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Consider the following logic circuit whose inputs are functions $f_1, f_2, f_3$ and output is $f$

Given that

$f_1(x,y,z) = \Sigma (0,1,3,5)$

$f_2(x,y,z) = \Sigma (6,7),$ and

$f(x,y,z) = \Sigma (1,4,5).$

$f_3$ is

1. $\Sigma (1,4,5)$
2. $\Sigma (6,7)$
3. $\Sigma (0,1,3,5)$
4. None of the above
edited | 3.3k views
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Here f2 should be : f2(x,y,z)=Σ(6,5) as given in original paper.

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Can you show the original paper? But that doesn't any way change the answer here as min-term 5 is in f3.
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Arjun sir great explanations
Thank you so much
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Great explanation Arjun sir! Understood easily ! Thanks :)
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what is the result of AND of Maxterm instead of minterms?
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Maxterm Means$: A+B+C$

AND of Maxterm is equal to  $(A+B+C). (\bar{A}+B+C)$ Product of Sum(POS)

$f = ((f_1f_2)'f_3')' = f_1f_2 + f_3$

In minimum sum of products form, AND of two expressions will contain the common terms. Since $f_1$ and $f_2$ don't have any common term, $f_1f_2$ is $0$ and hence $f = f3 =Σ(1, 4, 5).$
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here we dont have AND gate. We have NAND gate so f1 NAND(compl. of AND) f2 = {NULL}Compl. which mean f1 NAND f2 will have all the element. So, answer should be D
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Here f1 and f2 have $\sum (5)$ in common then why its that nothing in common ?

Here we have NAND - NAND Circuit, we can convert it to following AND - OR circuit. (As NAND is bubbled OR). Now it is easy to solve this question. F1 AND F2 = 0. SO whatever f3 is directly passed to output. So answer is A.

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sir how you convert NAND to AND and NOR to OR ..?

i got you answer but just only wants to know the hidden concept behind the your concept

and i think you should edit your answer "Here we have NAND - NAND Circuit to Here we have NAND - NOR Circuit"
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Here we don't have any nor gate its nand - nand realization which is equivalent to and-or realization.
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this should be the best answer

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