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The following expression was to be realized using 2-input AND and OR gates, but by mistake all 2-input AND gates were taken as 2-input NAND gates,

$$(a.b).c + (\bar a.c).d + (\bar b.c).d + a.d$$

What is the function finally realized ?

  1. $1$
  2. $\bar a + \bar b + \bar c + \bar d$
  3. $\bar a + b + \bar c + \bar d$
  4. $\bar a + \bar b + c + \bar d$
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I think 1 should be the answer

((not b nand c) nand d) + ((a nand b) nand c) + ((not a nand c) nand d)+ (a nand d)

reduces to

not a + b + not c + not b +not d
= 1  ?
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2 Answers

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Best answer

I have not indicated the k map properly but answer is 1

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How can you assume NOT gate is available and directly take complement for eg in (a'c) term you are writing it as (a'b)'  ??
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1 vote
1 vote
right answer is A
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