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24. Find the dual of: (A' + 0) • (B' + 1) = A' [ISC 2020]
Ans. Dual = A'•1 + B'• 0 = A'
25. State whether the following proposition is a tautology, contradiction or a contingency: [ISC 2020]
F = (P => Q) V (Q => ∼P)
Ans. F = (P => Q) V (Q => ∼P)
P Q ∼P P => Q Q => ∼P (P => Q) ∨ (Q => ∼P)
0 0 1 1 1 1
0 1 1 1 1 1
1 0 0 0 1 1
1 1 0 1 0 1
(using laws): (P' + Q) + (Q' + P') = 1 ( as Q + Q' = 1)
Hence, it is a TAUTOLOGY
26. Given the Boolean function F(A, B, C, D) = Σ(0, 1, 2, 3, 4, 5, 8, 9, 10, 11, 12, 13, 14). [ISC 2020]
Reduce the above expression by using a 4-variable Karnaugh map, showing the various groups (i.e., octals, quads and pairs).
Ans. F(A, B, C, D) = Σ (0, 1, 2, 3, 4, 5, 8, 9, 10, 11, 12, 13, 14)
C'.D' C'.D C.D C.D'
0 1 3 2
A'.B' 1 1 1 1
4 5 7 6
A'.B 1 1 0 0
12 13 15 14
A.B 1 1 0 1
8 9 11 10
A.B' 1 1 1 1
There are two octets and one quads:
Quad 1: (m +m +m +m +m +m +m +m ) = B'
1
0
2
10
11
9
3
8
Quad 2: (m +m +m +m +m +m +m +m ) = C'
12
8
1
5
9
4
13
0
Quad 3: (m +m +m +m ) = AD'
14
8
10
12
Hence, F(A, B, C, D) = B' + C' + AD'
27. Given the Boolean function F(A, B, C, D) = Σ(3, 4, 6, 9, 11, 12, 13, 14, 15). [ISC 2020]
Reduce the above expression by using a 4-variable Karnaugh map, showing the various groups (i.e., octals, quads and pairs).
Ans. C+D C+D' C'+D' C'+D
0 1 3 2
A+B 1 1 0 1
4 5 7 6
A+B' 0 1 1 0
12 13 15 14
A'+B' 0 0 0 0
8 9 11 10
A'+B 1 0 0 1
There are two quads and one pair:
Quad 1: (M M M M ) = B' + D
14
6
4
12
Quad 2: (M M M M ) = A' + D'
13
9
15
11
Pair: (M M ) = B + C' + D'
11
3
Hence, F(A, B, C, D) = ( B' + D) • (A' + D') • (B + C' + D')
28. Convert the following expression to its cardinal SOP form: [ISC 2020]
F(P,Q,R) = P'Q'R + P'QR + PQ'R' + PQR'
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