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17 votes

What is the equivalent Boolean expression in product-of-sums form for the Karnaugh map given in Fig

 

  1. $B\overline{D} + \overline{B}D$

  2. $(B + \overline{C} +D) (\overline{B} + C + \overline{D})$

  3. $(B + {D})(\overline{B} +\overline{ D})$

  4. $(B + \overline{D})(\overline{B} + {D})$

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(C) is correct option

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4 Answers

12 votes
12 votes
Best answer

Following two K-Maps are equivalent and represent the same boolean function.
While the first K-Map gives us the boolean expression in Sum-of-Product form, and the second K-Map gives us the same boolean function in Product-of-Sum form:

 

(i)
(ii)

 

(C) is correct option!

PS: If SOP is asked answer will be (A).
 

edited by
15 votes
15 votes

If all the empty cells are filled with '0' and then the POS expression is calculated as:

                                                                                                                               

CD\AB   00   01   11   10
00    0       0
01         0     0  
11       0     0  
10    0        0

   

Then the POS expression of f = (D+B)(D'+B') = BD' + B'D

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3 Comments

You are right
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So here, what will be correct answer?? A or C??
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The one in POS form
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5 votes
5 votes
First of all find the sop form then just complement it to get the corresponding pos form which comes out to be option c

4 Comments

Say , F = something in sop
taking complement,
F' = something in pos , yes that is right
but what we get in pos is F'
and what we need in pos is F.
So if we have  F = some minterms  i.e, sop
get F' = remaining minterms , that is also sop
that complement, we get F is pos
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This method doesn’t works! We have to find POS explicitly, complementing SOP won’t give correct POS here.
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no by complementing we get f complement which is in pos form
but we want f in pos form
so best method is to convert sop to pos using boolean laws
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3 votes
3 votes
Or Use the K-map get the Min terms s we get  in this question.#USING MIN-TERMS and convert them to MAX term Check for Options
as in using K map we get
F= D'B + B'D (Is would be easy till this part)
Now apply de-morgan's law
F'= (D + B') (D'+B)
solve this  we get
F'= DD' +DB + D'B' BB'

Apply de-morgan's again
F'' = (D'+B')(D+B)
Answer:

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