LTI: causality and stability

A Linear Time-Invariant (LTI) system is a system that obeys superposition (linear) and whose response to a delayed input is the same response delayed (time invariant). Its output for any input is given by convolution with its impulse response:

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

Causality

A system is causal if its output at time $t$ depends only on inputs at times $\le t$.

In the convolution, the term $x(t-\tau)$ for $\tau < 0$ is a future input value. For it never to be used, the weight $h(\tau)$ must be zero there:

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

Then $y(t) = \int_0^{\infty} h(\tau)\,x(t-\tau)\,d\tau$.

• Causal: $h(t) = e^{-2t}u(t)$, $h(t) = u(t) - u(t-1)$.
• Non-causal: $h(t) = e^{-|t|}$, $h(t) = \delta(t+2)$.
• Physically realizable real-time systems must be causal.

Stability

A system is BIBO stable if every bounded input, $|x(t)| \le M_x$, gives a bounded output.

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

So the output is bounded if

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

(the impulse response is absolutely integrable). This is also necessary: if the integral is infinite, the bounded input $x(t) = \mathrm{sgn}[h(-t)]$ gives $y(0) = \int|h(\tau)|d\tau = \infty$.

• Stable: $h(t) = e^{-2t}u(t)$, since $\int_0^\infty e^{-2t}dt = 0.5$.
• Unstable: $h(t) = u(t)$ (integrator), $h(t) = e^{2t}u(t)$.