Invertibility

An invertible system is one where you can uniquely recover the input from the output. Invertibility determines whether a communication channel can correct for distortion, whether an encryption transform can be decoded, and whether a sensor's output can be mapped back to the physical quantity it measures.

Core Concept

A system is invertible if distinct inputs always produce distinct outputs. Formally, if x1(t) is different from x2(t), then H{x1(t)} must differ from H{x2(t)}. If two different inputs produce the same output, you cannot tell them apart by observing the output, so inversion is impossible.

For LTI systems, invertibility translates to transfer functions. A system H(s) is invertible if there exists H_inv(s) such that H(s) * H_inv(s) = 1. This means H_inv(s) = 1/H(s). The inverse system has poles where H(s) had zeros and zeros where H(s) had poles.

Zeros of H(s) become poles of H_inv(s). If H(s) has zeros in the right-half plane, the inverse system H_inv(s) will have poles in the right-half plane, making the inverse system unstable. Such systems are called non-minimum phase and their inversion requires special handling.

Key Formula

Invertibility condition for LTI: H_inv(s) = 1/H(s) exists and is well-defined.

Cascade condition: H(s) * H_inv(s) = 1, so h(t) * h_inv(t) = delta(t).

For discrete LTI: H_inv(z) = 1/H(z). The system is invertible if H(z) has no zeros at z=0 or z=infinity that would make 1/H(z) improper.

Problem: Find the inverse system for H(s) = (s+2)/(s+5).

Step 1 — Apply invertibility formula:
  H_inv(s) = 1/H(s) = (s+5)/(s+2)

Step 2 — Check stability of inverse:
  H_inv(s) has pole at s = -2 (left-half plane -> stable)
  H_inv(s) has zero at s = -5

Step 3 — Verify cascade:
  H(s)*H_inv(s) = [(s+2)/(s+5)] * [(s+5)/(s+2)] = 1
  Cascade = 1 -> identity system -> input is recovered.

Step 4 — Check if H(s) has right-half plane zeros:
  Zero of H(s) at s = -2 (left-half plane) -> minimum phase
  Inverse is stable.

Final Answer: H_inv(s) = (s+5)/(s+2)
System is invertible and the inverse is BIBO stable.

Exam Tip: A system with a zero at the origin, like H(s) = s/(s+1), has H_inv(s) = (s+1)/s which contains an integrator pole at s=0. This makes the inverse system marginally stable, not BIBO stable. For GATE problems, always check both the existence of 1/H(s) and whether its poles are strictly in the left-half plane before claiming the inverse is realizable and stable.

Properties Summary

• Invertibility: distinct inputs map to distinct outputs; no two inputs produce the same output.
• LTI inverse: H_inv(s) = 1/H(s); poles of inverse are zeros of original system.
• Non-minimum phase: zeros of H(s) in right-half plane make H_inv(s) unstable.
• Non-invertible example: y(t) = x^2(t); both x(t)=+A and x(t)=-A produce same output.
• Non-invertible example: y[n] = 0 for all n (zero system); all inputs map to zero output.
• Delay system y(t)=x(t-T) is invertible; its inverse is an advance y(t)=x(t+T), which is non-causal.

Quick Revision

• Invertible: one-to-one mapping from input to output.
• LTI inverse system transfer function: H_inv = 1/H.
• Zeros of H become poles of H_inv.
• Right-half plane zeros in H make the inverse unstable (non-minimum phase system).
• The identity system has impulse response delta(t); H*H_inv must equal delta(t).
• A delay is invertible but its causal inverse (advance) is non-causal.
• Exam trap: students say every LTI system is invertible; a system with H(s)=0 or a system with repeated zeros collapses multiple inputs to the same output and cannot be inverted.