Q2GATE 2024MCQ

Simplified form of the Boolean function: $F(P, Q, R, S)=\bar{P} \bar{Q}+\bar{P} Q S+P \bar{Q} \bar{R} \bar{S}+P \bar{Q} R \bar{S}$ is

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📖 Explanation

Minimize function $=\bar{Q} \bar{S}+\bar{P} S$

Q3GATE 2019MCQ

The output expression for the Karnaugh map shown below is

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📖 Explanation

$Output=Q\bar{R}+S$

Q4GATE 2018MCQ

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

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📖 Explanation

Given, $f=m_0+m_2+m_3+m_5+m_7$ and $m_1=$ don't care

Q5GATE 2017MCQ

The Boolean expression $AB + A\bar{C} + BC$ simplifies to

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📖 Explanation

$BC+A\bar{C}$

Q6GATE 2016MCQ

The Boolean expression $\overline{(a+\bar{b}+c+\bar{d})+(b+\bar{c})}$ simplifies to

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📖 Explanation

$F=\overline{(a+\bar{b}+c+d)+(b+\bar{c})}$ $\;\;=\overline{(a+\bar{b}+c+d)}\cdot \overline{(b+\bar{c})}$ $\;\;=\bar{a}\cdot b\cdot \bar{c}\cdot d \cdot \bar{b} \cdot c$ $F=0$

Q7GATE 2015MCQ

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

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📖 Explanation

The SOP form of F is ( shown in K-map) So,POS form can be formed using '0' from the K-map. $POS=(A+\bar{B}+C)(\bar{A}+B+C)(\bar{A}+\bar{B}+C)$

Q8GATE 2015MCQ

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

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Q9GATE 2014MCQ

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

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📖 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 $f=(\bar{B}+C)\cdot (\bar{A}+C)\cdot (\bar{A}+\bar{B})\cdot (\bar{C}+D)$

Q10GATE 2014MCQ

Which of the following is an invalid state in an 8-4-2-1 Binary Coded Decimal counter

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Q11GATE 2012MCQ

In the sum of products function f(X,Y,Z)= $\sum$ (2,3,4,5), the prime implicants are

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📖 Explanation

$f(X,Y,Z)=\Sigma (2,3,4,5)$ $\therefore \;\; f(X,Y,Z)=X\bar{Y}+\bar{X}Y$

Q12GATE 2010MCQ

The following Karnaugh map represents a function F. Which of the following circuits is a realization of the above function

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📖 Explanation

From the figure, it is clear that, two NAND gates generate the $\bar{X}$ and $\bar{Y}$ and now two AND gates with i/p $\bar{X}$ and $\bar{Y}$ 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.

Q13GATE 2010MCQ

The following Karnaugh map represents a function F. A minimized form of the function F is

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📖 Explanation

$F=\bar{X}\bar{Y}+YZ$

Q14GATE 2004MCQ

The simplified form of the Boolean expression $Y = (\bar{A}BC+D)(\bar{A}D+\bar{B}\bar{C})$ can be written as

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📖 Explanation

$Y=(\bar{A}BC+D)(\bar{A}D+\bar{B}\bar{C})$ $\;\;=\bar{A}BCD+\bar{A}D+\bar{B}\bar{C}D$ $\;\;=(\bar{A}D+\bar{B}\bar{C}D)\;\;(A+1=1)$

Q15GATE 2003MCQ

The boolean expression $\bar{X}Y\bar{Z}+\bar{X}\bar{Y}Z+XY\bar{Z}+X\bar{Y}Z+XYZ$ can be simplified to

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📖 Explanation

By K-map The simplified for is $XY+Y\bar{Z}+\bar{Y}Z$

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