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I'm trying to get a good understanding of causality and I can't wrap my head around the first definition of it, as noted in this Wikipedia article:

Definition 1: A system mapping $x$ to $y$ is causal if and only if, for any pair of input signals $x_1(t)$ and $x_2(t)$ such that

$$x_{1}(t) = x_{2}(t), \quad \forall \ t \le t_{0},$$

the corresponding outputs satisfy

$$y_{1}(t) = y_{2}(t), \quad \forall \ t \le t_{0}.$$

Can anyone help explain this?

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  • $\begingroup$ This is not "the" definition of causality, but a corollary of the definition itself: a system is causal iff its output at time t does not depend on any future value of the input (i.e x(t+d) where d is positive). So your definition is a corollary of this fundamental definition. There are other corollaries that follow based on this definition by utilizing LTI, LCCDE or H(s) into it as provided in the below answer. $\endgroup$ – Fat32 Mar 16 '16 at 10:34
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So $x_1(t)$ and $x_2(t)$ are equal up to a certain time $t_0$, and diverge after that time. What this definition is saying is that, up to time $t_0$, the system will produce the exact same output, without regard for whether the input is actually $x_1(t)$ or $x_2(t)$.

If the system is non-causal, then it may "know" whether the input is actually $x_1(t)$ or $x_2(t)$, because it can "look ahead" in time beyond $t_0$, and it could modify its output accordingly, even before $t_0$.

As an example, consider a lightbulb connected to a battery and to a switch. The switch is initially on, and the lightbulb is also on. At time $t=0$, you decide whether you'll turn off the switch at time $t=60$ (one minute). The lightbulb, being causal, cannot know in advance what will happen, and will stay on at least up to $t=60$. A non-causal lightbulb will know in advance what will happen and may turn itself off at, say, $t=30<60$.

I hope this makes sense -- let me know if you need more clarification.

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  • $\begingroup$ Thank you for helping! I think I'm just having trouble understanding why that definition is necessary - what is it about that definition that defines a causal system? Because, according to my understanding, a causal system is just a system that only relies on present and past inputs - so i don't see where that definition fits in. $\endgroup$ – user20073 Mar 16 '16 at 5:31
  • $\begingroup$ @user20073: Those two definitions are just two different ways of saying the same thing. For a causal system the two signals in the wikipedia definition are identical as far as it can tell. That's why the output (up to the current time) must be the same. The reason why this is the case is of course because a causal system only knows about the past and the present, but not about the future. $\endgroup$ – Matt L. Mar 16 '16 at 8:44
  • $\begingroup$ @user20073 Matt is right: saying "up to time $t_0$, the system can't tell whether the input is $x_1(t)$ or $x_2(t)$" is just a different way of saying "the system cannot react to future inputs". $\endgroup$ – MBaz Mar 16 '16 at 13:46
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Causal System: cannot see into the future i.e. output at time n depends only on inputs up to time n. Formal definition: If $v(t) = x(t)$ for $t ≤ t_0$ then $H (v(t)) = H (x(t))% for t ≤ t0$. The following are equivalent:

  1. An LTI system is causal
  2. $h(t)$ is causal ⇔ $h(t) = 0$ for $t < 0 $
  3. $H(s)$ converges for $s = ∞$

Any right-sided sequence can be made causal by adding a delay.

Examples here: http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/assignments/MITRES_6_007S11_hw05_sol.pdf

http://home.npru.ac.th/sopapun/Solved_Problems.pdf

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  • $\begingroup$ He is talking about continuous time... (although being similar, better use "t" and H(s) ) $\endgroup$ – Fat32 Mar 16 '16 at 10:30

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