Consider the minterm list form of a Boolean function F given below.
- F(P, Q, R, S) = Σm(0, 2, 5, 7, 9, 11) + d(3, 8, 10, 12, 14)
Here, m denotes a minterm and d denotes a don’t care term. The number of essential prime implicants of the function F is _______ .
There are 3 prime implicant i.e., P’QS, Q’S’ and PQ’ and all are essential.
Because 0 and 2 are correct by only Q’S’, 5 and 7 are covered by only P’QS and 8 and 9 are covered by only PQ’.
Consider the Karnaugh map given below, where X represents “don’t care” and blank represents 0.
As all variables and their complements are available we can implement the function with only one NOR Gate.
Given f(w,x,y,z) = Σm(0,1,2,3,7,8,10) + Σd(5,6,11,15), where d represents the don’t-care condition in Karnaugh maps. Which of the following is a minimum product-of-sums (POS) form of f(w,x,y,z)?
K-Map for the function f is
Consider maxterms in K-map to represent function in product-of-sums (POS) form
f(w,x,y,z) = (w' + z')(x' + z)
Total 3 prime implicants are there.
Hence, option (A) matches.
Question 8 Explanation:
Let's check for option (C):
a'c' + ad' + abc' + c'd
Not equivalent to the K-map, we get in previous question.
Consider a Boolean function f (w, x, y, z). Suppose that exactly one of its inputs is allowed to change at a time. If the function happens to be true for two input vectors i1 = 〈w1, x1, y1, z1〉 and i2 = 〈w2, x2, y2, z2〉, we would like the function to remain true as the input changes from vectors i1 to i2 (i1 and i2 differ in exactly one bit position), without becoming false momentarily. Let f(w, x, y, z) = ∑(5, 7, 11, 12, 13, 15). Which of the following cube covers of f will ensure that the required property is satisfied?
Question 9 Explanation:
Static hazard is the situation where, when one input variable changes, the output changes momentarily before stabilizing to the correct value. The most commonly used method to eliminate static hazards is to add redundant logic (consensus terms in the logic expression).
f = X1 * X2 + X1' * X3
If (X1,X2,X3) = (1,1,1) then f=1 because X1 * X2 =1 X1' * X3 = 0.
Let the input is changed from 111 to 011 , then f = 1 because X1 * X2 = 0 X1' * X3 =1.
The output f will be momentarily 0 if AND gate X1 * X2 is faster than the AND gate X1' * X3.
This Hazard can be avoided by adding the term X2 * X3 (because X1 is in true form in first term and in complement form in the second term . So pick the fixed terms X2 and X3 from both terms) to f i.e f = X1 * X2 + X1' * X3 + X2 * X3
Option D is equivalent to f(w, x, y, z) = ∑(5,7,11,12,13,15)
Question 10 Explanation:
From given function 'f' we can draw,
There are two EPI,
A'C and AC'.
Question 11 Explanation:
⇒ w'y' + z'wx' + xyz'
Total 8 literals are there.
⇒ (z' + w')(z' + y')(w' + x')(x + z + w)
Total 9 literals are there.
Question 12 Explanation:
⇒ xz' + zx'
Question 13 Explanation:
Given k-map gives xy + xy + wz
⇒ wy + wz + xy
Question 15 Explanation:
Question 16 Explanation:
In sum of terms,any term is an implicant because it implies the function. So xz is an implicant and hence 'C' is the answer.
Question 17 Explanation:
Correct option is
Question 18 Explanation:
We can write this as
⇒ ABC + B'C' + A'C'
Question 19 Explanation:
⇒ y'z + xy
There are 19 questions to complete.