LTI: causality and time variance

LTI system

A linear time-invariant (LTI) system is one that is both

• linear: obeys superposition, $a x_1(t) + b x_2(t) \to a y_1(t) + b y_2(t)$, and
• time-invariant: a delay in the input gives the same delay in the output, $x(t - t_0) \to y(t - t_0)$.

An LTI system is completely described by its impulse response $h(t)$, and $y(t) = x(t) * h(t)$.

Causality

A system is causal if the output at any time $t_0$ depends only on present and past inputs ($t \le t_0$), never on future inputs. All real-time physical systems are causal.

• Causal: $y(t) = x(t-1)$, $y(t) = \int_{-\infty}^{t} x(\tau)d\tau$.
• Non-causal: $y(t) = x(t+1)$, $y(t) = x(-t)$ (at $t = -1$ it needs $x(1)$).
• For an LTI system: causal $\iff h(t) = 0$ for $t < 0$.

Time variance

A system is time-invariant if its behaviour does not change with time: if $x(t) \to y(t)$ then $x(t - t_0) \to y(t - t_0)$ for every $t_0$. Otherwise it is time-variant.

Test: find $y_1(t)$, the output for $x(t - t_0)$, and compare with $y(t - t_0)$.

Example 1: $y(t) = x(2t)$.

$$\begin{aligned} y_1(t) &= x(2t - t_0) \\ y(t - t_0) &= x(2(t - t_0)) = x(2t - 2t_0) \end{aligned}$$

They differ, so the system is time-variant.

Example 2: $y(t) = x(t)^2$: $y_1(t) = x(t - t_0)^2 = y(t - t_0)$, so it is time-invariant.