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Boolean algebra. Combinational and sequential circuits. Minimization. Number representations and computer arithmetic (fixed and floating point)

$$\scriptsize{\overset{{\large{\textbf{Mark Distribution in Previous GATE}}}}{\begin{array}{|c|c|c|c|c|c|c|c|}\hline
\textbf{Year}& \textbf{2022} & \textbf{2021-1}&\textbf{2021-2}&\textbf{2020}&\textbf{2019}&\textbf{2018}&\textbf{2017-1}&\textbf{2017-2}&\textbf{2016-1}&\textbf{2016-2}&\textbf{Minimum}&\textbf{Average}&\textbf{Maximum}
\\\hline\textbf{1 Mark Count} & 1&2&3&2&4&2&3&2&3&3&1&2.5&4
\\\hline\textbf{2 Marks Count} & 2&2&2&1&2&2&0&4&2&0&0&1.7&4
\\\hline\textbf{Total Marks} & 5&6&7&4&8&6&3&10&7&3&\bf{3}&\bf{5.9}&\bf{10}\\\hline
\end{array}}}$$

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Most answered questions in Digital Logic

44 votes
7 answers
31
GATE CSE 2015 Set 3 | Question: 43
The total number of prime implicants of the function $f(w, x, y, z) = \sum (0, 2, 4, 5, 6, 10)$ is __________
The total number of prime implicants of the function $f(w, x, y, z) = \sum (0, 2, 4, 5, 6, 10)$ is __________
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52 votes
7 answers
32
GATE CSE 2015 Set 1 | Question: 20
Consider a $4$-bit Johnson counter with an initial value of $0000.$ The counting sequence of this counter is $0, 1, 3, 7, 15, 14, 12, 8, 0$ $0, 1, 3, 5, 7, 9, 11, 13, 15, 0$ $0, 2, 4, 6, 8, 10, 12, 14, 0$ $0, 8, 12, 14, 15, 7, 3, 1, 0$
Consider a $4$-bit Johnson counter with an initial value of $0000.$ The counting sequence of this counter is $0, 1, 3, 7, 15, 14, 12, 8, 0$$0, 1, 3, 5, 7, 9, 11, 13, 15, ...
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24 votes
7 answers
33
GATE CSE 2011 | Question: 14
The simplified SOP (Sum of Product) from the Boolean expression $(P + \bar{Q} + \bar{R}) . (P + \bar{Q} + R) . (P + Q +\bar{R})$ is $(\bar{P}.Q+\bar{R})$ $(P+\bar{Q}.\bar{R})$ $(\bar{P}.Q+R)$ $(P.Q+R)$
The simplified SOP (Sum of Product) from the Boolean expression$$(P + \bar{Q} + \bar{R}) . (P + \bar{Q} + R) . (P + Q +\bar{R})$$ is $(\bar{P}.Q+\bar{R})$$(P+\bar{Q}.\bar...
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27 votes
7 answers
36
GATE CSE 2014 Set 1 | Question: 7
Consider the following Boolean expression for F: $F(P,Q,R,S)= PQ + \bar{P}QR + \bar{P}Q\bar{R}S$ The minimal sum$-$of$-$products form of $F$ is $PQ+QR+QS$ $P+Q+R+S$ $\bar{P} + \bar{Q}+ \bar{R}+ \bar{S}$ $\bar{P}R + \bar{R} \bar{P}S+P$
Consider the following Boolean expression for F: $F(P,Q,R,S)= PQ + \bar{P}QR + \bar{P}Q\bar{R}S$The minimal sum$-$of$-$products form of $F$ is$PQ+QR+QS$$P+Q+R+S$$\bar{P} ...
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31 votes
7 answers
37
GATE CSE 2013 | Question: 21
Which one of the following expressions does NOT represent exclusive NOR of $x$ and $y$? $xy + x′ y′$ $x\oplus y′$ $x′\oplus y$ $x′\oplus y′$
Which one of the following expressions does NOT represent exclusive NOR of $x$ and $y$?$xy + x′ y′$$x\oplus y′$$x′\oplus y$$x′\oplus y′$
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17 votes
6 answers
40
GATE CSE 2021 Set 2 | Question: 5
Which one of the following circuits implements the Boolean function given below? $f(x,y,z) = m_0+m_1+m_3+m_4+m_5+m_6$, where $m_i$ is the $i^{\text{th}}$ minterm.
Which one of the following circuits implements the Boolean function given below?$f(x,y,z) = m_0+m_1+m_3+m_4+m_5+m_6$, where $m_i$ is the $i^{\text{th}}$ minterm.
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0 votes
6 answers
43
UGC NET CSE | January 2017 | Part 2 | Question: 10
The hexadecimal equivalent of the octal number $2357$ is : $2EE$ $2FF$ $4EF$ $4FE$
The hexadecimal equivalent of the octal number $2357$ is :$2EE$$2FF$$4EF$$4FE$
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4 votes
6 answers
44
GATE ECE 2014
Above question modification. This is the actual question Boolean expression$:(x+y)(x+\bar{y})+\overline{{(x\bar{y}+\bar{x})}}$ $(A) x$ $(B)y$ $(C)xy$ $(D)x+y$
Above question modification.This is the actual questionBoolean expression$:(x+y)(x+\bar{y})+\overline{{(x\bar{y}+\bar{x})}}$$(A) x$ $(B)y$ $(C)xy$...
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23 votes
6 answers
46
GATE CSE 1990 | Question: 5-b
Show with the help of a block diagram how the Boolean function : $f=AB+BC+CA$ can be realised using only a $4:1$ multiplexer.
Show with the help of a block diagram how the Boolean function :$f=AB+BC+CA$can be realised using only a $4:1$ multiplexer.
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15 votes
6 answers
48
Simplified Boolean expression for A'BC+AB'C'+A'B'C'+AB'C+ABC
Simplified Boolean expression for A'BC+AB'C'+A'B'C'+AB'C+ABC A . AB B . B'C C . AB+(A'+AB')C D . AB'+BC+B'C'
Simplified Boolean expression for A'BC+AB'C'+A'B'C'+AB'C+ABCA . ABB . B'CC . AB+(A'+AB')CD . AB'+BC+B'C'
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10 votes
6 answers
49
ISRO2015-8
Minimum number of $2 \times 1$ multiplexers required to realize the following function, $f = \overline{A} \;\overline{B} C + \overline{A}\; \overline{B} \;\overline{C}$ Assume that inputs are available only in true form and Boolean a constant $1$ and $0$ are available. $1$ $2$ $3$ $7$
Minimum number of $2 \times 1$ multiplexers required to realize the following function, $f = \overline{A} \;\overline{B} C + \overline{A}\; \overline{B} \;\overline{C}$As...
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7 votes
6 answers
50
ISRO-2013-17
Two eight bit bytes $1100 0011$ and $0100 1100$ are added. What are the values of the overflow, carry and zero flags respectively, if the arithmetic unit of the CPU uses $2$'s complement form? $0, 1, 1'$ $1, 1, 0$ $1, 0, 1$ $0, 1, 0$
Two eight bit bytes $1100 0011$ and $0100 1100$ are added. What are the values of the overflow, carry and zero flags respectively, if the arithmetic unit of the CPU use...
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