Step 1: Understanding the Problem:
The question asks to find the 5th sample of the convolution of two discrete-time step sequences, $x[n]$ and the delayed step sequence $h[n]$.
Step 2: Key Formula or Approach:
The discrete-time convolution sum is defined as:
\[ y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty} x[k] h[n - k] \]
Step 3: Detailed Explanation:
• Given: $x[n] = u[n]$ and $h[n] = u[n-2]$.
• We need to compute $y[n]$ at $n = 5$:
\[ y[5] = \sum_{k=-\infty}^{\infty} u[k] u[5 - 2 - k] \]
\[ y[5] = \sum_{k=-\infty}^{\infty} u[k] u[3 - k] \]
• The unit step function $u[k]$ is defined as:
\[ u[k] = \begin{cases} 1, & k \geq 0 \\ 0, & k < 0 \end{cases} \]
• The term $u[3-k]$ is non-zero only when:
\[ 3 - k \geq 0 \implies k \leq 3 \]
• Therefore, both $u[k]$ and $u[3-k]$ are equal to 1 simultaneously when:
\[ 0 \leq k \leq 3 \]
• The sum simplifies to:
\[ y[5] = \sum_{k=0}^{3} (1 \times 1) = 1 + 1 + 1 + 1 = 4 \]
Step 4: Final Answer
Hence, the value of $y[5]$ is 4, which is option (B).