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Given a discrete-time LTI system defined by the impulse response $h[n]$, why is the output $y[n]$ of this system for an input $x[n]$ expressed as $$ y[n] = x[n] * h[n]=\sum\limits_{i=-\infty}^{\infty} x[i] h[n-i] $$

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  • $\begingroup$ What do you mean? Convolution is a mathematical operation, it is not used to "define" an LTI system. It can be used to easily find an LTI system's output to any input, but you can define an LTI system in multiple ways, for example $y(t)=3x(t)$. Can you also explain what do you think is the relationship between polynomial multiplication and convolution? $\endgroup$
    – MBaz
    Jun 8, 2015 at 23:58
  • $\begingroup$ I think the before signal is affecting to present signal. Also polynomial multiplication too. This fetures are common between relationship . $\endgroup$
    – gmotree
    Jun 9, 2015 at 0:18
  • $\begingroup$ @MBaz also convolution is the same correlation but it has differnce only associative. Also actually convolution calculation is very difficult. So we are using the frequency domain . It is much more easy to calculate $\endgroup$
    – gmotree
    Jun 9, 2015 at 0:32
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    $\begingroup$ @gmotree, LTI systems are defined by what LTI stands for: They're linear and time invariant. Convolution is the most general linear time invariant operation, and so every LTI system can be written as a convolution product. It's not even hard to prove. The relationship between polynomial multiplication and convolution is based on the same connection. Polynomial convolution with one of the two polynomials fixed is a linear operation. It is also shift invariant, because you can always factor out some x^k and factor it in afterwards. So the operation on the coefficient vector must be a convolution $\endgroup$
    – Jazzmaniac
    Jun 9, 2015 at 7:54
  • $\begingroup$ I think that this answer might answer the first part of your question. $\endgroup$
    – Matt L.
    Jun 9, 2015 at 15:42

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it takes more work to do this for continuous-time systems, so i'll just do it for discrete-time and we can all wave our arms to extend to continuous-time.

a general discrete-time input signal can be expressed as

$$ x[n] = \sum\limits_{i=-\infty}^{\infty} x[i] \delta[n-i] $$

where $\delta[n]$ is the Kronecker delta or "unit impulse function"

$$ \delta[n] \triangleq \begin{cases} 1, & \text{if }n = 0 \\ 0, & \text{if }n \ne 0 \end{cases} $$

so think of your input $x[n]$ as the sum of a bunch of individual impulse functions, each delayed to time $i$ and weighted by a constant which is equal to $x[i]$.

so, let's say we have a Linear and Time-Invariant discrete-time system, represented with operator, $\mathbb{T}$, and output:

$$ y[n] = \mathbb{T}\left\{ x[n] \right\} $$

and we input into that system a single unit impulse, $\delta[n]$. by definition, the "impulse response" of the LTI system is the output coming out when a unit impulse, $\delta[n]$, is input.

$$ h[n] \triangleq \mathbb{T}\left\{ \delta[n] \right\} $$

now, imagine a general input, $x[n]$ is input to $\mathbb{T}\{ \}$. then, because it's Linear:

$$ \begin{align} y[n] & = \mathbb{T}\left\{ x[n] \right\} \\ & = \mathbb{T}\left\{ \sum\limits_{i=-\infty}^{\infty} x[i] \delta[n-i] \right\} \\ & = \sum\limits_{i=-\infty}^{\infty} \mathbb{T}\left\{ x[i] \delta[n-i] \right\} \\ & = \sum\limits_{i=-\infty}^{\infty} x[i] \mathbb{T}\left\{ \delta[n-i] \right\} \\ \end{align} $$

and because it's Time Invariant,

$$ h[n-i] = \mathbb{T}\left\{ \delta[n-i] \right\} $$

then, putting the two together, you get

$$ \begin{align} y[n] & = \sum\limits_{i=-\infty}^{\infty} x[i] \mathbb{T}\left\{ \delta[n-i] \right\} \\ & = \sum\limits_{i=-\infty}^{\infty} x[i] h[n-i] \\ \end{align} $$

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  • $\begingroup$ (+1) For continuous-time systems, the sifting property can be used to express the input $x(t)$ as a "sum" of Dirac delta functions. $\endgroup$
    – mhdadk
    Nov 16, 2021 at 12:45

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