Timeline for How to prove these two definitions of the minimum phase transfer function are same?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 29, 2018 at 2:16 | vote | accept | Kim Jaewoo | ||
| Nov 26, 2018 at 12:10 | history | edited | Matt L. | CC BY-SA 4.0 | added 19 characters in body |
| Nov 26, 2018 at 9:42 | history | edited | Matt L. | CC BY-SA 4.0 | added 5 characters in body |
| Nov 26, 2018 at 9:34 | comment | added | Matt L. | @KimJaewoo: But this is equivalent to saying that the phase lag is minimum for a minimum-phase system, because the maximum phase lag is just the maximum range of the phase assuming that the phase at $\omega=0$ equals zero, which can always be achieved by choosing the sign of $H(\omega)$ accordingly. | |
| Nov 26, 2018 at 9:25 | comment | added | Kim Jaewoo | I was wrong about my inequality. The inequality I want to know is why $$\sup_{\omega} {\angle H_m(j\omega)}-\inf_{\omega} {\angle H_m(j\omega)}<\sup_{\omega} {\angle H(j\omega)}-\inf_{\omega} {\angle H(j\omega)}$$ | |
| Nov 26, 2018 at 8:05 | comment | added | Matt L. | I'm not sure I understand your comment, but your inequality seems wrong to me. It's just the minimum-phase system that has the minimum range of phase. But maybe I don't see what you mean. What is $H_m(j\omega)$? I assume you mean the phase response. | |
| Nov 26, 2018 at 1:36 | comment | added | Kim Jaewoo | I understood about the additional phase lag. I was wondering why $$\sup_{\omega}H_m(j\omega)-\inf_{\omega}H_m(j\omega)>\sup_{\omega}H(j\omega)-\inf_{\omega}H(j\omega) $$, because the phase angle value can be larger than the value when $\omega=0$ and be smaller than the value when $\omega=\infty$. Thanks. | |
| Nov 25, 2018 at 19:40 | history | answered | Matt L. | CC BY-SA 4.0 |