Animated Solution for Physics - Semiconductors: Identify the correct output signal $Y$ in the given combination of gates (as shown) for the given inputs $A$ and $B$.

Welcome to the fascinating world of digital electronics! In this problem, we are presented with a logic circuit consisting of three fundamental gates: an AND gate, a NOT gate, and an OR gate. Our mission is to determine the final output waveform, $Y$, given the specific timing diagrams for the two inputs, $A$ and $B$.

To tackle this, we need a systematic approach. Instead of trying to guess the output by looking at the squiggly lines, we will first translate the physical circuit into the language of mathematics: Boolean Algebra.

Let's trace the path of our signals.

First, look at the top gate. It's an AND gate taking both $A$ and $B$ as inputs. The output of an AND gate is true only if both inputs are true. Mathematically, we write this as a logical multiplication:

Next, look at the bottom path. The input $B$ is fed into a NOT gate (indicated by the triangle shape with a small bubble). A NOT gate simply inverts the signal. If it's a $1$, it becomes a $0$, and vice versa. We denote this with a bar over the variable:

Finally, these two intermediate signals, $Y_1$ and $Y_2$, are fed into an OR gate. An OR gate outputs true if at least one of its inputs is true. This is represented by logical addition. Therefore, our final output $Y$ is:

This single equation is the master key to our problem. It dictates exactly how the circuit will behave at any given moment.

Now, we turn our attention to the timing diagrams. The trick here is to break the continuous timeline into discrete intervals where the inputs $A$ and $B$ remain constant. Looking at the graphs, the natural intervals are $0-5$, $5-10$, $10-15$, and $15-20$.

Let's evaluate our master equation for each interval:

Interval $t\in (0,5)$: Looking at the graphs, input $A$ is high ($1$) and input $B$ is low ($0$). Let's plug these into our equation: So, the output $Y$ starts high.

Interval $t\in (5,10)$: In this segment, input $A$ drops to low ($0$), while input $B$ jumps to high ($1$). Substituting these values: The output $Y$ drops to low.

Interval $t\in (10,15)$: Here, both inputs $A$ and $B$ are high ($1$). Let's see what happens: The output $Y$ goes back to high.

Interval $t\in (15,20)$: Finally, both inputs $A$ and $B$ are low ($0$). The output $Y$ remains high.

We have successfully mapped out the entire behavior of the circuit! The output $Y$ is high ($1$) from $0$ to $5$, low ($0$) from $5$ to $10$, and high ($1$) continuously from $10$ to $20$.

When we compare our derived sequence ($1,0,1,1$) with the given options, we can clearly see that Option (c) perfectly matches our result.

Bonus Insight: If you are familiar with Boolean algebra theorems, you might notice that $A\cdot B+\overline{B}$ can be simplified using the distributive law to $A+\overline{B}$. If you evaluate this simpler expression for the same intervals, you will get the exact same waveform! This shows the beauty and consistency of digital logic.