Animated Solution for Physics - Semiconductor Electronics: To get output '1' at $R$, for the given logic gate circuit, the input values must be

Welcome! Let's decode this logic circuit to find out exactly what inputs $X$ and $Y$ will give us an output of $1$ at $R$. When faced with a complex web of logic gates, it is easy to feel overwhelmed. But don't worry, the trick is to divide and conquer.

Instead of trying to guess the inputs or drawing a massive truth table right away, we will break the circuit down into smaller, manageable pieces. We can do this by labeling the outputs of the intermediate gates.

Let's define two intermediate signals to make our lives easier. We will call the output of the top OR gate $P$, and the output of the bottom AND gate $Q$. By doing this, we can analyze the top and bottom branches independently before bringing them together at the final stage.

Let's analyze the top branch first. The input $X$ passes through a NOT gate before reaching the OR gate. This means the upper input to the OR gate is inverted, becoming $\bar{X}$. The lower input to this OR gate is directly connected to the input $Y$. Since an OR gate performs logical addition, the Boolean expression for our intermediate signal $P$ is simply:

Now, let's look at the bottom branch. The upper input to the AND gate is directly connected to the input $X$, and the lower input is directly connected to the input $Y$. An AND gate performs logical multiplication, so the expression for our second intermediate signal $Q$ is:

Finally, let's analyze the last stage. The intermediate signals $P$ and $Q$ feed into a NOR gate. A NOR gate is essentially an OR gate followed by a NOT gate. It performs an OR operation and then complements the result. Therefore, the final output $R$ is the complement of $P+Q$:

Let's substitute the expressions we found for $P$ and $Q$ into this equation to get the full picture:

Now comes the fun part: simplification! Look closely at the terms $Y+X\cdot Y$. According to the rules of Boolean algebra, this expression can be simplified dramatically. Think of it like factoring out $Y$: $Y(1+X)$. Since $1+X$ is always $1$ in Boolean logic, the whole term simplifies to just $Y$. This is known as the Absorption Law.

Applying this law, our expression for $R$ shrinks down to:

Next, we apply De Morgan's Law to break the long bar over the expression. Remember the rule: "break the line, change the sign". The OR operation ($+$) changes to an AND operation ($\cdot$), and the individual terms get complemented.

Since the double complement of $X$ is just $X$, we arrive at our beautifully simplified final expression:

We are given the condition that the final output $R$ must be $1$. Our simplified expression is an AND operation: $X\cdot \bar{Y}=1$.

For an AND gate to output a $1$, all of its inputs must be $1$. Therefore, $X$ must be $1$, and $\bar{Y}$ must be $1$. If $\bar{Y}=1$, then $Y$ must be $0$.

And there we have it! By systematically breaking down the circuit and using the elegant rules of Boolean algebra, we found the exact inputs: $X=1,Y=0$.