Step 1: Understanding the Concept:
We must derive the Boolean logic expression for the given circuit diagram and then apply the timing diagram inputs to find the output waveform. Step 2: Key Formula or Approach:
1. Trace the signal paths through the logic gates (NOT, AND, OR).
2. Determine the Boolean equation for $Y$ in terms of $A$ and $B$.
3. Sample the states of $A$ and $B$ at each time interval to evaluate $Y$. Step 3: Detailed Explanation:
Let's analyze the circuit diagram:
Input $A$ goes directly into the first input of an AND gate.
Input $B$ passes through a NOT gate. The output is $\overline{B}$.
This $\overline{B}$ goes into the second input of the AND gate.
The output of the AND gate is therefore $A \cdot \overline{B}$.
This AND output is fed into one input of an OR gate.
The original input $B$ is also routed directly into the second input of the OR gate.
The output $Y$ of the OR gate is:
$Y = (A \cdot \overline{B}) + B$.
Using Boolean algebra distributive law: $X + \overline{X}Z = X + Z$.
$Y = B + A\overline{B} = B + A$.
So the circuit effectively functions as an OR gate ($Y = A + B$).
Now, let's analyze the input timing diagrams:
For $t \in [0, 1]$: $A = 0$, $B = 0 \implies Y = 0 + 0 = 0$.
For $t \in [1, 2]$: $A = 1$, $B = 0 \implies Y = 1 + 0 = 1$.
For $t \in [2, 3]$: $A = 0$, $B = 1 \implies Y = 0 + 1 = 1$.
The output $Y$ must be $0$ for $[0,1]$, $1$ for $[1,2]$, and $1$ for $[2,3]$.
Looking at the provided graphical options, Graph 4 perfectly matches this sequence: it is low (0) up to $t=1$, and high (1) from $t=1$ to $t=3$. Step 4: Final Answer:
The correct output is represented by Graph 4.
