Applied Test Series

Question

The number of minterms in the function F is _______

 

 

Comments

On expanding F we will get:

=> (A+A’)B(C+C’)D + (A+A’)B’C(D+D’)

=> (AB + A’B)(CD + C’D) + (AB’ + A’B’)(CD + CD’)

=> ABCD + ABC’D + A’BCD + A’BC’D  + AB’CD + AB’CD’ + A’B’CD + A’B’CD’

=> m15 + m13 + m7 + m5 + m11 + m10 + m3 + m2

 

The total number of minterms is 8.

Answer 1

$F(A,B,C,D)=\overline{\overline {BD}.(\overline{\bar BC})}$

$F(A,B,C,D)=\overline {\overline{BD}}+\overline{\overline{\bar BC}}$

$F(A,B,C,D)=BD+\bar BC$

 Minterms are: $\sum m(2,3,5,7,10,11,13,15)$

So total $8$ minterms are present in the given function.