Timeline for Check whether a system has memory or not
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 17, 2021 at 2:24 | vote | accept | LM2357 | ||
| Oct 16, 2021 at 13:10 | history | edited | Matt L. |
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| Oct 16, 2021 at 12:51 | answer | added | Matt L. | timeline score: 1 | |
| Oct 15, 2021 at 16:24 | comment | added | LM2357 | Is it just the fact that the first system can take any input signal and the output would be the same but the second system does depend on our input signal (albeit only at a single point)? I guess, this is what differentiates the two and leads to one having memory and other being memoryless? I would appreciate if you can confirm my reasoning | |
| Oct 15, 2021 at 15:27 | comment | added | LM2357 | Yeah, so what kind of confuses me and seems a little contradicting is that the second system is also a constant value (it is well-defined). So, what actually is it that leads to both systems being in different categories i.e one having memory and the other memoryless | |
| Oct 15, 2021 at 13:21 | comment | added | Dilip Sarwate | Not only does the second system have memory but it also has foresight in that at times $t$ that are smaller than $t_0$, the system already knows that at a future time $t_0$, its input is going to be $x(t_0)$ which value the system is already outputting at time $t<t_0$. | |
| Oct 15, 2021 at 12:56 | comment | added | MBaz | I think you're on the right track; there is no fundamental difference between $x(t_0)$ and $K$ (as long as $x(t_0)$ is actually a well-defined value!). | |
| S Oct 15, 2021 at 12:53 | review | First questions | |||
| Oct 15, 2021 at 13:08 | |||||
| S Oct 15, 2021 at 12:53 | history | asked | LM2357 | CC BY-SA 4.0 |