Question

The SOP (sum of products) form of a Boolean function is $\sum$(0, 1, 3, 7, 11), where inputs are A,B,C,D (A is MSB, and D is LSB). The equivalent minimized expression of the function is

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

Answer 1

The correct option is D $(\overline{B}+C)(\overline{A}+C)(\overline{A}+\overline{B})(\overline{C}+D)$

The 4 variable Boolean function is given in canonical sum of products form as,
$f(A,B,C,D)=\sum (0,1,3,7,11)$
As the options are given in the simplified product of sum form, we first convert the given function in canonical product of sums form, as under $f(A,B,C,D)=\prod (2,4,5,6,8,9,10,12,13,14,15)$
Now by plotting the above function on a 4 variable K-map (maxterms map), obtain the simplified expression of the function

f = $(\overline{B}+C)(\overline{A}+C)(\overline{A}+\overline{B})(\overline{C}+D)$