I am trying to draw the logic gate of the following function using only two input nand gates and inverters A'C'D+AC'D'+AB'. So far I am stuck on trying create the AC'D' part. What I have drawn below would give me the some I think and not the product I am looking for.
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1\$\begingroup\$ Take a look at this question. The idea is the very same. \$\endgroup\$– Eugene Sh.Feb 27, 2019 at 14:49
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\$\begingroup\$ I think the problem is I am getting confused with the boolean algebra. Thanks @EugeneSh. \$\endgroup\$– Geralt of RiviaFeb 27, 2019 at 14:53
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1 Answer
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Simplify \$A'C'D+AC'D'+AB' ==> C'(A \oplus D)+AB' \$
You can draw \$(A \oplus D)\$using two-input NAND gates.
Let \$(A\oplus D)\$ = \$ X\$,
i.e, the equation becomes: $$C'X+AB' $$
Using De-Morgan's theorem, we can tweek the equation into:
$$\overline{\overline{(C'X)}.\overline{AB'}}$$
Now you can directly draw the circuit using NAND and NOT gates.
