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The function represented by the Karnaugh map given below is

  1. $A.B$
  2. $AB+BC+CA$
  3. $\overline{B \oplus C}$
  4. $A.BC$
asked in Digital Logic by Veteran (52.1k points)
edited by | 2.2k views
0
should the K-map have BC as 00, 01, 11, 10?
+1
yes. It must be a typo.
+7
it isn't typo it is tricky i would say . Just to confuse the people . if we rewrite it according to the standrard formula . asnwer is so simple as that . BUT min term m1 is 0 not 1 , this is a typo
0
what trick is that?? isnt it a typo?? If we consider normal kmap with 00,01,11,10 we get answer as C. now what trick are you talking about?
+2
Hey I found out slight mistake in this question added, I checked original 1998 question , the minterm value at  '001' is '0' ..so by clearing that mistake I solved further question..

    Here The k-map is not given in cyclic form/gray code form. so after converting it into cyclic code format we got two essential prime implicants ie B'C' and BC.
So
  SOP(Sum of Product)=B'C'+BC
                                        =(B xOR C) '
Answer : C
0
not clear....???
0
sir can uh please explain the meaning of typo in this question ??
0
I got

A' B' +B' C' +A' C + BC after solving the given c k-map is it correct or not can anyone clarify me ??
0
A\BC 00 01 11 10
0 1 1 1 0
1 1 0 1 0

BC+B'C'+A'B' OR BC+B'C'+A'C

plz correct if anything is wrong? 

0

meghna $m_1$ is $0$

0

Mk Utkarsh I tried it as given in question considering m1=1, if m1=0 then B(XNOR)C.

0
why m1 will 0? is there  mistake in question ?
0
I can't understand the question is right or wrong

1 Answer

+12 votes
Best answer

The given K-map is not standard as after "01" we have "10" and two variables are changing for consecutive column. This means it is not safe to merge adjacent $1s.$ By converting the K-map to standard form we get

which gives

$BC + \bar B\bar C = B \text{ XNOR } C = B \odot C$

This can be represented as negation of $\text{XOR} = \overline{B \oplus C}$

Option C is correct.

answered by Boss (45.1k points)
selected by
0
what about minterm formed by postion m0 and m1 ie A'B'
+2
This is typo minterm m1 is 0 not 1
+1
why can't we make a pair of 4??? if we have all the ones then we must have taken 8 pairs. why here is a problem with 4 ones?
+1

@Gurjot_Singh

  • Read the comments after the question
  • Closely obserb the given image and compare with the standard diagram used to represent K-Map.

This would solve your doubt

Answer:

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