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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+18 votes

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

- $\overline{Q+R}$
- $\overline{P+Q}$
- $\overline{P+R}$
- $\overline{P+Q+R}$

+2

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.

0

if you understood the above logic then similarly for the second also we will realise using the Or-and realization .

so at the end, we will get (pr+q) from first case and (pq+r) from the second case.

now this output will act as the input for the last NOR gate .

((pr+q) +(pq+r))' =which can be written as (r(p+1) +q(p+1))' ..

now , on simplifying this we will get (q+r)'

so at the end, we will get (pr+q) from first case and (pq+r) from the second case.

now this output will act as the input for the last NOR gate .

((pr+q) +(pq+r))' =which can be written as (r(p+1) +q(p+1))' ..

now , on simplifying this we will get (q+r)'

+21 votes

Best answer

$β\text{ Level 1:}$

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

$\text{ Level 2:}$

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

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

$\text{ Level 3:}$

$\\\overline{PR+QR+PQ+Q+R} =\overline{Q+R}\\ \therefore \text{ Answer: Option A} β$

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

$\text{ Level 2:}$

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

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

$\text{ Level 3:}$

$\\\overline{PR+QR+PQ+Q+R} =\overline{Q+R}\\ \therefore \text{ Answer: Option A} β$

+7 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.

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

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

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

[Q + R]'---------Answer.

+5 votes

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

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

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

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

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

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

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

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