Invertibility of LTI systems

A system is invertible if distinct inputs produce distinct outputs, so that the input can be recovered from the output. The system that recovers it is the inverse system.

x(t) --> [ h(t) ] --w(t)--> [ h1(t) ] --> y(t) = x(t)

Condition for LTI systems

Let the system have impulse response $h(t)$ and the inverse system $h_1(t)$. The cascade is an LTI system with impulse response $h(t) * h_1(t)$. For the cascade to be the identity system ($y(t) = x(t)$ for every $x$):

$$\begin{aligned} y(t) &= x(t) * \big[h(t) * h_1(t)\big] = x(t) \\ \Rightarrow\; h(t) * h_1(t) &= \delta(t) \end{aligned}$$

because $\delta(t)$ is the identity element of convolution: $x(t) * \delta(t) = x(t)$.

In the frequency domain (convolution property):

$$H(j\omega)\,H_1(j\omega) = 1 \;\Rightarrow\; H_1(j\omega) = \frac{1}{H(j\omega)}$$

So an LTI system is invertible only if $H(j\omega) \ne 0$ at all frequencies (no information destroyed). Similarly, in discrete time: $h[n] * h_1[n] = \delta[n]$.

Examples

1. Delay: $h(t) = \delta(t - t_0)$. Inverse: $h_1(t) = \delta(t + t_0)$, since $\delta(t - t_0) * \delta(t + t_0) = \delta(t)$.
2. Accumulator: $y[n] = \sum_{k=-\infty}^{n} x[k]$, $h[n] = u[n]$. Inverse: first difference $y[n] = x[n] - x[n-1]$, $h_1[n] = \delta[n] - \delta[n-1]$. Check: $u[n] - u[n-1] = \delta[n]$.
3. Not invertible: $y(t) = 0$, or an ideal LPF ($H = 0$ in stopband; those frequencies are lost).