Suppose $\mathcal{L}$ be invertible system with memory. Does $\mathcal{L}^{-1}$ have memory necessarily?

Intuitively I think the answer is "yes". There are many examples showing that. For instance $\mathcal{L}(x(t)) = x(t-2)$ and $\mathcal{L}(x(t)) = x(\frac t 3)$. Another example which seems problematic to me is $$\mathcal{L}(x(t)) = \int_{-\infty}^{t}x(\lambda)d\lambda$$The inverse is $$\mathcal{L}^{-1}(x(t)) = \frac{dx(t)}{dt}$$Does differentiator have memory? Of course the main question here is about memory of an invertible system which has memory. Note that here $\mathcal{L}$ can be nonlinear as well.

For clarity, I add some related definitions from Oppenheim's book:

Invertible system: A system is said to be invertible if distinct inputs lead to distinct outputs.

Causal system: A system is causal if the output at any time depends only on values of the input at the present time and in the past.

Memoryless system: A system is said to be memoryless if its output for each value of the independent variable at a given time is dependent only on the input at that same time.

  • $\begingroup$ Regarding the memory of the differentiator, see dsp.stackexchange.com/questions/58533/… $\endgroup$
    – MBaz
    Oct 9 '20 at 18:21
  • $\begingroup$ The inverse of a time delay is a time advance, right? Does the time advance have memory? $\endgroup$
    – MBaz
    Oct 9 '20 at 18:27
  • $\begingroup$ @MBaz Thanks. I've seen that. My question is more general. Also there are many answers in that link which really confuses me. $\endgroup$
    – S.H.W
    Oct 9 '20 at 18:33
  • $\begingroup$ @MBaz "A system is said to be memoryless if its output for each value of the independent variable at a given time is dependent only on the input at that same time." This is the definition which I'm using. So time advance have memory. $\endgroup$
    – S.H.W
    Oct 9 '20 at 18:35
  • $\begingroup$ Well, I don't agree with that definition. Seeing into the future is not "memory", it is "non-causality". $\endgroup$
    – MBaz
    Oct 9 '20 at 18:37

For time-based systems, I understand that it is difficult to imagine a memory of the future. But for general systems, $-t$ and $t$ are just left and right (think of a spatial system). Other discussions are in LTI system $y(t)=x(t−T)$ with or without memory, What is a memory less system?, or A question about the concept of the time.

By definition of invertibility, $\mathcal{L}^{-1}$ is such that $\mathcal{L}^{-1}( \mathcal{L}(x))=x$. But also that $\mathcal{L}( \mathcal{L}^{-1}(x))=x$ (by the way, derivatives and integrals are not inverses). Let us suppose the converse: $ \mathcal{L}^{-1}$ has no memory. Hence $\mathcal{L}^{-1}(x[n]))$ can only use the present state, and $\mathcal{L}$ as well to yield $(x[n])$.

So, if $\mathcal{L}^{-1}$ is memoryless, $\mathcal{L}$ is memoryless as well. By contraposition, the converve is true

  • $\begingroup$ Thanks. I don't understand why $\mathcal{L}$ has to have memory in order to $\mathcal{L}( \mathcal{L}^{-1}(x[n]))=x[n]$ holds. Would you elaborate, please? $\endgroup$
    – S.H.W
    Oct 11 '20 at 20:59
  • $\begingroup$ I used an argument based on logic. It is not constructive in the common sense. I suppose the converse. It entails that the initial hypothesis on $\mathcal{L}$ cannot be verified. Hence, my initial supposition is false $\endgroup$ Oct 12 '20 at 21:30
  • 1
    $\begingroup$ I see. I don't understand the part "and $\mathcal{L}$ as well". Why $ \mathcal{L} $ can only use the present state to yield $x[n]$? Maybe it uses future or past values of the input as well and still yields $x[n]$. $\endgroup$
    – S.H.W
    Oct 12 '20 at 21:39
  • $\begingroup$ Because (from my hypothesis) $\mathcal{L}^{-1}$ has no memory. So $\mathcal{L}^{-1}(x)$ can only use the present state. Hence, $\mathcal{L}$ is only given sometimes related to the present state $\endgroup$ Oct 12 '20 at 21:46
  • $\begingroup$ It's a real shame for me that I don't get this part of your answer still. Anyway, thank you so much. $\endgroup$
    – S.H.W
    Oct 12 '20 at 23:47

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