The circuit diagram shows a logic combination | vedclass

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(C) $1$. The circuit consists of an $AND$ gate, a $NOR$ gate, a $NOT$ gate, and a $NOR$ gate as the final output stage.$2$. Let the output of the $AND

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$0, 1, 0$

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(C) $1$. The circuit consists of an $AND$ gate, a $NOR$ gate, a $NOT$ gate, and a $NOR$ gate as the final output stage.$2$. Let the output of the $AND$ gate be $X$. $X = P \cdot Q$.$3$. Let the output of the $NOR$ gate (connected to $R$ and $S$) be $N$. $N = \overline{R+S}$.$4$. This output $N$ passes through a $NOT$ gate, so the input to the final $NOR$ gate is $Y = \overline{N} = R+S$.$5$. The final output $Z$ is the $NOR$ of $X$ and $Y$, so $Z = \overline{X+Y}$.$6$. Given new inputs: $P=0, Q=1, R=0, S=1$.$7$. Calculate $X$: $X = P \cdot Q = 0 \cdot 1 = 0$.$8$. Calculate $Y$: $Y = R+S = 0+1 = 1$.$9$. Calculate $Z$: $Z = \overline{X+Y} = \overline{0+1} = \overline{1} = 0$.$10$. Thus, the new states are $X=0, Y=1, Z=0$.

The circuit diagram shows a logic combination. Initially, for inputs $P, Q, R$ and $S$ all at state $1$, the outputs are $X=1, Y=1$ and $Z=0$. When inputs $P$ and $R$ change to state $0$ while inputs $Q$ and $S$ remain at state $1$, what are the new states of outputs $X, Y$ and $Z$?

Similar Questions

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