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The output of an LTI system is the convolution of the input with the system’s impulse response. When \( h[n] \) is nonzero for \( n = 0,1,2 \), the output becomes a weighted sum of the current and two past input values.
Updated On: Apr 16, 2025
- \( x[n] + x[n-1] + x[n-2] \)
- \( x[n-1] + x[n] + x[n+1] \)
- \( x[n] + x[n+1] + x[n+2] \)
- \( x[n+1] + x[n+2] + x[n+3] \)
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The Correct Option is A
Solution and Explanation
The system's impulse response \( h[n] = \mathcal{S}\{\delta[n]\} \) is:
\[
h[n] =
\begin{cases}
1, & n = 0, 1, 2 \\
0, & \text{otherwise}
\end{cases}
\]
This means the system performs convolution:
\[
y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty} x[k] h[n - k]
\]
Since \( h[n] = \delta[n] + \delta[n-1] + \delta[n-2] \), we get:
\[
y[n] = x[n] + x[n-1] + x[n-2]
\]
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