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+12 votes
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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. $\overline{P+Q+R}$
asked in Digital Logic by Veteran (101k points)
edited | 2.2k 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.

4 Answers

+14 votes
Best answer
$​\text{ Level 1:}$
$(\overline{P + Q}) (\overline{Q + R}) (\overline{P + R}) (\overline{Q + R})$

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

$\text{ Level 3:}$
$\\\overline{PR+QR+PQ+Q+R} =\overline{Q+R}\\ \therefore \text{ Answer: Option A} ​$
answered by Active (4.2k points)
selected
0
can you please give simplification of level 3.?
0
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
0
Good Explanation
+6 votes
[((P'Q') + (Q'R'))' + (P'R') + (Q'R'))']'

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

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

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

[Q + R]'---------Answer.
answered by Boss (19.7k points)
+3 votes
$f=(\overline{(\overline{(\overline{P+Q})+(\overline{Q+R})})+(\overline{(\overline{P+R})+(\overline{Q+R})})})$
Using De-Morgan's law
$\rightarrow(\overline{(\overline{\overline{PQ}+\overline{QR}})+(\overline{\overline{PR}+\overline{QR}})})$
$\rightarrow(\overline{(P+Q)(Q+R)+(P+R)(Q+R)})$
$\rightarrow(\overline{(PQ+PR+Q+QR)+(PQ+PR+R+QR)})$
Q + QR = Q and R + QR = R [absorption law]
$\rightarrow(\overline{(PQ+PR+Q)+(PQ+PR+R)})$
Q + PQ = Q and R + PR = R
$\rightarrow(\overline{PR+R+PQ+Q})$
$\rightarrow(\overline{Q+R})$
answered by Active (2k points)
–1 vote
answer - A

use de morgan's law
answered by Loyal (9k points)


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