- #1

optrix

- 33

- 0

In particular, I have:

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

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In summary, to calculate the time taken for the output to fall below 1% of its initial value for a discrete-time LTI system with unit impulse response function h[n], one can determine n by solving the inequality h[n] < 0.01 · h[0] or using the formula n < \log_\alpha \left( \frac{0.01\alpha(1-\alpha-\alpha^2)}{1-\alpha^2} \right), where \alpha is the parameter in the function. This can be done algebraically instead of using a simulation.

- #1

optrix

- 33

- 0

In particular, I have:

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

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- #2

berkeman

Mentor

- 68,088

- 21,649

optrix said:

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

optrix

- 33

- 0

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.

- #4

falbani

- 8

- 0

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

h[n] < 0.01 · h[0]

- #5

optrix

- 33

- 0

falbani said:

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]

A Discrete LTI (Linear Time-Invariant) filter impulse response is the output of a discrete-time LTI filter when the input is an impulse signal. It represents the relationship between the input and output of the filter and is typically represented by a sequence of numbers.

The impulse response of a Discrete LTI filter can be calculated by taking the inverse Fourier transform of the frequency response of the filter. This can also be done by using the filter coefficients and recursively applying the filter equation to an input impulse signal.

The impulse response of a Discrete LTI filter is important because it provides a complete description of the filter's behavior. It allows us to analyze the filter's frequency response, stability, and other properties. It also serves as a basis for designing and implementing the filter in different applications.

No, the impulse response of a Discrete LTI filter is time-invariant, meaning it does not change with time. This property is a fundamental characteristic of LTI systems and allows us to analyze and design filters using mathematical tools such as Fourier transforms and convolution.

The impulse response of a Discrete LTI filter can be used in various signal processing applications such as filtering, signal reconstruction, and system identification. It allows us to manipulate and analyze signals in both time and frequency domains, making it a powerful tool in signal processing.

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