I have been trying to figure this out for a while now. Everywhere I have looked I could easily find examples of invertible LTI systems, but I could not find any counterexamples. Can anybody shed some light on this for me?

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    $\begingroup$ What do you mean by invertible? Is this continuous or discrete? $\endgroup$ – copper.hat Nov 11 '19 at 5:07
  • $\begingroup$ i would say that, if stability remains a requirement for an LTI system to be invertable, then any continuous-time LTI system, $H(s)$, with any zeros in the right half-plane is not invertable. nor would any discrete-time LTI system, $H(z)$, having zeros outside the unit circle be invertable. $\endgroup$ – robert bristow-johnson Nov 11 '19 at 19:12
  • $\begingroup$ @copper.hat: They probably mean a system that can transform the output back into the input. $\endgroup$ – user541686 Nov 11 '19 at 22:50

You need to define what you mean by "invertible". Do you mean invertible by a causal and stable system? If yes, then any system that is not minimum-phase is not invertible (because the inverse system can't be causal and stable).

Example of a system that cannot be inverted by a causal and stable system: a simple delay $y(t)=x(t-T)$, $T>0$, could only be inverted by a non-causal system.

If by "invertible" you mean a system that can be inverted by a system that is possibly not causal and/or not stable, then it's still straightforward to find systems that can't be inverted according to this broader criterion: just take a system with one or more zeros in its frequency response. The information about the input signal at the frequencies where the system's frequency response is zero cannot be recovered by any other system.

Example of a system that cannot be inverted by any system: an ideal low-pass filter cannot be inverted by any system, because any information above the filter's cut-off frequency is lost and cannot be recovered.

EDIT: the latter definition of invertibility (where causality and stability of the inverse system are ignored) is just the definition of an injective mapping. In terms of systems this means that no two distinct input signals generate the same output signal. This category is what is also described in Laurent Duval's answer. Note, however, that in practice we usually prefer inverse systems that can also be realized, i.e., we require stability and causality, and then the first definition of invertibility given above is appropriate.


A necessary condition for invertibility is that any output has only one possible input (or injectivity, as proposed in comments). Since we are looking at counterexamples, we can look at when this condition is not satisfied.

The null system, that turns every signal into a zero flat line, is not invertible, but a bit trivial.

A system that computes a discrete derivative, eg with impulse response $[1 \, -1]$, is not invertible in that sense, as for any signal $s[n]$, and constant $c$, all signals $s[n]+c$ have the same output discrete derivative.

On the theoretical side, since LTI systems are characterized by their eigenvectors/eigenvalues, an LTI system with a zero eigenvalue lacks one necessary condition for being invertible.

Matt's answer (and his former one How to determine if the system is invertible) is more detailed in terms of "realizability".

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    $\begingroup$ I'm sure you know this, but: what you're describing is actually just injectivity. For invertibility you also need surjectivity, though for linear ℝⁿ-endomorphisms that already follows from injectivity. $\endgroup$ – leftaroundabout Nov 11 '19 at 10:30
  • $\begingroup$ You are very right. I made classical the shortcut that an injective function is bijective on its image Is every injective function invertible? $\endgroup$ – Laurent Duval Nov 11 '19 at 10:38

Whether LTI or not all systems are invertible if

unique (distinct) inputs produce unique (distinct) outputs

Causality and stability are later concerns for making sense of the obtained inverse system.

For example the inverse to the delay system

$$y[n] = x[n-d] $$


$$y[n] = x[n+d] $$

Which is clearly noncausal for $d > 0$, and is not realizable in real-time processing. But this system will be perfectly implemented in on offline audio processing console. Furthermore it can even be implemented in real-time image processing where the index is not time but space and causality does not restrict realizbility.

If you are into stable realizations (which makes sense), then your inverse system should include the unit circle in its Fourier transform. One simple conclusion for FIR filters is that if the forward filters frequency responses include zeros on the unit circle then the inverse filter will be unstable.

The following non-LTI system

$$y[n] = g[n] x[n]$$

is invertible iff $g[n] \neq 0$ for all $n$. Otherwise it's not invertible if for a set of $n$ $g[n] = 0$.

The nonlinear system

$$y[n] = x[n]^2 $$

is not invertible, whereas the nonlinear system

$$ y[n] = x[n]^3 $$

is invertible.

For LTI systems with rational transfer functions

$$ H(Z) = \frac{B(z)}{A(z)} $$

the inverse system will be

$$H_i(z) = \frac{1}{H(z)} = \frac{A(z)}{B(z)} $$

Poles and zeros are replaced and stability and causality of inverse system is analysed based on these new poles.


In addition to all the answers that are correct in a mathematical sense, in a practical sense, a system whose frequency response goes below some finite but small-enough value will not be usefully invertable, even if a simple mathematical analysis would suggest that it is.

In frequency-domain terms, the frequency response of a system's inverse will have gain equal to the reciprocal of the system gain at each frequency point. In practical terms, once this gain gets too great, all you will be doing is amplifying noise rather than actually contributing to a reasonable estimate of the input signal.

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    $\begingroup$ and these kind of considerations open a door of discussions into the huge body of system inversion... $\endgroup$ – Fat32 Nov 11 '19 at 20:29

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