---
title: "Invertibility of LTI systems"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:lti-invertibility-condition"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/lti-invertibility-condition
site: BCT Engineering
---

# Invertibility of LTI systems

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2078 Baisakh, 5 marks
- Video: [Invertible LTI system](https://www.youtube.com/watch?v=V20LmLEcHlE) — Shrenik Jain
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/lti-invertibility-condition
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

Explain invertibility of LTI system and derive the necessary condition for two LTI systems to become inverse of each other.

## Short answer

An LTI system is invertible if distinct inputs give distinct outputs, so an inverse recovers the input. Cascade identity requires h(t)*h1(t)=delta(t) or H(jw)H1(jw)=1. Hence invertible iff H(jw) is nonzero at all frequencies.

## Answer

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**.

```text
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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. When is a system invertible?
   A. When its output equals its input
   B. When it is causal
   C. When it has finite memory
   D. When distinct inputs give distinct outputs, so the input can be recovered
   Answer: D. Invertibility means the input can be uniquely recovered from the output. (Key point: Invertibility meaning)

2. For an LTI system h(t) and its inverse h1(t) in cascade, what must hold?
   A. h(t)*h1(t) = δ(t)
   B. h(t)*h1(t) = u(t)
   C. h(t)h1(t) = 1
   D. h(t)+h1(t) = δ(t)
   Answer: A. The cascade must be the identity, and δ(t) is the identity for convolution. (Key point: Condition in time domain)

3. What is the inverse's frequency response?
   A. H1(jω) = H(jω)
   B. H1(jω) = 1/H(jω)
   C. H1(jω) = -H(jω)
   D. H1(jω) = H(-jω)
   Answer: B. Convolution becomes multiplication, so H·H1 = 1. (Key point: Frequency domain)

4. Which system is not invertible?
   A. Pure delay δ(t-t0)
   B. Accumulator
   C. Ideal LPF, since stopband frequencies are lost
   D. Identity system
   Answer: C. H=0 in the stopband destroys information that cannot be recovered. (Key point: Non-invertible example)

## Related in Signals and Systems

- [Short note: Invertibility of LTI system](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Ashort-note-invertibility) — Signals and Systems, chapter 5
- [Causality condition for CT LTI system](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Acausal-ct-lti-condition) — Signals and Systems, chapter 5
- [Condition for LTI system to be causal](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Alti-causality-condition) — Signals and Systems, chapter 6, asked 3×
- [Cascade of two CT LTI systems](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Acascade-impulse-response) — Signals and Systems, chapter 5
- [LTI properties; BIBO stability condition](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Alti-bibo-stability) — Signals and Systems, chapter 6
- [Properties of DT LTI systems](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adt-lti-properties) — Signals and Systems, chapter 6, asked 3×

## Related in other subjects

- [LTI system and closed loop in modern systems](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alti-define-cl-pronounced) — Control Systems, chapter 1
- [GFS vs other file systems](https://www.bctengineer.com/question.md?key=semester_8%2Fbig_data_technologies%3Agfs-differ-other-fs) — Big Data Technologies, chapter 2
- [Condition for matrix inverse to exist](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Acondition-for-matrix-inverse-to-exist) — Mathematics III, chapter 1, asked 6×
- [Condition for matrix inverse to exist](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Acondition-for-matrix-inverse-to-exist-two) — Mathematics III, chapter 1, asked 6×