what i know is that for a system to be invertibel it should be one-one , but I am confused that if i am given a transfer function of a LTI system how can I prove or verify if it is invertible.

Another query is that can a system has more than one inverse systems , sometime before I used to think that inverse pairs are unique but just solving an example of finding inverse of differentiation as LTI system i came to result that it has got two inverses




3 Answers 3


In general LTI System is invertible if it has neither zeros nor poles in the Fourier Domain (Its spectrum). The way to prove it is to calculate the Fourier Transform of its Impulse Response.

The intuition is simple, if it has no zeros in the frequency domain one could calculate its inverse (Element wise inverse) in the frequency domain.

Few remarks for the practice world:

  1. If for a certain system we know the input signals are limited to certain band (Or zones) in frequency domain it is enough the LTI system has no zeros in this zone only.
  2. In practice we have noise hence even values which are not mathematically zero but close to 0 (Or even low relative to the gain of the LTI system, Say ~-60 [dB]) are making the system not invertible.

Look for many question in this community about Deconvolution and Inverse Problems.

  • 1
    $\begingroup$ Even systems with no zeros on the the imaginary axis might be not invertible by a causal and stable system. $\endgroup$
    – Matt L.
    Jun 25, 2019 at 13:15
  • $\begingroup$ He didn't mention casual or stable system to this matter. The OP only mentioned LTI. $\endgroup$
    – Royi
    Jun 25, 2019 at 14:31
  • $\begingroup$ That's true, but an inverse system that is unstable is generally not very useful, that's what I tried to explain in my answer. BTW, wasn't me downvoting you. $\endgroup$
    – Matt L.
    Jun 25, 2019 at 15:07
  • $\begingroup$ Of course you are right. I know it wasn't you :-). By the way it was me for you +1 :-). Regarding general and real life, I'd say that usually you care about small part of the domain and then you have much more freedom. Text book wise, you're completely right. $\endgroup$
    – Royi
    Jun 25, 2019 at 15:13
  • $\begingroup$ I know what you mean, but if we say a system is invertible we don't want to make restrictions with regard to the input signal. I addressed the example of the differentiator in my answer, which is strictly speaking not invertible, unless we don't care about the constant. So one could say it's invertible for zero mean signals. That's probably what you mean. $\endgroup$
    – Matt L.
    Jun 25, 2019 at 15:17

You have to be clear what you mean by "invertible". Commonly, you want the inverse system to be causal and stable, and that puts certain restrictions on the original system.

In the case of systems with rational transfer functions, you just have to look at the zeros of the transfer function, because they become the poles of the inverse system. If all zeros of the original system lie in the left half plane (for continuous-time systems), or inside the unit circle (for discrete-time systems), then the system is invertible, and its inverse system is causal and stable. Such systems are called minimum-phase systems. Minimum-phase systems can be inverted by causal and stable systems. Note that zeros on the imaginary axis (for continuous-time systems) or on the unit circle (for discrete-time systems) are not allowed. Sometimes such systems are called strictly minimum-phase.

The differentiator mentioned in your question is an example of a system that cannot be inverted. Since the differentiator has a zero at DC you cannot recover the DC value of the input signal. This is equivalent to the fact that the integral of a function is determined up to a constant. Of course, in practice one might be happy with recovering the signal up to an unknown constant, and in this case an integrator with impulse response $h(t)=u(t)$ is a good choice. Note that all systems with impulse response $h(t)=u(t)+C$ with some constant $C$ could be used, and the other one you found is just $-u(-t)=u(t)-1$. But only the one with $C=0$ is causal, and none of them is stable.

  • $\begingroup$ If I don't pose any practical requirement of both system and its inverse being causal and stable then how can i state using mathematical manipulations that a particular system is invertible or not.Consider the point that you mentioned with differentiator now i have read that for inverse pairs u*i=impulse and i was able to do so for both u(t) and -u(-t) ;I didn't thought abut point of DC value you mentioned. $\endgroup$
    – Buzz bee
    Jun 25, 2019 at 17:41
  • $\begingroup$ What I am seeking for is an algorithm (as I have just started) to proceed for proving if system is invertible when given system is complex.For learning purpose I don't wish to consider that both systems has to be realizable ( causal and stable).Will you plz. help for it because before your answer I was readily sure that differentiator is invertible system. $\endgroup$
    – Buzz bee
    Jun 25, 2019 at 17:45

The term LTI system is a bit broad, so perhaps restricting ourselves to single input-single output systems makes sense. Let's just look at $s$ (Laplace) for now. The $Z$ transform follows in a straightforward way.

Also, if the system is not stable, a practical inverse filter will not recover the input. Convolution commutes so following a stable filter with an unstable inverse is equally fraught.

We might also ignore situations where one can apply filters in both time directions. Both filters are causal or anti causal.

If you have the transfer function $H(s)$,then $G(s)$ such that $G(s)H(s)=1$ constitutes an invertible system. $$ H(s) =\frac{a_0+s a_1 + s^2 a_2 + \dots a_{N-1} s^{N-1}+ a_{N} s^N} {b_0+s b_1 + s^2 b_2 + \dots b_{M-1} s^{M-1}+ b_{M} s^M}= \frac{\prod_{k=0}^{N-1} (s-\lambda_k)}{\prod_{n=0}^{M-1}(s-\gamma_n)} $$ so $$ G(s)=\frac{b_0+s b_1 + s^2 b_2 + \dots b_{M-1} s^{M-1}+ b_{M} s^M} {a_0+s a_1 + s^2 a_2 + \dots a_{N-1} s^{N-1}+ a_{N} s^N}= \frac{\prod_{n=0}^{M-1}(s-\gamma_n)}{\prod_{k=0}^{N-1} (s-\lambda_k)}. $$ but as a practical matter, if $H(s)$ is stable (and causal), $G(s)$ should also be stable (and causal) which brings us to minimum phase filters.

So while the poles $\gamma_n$ need to be stable (left side of the s plane) for $H(s)$ to be stable, the zeros of $H(s)$ , $\lambda_k$ also need to be in the left side of the s plane for $G(s)$ to be stable. For the discrete time case, the same is true of the poles and zeros of the Z transform, they need to be interior to the unit circle. If we have anticausal filter, all the poles and zeros are outside the unit circle, which explains, you dual solutions.

Another approach, this time for discrete time systems, a filter can be expressed in terms of the impulse response $h[0] h[1] \dots h[N-1]$.

A filter can be expressed as a (possibly infinite) matrix $H$ vector $\mathbf{x}$ product: $$ \mathbf{y}=\left| \begin{array}{cccccc} h[0] & 0 & 0 & 0 & 0 & \dots\\ h[1] & h[0] & 0 & 0 & 0 & \dots \\ h[2] & h[1] & h[0] & 0 & 0 & \dots \\ \vdots & \ddots & \ddots & \ddots & \dots & \\ h[N-1] & \dots & \\ 0 & \ddots & \ddots & \ddots & \ddots && \end{array}\right| \mathbf{x} $$ If $H$ has an inverse , $\mathbf{x}$ can be recovered from $\mathbf{y}$ one obvious condition is $h[0]\ne 0$.

as an example

 h= [ 1 1 1 1 0 0 0 0]
h =
     1     1     1     1     0     0     0     0
>> H=toeplitz(h,[h(1) zeros(1,length(h)-1)])
H =
     1     0     0     0     0     0     0     0
     1     1     0     0     0     0     0     0
     1     1     1     0     0     0     0     0
     1     1     1     1     0     0     0     0
     0     1     1     1     1     0     0     0
     0     0     1     1     1     1     0     0
     0     0     0     1     1     1     1     0
     0     0     0     0     1     1     1     1
>> inv(H)
ans =
     1     0     0     0     0     0     0     0
    -1     1     0     0     0     0     0     0
     0    -1     1     0     0     0     0     0
     0     0    -1     1     0     0     0     0
     1     0     0    -1     1     0     0     0
    -1     1     0     0    -1     1     0     0
     0    -1     1     0     0    -1     1     0
     0     0    -1     1     0     0    -1     1

in this case, the inverse is also Toeplitz, thus corresponding to a FIR filter. Toeplitz matrices don’t generally have Toeplitz inverses.

  • $\begingroup$ Out of curiosity: I know these arguments for LTI systems that have rational transfer functions. This is what we treat in standard signal and systems literature. But can we say anything about LTI systems that do not have a rational transfer function? I guess not, right? $\endgroup$
    – Florian
    Jun 25, 2019 at 14:24
  • $\begingroup$ I'm not sure the multiplication being equal to 1 means invertible. In some cases one system could have a zero the other system can vanish. Hence their multiplication is one yet cascading them won't invert the signal. $\endgroup$
    – Royi
    Jun 25, 2019 at 14:33
  • 1
    $\begingroup$ @Florian, LTI that doesn't have a rational transfer function, like a delay in continuous time, which has an infinite number of terms. There are rational approximations that mimic infinite order systems. For delay, advance would be the inverse but that isn't causal. $\endgroup$
    – user28715
    Jun 25, 2019 at 14:46
  • $\begingroup$ @Royi, so when is a composition of minimum phase transfer functions, equal 1, not invertible? $\endgroup$
    – user28715
    Jun 25, 2019 at 14:55
  • $\begingroup$ I don't think you mention you talk about minimum phase transfer functions. You write: "If you have the transfer function $H(s)$,then $G(s)$ such that $G(s)H(s)=1$ constitutes an invertible system.". Then giving a trivial case to disprove this is easy. Create a polynomial with zero at $ {\omega}_{0} $ and create a rational system with pole at $ {\omega}_{0} $. Minimum phase is the way to exclude such cases. But you don't mention this. $\endgroup$
    – Royi
    Jun 25, 2019 at 15:04

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