GATE CSE 1993 | Question: 6.1

Question

Identify the logic function performed by the circuit shown in figure.

 

• A. exclusive OR
• B. exclusive NOR
• C. NAND
• D. NOR
• E. None of the above

Comments

Simplify the given circuit:

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there is one more easy method.

shaded region works as Two level OR And ;

now you need to do :

[x+(x'y')] And [y +(x'y')] = [x+y'] And [y+ x']= xy+x'y'

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@anu shaded portion is same as in the comment posted by me?

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NOR-NOR Realization is same as OR-AND realization.

Answer 1

$(x+x'y') .(y+x'y') = (x+y')(x'+y) = xy + x'y' = \text{Exclusive-NOR}$

Comments

johnson counter ???????????

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HOW EXCLUSIVE NOR PLZ SIR TELL THIS STEP BY STEP?

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@Arjun sir please modify this answer as question demands for identifying the logic function based on circuit and in answer part explanation for 6.2 and 6.3 is given which is not required.

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For EX-NOR of 2 boolean variable if we simply replace the variables with their complement , the value remains same .

F = a.b + a’.b’ – > replace a with a’ and b with b’ – > a’.b’ + ab

We will be using this theory here . Now , look at the 3 XNOR gates from right . If we simply imagine that the ‘NOT’ from first 2 XNOR and is taken and made as a NOT gate to the inputs of the final XNOR gate then the final XNOR doesn’t change and the other 2 XNOR becomes XOR .

then the boolean formula becomes something like ,

XNOR(XOR(x,XNOR(x,y)), XOR(y,XNOR(x,y)))

= NOT(XOR(XOR(x,XNOR(x,y)) , XOR(y,XNOR(x,y))))

= NOT(XOR(x,XOR(x,y),y,XOR(x,y)))

= XNOR(x,y)

Answer 2

ANS: B

( (x+(x+y)')' + (y+(x+y)')' )' =xy +x'y'