# User talk:مبتدئ

{{unblock|reason given for blocking me is user name. None of the criterias listed in the section called inappropriate usernames applies to me. Please unblaock! Best Regards مبتدئ (talk) 23:03, 2 October 2008 (UTC)}}

Are you sure this is the username you want? Bear in mind that, per [1], many users of English Wikipedia will find your name unreadable on their screens. I see it as %D9%85%D8%A8%D8%AA%D8%AF%D8%A6. -FisherQueen (talk · contribs) 23:12, 2 October 2008 (UTC)
This user has SUL and is an established user on other wikis - see sul:مبتدئ. I've helped them set up an English signature over IRC and believe that this satisfied the non-latin requirements in WP:U I'm inclined to unblock this user. east718 // talk // email // 23:16, 2 October 2008 (UTC)

test [[user:مبتدئ|مبتدئ]] ([[user talk:مبتدئ|talk]]) (talk) 23:16, 2 October 2008 (UTC)

It's fine with me if this is the username this user wants; it isn't against the rules. For what it's worth, I see the name correctly in the page title and in her signatures, but incorrectly in the template. -FisherQueen (talk · contribs) 23:19, 2 October 2008 (UTC)
Cool, I've unblocked them. I've noticed that sort of voodoo with the unblock template before, I guess it's just some background escaping we have going on in the template code or something. Welcome to the English Wikipedia, مبتدئ! east718 // talk // email // 23:23, 2 October 2008 (UTC)

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test مبتدئ (talk) 23:24, 2 October 2008 (UTC)

## Minimum phase

Hi, I have partially reverted your edits to minimum phase as they appear to contradict a textbook that I have. I added a reference to this source. Also, could you clarify and reference your discussion about minimum phase in nonlinear systems? I am not familiar with this terminology and would appreciate a reference for this. Thanks, and welcome to English Wikipedia! --Zvika (talk) 07:42, 7 December 2008 (UTC)

Hi zvika
can you clarify in which extent what i wrote contradicts a books that you have? may be i m wrong i dont know. As far as i know minimum phased systems means (roughly speaking) that the inverse system is stable. And as you know a system is also stable when it has a pole at 0 (2 poles at zero means instable). Now in the case of linear systems where the transfer functions are expressed in terms of polinomials this means you can write the numerator as s*(s-a)*(s-b)*..... . You can easly construct a system which is not proper and fullfill this req. It is clear that non proper systems do not make any physical sens. Concerning the zerp dynamic: it is the dynamic remaning in the system and which is not observable at the output. For a fully observable system you will see the hole dynamic. For an only partially observable system some dynamic may be hidden. As you know the concept and poles is not defined for nonlinear systems. But there is an extension which extend many of the concepts of linear systems to nonlinear ones. In the case of zeros it is the zerodynamic. The connection is that if you linearize your zero dynamic and find its poles this will be the same as the zeros of the linearized system.
best regards مبتدئ (talk) 19:53, 7 December 2008 (UTC)

Thanks for the detailed response. The book I cited (Kailath) says: "A LTI system ${\displaystyle H(z)}$ is called minimum-phase if both it and its inverse, ${\displaystyle H^{-1}(z),}$ are stable and causal." (Kailath et al., Linear Estimation, p.193.) This is exactly the definition currently given in the article, from which you removed the "causal" part, which I then restored.
All the best, --Zvika (talk) 07:35, 8 December 2008 (UTC)
Sorry to butt in here, but I thought I might have something to add. It's better to define minimum phase in terms of stable poles and "stable zeros." If a system has stable poles AND stable zeros, then it MUST be causal. That is, causality is a CONSEQUENCE of being minimum phase. Consider the system...
${\displaystyle {\frac {1}{s+5}}}$
The system is clearly stable because all of its poles are negative. However, it has a zero at ${\displaystyle s=\infty }$, and so it is not minimum phase. As a consequence, the inverse system:
${\displaystyle s+5}$
is not causal. Note that the inverse system has a pole at ${\displaystyle s=\infty }$, which shows that it's not stable either. And so the casual/stable semantics go together. Put another way, we have to have a pole for every zero and vice versa for the system to be minimum phase. —TedPavlic | (talk) 00:14, 30 December 2008 (UTC)
Likewise, for discrete-time systems, you'll always get it right if you put it in terms of stable poles (inside the unity circle) and "stable zeros" (inside the unit circle). If you have an unmatched pole or zero, then one of the systems is guaranteed to have an unstable pole. —TedPavlic | (talk) 00:17, 30 December 2008 (UTC)
Further, you should keep in mind that the phrase minimum phase has less meaning when there are more poles than zeros or vice versa. When we say that a system is minimum phase, that means that of all of the transfer functions with identical magnitude response curves, this transfer function has the smallest phase shift. For example, the transfer functions
${\displaystyle {\frac {s+10}{s+5}}}$ and ${\displaystyle {\frac {s-10}{s+5}}}$
are both stable and both have the same magnitude response. However, the first transfer function has less of a contribution to the phase shift than the second one. The first transfer function has no phase shift at ${\displaystyle \omega =0}$ and no phase shift as ${\displaystyle \omega \to \infty }$. However, the second transfer function has a ${\displaystyle -180^{\circ }}$ phase shift at ${\displaystyle \omega =0}$ and no phase shift as ${\displaystyle \omega \to \infty }$. Hence, the second transfer function has the maximum phase contribution and the first transfer function has the minimum phase contribution. These terms really only make sense when you have the same number of poles and zeros. Sure, you could apply the same reasoning when you have several poles and zeros but one fewer zero than pole, but that's not as useful. —TedPavlic | (talk) 13:35, 30 December 2008 (UTC)
One last try at something that might help. Consider simple unity-gain negative feedback around a minimum phase system (i.e., a system that has the same number of zeros and poles and has all zeros and poles in the LHP). By increasing the error gain, you can be sure that the closed-loop poles will eventually all be LHP. This is one of the reasons why control is much easier with minimum-phase systems. With maximum-phase systems, you get strange delayed phenomena. For example, you might start a process that will eventually lead to a higher output, but immediately after applying your control, the output gets much lower. This kind of delayed response can drive a system to instability. If you consider your root locus, the instability coincides with the closed loop poles moving out of the LHP into the RHP to meet up with the non-minimum phase zeros. Using this example, can you see why the designation of minimum-phase makes the most sense when you have an equal number of poles and zeros (i.e., when both the system and its inverse are causal)? —TedPavlic | (talk) 13:52, 30 December 2008 (UTC)

## Nonlinear observability

A little while ago, you left a comment on my talk page about nonlinear observability. I have finally responded there. Could you take a look and give me your response? Thanks. —TedPavlic | (talk) 23:59, 29 December 2008 (UTC)

You once asked me to add some content on nonlinear observability. With help of Professor Drakunov, the state observer page has been enhanced to include a section on state observers for non-linear systems. The section on sliding mode observers goes into a lot of detail. Additionally, a while ago I added a sliding mode observer example to the sliding mode control page. I recommend you give those pages a look. —TedPavlic (talk/contrib/@) 23:03, 6 July 2009 (UTC)