The impulse response. h[n] of a linear time invariant system is given ...
The given impulse response of a system
h[n] = u[n + 3] + u[n – 2] – 2u[n – 7]
The given system equation is non-causal for any value of n, since output depends upon the future input.
The given system is stable.
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The impulse response. h[n] of a linear time invariant system is given ...
Answer:
To determine the stability and causality of the given system, we need to analyze the impulse response h[n] = u[n-3] - u[n-7].
Stability:
A system is considered stable if its impulse response is absolutely summable, i.e., the sum of the magnitudes of all the samples is finite. In other words, if ∑|h[n]| < ∞,="" then="" the="" system="" is="" />
Let's calculate the sum of the magnitudes of all the samples in h[n]:
∑|h[n]| = ∑|u[n-3] - u[n-7]|
Considering the given impulse response, we can observe that only three samples have non-zero values: n = 3, n = 4, and n = 5. For all other values of n, h[n] = 0.
∑|h[n]| = |1| + |1| + |1| = 3 < />
Since the sum of the magnitudes of all the samples is finite, the system is stable.
Causality:
A system is considered causal if the output depends only on present and past inputs, i.e., for n < 0,="" the="" output="" is="" zero.="" in="" other="" words,="" if="" h[n]="0" for="" n="" />< 0,="" then="" the="" system="" is="" />
Let's analyze the given impulse response h[n]:
h[n] = u[n-3] - u[n-7]
For n < 3,="" both="" u[n-3]="" and="" u[n-7]="" are="" zero,="" so="" h[n]="0." therefore,="" the="" system="" is="" />
Conclusion:
Based on the analysis, we can conclude that the given system is stable (option 'A') and causal.