To find the output \( y[n] \) of an LTI system with input \( x[n] = u[n] - u[n-3] \) and impulse response \( h[n] \), you would use the **convolution sum** formula:
\[
y[n] = (x * h)[n] = \sum_{k=-\infty}^{\infty} x[k] \cdot h[n-k]
\]
Here’s a step-by-step outline of how to proceed:
### 1. **Understand the Input Signal:**
The input signal \( x[n] = u[n] - u[n-3] \) represents a finite-duration pulse. It is a discrete signal that equals 1 for \( n = 0, 1, 2 \), and 0 otherwise. Mathematically:
\[
x[n] = \begin{cases}
1 & \text{for } n = 0, 1, 2 \\
0 & \text{otherwise}
\end{cases}
\]
### 2. **Impulse Response Representation:**
You need to know the specific form of the impulse response \( h[n] \) from the figure provided. The impulse response will determine how the system reacts to each input value. Let’s assume \( h[n] \) is known (you should substitute its values accordingly based on the figure provided).
### 3. **Convolution Calculation:**
With \( x[n] \) and \( h[n] \) defined, perform the convolution sum:
\[
y[n] = \sum_{k=-\infty}^{\infty} x[k] \cdot h[n-k]
\]
Since \( x[k] \) is nonzero only for \( k = 0, 1, 2 \), the sum reduces to:
\[
y[n] = x[0] \cdot h[n] + x[1] \cdot h[n-1] + x[2] \cdot h[n-2]
\]
This simplifies further since \( x[k] = 1 \) for \( k = 0, 1, 2 \), so:
\[
y[n] = h[n] + h[n-1] + h[n-2]
\]
### 4. **Final Output:**
The final output \( y[n] \) is the sum of the impulse response shifted by 0, 1, and 2 time steps. You can calculate this for each value of \( n \) based on the known form of \( h[n] \).
If you provide the figure for \( h[n] \), I can help with the exact output.