Question

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

Answer 1

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

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.