Question:
Two discrete-time linear time-invariant systems with impulse responses
\(h_1[n] = \delta[n-1] + \delta[n+1]\) and \(h_2[n] = \delta[n] + \delta[n-1]\)
are connected in cascade. The impulse response of the cascaded system is
Two discrete-time linear time-invariant systems with impulse responses
\(h_1[n] = \delta[n-1] + \delta[n+1]\) and \(h_2[n] = \delta[n] + \delta[n-1]\)
are connected in cascade. The impulse response of the cascaded system is
Show Hint
For cascaded LTI systems, always convolve impulse responses. Each shifted delta simply shifts the output.
Updated On: Dec 29, 2025
- \( \delta[n-2] + \delta[n+1] \)
- \( \delta[n-1]\delta[n] + \delta[n+1]\delta[n-1] \)
- \( \delta[n-2] + \delta[n-1] + \delta[n] + \delta[n+1] \)
- \( \delta[n]\delta[n-1] + \delta[n-2]\delta[n+1] \)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Recall that cascaded LTI systems use convolution.
\[
h[n] = h_1[n] h_2[n].
\]
Step 2: Expand the convolution.
\[
h_1[n] = \delta[n-1] + \delta[n+1],
h_2[n] = \delta[n] + \delta[n-1].
\]
Convolving term-by-term:
\[
\delta[n-1] \delta[n] = \delta[n-1],
\delta[n-1] \delta[n-1] = \delta[n-2],
\delta[n+1] \delta[n] = \delta[n+1],
\delta[n+1] \delta[n-1] = \delta[n].
\]
Step 3: Add all results.
\[
h[n] = \delta[n-2] + \delta[n-1] + \delta[n] + \delta[n+1].
\]
Step 4: Conclusion.
The cascaded impulse response is
\[
\delta[n-2] + \delta[n-1] + \delta[n] + \delta[n+1].
\]
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