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Recent questions tagged combinational-circuits

2 votes
1 answer
2
GATE ECE 2010 | Question: 11
Match the logic gates in Column A with their equivalents in Column B. $\text{P-2, Q-4, R-1, S-3}$ $\text{P-4, Q-2, R-1, S-3}$ $\text{P-2, Q-4, R-3, S-1}$ $\text{P-4, Q-2, R-3, S-1}$
Match the logic gates in Column A with their equivalents in Column B.$\text{P-2, Q-4, R-1, S-3}$$\text{P-4, Q-2, R-1, S-3}$$\text{P-2, Q-4, R-3, S-1}$$\text{P-4, Q-2, R-3...
2 votes
1 answer
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GATE ECE 2010 | Question: 12
For the output $\text{F}$ to be $1$ in the logic circuit shown, the input combination should be $\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0$ $\text{A = 1, B = 0, C = 0}$ $\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=0$ $\text{A = 0, B = 0, C = 1}$
For the output $\text{F}$ to be $1$ in the logic circuit shown, the input combination should be$\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0$$\text{A = 1, B = 0, C = 0}$$\ma...
2 votes
1 answer
4
GATE ECE 2010 | Question: 39
The Boolean function realized by the logic circuit shown is $\text{F} = \sum_{\text{m}} (0, 1, 3, 5, 9, 10, 14)$ $\text{F}=\sum_{\text{m}}(2,3,5,7,8,12,13)$ $\text{F}=\sum_{\text{m}}(1,2,4,5,11, 14,15)$ $\text{F}= \sum_{\text{m}}(2,3,5,7,8,9,12)$
The Boolean function realized by the logic circuit shown is$\text{F} = \sum_{\text{m}} (0, 1, 3, 5, 9, 10, 14)$$\text{F}=\sum_{\text{m}}(2,3,5,7,8,12,13)$$\text{F}=\sum_{...
0 votes
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GATE ECE 2021 | Question: 12
​​​​​​Addressing of a $32K\:\times\:16$ memory is realized using a single decoder. The minimum number of $\text{AND}$ gates required for the decoder is $2^{8}$ $2^{32}$ $2^{15}$ $2^{19}$
​​​​​​Addressing of a $32K\:\times\:16$ memory is realized using a single decoder. The minimum number of $\text{AND}$ gates required for the decoder is$2^{8}$...
0 votes
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10
GATE ECE 2016 Set 3 | Question: 17
The logic functionality realized by the circuit shown below is $OR$ $XOR$ $NAND$ $AND$
The logic functionality realized by the circuit shown below is $OR$$XOR$$NAND$$AND$
0 votes
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11
GATE ECE 2016 Set 3 | Question: 18
The minimum number of $2$-input $NAND$ gates required to implement a $2$-input $XOR$ gate is $4$ $5$ $6$ $7$
The minimum number of $2$-input $NAND$ gates required to implement a $2$-input $XOR$ gate is$4$$5$$6$$7$
0 votes
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14
GATE ECE 2016 Set 1 | Question: 43
The functionality implemented by the circuit below is $2$-to-$1$ multiplexer $4$-to-$1$ multiplexer $7$-to-$1$ multiplexer $6$-to-$1$ multiplexer
The functionality implemented by the circuit below is$2$-to-$1$ multiplexer $4$-to-$1$ multiplexer$7$-to-$1$ multiplexer$6$-to-$1$ multiplexer
1 votes
0 answers
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GATE ECE 2015 Set 2 | Question: 14
In the figure shown, the output ܻ$Y = AB + \overline{C}\:\:\overline{D}$ is required to be ܻ The gates $G1$ and $G2$ must be, respectively, NOR, OR OR, NAND NAND, OR AND,NAND
In the figure shown, the output ܻ$Y = AB + \overline{C}\:\:\overline{D}$ is required to be ܻ The gates $G1$ and $G2$ must be, respectively,NOR, OROR, NANDNAND, ORAND,NA...
1 votes
0 answers
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GATE ECE 2014 Set 4 | Question: 15
In the circuit shown in the figure, if $C=0$, the expression for $Y$ is $Y=A \overline{B} + \overline{A}B$ $Y=A+B$ $Y=\overline{A} + \overline{B}$ $Y=A \: B$
In the circuit shown in the figure, if $C=0$, the expression for $Y$ is$Y=A \overline{B} + \overline{A}B$$Y=A+B$$Y=\overline{A} + \overline{B}$$Y=A \: B$
1 votes
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23
GATE ECE 2014 Set 3 | Question: 42
If $X$ and $Y$ are inputs and the Difference $(D=X-Y)$ and the Borrow $(B)$ are the outputs, which one of the following diagrams implements a half-subtractor?
If $X$ and $Y$ are inputs and the Difference $(D=X-Y)$ and the Borrow $(B)$ are the outputs, which one of the following diagrams implements a half-subtractor?
0 votes
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29
GATE ECE 2018 | Question: 8
The logic function $f(X, Y)$ realized by the given circuit is NOR AND NAND XOR
The logic function $f(X, Y)$ realized by the given circuit is NORANDNANDXOR
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