1.7k views

The Boolean function in sum of products form where K-map is given below (figure) is _______ edited | 1.7k views
+2
simple approach:

1) just draw truth table for 3 variables(0 to 7)

2) now set function value acoording to given kmap

3) minimize function now using kmap
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Someone, please post a proper solution.

Answer - $ABC + B'C' + A'C'$

Expand this $K$ map of $2$ variables $($$4 cells) to K map of three variable ($$8$ cells$)$

Entries which are non zero are: $A'B'C', AB'C', A'BC'$ and $ABC$

Minimize $SOP$ expression using that $K$ map.

edited
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You missed an entry AB'C'
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A'B'C' + A'B C' + ABC = A'C'(B' + B) + ABC = A'C' + ABC
+1
The given K-map has an entry B'C'. So, when you expand with A, it must be AB'C' + A'B'C'.
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I agree I'll edit the answer

finally it comes out as ABC + B'C' + A'C'
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@Arjun sir is this ans correct?
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how we can  expand this K map of 2 variables (4 cells) to K map of three variable (8 cells)?????
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how to convert 2 variable k map to 4 variable k map?
+16

Convert 2 variable k map to 3 variable k map like this. look b and c at 0,0 fill 1 in new table, at 0,1 fill 0, at 1,0 fill A', at 1,1 fill A

+1
This guy has everywhere posted solutions with half or no explanations and even the solutions get selected as best one.
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What he has explained is correct.
If you don't know what he is saying, just check how to construct a characteristic table of a flip flop. You will get the answer.
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@Peeyush Pandey

Nice approach!!

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Is this thing valid?

$\mathbf{A'(BC+B'C') = A'}$

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make a characteristic table for 3 variables and check all 8 combinations, if L.H.S. matches R.H.S for every combination then it is valid else not.
B'C'+ BC'A'+ ABC = C'(B' + BA') + ABC = C'(A'+ B') + ABC = A'C'+ B'C'+ ABC
by
edited by
+2
sir, this is variable entrant map??? right
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Yes, this is directly from variable entrant map.

Alternate Method:-

 C B 0 1 0 1 0 1 A’ A

Step1: (Mark all the variables in Cell as 0)

 C B 0 1 0 1 0 1 0 0

f0=B’C’  ---- i

Step2:  Minterm for variable A’. (Mark minterm obtained for 1 as Don’t Care and target variable as 1)

 C B 0 1 0 X 0 1 1 0

fA’=A’C’ --- ii

Step2:  Minterm for variable A. (Mark minterm obtained for 1 as Don’t Care and target variable as 1)

 C B 0 1 0 X 0 1 0 1

fA=ABC ---iii

f=ABC+A’C’+B’C’ (Ans)

+1 vote

### Variable Entered Map:

 $B$ $C$ $f$ $0$ $0$ $1$ $0$ $1$ $0$ $1$ $0$ $A'$ $1$ $1$ $A$

$\downarrow$

 $A$ $B$ $C$ $f$ $0$ $0$ $0$ $1$ $0$ $0$ $1$ $0$ $0$ $1$ $0$ $1$ $0$ $1$ $1$ $0$ $1$ $0$ $0$ $1$ $1$ $0$ $1$ $0$ $1$ $1$ $0$ $0$ $1$ $1$ $1$ $1$

$f=\Sigma (0,2,4,7)$

$f=A'B'C'+A'BC'+AB'C'+ABC$

(b!c!) +ab +ac!        ?

Here's a very simplified approach. All you have to do is observe.

The function $F$ is true when B and C are 0.

$A$ is true, when B and C are 1.

$\overline{A}$ is true, when B is 1 and C is 0.

Make K-map of three variables. (8 cells)

Where B and C are 0, put 1 in that cell irrespective of what A is.

Where B and C are 1, put 1 in the cell against A.

Where BC = 10, put 1 in the cell against $\overline{A}$, ie, where A is 0.

This is the K-map you'll get. $F = ABC+\overline{B} \overline{C}+\overline{A} \overline{C}$