i have a discrete-time LTI system $L$ that takes input signal $x[n]$ and gives the output signal $y[n]$. since $L$ is LTI, $y[n]$ can be derived as a sum of shifted and scaled impulse responses of $L$. but why is that a useful fact and can you show a practical example where this saves us work?

this is my attempted answer and i would like to know if it's correct: suppose $L$ is an accumulator,

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

then by LTI properties, this can be rewritten as a sum of shifted and scaled impulse responses:

$y[n] = \sum_{k=-\infty}{\infty}h(n-k)$

where $h(n-k)$ is the impulse response function.

what does this gain us, intuitively? it seems that this means that if we could only measure the response of $y(n)$ to a single (unit) impulse, we could derive through shifting, scaling and summation $L$'s output to any arbitrary input signal $x[n]$. so we wouldn't have to measure $L$'s output to a more complicated signal, a single pulse would be enough. is this right?

if so how would it work in this example? suppose the input signal $x_0[n]$ was:

$x_0[0] = 1, x_0[k] = 0, \forall k \neq n$

then clearly $L[x_0[n]] = 1$. how can we use this to derive the response to a new input signal $x'[n]$:

$x'[0] = 1, x'[2] = 1, x'[3] = 2$

the result should be 4. how do you formally derive this answer using only the system response to $x_0$?

  • $\begingroup$ @MBaz: those answers are not detailed enough and don't use a concrete real system as example like the accumulator. i'd like to see an analysis of the accumulator LTI $\endgroup$ – mvd Dec 4 '16 at 23:49
  • $\begingroup$ Perhaps if you wrote down EXPLICITLY what the impulse response of the accumulator system is instead of pushing back at MBaz, you will see the connection. I too vote to close this as a duplicate. $\endgroup$ – Dilip Sarwate Dec 4 '16 at 23:59

A sum of shifted and scaled impulse responses would look more like

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

The impulse response of an accumulator is

$ h[n] = u[n] $

where $u[n]$ is the unit step function,

$u[n] = \begin{cases} 1 & n \ge 0 \\ 0 & \mathrm{otherwise} \end{cases}$

so for your example case of

$ x[n] = \{1, 1, 2\} $

The result will be

$ \begin{align} y[0] &= \sum_{k=-\infty}^0 x[k] \cdot h[-k] \\ &= 1 \cdot 1 + 1 \cdot 0 + 2 \cdot 0 \\ &= 1 \\ y[2] &= \sum_{k=-\infty}^2 x[k] \cdot h[2-k] \\ &= 1 \cdot 1 + 1 \cdot 1 + 2 \cdot 1 \\ &= 4 \end{align}$

and hopefuly you can see how $y[1]$ will be 2, and $y[n], n> 2$ will be 4, as expected.

Your equations are incorrect for the sum of scaled and shifted impulse responses, and for your first input signal $x_0$.

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