LTI system: linearity, stability, TI, causality

A Linear Time-Invariant (LTI) system is a system that is both linear (obeys superposition) and time invariant (its behaviour does not change with time). An LTI system is completely described by its impulse response $h(t)$, and its output for any input is the convolution $y(t) = x(t) * h(t)$.

(a) Linearity

A system is linear if it satisfies superposition = additivity + homogeneity:

$$a\,x_1(t) + b\,x_2(t) \;\rightarrow\; a\,y_1(t) + b\,y_2(t)$$

• Additivity: $x_1 + x_2 \rightarrow y_1 + y_2$.
• Homogeneity (scaling): $a x \rightarrow a y$.
• A consequence: zero input gives zero output.
• Examples: $y(t) = 3x(t)$, $y(t) = \frac{dx}{dt}$ are linear; $y(t) = x^2(t)$ and $y(t) = 2x(t) + 1$ are not.
• Convolution itself is a linear operation, so an LTI system is linear in its input.

(b) Stability

A system is BIBO stable (bounded-input bounded-output) if every bounded input $|x(t)| \le M_x < \infty$ gives a bounded output $|y(t)| \le M_y < \infty$.

For an LTI system:

$$|y(t)| = \left|\int h(\tau)x(t-\tau)d\tau\right| \le M_x \int_{-\infty}^{\infty}|h(\tau)|\,d\tau$$

So the condition is that the impulse response is absolutely integrable:

$$\int_{-\infty}^{\infty} |h(t)|\,dt < \infty$$

• $h(t) = e^{-2t}u(t)$: area $= 1/2$, stable.
• $h(t) = u(t)$ (integrator): area infinite, unstable.

(c) Time invariance

A system is time invariant if a time shift of the input causes the same time shift of the output:

$$x(t) \rightarrow y(t) \;\Rightarrow\; x(t - t_0) \rightarrow y(t - t_0)$$

• Test: find the output for the delayed input, then compare with $y(t-t_0)$.
• $y(t) = x(t-2)$ is time invariant.
• $y(t) = t\,x(t)$ is time varying, because the delayed input gives $t\,x(t-t_0)$ but $y(t-t_0) = (t-t_0)x(t-t_0)$.
• Systems with constant coefficients are time invariant.

(d) Causality

A system is causal if its output at time $t$ depends only on present and past inputs ($\tau \le t$), not on future inputs.

For an LTI system, $y(t) = \int h(\tau)x(t-\tau)d\tau$. To avoid using future values $x(t-\tau)$ with $\tau < 0$:

$$h(t) = 0 \quad \text{for } t < 0$$

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

• $h(t) = e^{-t}u(t)$ is causal.
• $h(t) = e^{-|t|}$ is non-causal (it is stable, though).