All three, got a quiz tomorrow morning and I understand almost none of it.
Fight smart, not fair.Okay, I've got my notes and textbook from last semester out. Why don't you ask a specific question, like something from a homework that's likely to be on said quiz that you don't understand. Then we can work from there (considering I haven't exactly prepared a lecture on this).
In general a root locus plot is plot of the roots of a characteritic equation for a system as a "function" of your system gain (usually a value K). Function is in quotes only because it is not a simple thing to plot.
Fear the Gothilolions! | Anime listHow do I figure out the difference between the zeroes, the poles, and what do I do with them would be the first step.
Bed time, see you in the morning. I need a lesson on pretty much the whole thing though.
edited 25th Feb '10 8:44:55 PM by Deboss
Fight smart, not fair.You've got an open-loop transfer function, correct?
Zeroes are the values which would make that function go to zero. So what values of "s" would make the numerator of your transfer function be zero.
Poles are essentially the same thing for the denominator. They make the transfer function go to infinity, which is caused by dividing by zero.
You plot these points on the real-imaginary axis. This is important because even something of the form s^2 + 4 has a corresponding pole/zero (in this case, +/-2i).
Once we have these points plotted,the next step is to graph the root locus itself, as stated earlier, the root locus is a plot of the roots of the characteristic equatin as a function of some parameter, usually a gain K. Plotting these can get... messy. I actually never had to do it by hand, and doing it by hand is a long and complicated process. We just stuck the transfer function into Matlab and made it do the hard part.
My prof did give us a few general rules though:
- Loci alwyas start at poles.
- A loci will always terminate at an open loop zero as K -> inf. A better way to phrase that I think, would be, if you have an O.L. zero, a loci will terminate at it. A loci can go off to any asymtopic value though.
- A loci will alwyas exist to the left of an odd # pole or zero on the real axis. In essence, if you go to the most right hand pole/zero, that becomes pole number 1. A loci will either be coming out of or going into (depending on whether it is a pole or a zero) that point's left side.
Yay wall of text!
EDIT
Wait, don't sleep, I just finished my wall of text!
edited 25th Feb '10 8:57:11 PM by zzzdragon
Fear the Gothilolions! | Anime listFourier question: How can you tell without looking at a signal's graph if it's periodic or aperiodic? Because if it's periodic, I have to use the Fourier series to get the frequency domain equivalent but if it's aperiodic, I have to use a Fourier transform to do the same.
My FF.net accountAre you just looking at raw data? I can't think of any way to figure out the periodicity of that short of trial and error or graphing it and looking for patterns.
If you've actually got the signal's function to work with, that's a different matter. I almost always did when I was learning about Fourier transforms — but I learned about them in a math context, not a signal-processing context.
edited 25th Feb '10 9:42:53 PM by Nornagest
I will keep my soul in a place out of sight, Far off, where the pulse of it is not heard.Got it, I'll look again in the morning to see if I'll remember any of it.
Fight smart, not fair.I forgot to mention this, but it's kind of important. The system is considered stable whenever the value of the roots of the characteristic equation are negative. So any K that makes causes the value of the loci to become positive causes the system to become unstable.
Graphically, your loci curve may arc around and cross the imaginary axis onto the right hand side. The values of K that cause it to cross and remain on the positive side are the values of K that cause the system to become unstable.
edited 26th Feb '10 5:16:35 AM by zzzdragon
Fear the Gothilolions! | Anime listI made it to State level for Windows Network Administration in BPA, guys!
...anyone wanna help me with how best to study for it? Last year, the test mostly covered OLD server, like I Ps. Speaking of, somebody want to explain those to me? How do you know which is a Class A, Class B, and so on? And how do you tell a fake IP apart from a real one?
be warned, user is known affectionately as The Sneakiest BastardThere isn't such a thing as Class A, B or C anymore. Not for a long while, ever since CIDR (Classless Inter-Domain Routing), which was introduced in 1993.
Historically, though, Class A was an IP address with a zero as the first bit, and thus a first octet (byte) of 127 or lower (and not zero). These were the largest networks handed out; there were 126 of them, and each one had three octets for local allocation — sixteen million potential addresses behind it, though in actual fact subnetworking wastes a lot of them.
Class B had a first bit of 1 and a second bit of 0, and used first octets between 128 and 191. The second octet was also part of the network number, leaving two for local allocation — sixteen bits, or just over sixty-five thousand local addresses. There were sixteen thousand-odd possible class B networks.
Class C has a first and second bit of 1 and a third bit of zero, and covered the first octet range of 192-223. Three octets (bytes) were the network number, leaving only one octet for the local addresses. This gave 253 local addresses to use. There were two million-odd possible class C networks.
This scheme wasted a lot of space, and was eventually abandoned. People still use the terms, when talking about sizes of allocations, but really these days you should specify the number of network bits with a slash in front: /8, /16 and /24 match up to classes A, B and C.
The astute will notice that classes A, B and C didn't actually cover the entire address space. The rest was used for other purposes, such as multicast.
Companies or institutions who've been on the net for a long time may still have allocations that conform to the old scheme; for instance, the university I work at uses 128.125.x.x for much of their stuff, which is a Class B under the old rules (first octet between 128 and 191).
Addendum: This document shows the current top-level I Pv 4 allocations. You'll notice that a fair few US megacorporations still have Class A-sized allocations of their own, though less than they used to. Lots of American defense networks, too.
edited 26th Feb '10 11:46:32 PM by Morven
A brighter future for a darker age.(stares at Morven) ...Please marry me, like, right now. Thanks for the help- the tests last year talked about a lot of older stuff like Kerberos authentication and I Ps a lot, so I figured I'd need help.
(curls up with the Addendum- well, there goes the "more than 3 hours of sleep tonight" plan)
be warned, user is known affectionately as The Sneakiest BastardAny idea how to do the anti-derivative of (3x^2)^2? Where I got:
Anti-derivative of (3x^2)^2
= anti-derivative of the whole, divided by the anti-derivative of its inside?
{[(3x^2)^3]÷3} ÷ x^3
= 27x^6 ÷ 3 ÷ x^3
= 9x^3
According to my notes, it's actually....something else. Not having a math refresher included with this course really didn't help.
edited 1st Mar '10 8:46:28 PM by Barcode711
Worshipper of Ahura Mazda, as proclaimed by Zoroadster http://twitter.com/bpglobalprYou mean (3x2)2? That equals 9x4, so just calculate the antiderivative of that. 9/5 * x5, I believe.
[1] This facsimile operated in part by synAC.@Barcode: You can't do that. Tzetze's method is the way to do it.
edited 1st Mar '10 8:49:37 PM by Ironeye
I'm bad, and that's good. I will never be good, and that's not bad. There's no one I'd rather be than me.^^Notes suggested the same. Got the right answer. Not exactly sure why I can't do it the original way, however. Guess I'll ask tomorrow, though I imagine it's not important, I want to know how to do it damnit
Thanks.
^Oh, really? Scratch that, then. Again, thanks.
edited 1st Mar '10 8:51:56 PM by Barcode711
Worshipper of Ahura Mazda, as proclaimed by Zoroadster http://twitter.com/bpglobalprBasically it looks like you're overcomplicating things. If you can simplify the expression before taking the antiderivative, do so. I think.
[1] This facsimile operated in part by synAC.Man, I seem to keep missing your posts addressed to me.
It is mostly grammar I am having trouble with, I can not remember all of the rules.
Any part in particular?
[1] This facsimile operated in part by synAC.All of it.
That's pretty vague. Adjectives go after nouns but numbers go before, adjectives have to agree with nouns in gender and pluralization, the conjugation...
[1] This facsimile operated in part by synAC.I can not even form a complete sentence. I have no idea hw I managed to pass the first class.
Was it Mapi that was the EE? Yeah, we got to the EE section of the FE exam review, Holy Fucking Shit. Damn, I remember almost none of that and the booklet doesn't help, any advice?
Fight smart, not fair.Doing homework right now...
Who assists Headquarters USAF with reviewing each Air Force specialty for awarding or adjusting SRB?
Any takers?
Which SRB is that? My dictionary thing is spitting several dozen possibilities out.
Fight smart, not fair.
@Deboss
What specifically are you having problems with? How to make, how to read? Where stuff is stable?
edited 25th Feb '10 7:20:51 PM by zzzdragon
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