C'.I + A'.F'.C.I' + A'.F'.C.I + A'.F.C'.I + A'.F.C.I' + A'.F.C.I + A.F'.C'.I' + A.F'.C'.I + A.F'.C.I' + A.F'.C.I

                 In the cardinal form, the expression is: Σ(1, 2, 3, 5, 6, 7, 8, 9, 10, 11)







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                                                                                                      Boolean Algebra   37
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