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What is the boolean expression for the output $f$ of the combinational logic circuit of NOR gates given below?

1. $\overline{Q+R}$
2. $\overline{P+Q}$
3. $\overline{P+R}$
4. $\overline{P+Q+R}$
edited | 2.4k views
+1

The question is the realization of OR-AND(Product of Sum) using NOR gates.

This is the realization of (P+Q)(Q+R) using NOR gates.

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Give full explanation

$​\text{ Level 1:}$
$(\overline{P + Q}) (\overline{Q + R}) (\overline{P + R}) (\overline{Q + R})$

$\text{ Level 2:}$
$\overline{(\overline{P + Q})+(\overline{Q + R})} = (P+Q)(Q+R)=PQ+PR+Q+QR$
$\overline{(\overline{P + R}) (\overline{Q + R})} = (P+R)(Q+R)=PQ+R+QR+PR$

$\text{ Level 3:}$
$\\\overline{PR+QR+PQ+Q+R} =\overline{Q+R}\\ \therefore \text{ Answer: Option A} ​$
selected
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can you please give simplification of level 3.?
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PR+QR+PQ+Q+R -->

Take Q common in Q and QP. We get Q(1+P) = Q

Similarly, R common in PR and QR. We get Q and P and the result.
+1
we can simply replace nor nor ckt to or and ... it will make easy ckt
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Good Explanation
[((P'Q') + (Q'R'))' + (P'R') + (Q'R'))']'

[(Q'(P' + R'))' + (R'(P' + Q'))']'

[(Q + PR + R + PQ)]'

[Q(1 + P) + R(1 + P)]'

$f=(\overline{(\overline{(\overline{P+Q})+(\overline{Q+R})})+(\overline{(\overline{P+R})+(\overline{Q+R})})})$

Using De-Morgan's law: $\overline{\overline{(P+Q)}+\overline{(Q+R)}}=\overline{\overline{(P+Q)}}.\overline{\overline{(Q+R)}}$ and $\overline{\overline{(P+R)}+\overline{(Q+R)}}=\overline{\overline{(P+R)}}.\overline{\overline{(Q+R)}}$

$\rightarrow(\overline{(P+Q).(Q+R)+(P+R).(Q+R)})$
$\rightarrow(\overline{(PQ+PR+Q+QR)+(PQ+PR+R+QR)})$

Q + QR = Q and R + QR = R [absorption law]

$\rightarrow(\overline{(PQ+PR+Q)+(PQ+PR+R)})$

Q + PQ = Q and R + PR = R

$\rightarrow(\overline{PR+R+PQ+Q})$
$\rightarrow(\overline{Q+R})$
edited
–1 vote

use de morgan's law

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