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+16 votes

In the Karnaugh map shown below, $X$ denotes a don’t care term. What is the minimal form of the function represented by the Karnaugh map?

  1. $\bar{b}.\bar{d} + \bar{a}.\bar{d}$

  2. $\bar{a}.\bar{b} + \bar{b}.\bar{d} + \bar{a}.b.\bar{d}$

  3. $\bar{b}.\bar{d} + \bar{a}.b.\bar{d}$

  4. $\bar{a}.\bar{b} + \bar{b}.\bar{d} + \bar{a}.\bar{d}$

asked in Digital Logic by Veteran (59.9k points)
edited by | 1.4k views

4 Answers

+19 votes
Best answer

$2$ quads are getting formed:                                                            

Value for first one is $a'd'$ and value for 2$^{nd}$ one is $b'd'.$

Answer is Option A.

answered by Junior (617 points)
edited by
Also, only 2 essential PI are formed and they cover all the minterms.The size of each EPI is of $2^2$ so final minimum expression must have exactly 2 terms and in each term you'll have only $4-2=2$ variables.Only (A) satisfies.
+3 votes
m0 ,m1,m8 ,m9 form one quad a'd'

m0,m2,m8,m10 form one quad b'd'

So f=a'd'+b'd'

Ans is a
answered by Boss (32.1k points)
hi, why we are not taking m0,m4,m12,m8 means with dont care
+1 vote
answer - A
answered by Loyal (8.7k points)
0 votes

X denotes a don’t care term ,means we can take them whenever necessary.Here, if we take 1 don't care,it is sufficient (m10)

m0 ,m1,m8 ,m9 -> a'd'

m0,m2,m8,m10 -> b'd'

So, Option A is correct

answered by Active (3.8k points)

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