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 $$ y[n] = x[n] * h[n]=\sum\limits_{i=-\infty}^{\infty} x[i] h[n-i] $$

  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented Jun 9, 2015 at 0:32
  • 2
    $\begingroup$ @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
    Commented Jun 9, 2015 at 7:54
  • $\begingroup$ I think that this answer might answer the first part of your question. $\endgroup$
    – Matt L.
    Commented Jun 9, 2015 at 15:42

2 Answers 2


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

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

  • $\begingroup$ (+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
    Commented Nov 16, 2021 at 12:45

To complement @robert bristow-johnson's answer, the continuous-time convolution operation naturally arises when attempting to find the output of an LTI system as the solution to an ordinary differential equation (ODE).

Without loss of generality, I will only consider a single-input single-output (SISO) LTI system. A similar derivation applies to the multiple-input multiple output case. Consider the state equation of a SISO LTI system, $$ \frac{\text{d}x}{\text{d}t}(t) = ax(t) + bu(t) \tag{1} \label{st_eq} $$ where

  • $x(t)$ is a time-varying state variable,
  • $\frac{\text{d}x}{\text{d}t}(t)$ is its time derivative, which is a function of time,
  • $u(t)$ is an input signal, and
  • $a$ and $b$ are time-invariant constants.

Note that $\eqref{st_eq}$ represents a first-order ODE. The corresponding observation (or output) equation can be defined as $$ y(t) = cx(t) + du(t) \tag{2} \label{obsv_eq} $$

To determine the output $y(t)$ given an input $u(t)$, we need to determine the function $x(t)$ that satisfies $\eqref{st_eq}$. Let the function that satisfies $\eqref{st_eq}$ be represented as $x^*(t)$. We then substitute $x^*(t)$ in place of $x(t)$ in $\eqref{obsv_eq}$ to determine the output $y(t)$.

To determine the function $x(t)$ that satisfies $\eqref{st_eq}$, we proceed as follows. First, subtract $ax(t)$ from both sides of $\eqref{st_eq}$ to get $$ \frac{\text{d}x}{\text{d}t}(t) - ax(t) = bu(t) $$ Next, multiply both sides of this equation by $e^{-at}$ to get $$ e^{-at}\frac{\text{d}x}{\text{d}t}(t) - e^{-at}ax(t) = e^{-at}bu(t) $$ Now note that the left-hand side of this equation represents the result of a product rule differentiation. That is, if we let $u(t) = x(t)$ and $v(t) = e^{-at}$, then \begin{align} \frac{\text{d}}{\text{d}t}(u(t) \cdot v(t)) &= \frac{\text{d}u}{\text{d}t}(t) \cdot v(t) + \frac{\text{d}v}{\text{d}t}(t) \cdot u(t) \\ &= \frac{\text{d}x}{\text{d}t}(t) \cdot e^{-at} + (-ae^{-at}) \cdot x(t) \\ &= e^{-at}\frac{\text{d}x}{\text{d}t}(t) - e^{-at}ax(t) \\ &= \frac{\text{d}}{\text{d}t}(x(t) \cdot e^{-at}) \end{align} Therefore, \begin{align} e^{-at}\frac{\text{d}x}{\text{d}t}(t) - e^{-at}ax(t) &= e^{-at}bu(t) \\ \frac{\text{d}}{\text{d}t}(x(t) \cdot e^{-at}) &= e^{-at}bu(t) \end{align} This is another first-order ODE, but it can now be solved using a simple integration. We will use the first fundamental theorem of calculus to do this, \begin{align} \int_{\tau}^t \frac{\text{d}}{\text{d}t}(x(\lambda) \cdot e^{-a\lambda}) \, \text{d}\lambda &= \int_{\tau}^t e^{-a\lambda}bu(\lambda) \, \text{d}\lambda \\ x(t) \cdot e^{-at} - x(\tau) \cdot e^{-a\tau} &= \int_{\tau}^t e^{-a\lambda}bu(\lambda) \, \text{d}\lambda \end{align} where the lower-limit $\tau$ is used to determine which antiderivative is chosen. We then move the $x(\tau) \cdot e^{-a\tau}$ term to the right to get $$ x(t) \cdot e^{-at} = x(\tau) \cdot e^{-a\tau} + \int_{\tau}^t e^{-a\lambda}bu(\lambda) \, \text{d}\lambda $$ and then multiply both sides by $e^{at}$ to get \begin{align} x(t) &= e^{at} \cdot x(\tau) \cdot e^{-a\tau} + e^{at} \cdot \int_{\tau}^t e^{-a\lambda}bu(\lambda) \, \text{d}\lambda \\ x(t) &= x(\tau) \cdot e^{a(t - \tau)} + \int_{\tau}^t e^{a(t - \lambda)}bu(\lambda) \, \text{d}\lambda \end{align} Out of interest, note that $$ x(t) = \underbrace{x(\tau) \cdot e^{a(t - \tau)}}_{\text{complementary function}} + \underbrace{\int_{\tau}^t e^{a(t - \lambda)}bu(\lambda) \, \text{d}\lambda}_{\text{particular integral}} $$ Unfortunately, this expression for $x(t)$ is still ambiguous, as we have not yet specified a value for $\tau$. We do this using an initial condition. For example, suppose we know that $x(\tau) = x_0$ for $\tau = 0$. Then, we can substitute $\tau = 0$ into the equation above to get \begin{align} x(t) &= x(0) \cdot e^{at} + \int_{0}^t e^{a(t - \lambda)}bu(\lambda) \, \text{d}\lambda \\ x(t) &= x_0 \cdot e^{at} + \int_{0}^t e^{a(t - \lambda)}bu(\lambda) \, \text{d}\lambda \end{align} We now have an exact expression for $x(t)$. We can now substitute this expression into the observation equation in $\eqref{obsv_eq}$ to determine an expression for the output $y(t)$, \begin{align} y(t) &= cx(t) + du(t) \\ y(t) &= c\left[x_0 \cdot e^{at} + \int_{0}^t e^{a(t - \lambda)}bu(\lambda) \, \text{d}\lambda\right] + du(t) \end{align} It is now possible to express the output $y(t)$ as the convolution of two functions. To see this, suppose that $c = 1$ and $d = 0$, such that the equation above simplifies to $$ y(t) = x_0 \cdot e^{at} + \int_{0}^t e^{a(t - \lambda)}bu(\lambda) \, \text{d}\lambda $$ Next, suppose that $x_0 = 0$, such that $$ y(t) = \int_{0}^t e^{a(t - \lambda)}bu(\lambda) \, \text{d}\lambda $$ If $u(\lambda) \neq 0$ when $t \in [0,t]$ and $u(\lambda) = 0$ otherwise, then this expression for $y(t)$ represents the convolution of the functions $h(t) = b \cdot e^{at}$ and $u(t)$, \begin{align} y(t) &= \int_{0}^t e^{a(t - \lambda)}bu(\lambda) \, \text{d}\lambda \\ &= (b \cdot e^{at}) * u(t) \\ &= h(t) * u(t) \end{align} where $*$ represents the convolution operation. This means that the function $h(t) = b \cdot e^{at}$ is the impulse response of this LTI system when the input is $u(t)$, the output is $x(t)$ (the $c = 1$ and $d = 0$ assumptions above), and the initial state is $x(0) = 0$.

  • $\begingroup$ Nice answer! By the way, this explains why the impulse response of a first order system (your example) is exactly $be^{at}$. Your answer also explains why we need to consider zero initial conditions to take the transfer function: $H(s)=Y(s)/U(s)$ (convolution is multiplication on frequency domain) only if the term $x_0\cdot e^{at}$ disapear. $\endgroup$ Commented Aug 9, 2023 at 0:51

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