Timeline for What does it mean for a transfer function matrix to be real, rational, and stable?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 15, 2022 at 20:43 | vote | accept | mhdadk | ||
| Sep 15, 2022 at 20:21 | comment | added | Matt L. | See my answer below. | |
| Sep 15, 2022 at 20:14 | answer | added | Matt L. | timeline score: 2 | |
| Sep 15, 2022 at 18:56 | comment | added | mhdadk | @MattL. that makes sense. It seems then that my understanding of "real" is wrong here. Any idea what it could mean instead? I'm thinking that it means that all the poles of each $G_{ij}(s)$ are real, but I'm not sure if that's right. | |
| Sep 15, 2022 at 18:50 | comment | added | Matt L. | No, but how could it be real for any complex-valued argument? | |
| Sep 15, 2022 at 18:13 | comment | added | mhdadk | @MattL. No I meant $s_0 \in \mathbb C$. My understanding is that a transfer function matrix $\mathbf G(s)$ can be defined as the function $\mathbf G : \mathbb C \to \mathbb C^{m \times n}$. Is my understanding wrong? | |
| Sep 15, 2022 at 18:09 | comment | added | Matt L. | Don't you mean $s_0\in\mathbb{R}$? | |
| Sep 15, 2022 at 17:59 | history | asked | mhdadk | CC BY-SA 4.0 |