Intuition behind impulse response terms in convolution We're learning about convolution in my signals and systems class right now. I have been able to do all of the problems by simply working out the respective sum/integral, but I'm still having trouble gaining the intuition behind it.

Consider the following example. Let $x[n]$ be a discrete-time signal and input it into some LTI system with impulse response $h[n]$. Then,

$$
y[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k].
$$

Let's plug in some values to make this more concrete. Suppose that we want to compute $y[6]$. Well,

$$
y[6] = \sum_{k=-\infty}^{\infty} x[k] h[6-k] = \cdots + x[4]h[2] + x[5]h[1] + x[6]h[0] + x[7]h[-1] + \cdots.
$$

I understand the shifting, but I feel as if the multiplications should be in a different order. Namely, why are we multiplying $x[7]$ by $h[-1]$. I feel as if we should be multiplying it by $h[1]$, since we've essentially shifted everything to the right by $6$ units, to $6$ is the new $0$, which would mean that $7$ is the new $1$ (under the shifting). I have it backwards, and understand why when I work out the math, but why?

More concretely, my question is the following:

With respect to the above example, what exactly is the meaning of $h[2]$? $h[-1]$? $h[k]$ in general?
The coefficient $h[n]$ is the value of the system's response at time $n$ when the input signal was an impulse at $n=0$. Obviously, that's why we call $h[n]$ the impulse response. From this you can see that for a system to be causal, $h[n]$ must be zero for $n&lt;0$, otherwise the system would "know" in advance that an impulse will come at $n=0$.

Note that any discrete-time signal $x[n]$ can be written as a sum of unit impulses:

$$x[n]=\sum_{k=-\infty}^{\infty}x[k]\delta[n-k]\tag{1}$$

Since the response to a shifted impulse $\delta[n-k]$ is $h[n-k]$, and since the system is linear and time-invariant, from $(1)$ the output signal must be given by

$$y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]\tag{2}$$

which is of course the discrete-time convolution. Consequently, positive indices of $h[n]$ correspond to the memory of the system, that's why $h[1]$ is multiplied with the past input sample $x[n-1]$, etc.