The impulse response h[n] of an LTI system is

h[n] = u[n + 3] + u[n - 2] - 2u[n - 7].

Then the system is:

1. Stable

2. Casual

3. Unstable

4. Not casual.

Which of these are correct?

This question was previously asked in
ESE Electronics 2010 Paper 2: Official Paper
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  1. 1 and 2 only
  2. 2 and 3 only
  3. 3 and 4 only
  4. 1 and 4 only

Answer (Detailed Solution Below)

Option 4 : 1 and 4 only
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Detailed Solution

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Concept:

  • LTI system is causal when h[n] = 0 for n < 0
  • LTI system is stable when \(\mathop \sum \nolimits_{n = - \infty }^\infty \left| {h\left[ n \right]} \right| < \infty \) i.e. It should be absolutely summable.


Calculation:

h[n] = u(n + 3) + u(n – 2) – 2 u(n – 7)

F1 Neha Shraddha 31 08.2021 D1 Correction D

∵ h[n] ≠ 0     n < 0 ….. system is Non causal

\(\mathop \sum \nolimits_{n = - \infty }^\infty h\left( n \right) < \infty \) i.e. given impulse response is summable hence it is stable.

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