In other words, what does the impulse response tell us about the characteristics of the system? For simplicity, let's assume we deconvolve a discrete output and a discrete input to obtain an impulse response that has n points. Does the amplitude of the first (or second, or third...) point indicate something? Does the length n tell us anything?


See this question for a long description of the impulse and frequency responses of a system.

In short, the impulse response of a discrete-time LTI system tells you something very specific. If the system has impulse response $h[n]$ and you apply an input signal of the following form:

$$ x[n] = \delta[n] = \begin{cases} 1, && n = 0 \\ 0, && \text{otherwise} \end{cases} $$

then the output of the system will be exactly $h[n]$. This input signal $\delta[n]$ is a discrete-time impulse, known as a Kronecker delta function. To your question specifically, the value of the first point in the impulse response indicates what the system's output will be at time instant $n = 0$ when you input an impulse. Likewise, the second value indicates what the system output will be at time instant $n = 1$, and so on.

This becomes a powerful tool for analysis when you combine it with the properties of linearity and time invariance that LTI systems have. You can decompose any discrete-time signal into a sum of multiple discrete-time impulses, all scaled and shifted in time in different ways. Since the system is linear and time-invariant, scaling and shifting the input signal has the same effect on the output, so if you input a signal of the form:

$$ a\delta[n-k] = \begin{cases} a, && n = k \\ 0, && \text{otherwise} \end{cases} $$

then the output of the system will be $ah[n-k]$; the scaling and time shift on the input just cause likewise scaling and time shifting on the output. Since for any discrete-time input signal, you can break it down into a sum of scaled and time-shifted impulses, you can do the same to the system's output; that output will just be a sum of correspondingly scaled and time-shifted impulse responses. That is, the system's output $y[n]$ for input signal $x[n]$ of length $K$ samples can be written as:

$$ y[n] = \sum_{k=0}^{K-1} x[k]h[n-k] $$

For each point $k \in [0, 1, \ldots , K-1]$, we take the corresponding value of the input signal and use it to scale a copy of the impulse response shifted by $k$ samples. The resulting output is the sum of all of these impulse responses. This is the discrete-time convolution sum.

  • $\begingroup$ Thanks Jason. I actually posted this question after reading your awesome summary in another post about impulse and frequency response. I was just looking for some plain (formula free) explanation; i.e. if I have a specific physical system such as a ball in a tube I give the ball an instantaneous kick (!), how would the resulting movement of the ball correlate with the systems impulse response? I hope this can be considered an LTI system, of course. $\endgroup$ – Xe M Sep 14 '17 at 13:37
  • $\begingroup$ If your instantaneous kick could be well-modeled by an impulse, then the movement of the ball would closely match its impulse response (provided the system was LTI). $\endgroup$ – Jason R Sep 14 '17 at 16:03
  • $\begingroup$ Ok, that would make sense. I assume that the impulse response, if an exponential decay, would be proportional to the ball's velocity over time, yes? Last thing, do we expect the integral of a discrete impulse response (area under curve) to be 1, or in an LTI system this is not a requirement? Thanks $\endgroup$ – Xe M Sep 14 '17 at 17:11
  • $\begingroup$ If the output of the system that you're measuring $y[n]$ is the position of the ball versus time, then the impulse response instead would represent its position versus time, provided that you smacked it hard enough to instantaneously give it an initial displacement. If the impulse response was an exponential decay, then you would expect its position (not velocity) to have that decay behavior. With that said, the derivative of a decaying exponential is also a decaying exponential (with opposite sign), so its velocity would decay exponentially as well. $\endgroup$ – Jason R Sep 14 '17 at 18:09
  • $\begingroup$ And no, there is no requirement for the area underneath the impulse response to be 1 or any specific value. $\endgroup$ – Jason R Sep 14 '17 at 18:09

The impulse response of an LTI (linear time invariant) system has the same physical meaning of the differential equation of a general dynamical system has.

So every output can be computed from the input and the impulse response convolution, without having to solve the differential equation. The frequency charactheristics can also be deduced from the Fourier transform of the impulse response.

Several properties of the (LTI) systems can be seen in the impulse reponse (and Fourier transform of it) expressions, such as slow system, fast system, wide band narrow band systems, overshoot, oscillations etc...

  • $\begingroup$ Thanks Fat32. That's very interesting! How do you determine how fast a system is (and other properties) from the impulse response and its Fourier transform? Any good documentation on this? By the way, can general dynamic systems be LTI systems as well? Such as flow of water in a pipe? $\endgroup$ – Xe M Sep 14 '17 at 13:40
  • $\begingroup$ For a beginning treatment look from any Signals and Systems book the chapter on Transform and Time domain analysis of LTI systems. For example Ch6 of oppenheim's book will provide you a lot of insights into this. $\endgroup$ – Fat32 Sep 14 '17 at 13:59
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    $\begingroup$ Ok, thanks for this. Two last questions before getting myself deep into studying: Does the impulse response have any units of measurement? If not, is it proportional to any physical characteristic of the system? $\endgroup$ – Xe M Sep 14 '17 at 16:48
  • $\begingroup$ @XeM you have two answers now, If they provided useful (and correct) information to you then you can upvote them and select one of them as the answer. Otherwise the question hangs and rings in the community list as being unanswered etc... $\endgroup$ – Fat32 Sep 14 '17 at 16:54
  • $\begingroup$ But I am still asking follow up questions! $\endgroup$ – Xe M Sep 14 '17 at 17:47

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