Define the LTI system $\mathcal{H} : x\mapsto y$

Define the convolution for continuous-time system : $$ (x*h)(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)\;\text{d}\tau $$ The initial rest condition states that :

An LTI system with an input signal $x(t)$ is causal if and only if $x(t)=0$, $\forall t<0$

Now I have noticed that in some textbooks, the definition of convolution would have $h(\tau)$ instead of $x(\tau)$ and $x(t-\tau)$ instead of $h(t-\tau)$ (since $x*h=h*x$) but the initial rest condition now becomes :

An LTI system with an impulse response $h(t)$ is causal if and only if $h(t)=0$, $\forall t<0$

My question is :

Why does commutativity allow that for a LTI system to be causal then in one definition $x(t)=0$, $\forall t<0$ while in another definition $h(t)=0$, $\forall t<0$ knowing that $x$ and $h$ are two different terms?


1 Answer 1


It should be clear that a property of a system, such as causality, cannot be determined by looking at its input signals. For a linear time-invariant system, it is its impulse response $h(t)$ from which properties such as causality or stability can be determined.

Only the second definition in the question is correct: a causal LTI system has an impulse response $h(t)$ which equals zero for $t<0$. This implies that the system's output at time $t_0$ only depends on values of the input signal $x(t)$ for $t\le t_0$, and not on future values $x(t)$, $t>t_0$.

The output signal of a causal LTI system can be written as


An initial-rest condition just means that if the input $x(t)$ is zero for $t<t_0$ then the output $y(t)$ must also be zero for $t<t_0$.

  • $\begingroup$ So if we say something like "For any $t_{0}$ and any input $x(t)$ such that $x(t)=0$ for $t<t_{0},$ the corresponding output $y(t)$ must also be zero for $t<t_{0}"$ is only true if we know that the system is LTI causal? $\endgroup$ Feb 28, 2021 at 17:12
  • $\begingroup$ Alright, got it thank you! $\endgroup$ Feb 28, 2021 at 17:15
  • $\begingroup$ @xXACEXx: I correct my previous comment: the condition given in your comment is valid for causal linear systems. There are non-linear systems that satisfy this requirement but are not causal, and there are causal non-linear systems that do not satisfy this condition. $\endgroup$
    – Matt L.
    Mar 1, 2021 at 14:23
  • $\begingroup$ So if we prove that $h(t)=0$ when $t<0$, then we would be proving causality for LTI systems so then the statement in my previous comment would be valid. right? Because I am confusing between the necessary condition and the sufficient condition for the causality of LTI systems and how they are related. $\endgroup$ Mar 1, 2021 at 14:30
  • 1
    $\begingroup$ @xXACEXx: Yes, sure, as written. $\endgroup$
    – Matt L.
    Mar 6, 2021 at 18:18

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