- #1

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In particular, I have:

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

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- Thread starter optrix
- Start date

- #1

- 32

- 0

In particular, I have:

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

- #2

berkeman

Mentor

- 59,682

- 9,838

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

- 8

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

h[n] < 0.01 · h[0]

- #5

- 32

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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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