Discrete LTI filter impulse response

[optrix]

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:

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

 

[berkeman]

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.

 

[optrix]

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.

 

[falbani]

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]

 

[optrix]

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 $$\alpha<1$$

I managed to get the inequality:

$$n < \frac{ln\left( \frac{0.01\alpha(1-\alpha-\alpha^2)}{1-\alpha^2} \right)}{ln\alpha}$$

Is this right?

Edit this could also be written as:

$$n < \log_\alpha \left( \frac{0.01\alpha(1-\alpha-\alpha^2)}{1-\alpha^2} \right)$$