# 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:
$F=\overline{(a+\bar{b}+c+d)+(b+\bar{c})}$
$\;\;=\overline{(a+\bar{b}+c+d)}\cdot \overline{(b+\bar{c})}$
$\;\;=\bar{a}\cdot b\cdot \bar{c}\cdot d \cdot \bar{b} \cdot c$
$F=0$
 Question 6
The output expression for the Karnaugh map shown below is
 A $A+\bar{B}$ B $A+\bar{C}$ C $\bar{A}+\bar{C}$ D $\bar{A}+C$
GATE EE 2016-SET-2   Digital Electronics
Question 6 Explanation:

$F=A+\bar{C}$
 Question 7
Consider the following Sum of Products expression, F.
$F=ABC+\bar{A}\bar{B}C+A\bar{B}C+\bar{A}BC+\bar{A}\bar{B}\bar{C}$
The equivalent Product of Sums expression is
 A $F=(A+\bar{B}+C)(\bar{A}+B+C)(\bar{A}+\bar{B}+C)$ B $F=(A+\bar{B}+\bar{C})(A+B+C)(\bar{A}+\bar{B}+\bar{C})$ C $F=(A+\bar{B}+\bar{C})(A+\bar{B}+\bar{C})(A+B+C)$ D $F=(A+\bar{B}+\bar{C})(A+B+\bar{C})(A+B+C)$
GATE EE 2015-SET-2   Digital Electronics
Question 7 Explanation:
The SOP form of F is ( shown in K-map)

So,POS form can be formed using '0' from the K-map.
$POS=(A+\bar{B}+C)(\bar{A}+B+C)(\bar{A}+\bar{B}+C)$
 Question 8
f(A,B,C,D)=$\Pi$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
 A $(A+\bar{C}+D)(\bar{A}+B+D)$ B $(A\bar{C}D)+(\bar{A}BD)$ C $\bar{A}C\bar{D}+A\bar{B}C\bar{D}+A\bar{B}\bar{C}\bar{D}$ D $(B+\bar{C}+D)(A+\bar{B}+\bar{C}+D)(\bar{A}+B+C+D)$
GATE EE 2015-SET-1   Digital Electronics
 Question 9
The SOP (sum of products) form of a Boolean function is $\Sigma$(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
 A $(\bar{B}+C)(\bar{A}+C)(\bar{A}+\bar{B})(\bar{C}+D)$ B $(\bar{B}+C)(\bar{A}+C)(\bar{A}+\bar{C})(\bar{C}+D)$ C $(\bar{B}+C)(\bar{A}+C)(\bar{A}+\bar{B})(\bar{C}+\bar{D})$ D $(\bar{B}+C)(A+\bar{B})(\bar{A}+\bar{B})(\bar{C}+D)$
GATE EE 2014-SET-2   Digital Electronics
Question 9 Explanation:
The 4 variable Boolean function is given in canonical sum of product form as,
$f(A,B,C,D)=\Sigma (0,1,3,7,11)$
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
$f(A,B,C,D)=\Pi (2,4,5,6,8,9,10,12,13,14,15)$
Now by plotting the above function on a 4 variable K-map (Maxterm map), we obtain the simplified expression of the function

$f=(\bar{B}+C)\cdot (\bar{A}+C)\cdot (\bar{A}+\bar{B})\cdot (\bar{C}+D)$
 Question 10
Which of the following is an invalid state in an 8-4-2-1 Binary Coded Decimal counter
 A 1 0 0 0 B 1 0 0 1 C 0 0 1 1 D 1 1 0 0
GATE EE 2014-SET-2   Digital Electronics
Question 10 Explanation:
Binary coded decimal counter counts from 0 to 9. So, 1100 is an initial state i.e. 12.
There are 10 questions to complete.