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+19 votes
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Consider the following expression

$a\bar d + \bar a\bar c + b\bar cd$

Which of the following Karnaugh Maps correctly represents the expression?



  1.  

  2.  

in Digital Logic by Boss (16.3k points)
edited by | 1.1k views

2 Answers

+3 votes
Best answer

$a\bar d + \bar a \bar c+b \bar c d = \overset{m_8}{a\bar b \bar c \bar d} + \overset{m_{10}}{a\bar b  c \bar d} +\overset{m_{12}}{a b \bar c \bar d} + \overset{m_{14}}{a b  c \bar d} $

$\qquad + \overset{m_0}{\bar a \bar b \bar c \bar d} + \overset{m_4}{\bar a  b \bar c \bar d}+ \overset{m_1}{\bar a \bar b \bar c  d}+ \overset{m_5}{\bar a  b \bar c  d}$

$\qquad + \overset{m_5}{\bar ab \bar c d}+ \overset{m_{13}}{ab\bar c d}$

When we minimize a K-map, we can assume either $0$ or $1$ for don't cares. But here they have asked for the expression represented by the K-map. So we can consider $X$ as $1$ and not as a don't care. Also the given expression is equivalent to the above K-map but not the minimal one. Minimal expression will be $\bar a \bar c + b\bar c+ a\bar d.$

Hence, Option A. 

by Veteran (418k points)
selected by
+10 votes

Answer is a.

by Active (5k points)
edited by
0
But we can still derive this expression using (c) and (d) as well right? They didn't ask for a minimal don't care set of K-map or something.

Please explain how are you being confident of the option (a)?
+1
Even the given expression is not the minimal for option A.

d can be removed from bc'd.
0

@Sunit Acharya 

I think the expression should be able to cover every minterm . And it does not cover every minterm from the kmap c and d . 

0
here X represents 1 or don't care
Answer:

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