GATE CSE 2010 | Question: 31

Question

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

• A. $\overline{Q+R}$
• B. $\overline{P+Q}$
• C. $\overline{P+R}$
• D. $\overline{P+Q+R}$

Comments

Try these one too

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1st line ??

Kse??

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 Bharti goyal by using distribution law.

Answer 1

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

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

$\text{ Level 3:}$
$\\\overline{PR+QR+PQ+Q+R} =\overline{Q+R}\\ \therefore \text{ Answer: Option A} ​$

Comments

Good Explanation

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${\color{Purple} {\implies \text{NOR - NOR Realization is equal to OR - AND  Realization.}}}$

${\color{Teal} {\implies \text{NAND - NAND Realization is equal to AND - OR  Realization.}}}$

So, we can easily solve and get the simplified expression $ = \overline{Q + R}$

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Yes

Answer 2

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

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

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

Comments

Nicely explained :) tnx

Answer 3

[((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.

Answer 4

Hope you find this helpful... 

Comments

Yes