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)
37
Boolean Algebra 37

