To implement y=abcd using only two input nand gates, minimum no. Of requirement of gate is
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Answer:
Explanation:
To implement y=ab + cd using only two input nand gates, minimum no. Of requirement of gate is
3 NAND GATE
' - representing -ve
(A . B)' = A' + B'
( C . D) ' = C' + D'
((A' + B').(C' + D'))' = AB + CD
A B C D Y = AB + CD
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
0 1 0 1 0
0 1 1 0 0
0 1 1 1 1
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1
A B NAND(AB) (A.B)'
0 0 1
0 0 1
0 0 1
0 0 1
0 1 1
0 1 1
0 1 1
0 1 1
1 0 1
1 0 1
1 0 1
1 0 1
1 1 0
1 1 0
1 1 0
1 1 0
C D NAND(CD) (C.D)'
0 0 1
0 1 1
1 0 1
1 1 0
0 0 1
0 1 1
1 0 1
1 1 0
0 0 1
0 1 1
1 0 1
1 1 0
0 0 1
0 1 1
1 0 1
1 1 0
NAND(AB) NAND(CD) NAND
(A.B)' (C.D)' ((A.B)' .(C.D)')'
1 1 0
1 1 0
1 1 0
1 0 1
1 1 0
1 1 0
1 1 0
1 0 1
1 1 0
1 1 0
1 1 0
1 0 1
0 1 1
0 1 1
0 1 1
0 0 1