The output from a NAND gate having inputs A a | vedclass

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(D) $NAND$ gate produces a low output $(0)$ only when both inputs are high $(1)$. Otherwise, it produces a high output $(1)$. The truth table for a $N

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(D) $NAND$ gate produces a low output $(0)$ only when both inputs are high $(1)$. Otherwise, it produces a high output $(1)$. The truth table for a $NAND$ gate is:| $A$ | $B$ | $Y = \overline{A \cdot B}$ ||---|---|---|| $0$ | $0$ | $1$ || $0$ | $1$ | $1$ || $1$ | $0$ | $1$ || $1$ | $1$ | $0$ |By analyzing the input waveforms for $A$ and $B$ at different time intervals:$1$. For the first interval, $A=1, B=1$, so $Y=0$.$2$. For the second interval, $A=0, B=0$, so $Y=1$.$3$. For the third interval, $A=0, B=1$, so $Y=1$.$4$. For the fourth interval, $A=1, B=0$, so $Y=1$.$5$. For the fifth interval, $A=1, B=1$, so $Y=0$.$6$. For the sixth interval, $A=0, B=0$, so $Y=1$.$7$. For the seventh interval, $A=0, B=1$, so $Y=1$.Comparing this sequence $(0, 1, 1, 1, 0, 1, 1)$ with the given options, the waveform in image $D$ matches this output.

The output from a $NAND$ gate having inputs $A$ and $B$ given below will be:

Similar Questions

The following figure shows the variation of inputs $A$ and $B$ with time. If these inputs are given to a $NAND$ gate,select the correct output signal $Y$ from the given options.

The given circuit works as:

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The truth table given below is for ($A$ and $B$ are the inputs,$Y$ is the output).
$A$ $B$ $Y$
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The following figure shows a logic gate circuit with two inputs $A$ and $B$ and the output $Y.$ The voltage waveforms of $A, B$ and the output $Y$ are as given in the diagram. The logic gate is $:-$

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