# An invertible system with memory

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.

• Regarding the memory of the differentiator, see dsp.stackexchange.com/questions/58533/… – MBaz Oct 9 '20 at 18:21
• The inverse of a time delay is a time advance, right? Does the time advance have memory? – MBaz Oct 9 '20 at 18:27
• @MBaz Thanks. I've seen that. My question is more general. Also there are many answers in that link which really confuses me. – S.H.W Oct 9 '20 at 18:33
• @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. – S.H.W Oct 9 '20 at 18:35
• Well, I don't agree with that definition. Seeing into the future is not "memory", it is "non-causality". – 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
• 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? – S.H.W Oct 11 '20 at 20:59
• 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 – Laurent Duval Oct 12 '20 at 21:30
• 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 $x[n]$. – S.H.W Oct 12 '20 at 21:39
• 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 – Laurent Duval Oct 12 '20 at 21:46