# 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.