Discrete LTI filter impulse response

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Discussion Overview

The discussion revolves around calculating the time taken for the output of a discrete-time LTI system to fall below 1% of its initial value after a unit impulse is applied. Participants explore methods for both algebraic determination and simulation approaches, focusing on the properties of the impulse response function h[n].

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks how to calculate the time for the output to fall below 1% of its initial value using the impulse response function h[n].
  • Another participant suggests running a simulation, mentioning prior experience with Excel for this purpose.
  • A participant expresses a desire to know the algebraic method for determining the time instead of relying on simulation.
  • Concerns are raised about the decay of h[n], with one participant stating that if h[n] does not decay, it will never fall below 1% of its initial value.
  • For impulse responses that do decay, a participant proposes solving the inequality h[n] < 0.01 · h[0] to find n.
  • A correction is made regarding the form of h[n], which is adjusted to h[n]=(1-\alpha^2)\alpha^{n-1}u[n-1]-\alpha \delta[n-1], indicating that it will converge for α<1.
  • A participant presents an inequality for n based on the corrected h[n], seeking validation for the expression derived.

Areas of Agreement / Disagreement

Participants express differing views on the decay properties of h[n], with some asserting that certain forms do not decay while others provide a corrected form that does. The discussion remains unresolved regarding the validity of the derived inequality and the method for calculating the time.

Contextual Notes

The discussion includes assumptions about the decay of the impulse response and the conditions under which it converges, which are not fully explored or resolved.

optrix
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If I have the unit impulse response function for a discrete-time LTI system (Unit sequence response?), h[n], how can I calculate the time taken for the output to fall below 1% of its initial value, after a unit impulse is applied to the input?

In particular, I have:

h[n]=(\alpha ^{-1}-\alpha )u[n-1]-\alpha \delta [n-1]
 
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optrix said:
If I have the unit impulse response function for a discrete-time LTI system (Unit sequence response?), h[n], how can I calculate the time taken for the output to fall below 1% of its initial value, after a unit impulse is applied to the input?

In particular, I have:

h[n]=(\alpha ^{-1}-\alpha )u[n-1]-\alpha \delta [n-1]

Can you just run a simulation? I've used Excel for that before.
 
berkeman said:
Can you just run a simulation? I've used Excel for that before.

I could do, but I would like to know the method for determining it algebraically.
 
That h[n] doesn't seem to decay, so it will never fall bellow 1% of its initial value.

But, for others h[n] which do decay, I will solve for n this simple inequality:

h[n] < 0.01 · h[0]
 
falbani said:
That h[n] doesn't seem to decay, so it will never fall bellow 1% of its initial value.

But, for others h[n] which do decay, I will solve for n this simple inequality:

h[n] < 0.01 · h[0]

Thank you, it was meant to be:

h[n]=(1-\alpha ^2)\alpha ^{n-1}u[n-1]-\alpha \delta [n-1]

In which case it will converge for \alpha&lt;1

I managed to get the inequality:

n &lt; \frac{ln\left( \frac{0.01\alpha(1-\alpha-\alpha^2)}{1-\alpha^2} \right)}{ln\alpha}

Is this right?

Edit this could also be written as:

n &lt; \log_\alpha \left( \frac{0.01\alpha(1-\alpha-\alpha^2)}{1-\alpha^2} \right)
 

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