Initial rest condition applied on $x(t)$ vs $h(t)$

Question

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?

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

$$y(t)=\int_{0}^{\infty}h(\tau)x(t-\tau)d\tau=\int_{-\infty}^tx(\tau)h(t-\tau)d\tau$$

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 $t<t_0$.