Discrete LTI filter impulse response

  • Thread starter optrix
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  • #1
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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:

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

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  • #2
berkeman
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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:

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

Can you just run a simulation? I've used Excel for that before.
 
  • #3
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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.
 
  • #4
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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]
 
  • #5
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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:

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

In which case it will converge for [tex]\alpha<1[/tex]

I managed to get the inequality:

[tex]n < \frac{ln\left( \frac{0.01\alpha(1-\alpha-\alpha^2)}{1-\alpha^2} \right)}{ln\alpha} [/tex]

Is this right?

Edit this could also be written as:

[tex]n < \log_\alpha \left( \frac{0.01\alpha(1-\alpha-\alpha^2)}{1-\alpha^2} \right)[/tex]
 

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