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Let x[n] = e-2n u[n] be the input to a system. Which of the following impulse responses gives the bounded output for this input?
  • a)
     h[n] = n
  • b)
     h[n] = 1/n
  • c)
     h[n] = 1/n2
  • d)
     Both 2 & 3
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Let x[n] = e-2nu[n] be the input to a system. Which of the following i...
A system is said to be BIBO stable if it’s impulse response is absolutely summable,

Out of all the responses only  is absolutely summable.

Is not absolutely summable it is divergent.
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Most Upvoted Answer
Let x[n] = e-2nu[n] be the input to a system. Which of the following i...
D) Both 2 and 3.

To determine if the output of a system is bounded, we need to check if the convolution sum is bounded. The convolution sum is given by:

y[n] = ∑(x[k]h[n-k])

For the given input x[n] = e^(-2n)u[n], we can substitute it into the convolution sum:

y[n] = ∑(e^(-2k)u[k]h[n-k])

Let's consider each option:

a) h[n] = n
Substituting this into the convolution sum:

y[n] = ∑(e^(-2k)u[k](n - k))

The sum is not bounded because the term (n - k) is unbounded.

b) h[n] = 1/n
Substituting this into the convolution sum:

y[n] = ∑(e^(-2k)u[k](1/n)(1/(n-k)))

The sum is not bounded because the term 1/(n-k) is unbounded.

c) h[n] = 1/n^2
Substituting this into the convolution sum:

y[n] = ∑(e^(-2k)u[k](1/n^2)(1/(n-k)))

The sum is bounded because both (1/n^2) and (1/(n-k)) are bounded.

Therefore, the impulse response h[n] = 1/n^2 gives the bounded output for the given input x[n] = e^(-2n)u[n].
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