+1 vote
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| 209 views
+1
it is simply, ∑m(1,3,5,6) -----> just cancel the bubbles...

evaluate each and every option, which is equal to the above function, that is the answer.

( Option A is the answer )
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why I got  (Y+Z) (X'+ Y'+Z') (X+Y'+Z)  dunno

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OOh  both (X+Y'+Z)  and (X+Z) = 010  = 2

got it !

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I have doubt here :-

Since the last gate is a Nand gate can't i write:-

F = (X+Y+Z')(X+Y'+Z')(X'+Y+Z')(X'+Y'+Z)

1             3              5             6

Why its wrong ?
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then you have to apply NAND gate on ∏(1,3,5,6).
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Ohk so the answer should be F' right ?

I need to take complement of above F for getting answer isn't it ?
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yes, after getting result of your approach, you should complement it !
+1
Inside the box, there is one level of AND gates and bubbled o/p make them equivalent to NAND gate. And 2nd level is also NAND gate. So it's a simple NAND- NAND circuit. Or you can cancel out the bubbles and view it as AND-OR circuit.
So $F= \Sigma(1,3,5,6) = \Pi(0,2,4,7)$.
Now simplify the obtained POS.
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I got one doubt i get that how it became Σ(1,3,5,6) but we are answering Π(0,2,4,7) which is the complement of the above function ?

What i get so far is :- F = (1'.3'.5'.6')' = 1 + 3 + 5 + 6 = Σ(1,3,5,6) ---> This is the actual F but  Π(0,2,4,7) is F' na
+2

i hope, you confused !

if F(X,Y,Z) = ∑m (1,3,5,6), then

F(X,Y,Z) = ∑m (1,3,5,6) = πM (0,2,4,7) --------> Those are equal, just we represent in POS to SOP form.

complement of F(X,Y,Z) = ∑m (0,2,4,7) = πM (1,3,5,6)

bubbled nor gate means AND gate so in the diagram max terms are there so find kmap for minterms
by (185 points)