2
$\begingroup$

Why do we always characterize a LTI system by its impulse response and not by another response, like the step response? What does the impulse response have that is so special?

$\endgroup$
  • $\begingroup$ possible duplicate of What is meant by a system's "impulse response" and "frequency response?" $\endgroup$ – Jason R Feb 3 '14 at 17:48
  • $\begingroup$ While I do not think it is a duplicate, I do think that the linked question answers this one as well. Welcome to DSP.se, and feel free to ask questions, but please put some effort in to finding the answer for yourself before asking the question. Not only does that make the question more specific, it also gets you better answers that focus on your problems specifically. $\endgroup$ – penelope Feb 4 '14 at 20:02
4
$\begingroup$

Let's try a simple example. A discrete-time LTI system has unit pulse response $$h[0] = 1,\quad h[1] = 1,\quad h[n] = 0 ~~\text{for all}~ n \geq 2. \tag{1}$$ The response $y$ of this system to a signal $x$ given by $$x[0] = 1,\quad x[1] = -1,\quad x[2] = 1,\quad x[n] = 0 ~~\text{for all}~ n \geq 3 \tag{2}$$ is readily calculated via the convolution formula, and we get $$y[0] = 1, \quad y[1] = 0, \quad y[2] = 0, \quad y[3] = 1,\quad y[n] = 0 ~~\text{for all}~ n \geq 4.\tag{3}$$

============================================================

The unit-step response $g$ of the same LTI system is easily found to be $$g[0] = 1,\quad g[n] = 2 ~~\text{for all}~ n \geq 1.\tag{4}$$

So, suppose that you are given only the step response $g$ in $(4)$ but not the pulse response in $(1)$. How will you compute the response of this LTI system to input $x$? It is known that the answer is $(3)$, but you are asked to figure out how to get from $(2)$ and $(4)$ to $(3)$.

$\endgroup$
2
$\begingroup$

The math for analyzing the behavior of applying an arbitrary signal to an LTI system (filtering, etc.) is usually more compact in description when using the impulse response rather than the step response.

$\endgroup$
0
$\begingroup$

any model response must be from a model input that contains non-zero energy at all frequencies (or virtually all). both the impulse and the step do that, but i think the reason why the impulse response is preferred over the step response is that the Fourier transform, Laplace transform, Z transform, DTFT, DFT (whatever) of an impulse is the number 1 (for all frequencies $\omega$ or all $s$ or $k$ or whatever). if you end up calculating the transfer function by dividing the transform of the output by the transform of the input, it's pretty easy to divide by 1.

$\endgroup$
0
$\begingroup$

Given the response of an LTI system to an impulse input, it is then easy to get the corresponding output to an arbitrary input. An arbitrary input can be seen as the sum of weighted and time-delayed impulses. Because the systems is Linear Time Invariant (LTI) the output is simply obtained summing the weighted and time delayed versions of the impulse response. The weighting and time delay factors are exactly the same as those used in obtain the input signal from an impulse.

$\endgroup$
  • $\begingroup$ What exactly does "The weighting and time delay factors are exactly the same as those used in obtain the input signal from an impulse." mean? (emphasis added) One obtains the input signal from an impulse? $\endgroup$ – Dilip Sarwate Feb 4 '14 at 21:42
  • $\begingroup$ @DilipSarwate Take your example at $x[1]=-1$. In this case the impulse is scaled by -1 and is also delayed by 1 sample. This produces the impulse response which is weighted by -1 and delayed by 1 sample (The same scaling and time delay used on the impulse at $x[1]$. The total output is obtained by summing the responses to each of the inputs i.e. $x[0]$, $x[1]$, $x[2]$ etc. Is that any clearer. $\endgroup$ – David Feb 5 '14 at 13:49
  • $\begingroup$ Sorry, no clearer. You correctly say, in your comment, that the output is the sum of scaled and time-delayed copies of the impulse response. No arguments about that. But what I was asking about is the claim in your answer that says that the input signal is obtained from an impulse. How does the impulse (or the impulse response) know which input signal I am thinking of today especially since it is different from the one that I was thinking of yesterday?? $\endgroup$ – Dilip Sarwate Feb 5 '14 at 14:14
  • $\begingroup$ @DilipSarwate The impulse doesn't know anything. It just a mathematical tool. It just allows you to get the response to an arbitrary input by constructing it from scaled and delayed versions of the impulse. The corresponding output is obtained by scaling and shifting the impulse response. I give a more mathematical derivation in my answer here. You can also look at it like a Green's function in Diff. equations. From the Greens function you can get the response to any input. $\endgroup$ – David Feb 5 '14 at 14:33
0
$\begingroup$

Let

  • $\mathcal H$ be a discrete-time LTI system whose impulse response is $h := \mathcal H (\delta)$, where $\delta$ is the unit impulse.
  • $x$ be a finite discrete-time signal of length $N$ whose value at $k$ is $x_k$.
  • $\mathcal D$ be the delay operator.

Hence, signal $x$ can be written as a superposition of delayed and scaled unit impulses

$$x = x_0 \delta + x_1 \mathcal D (\delta) + x_2 \mathcal D^2 (\delta) + \cdots + x_{N-1} \mathcal D^{N-1} (\delta) = \sum_{k=0}^{N-1} x_k \mathcal D^k (\delta)$$

Feeding $x$ into $\mathcal H$, we obtain

$$y := \mathcal H (x) = \mathcal H \left(\sum_{k=0}^{N-1} x_k \mathcal D^k (\delta)\right)$$

Since $\mathcal H$ is linear,

$$y = \mathcal H \left(\sum_{k=0}^{N-1} x_k \mathcal D^k (\delta)\right) = \sum_{k=0}^{N-1} x_k (\mathcal H \circ \mathcal D^k) (\delta)$$

Since $\mathcal H$ is not merely linear but also time-invariant, operators $\mathcal H$ and $\mathcal D$ do commute and, thus,

$$y = \sum_{k=0}^{N-1} x_k (\mathcal H \circ \mathcal D^k) (\delta) = \sum_{k=0}^{N-1} x_k (\mathcal D^k \circ \mathcal H) (\delta) = \sum_{k=0}^{N-1} x_k \mathcal D^k (\mathcal H (\delta)) = \sum_{k=0}^{N-1} x_k \mathcal D^k (h)$$

Hence, the output signal $y$ is the superposition of delayed and scaled versions of the impulse response $h$. What does $h$ have that is so special?

Note that a discrete-time signal $x$ is usually defined by

  • its value $x_0$ at $0$
  • its value $x_1$ at $1$
  • its value $x_2$ at $2$

and so forth. Hence, borrowing from linear algebra, the canonical basis for $\mathbb R^{\mathbb Z}$ is formed by delayed versions of $\delta$. From linearity, knowing the system's response to each element of the canonical basis fully characterizes the system. Unfortunately, the canonical basis for $\mathbb R^{\mathbb Z}$ has infinitely many elements. Fortunately, from time-invariance, knowing the system's response to $\delta$ is enough to fully characterize the system, as the system's response to, say, $\mathcal D (\delta)$ is simply $\mathcal D (h)$.

$\endgroup$
  • $\begingroup$ This is fully correct and nicely presented. However I'm not sure if it is what the OP really asked. His concern was more why we pick the IR over other equivalent characterisations. Maybe you can add something about the existence conditions of the impulse response versus those of the step response or the frequency response. Then you'll get my upvote for relevance. That said, it's nice to see a fellow math aficionado here! $\endgroup$ – Jazzmaniac Oct 22 '16 at 14:39
  • $\begingroup$ @Jazzmaniac I think I answered the OP's question. The "natural" way of dealing with discrete-time signals is in terms of shifted and scaled impulses, so much so that that is how one sketches them. Thus, using the impulse response is also natural. Of course, if one has piecewise constant or piecewise affine signals, then one would write them in terms of step functions or ramp functions, respectively. Also, sinusoidal signals can be more efficiently written in the Fourier basis. However, most signals have no such structure and, thus, the use of impulses is "natural". $\endgroup$ – Rodrigo de Azevedo Oct 22 '16 at 15:36
  • $\begingroup$ I don't disagree. However I believe that the OP already knew this and asked specifically for a deeper reason. I might be wrong, and like I said, your answer is perfectly correct. However, it doesn't really add anything to the accepted answer linked by Jason R above. $\endgroup$ – Jazzmaniac Oct 22 '16 at 17:49
0
$\begingroup$

The impulse response, the step response, and the frequency response all come from the "linear" nature of a LTI system. Therefore, which one is preferred is actually determined by which decomposition manner is preferred. Generally, we are familiar with the $\delta(t)$ sampling concept, and the $\delta[n]$ decomposition is more direct, thus the impulse response is more commonly used than the step response.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.