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

• A. \( \delta[n-2] + \delta[n+1] \)
• B. \( \delta[n-1]\delta[n] + \delta[n+1]\delta[n-1] \)
• C. \( \delta[n-2] + \delta[n-1] + \delta[n] + \delta[n+1] \)
• D. \( \delta[n]\delta[n-1] + \delta[n-2]\delta[n+1] \)

Hint:

For cascaded LTI systems, always convolve impulse responses. Each shifted delta simply shifts the output.

Answer 1

The Correct Option is C

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 \[ \delta[n-2] + \delta[n-1] + \delta[n] + \delta[n+1]. \]