Previous Year Questions- Boolean Algebra and Minimization | Analog and Digital Electronics - Electrical Engineering (EE) PDF Download

Q1: Simplified form of the Boolean function:
is (2024)
(a)
(b)
(c)
(d)
Ans: (a)
Sol: Minimize function =

Q2: The output expression for the Karnaugh map shown below is (2019)
(a)
(b)
(c) $QR+S$QR + S
(d)
Ans: (a)
Sol:
Output =

Q3: Digital input signals A, B, C with A as the MSB and C as the LSB are used to realize the Boolean function
F = m₀ + m₂ + m₃ + m₅ + m₇, where m_i
denotes the i^th minterm. In addition, F has a don't care for m₁. The simplified expression for F is given by  (2018)
(a)
(b)
(c)
(d)
Ans: (b)
Sol: Given,  f = m₀ + m₂ + m₃ + m₅ + m₇ and m₁ = don't care

Q4: The output expression for the Karnaugh map shown below is (SET-1(2017))
(a)
(b)
(c)
(d)
Ans: (d)
Sol: 
Q5: The Boolean expression  simplifies to  (SET-1  (2017))
(a)
(b)
(c)
(d) AB + BC
Ans: (a)
Sol: 
Q6: The Boolean expression simplifies to (SET-2 (2016))
(a) 1
(b)
(c) a.b
(d) 0
Ans: (d)
Sol: F = 0

Q7: The output expression for the Karnaugh map shown below is (SET-2 (2016))
(a)
(b)
(c)
(d)
Ans: (b)
Sol: 
Q8: Consider the following Sum of Products expression, F. The equivalent Product of Sums expression is (SET-2 (2015))
(a)
(b)
(c)
(d)
Ans: (a)
Sol: The SOP form of F is ( shown in K-map)
So,POS form can be formed using '0' from the K-map.

Q9: f(A, B, C, D) = Π M(0, 1, 3, 4, 5, 7, 9, 11, 12, 13, 14, 15) is a maxterm representation of a Boolean function f(A, B, C, D) where A is the MSB and D is the LSB. The equivalent minimized representation of this function is (SET-1 (2015))
(a)
(b)
(c)
(d)
Ans: (a)

Q10: 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 (SET-2  (2014))
(a)
(b)
(c)
(d)
Ans: (a)
Sol: The 4 variable Boolean function is given in canonical sum of product form as,
 f(A, B, C, D) = Σ(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 sum form, as under
 f(A, B, C, D) = Π(2, 4, 5, 6, 8, 9, 10, 12, 13, 14, 15)
Now by plotting the above function on a 4 variable K-map (Maxterm map), we obtain the simplified expression of the function

Q11: Which of the following is an invalid state in an 8-4-2-1 Binary Coded Decimal counter (SET-2 (2014))
(a) 1 0 0 0
(b) 1 0 0 1
(c) 0 0 1 1
(d) 1 1 0 0
Ans: (d)
Sol: Binary coded decimal counter counts from 0 to 9. So, 1100 is an initial state i.e. 12.

Q12: In the sum of products function f(X, Y, Z) = ∑(2, 3, 4, 5), the prime implicants are  (2012)
(a)
(b)
(c)
(d)
Ans: (a)
Sol: f(X, Y, Z) = Σ(2, 3, 4, 5)

Q13: The following Karnaugh map represents a function F.
Which of the following circuits is a realization of the above function (2010)
(a) (b) (c) (d) Ans: (d)
Sol: From the figure, it is clear that, two NAND gates generate the  and now two AND gates with i/p  and inputs Y and Z is used to generate two terms of SOP form and OR gate is used to sum them and generate th F.

Q14: The following Karnaugh map represents a function F.
A minimized form of the function F is  (2010)
(a)
(b)
(c)
(d)
Ans: (b)
Sol:

Q15: The simplified form of the Boolean expression  can be written as (2004)
(a)
(b)
(c)
(d)
Ans: (a)
Sol: 
Q16: The boolean expression can be simplified to (2003)
(a)
(b)
(c)
(d)
Ans: (b)
Sol: By K-map
The simplified for is.