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Short note: Invertibility of LTI system

Short answer

LTI system is invertible if there exists h_inv with h*h_inv=δ, equivalently H(ω)≠0 for all ω and H_inv=1/H. Examples: delay inverse advance, integrator inverse differentiator. Not invertible if information lost, e.g. ideal LPF.

Write a short note on invertibility of LTI system.

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

x(t) --> [ h(t) ] --> y(t) --> [ h_inv(t) ] --> x(t)

Condition for an LTI system: the system h(t)h(t) is invertible if there is an LTI system hinv(t)h_{inv}(t) such that the cascade gives the identity system:

h(t)∗hinv(t)=δ(t)h(t) * h_{inv}(t) = \delta(t)

In the frequency domain: H(ω) Hinv(ω)=1H(\omega)\,H_{inv}(\omega) = 1, so Hinv(ω)=1/H(ω)H_{inv}(\omega) = 1/H(\omega). This needs H(ω)≠0H(\omega) \ne 0 at every frequency.

Examples:

  • y(t)=x(t−t0)y(t) = x(t - t_0), h(t)=δ(t−t0)h(t) = \delta(t-t_0): inverse hinv(t)=δ(t+t0)h_{inv}(t) = \delta(t+t_0).
  • Integrator h(t)=u(t)h(t) = u(t): inverse is the differentiator, y(t)=dxdty(t) = \frac{dx}{dt}.
  • y[n]=x[n]−x[n−1]y[n] = x[n] - x[n-1]: inverse is the accumulator y[n]=∑k=−∞nx[k]y[n] = \sum_{k=-\infty}^{n} x[k].
  • Not invertible: y(t)=0y(t) = 0, y(t)=x2(t)y(t) = x^2(t) (sign is lost), an ideal LPF (high frequencies are lost).

Use: channel equalization in communication, deconvolution, decoding.

✓ Answer checked by Claude Opus 5.5 (high effort)

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