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NAND is logically complete. Use only NAND gates to constructgate-level circuits that compute the

following.
a. NOT
b. AND
c. OR

1 Answer

6 votes

Answer:

Hi, for this exercise we have two laws to bear in mind:

Morgan's laws

NOT(А).NOT(В) = NOT(A) + NOT (B)

NOT(A) + NOT (B) = NOT(А).NOT(В)

And the table of the Nand

INPUT OUTPUT

A B A NAND B

0 0 1

0 1 1

1 0 1

1 1 0

Let's start!

a.

Input OUTPUT

A A A NAND A

1 1 0

0 0 1

b.

Input OUTPUT

A B (A NAND B ) NAND (A NAND B )

0 0 0

0 1 0

1 0 0

1 1 1

C.

Input OUTPUT

A B (A NAND A ) NAND (B NAND B )

0 0 0

0 1 1

1 0 1

1 1 1

Step-by-step explanation:

In the first one, we only need one input in this case A and comparing with the truth table we have the not gate

In the second case, we have to negate the AND an as we know how to build a not, we only have to make a nand in the two inputs (A, B) and the make another nand with that output.

In the third case we have that the OR is A + B and we know in base of the morgan's law that:

A + B = NOT(NOT(А).NOT(В))

So, we have to negate the two inputs and after make nand with the two inputs negated.

I hope it's help you.

User Abhishek Dhiman
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