GATE IT 2008 | Question: 9

Question

What Boolean function does the circuit below realize?

• A. $xz + \bar{x}\bar{z}$
• B. $x\bar{z} + \bar{x}{z}$
• C. $\bar{x}\bar{y} + {y}{z}$
• D. $xy + \bar{y}\bar{z}$ 

Comments

@Shaik Masthan sir so can we say that it is an ambiguous question? 

 

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here you just have to create kmap for minterms (0,2,5,7) and take complement of there as nor is used
Or
we can do kmap for Maxterms(0,2,5,7) and we will get the answer
as B) x xNor z

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here we consider Z as msb or X as msb output will be same that's why numbering not given

Answer 1

Answer: B

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We have to note that, in the question it is not mentioned whether we are realizing F(X,Y,Z) or F(Z,Y,X). 

Let consider we are realizing F(X,Y,Z).

$F(X,Y,Z) =\overline{(O_0+O_2+O_5+O_7)}$

$F(X,Y,Z) =\overline{(O_{000}+O_{010}+O_{101}+O_{111})}$ (in binary format)

$F(X,Y,Z) =\overline{(\bar X.\bar Y. \bar Z+\bar X.Y.\bar Z+X.\bar Y.Z+X.Y.Z)}$

$F(X,Y,Z) =\overline{(\bar X.\bar Z.(\bar Y+Y))+X.Z.(\bar Y+Y))}$

$F(X,Y,Z) = \overline{(\bar X \bar Z + XZ)} = X\bar Z + \bar XZ$

 

Let consider we are realizing F(Z,Y,X).

$F(Z,Y,X) =\overline{(O_0+O_2+O_5+O_7)}$

$F(Z,Y,X) =\overline{(O_{000}+O_{010}+O_{101}+O_{111})}$ (in binary format)

$F(Z,Y,X) =\overline{(\bar Z.\bar Y. \bar X+\bar Z.Y.\bar X+Z.\bar Y.X+Z.Y.X)}$

$F(Z,Y,X) =\overline{(\bar Z.\bar X.(\bar Y+Y))+Z.X.(\bar Y+Y))}$

$F(Z,Y,X) = \overline{(\bar Z \bar X + XZ)} = Z\bar X + \bar ZX = X\bar Z + \bar XZ$ 

 

In both cases, answer is same.

Comments

are we considering 'x' as MSB or 'z' as MSB?

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consider x as MSB .
 even if you consider z as MSB answer will be same..

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Those which are connected to the NOR gate. We have to consider them only.

Answer 2

Decoders have minterms directly , so we need not minimize the function.

The given function takes m0,m2,m5,m7 and nor (NOR GATE IN THE CIRCUIT IS SHOWN) them,  that is the function will implement minterms m1,m3,m4,m7.

Drawing the kmap and solving will give option B

Comments

Decodes gives min terms i.e at those values function will be 1. Connecting NOR gate will take the complement of the function. Like first it was giving min terms, now it will give 1 for max terms. Though it what yo said will work but it is incorrect to say as min terms.  "implement minterms m1,m3,m4,m7" These are max terms and we will fill 0 in this place in Kmap.

PS: Answer won't change I am just telling the terminology.

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m1, m3, m4, m7 will be minterms only because m0,m2,m5,m7 are max terms then others are minterms...

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Mind blown

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The given function takes m0,m2,m5,m7 and nor (NOR GATE IN THE CIRCUIT IS SHOWN) them,  that is the function will implement minterms m1,m3,m4 the last one should be m6