GATE EE 2014 Digital Electronics Q 2 - Grad Plus

# GATE EE 2014 Digital Electronics Q 2

The SOP (sum of products) form of a Boolean function is Σ(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) \left(\overline B+C\right)\;\left(\overline A+C\right)\;\left(\overline A+\overline B\right)\;\left(\overline C+D\right)

b) \left(\overline B+C\right)\;\left(\overline A+C\right)\;\left(\overline A+\overline C\right)\;\left(\overline C+D\right)

c) \left(\overline B+C\right)\;\left(\overline A+C\right)\;\left(\overline A+\overline C\right)\;\left(\overline C+\overline D\right)

d) \left(\overline B+C\right)\;\left(A+\overline B\right)\;\left(\overline A+\overline B\right)\;\left(\overline C+D\right)

Ans: (a)

Explanation

The 4 variable Boolean function is given in canonical (SOP) sum of products form as, f(A, B, C,D)=Σ(0, 1, 3, 7, 11)

But, the options are given in the simplified (POS) product of sums form. So, convert the given function in canonical product of sums form,

f(A, B, C, D)= Π(2, 4, 5, 6, 8, 9, 10, 12, 13, 14, 15)

Plotting the above function on a 4 variable K-map (Maxterms map), we obtain the simplified expression of the function

f=\left(\overline B+C\right)\cdot\left(\overline A+C\right)\cdot\left(\overline A+\overline B\right)\cdot\left(\overline C+D\right)
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