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57 votes
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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
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4 Answers

Best answer
97 votes
97 votes
$f = ((f_1f_2)'f_3')' = f_1f_2 + f_3$

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 $f = f_3 =Σ(1, 4, 5).$

Correct Answer: $A$
edited by
44 votes
44 votes

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.

13 votes
13 votes

AND of two minterm expression gives common terms.

$f1\ AND\ f2\equiv f1\cap f2.$

So, $f1\ NAND\ f2\equiv$${(f1\cap f2)'}$

$f1\cap f2=\{0,1,3,5\}\cap\{6,7\}=\phi$ 

And, we have,

$f=[(f1\cap f2)' \cap (f3)']'=f1 \cap f2 \bigcup f3 = \{1,4,5\}$

But, $f1 \cap f2 = \phi$

$\Rightarrow \phi\ U  f3 = \{1,4,5\}$

$ \Rightarrow f3 =\{1,4,5\}$

So, Correct Answer is (A).

Answer:

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