Question 1 |
The output expression for the Karnaugh map shown below is
Q\bar{R}+S | |
Q\bar{R}+\bar{S} | |
QR+S | |
QR+\bar{S} |
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
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
\bar{A}\bar{C}+\bar{B}C+AC | |
\bar{A}+C | |
\bar{C}+A | |
\bar{A}C+BC+A\bar{C} |
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
B\bar{D}+BCD | |
B\bar{D}+AB | |
B\bar{D}+ABC | |
B\bar{D}+ABC |
Question 3 Explanation:
ABC+B\bar{D}
Question 4 |
The Boolean expression AB + A\bar{C} + BC simplifies to
BC+A\bar{C} | |
AB+A\bar{C}+B | |
AB+A\bar{C} | |
AB+BC |
Question 4 Explanation:
BC+A\bar{C}
Question 5 |
The Boolean expression \overline{(a+\bar{b}+c+\bar{d})+(b+\bar{c})} simplifies to
1 | |
\overline{a.b} | |
a.b | |
0 |
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
\;\;=\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+\bar{B} | |
A+\bar{C} | |
\bar{A}+\bar{C} | |
\bar{A}+C |
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
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
F=(A+\bar{B}+C)(\bar{A}+B+C)(\bar{A}+\bar{B}+C) | |
F=(A+\bar{B}+\bar{C})(A+B+C)(\bar{A}+\bar{B}+\bar{C}) | |
F=(A+\bar{B}+\bar{C})(A+\bar{B}+\bar{C})(A+B+C) | |
F=(A+\bar{B}+\bar{C})(A+B+\bar{C})(A+B+C) |
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)
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)=\PiM(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+\bar{C}+D)(\bar{A}+B+D) | |
(A\bar{C}D)+(\bar{A}BD) | |
\bar{A}C\bar{D}+A\bar{B}C\bar{D}+A\bar{B}\bar{C}\bar{D} | |
(B+\bar{C}+D)(A+\bar{B}+\bar{C}+D)(\bar{A}+B+C+D) |
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
(\bar{B}+C)(\bar{A}+C)(\bar{A}+\bar{B})(\bar{C}+D) | |
(\bar{B}+C)(\bar{A}+C)(\bar{A}+\bar{C})(\bar{C}+D) | |
(\bar{B}+C)(\bar{A}+C)(\bar{A}+\bar{B})(\bar{C}+\bar{D}) | |
(\bar{B}+C)(A+\bar{B})(\bar{A}+\bar{B})(\bar{C}+D) |
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)
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
1 0 0 0 | |
1 0 0 1 | |
0 0 1 1 | |
1 1 0 0 |
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.
