SOLUTIONS:
1. a. F (A,B,C) = A’ B’ C’ + A’ B’ C + A B’ C’ + A B’ C + A B C’ + A B C
[Distributive]
= A’B’ (C’ + C) + AB’ (C’ + C) + AB (C’ + C) [ Complementarity]
= A’B’ + AB’ + AB [Idempotency]
= A’B’ + AB’ + AB’ +AB [ Distributive]
= B’ (A’ + A) + A (B’ + B)
[ Complementarity]
= B’ + A
b. F(A,B) = (A’+B) ( A+B) [Distributive]
= AA’ + A’B+ AB+ BB [Complementarity ]
= A’B+AB+BB [Distributive]
= B ( A’+A+1)
[Complementarity]
=B.1
[ identity ]
=B
2. a.
a’b’ + a’c’ + bc’ [Complementarity & Distributive]
= a’b’ + (a’c’b’+a’c’b) + bc’ [combining]
= (a’b’ + a’b’c’) + (bc’a’ + bc’)[ commutativity & associativity]
= a’b’ + bc’[covering]
b. (a’+ b)(a’+ c’)(b’+ c’) [Associative]
= (a’+ b)(b’+ c’)(a’+ c’)[commutativity]
= (a’+ b)(b’+ c’)[consensus]
3.
a) Truth Table
b) SOP
Y0 =
(A’B’C’D)+(A’B’CD’)+(A’BC’D’)+(A’BCD)+(AB’C’D’)+(AB’CD)+(ABC’D)+(A
BCD’)
Y1=
(A’B’CD)+(A’BC’D)+(A’BCD’)+(A’BCD)+(AB’C’D)+(AB’CD’)+(AB’CD)+(AB
C’D’)+(ABC’D)+(ABCD’)
Y2 = ABCD
c) POS
Y0=
(A+B+C+D)(A+B+C’+D’)(A+B’+C+D’)(A+B’+C’+D)(A’+B+C+D’)(A’+B+C’+
D)(A’+B’+C+D)(A’+B’+C’+D’)
Y1=
(A+B+C+D)(A+B+C+D’)(A+B+C’+D)(A+B’+C+D)(A’+B+C+D)(A’+B’+C’+D’
)
Y2 = πM (0,1,2,3,4,5,6,7,8,9,10,11,12,13,14)
4. a.
The simplified function is:
F = B’D + A’B’C’ + ACD + ABD’
b.
5. a.
F(A,B,C,D) = (A + B’ + C’)(A + B + C)
b.
F(A,B,C,D) = D (A’ + C’)
6. a.
Since the universal gates {AND, OR, NOT can be constructed from
the NAND gate, it is universal.
b. The given function is universal.
7. a.
Hence, the result is proved.
The same result will be obtained for Expansion for y and Expansion for z
(Try it!!!)
b.
Thus, proved. Same result will be obtained for Expansion for y.
8. f (a, b, c, d) = Sm(0,1, 2, 3, 4, 5, 7,8,12) + S d(10,11)