Timeline for An invertible system with memory
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2020 at 23:47 | comment | added | S.H.W | It's a real shame for me that I don't get this part of your answer still. Anyway, thank you so much. | |
| Oct 12, 2020 at 23:45 | vote | accept | S.H.W | ||
| Oct 12, 2020 at 21:46 | comment | added | Laurent Duval | 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 | |
| Oct 12, 2020 at 21:39 | comment | added | S.H.W | 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 $x[n]$. | |
| Oct 12, 2020 at 21:30 | comment | added | Laurent Duval | 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 | |
| Oct 11, 2020 at 20:59 | comment | added | S.H.W | 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? | |
| Oct 11, 2020 at 20:19 | history | answered | Laurent Duval | CC BY-SA 4.0 |