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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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