Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 20 (Q.No. 48)
48.
The minimum number of 2 input NAND gates required to implement Boolean function A B C is A, B, C are available is
Discussion:
2 comments Page 1 of 1.
Jane said:
8 years ago
One NAND gate to get bar B.
Two NAND gate to get AC.
Two NAND gate to get required result.
Total NAND gate required = 1+2+2 = 5.
Two NAND gate to get AC.
Two NAND gate to get required result.
Total NAND gate required = 1+2+2 = 5.
(1)
Arsh said:
7 years ago
One NAND gate to get bar B.
Second NAND gate to get bar[A barB C].
Third NAND gate to get A bar B C.
(Double inversion)
Second NAND gate to get bar[A barB C].
Third NAND gate to get A bar B C.
(Double inversion)
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