Boolean Algebra and Minimization


Question 1
The output expression for the Karnaugh map shown below is
A
Q\bar{R}+S
B
Q\bar{R}+\bar{S}
C
QR+S
D
QR+\bar{S}
GATE EE 2019   Digital Electronics
Question 1 Explanation: 


Output=Q\bar{R}+S
Question 2
Digital input signals A,B,C with A as the MSB and C as the LSB are used to realize the Boolean function
F=m_{0}+m_{2}+m_{3}+m_{5}+m_{7}, \; where \; m_{i}
denotes the i^{th} minterm. In addition, F has a don't care for m_1. The simplified expression for F is given by
A
\bar{A}\bar{C}+\bar{B}C+AC
B
\bar{A}+C
C
\bar{C}+A
D
\bar{A}C+BC+A\bar{C}
GATE EE 2018   Digital Electronics
Question 2 Explanation: 
Given, f=m_0+m_2+m_3+m_5+m_7 and m_1= don't care


Question 3
The output expression for the Karnaugh map shown below is
A
B\bar{D}+BCD
B
B\bar{D}+AB
C
B\bar{D}+ABC
D
B\bar{D}+ABC
GATE EE 2017-SET-1   Digital Electronics
Question 3 Explanation: 


ABC+B\bar{D}
Question 4
The Boolean expression AB + A\bar{C} + BC simplifies to
A
BC+A\bar{C}
B
AB+A\bar{C}+B
C
AB+A\bar{C}
D
AB+BC
GATE EE 2017-SET-1   Digital Electronics
Question 4 Explanation: 


BC+A\bar{C}
Question 5
The Boolean expression \overline{(a+\bar{b}+c+\bar{d})+(b+\bar{c})} simplifies to
A
1
B
\overline{a.b}
C
a.b
D
0
GATE EE 2016-SET-2   Digital Electronics
Question 5 Explanation: 
F=\overline{(a+\bar{b}+c+d)+(b+\bar{c})}
\;\;=\overline{(a+\bar{b}+c+d)}\cdot \overline{(b+\bar{c})}
\;\;=\bar{a}\cdot b\cdot \bar{c}\cdot d \cdot \bar{b} \cdot c
F=0


There are 5 questions to complete.