# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

6 Pages

### 5.10A-C,5.11

Course: EE 4580, Fall 2010
School: LSU
Rating:

#### Document Preview

Sorry, a summary is not available for this document. Register and Upgrade to Premier to view the entire document.

Register Now

#### Unformatted Document Excerpt

Coursehero >> Louisiana >> LSU >> EE 4580

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
There is no excerpt for this document.
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

LSU - EE - 4580
EEE 480HW # 6 SOLUTIONSProblem 5.21The root-locus asymptotes for pure-gain compensation are at 2:5. Since the required closed-loop polelocations are 1 j , we need to introduce a lag compensator. Observe that asymptotes at 1 are obtained forpoles at 0
LSU - EE - 4580
LSU - EE - 4580
Solution to Homework 6 of EE4002 Spring 2004
LSU - EE - 4580
Problem 6.25Using the MATLAB command[GM PM]=margin(sys) we findGM = 1.57 and PM = 100.55From the Bode plot it can be observedthat the gain margin is so small that aslight increase in gain may result inunstable closed loop system. The phase onthe o
LSU - EE - 4580
EEE 480HW # 8 SOLUTIONSNOTE: Ballpark computations relating bandwidth (wBW ) and crossover frequency (wGC ).In classical compensator designs, it is often useful to convert closed-loop bandwidth specications to crossover specpications. The closed-loop
LSU - EE - 4580
Problem 6Problem 6.25Using theMATLABcommand[GM PM]=margin(sys)we findGM = 1.57 andPM = 100.55From the Bodeplot it can beobserved thatthe gainmargin is sosmall that aslight increasein gain mayresult inunstableclosed loopsystem. Thepha
LSU - EE - 4580
LSU - EE - 4580
LSU - EE - 4580
Course OutlineWeek28 JulyIntroduction204 Aug.Frequency Domain Modelling311 Aug.Block Diagrams418 Aug.System Response525 Aug.Feedback System Characteristics601 Sept.Root Locus708 Sept.Root Locus 2815 Sept.Bode Plots9Dr. Stefan B.
LSU - EE - 4580
GAIN AND PHASE MARGIN-The gain margin is defined as the change in open loop gain required to make the systemunstable. Systems with greater gain margins can withstand greater changes in systemparameters before becoming unstable in closed loop.-The phas
LSU - EE - 4580
Course OutlineWeekDateContent128 JulyIntroductionAssignment notes2Block Diagrams418 Aug.System Response525 Aug.Feedback System Characteristics601 Sept.Root Locus708 Sept.Root Locus 215 Sept.Bode Plots9Dr. Stefan B. WilliamsFreque
LSU - EE - 4580
Course ObjectivesMech 3800 : System ControlIntroduction To introduce the methods used for theanalysis and design of linear time invariantfeedback control systemsDr. Stefan B. WilliamsDr. Stefan B. WilliamsAdministrative DetailsAdministrative Deta
LSU - EE - 4580
Course OutlineWeekContent128 JulyIntroduction2Mech 3800 : System ControlRoot LocusDateAssignment notes04 Aug.Frequency Domain Modelling311 Aug.Block Diagrams418 Aug.System Response525 Aug.Feedback System Characteristics01 Sept.Root
LSU - EE - 4580
Angles of departure and arrival Looking at the open loop poles and zerosbelow, the root locus starts at the poles andends at zeros but at what angles?j Taking a point () close to a complex pole.Assuming this point lies on the root locus, theangles
LSU - EE - 4580
Course OutlineWeek28 JulyIntroduction204 Aug.Frequency Domain Modelling311 Aug.Block Diagrams418 Aug.System Response525 Aug.Feedback System Characteristics601 Sept.Root Locus708 Sept.Root Locus 2815 Sept.Bode Plots9Dr. Stefan B.
LSU - EE - 4580
Root-Locus DesignThe root-locus can be used to determine the value of the loop gain K , which results in asatisfactory closed-loop behavior. This is called the proportional compensator orproportional controller and provides gradual response to deviatio
LSU - EE - 4580
S Plane ControlXia HongMay 14, 2004Books: Dorf and Bishop (1998) Modern ControlSystems Addison and Wesley Franklin, Powell and Emami-Naeini (1994)Feedback Control of Dynamic Systems Addison and Wesley. Kuo (1995) Automatic Control Systems Wiley.
LSU - EE - 4580
Routh-Hurwitz Criterion: Special CasesZero only in the rst columnWhen forming the Routh table, replace thezero with a small number and evaluate thest column for positive or negative values of .Problem: Determine the stability of the closedloop transf
LSU - EE - 4580
Stability Design via Routh-HurwitzGiven the system below nd the range of thegain K that will lead to stability, unstabilityand marginal stability.R(s)+E(s)KC(s)s(s+7)(s+11)Closing the loop:KT (s) = 3s + 18s2 + 77s + K(1)1s3177s218Ks1
LSU - EE - 4580
Root locus techniquesj5K=504321K=0K=255R(s)+E(s)KK=00C(s)s(s+10)K=50The root locus show the changes in the transient response as the gain ,K , is varied.K &lt; 25 real poles, overdampedK = 25 multiple poles, critically dampedK &gt; 25 Com
LSU - EE - 4580
Sketching the root locusWe can see that the root locus can be plotted by locating the points in the s-plane forwhich the angles add up to an odd multiple of180o. We can now construct rules for sketching a root locus, plotting any point of interestif r
LSU - EE - 4580
Rening the root locus sketchWe have covered the rules for a rapid sketchof the root locus. For full use of the sketch weneed to calculate the points of interest of thesketch, point on the real axis where the rootlocus enters or leaves the complex pla
LSU - EE - 4580
Angles of departure and arrival Looking at the open loop poles and zerosbelow, the root locus starts at the poles andends at zeros but at what angles?j Taking a point () close to a complex pole.Assuming this point lies on the root locus, theangles
LSU - EE - 4580
Designing via root locus We have seen how a root locus graphicallydisplays stability information, in addition rootlocus can also display the transient response ofa system.jclosed loop poleopen loop poleA and B represent the actual and desired tran
LSU - EE - 4580
Improving transient response Again, we will study two techniques, placinga pure dierentiator in the forward path, andplacing a zero and more distant pole.Ideal derivative compensation (PD)The transient response of a system can be selected by choosing
LSU - EE - 4580
Types of electrical noises(i) Johnson NoiseJohnson noise occurs in any device that dissipates power, such as resistor,hence it is very common. Principle of equipartition of energy states: For asystem in thermal equilibrium with its surroundings : The
LSU - EE - 4580
SignalsSignal a transmitted effect conveying a messageA essential characteristic of a signal is that of change, since it must becapable of carrying information.The change must be partly unpredictable.It is also necessary to discuss the sinusoidal or
LSU - EE - 4580
EE4580 Test Fall 2004Suppose that the plant model is given byG(s) =10(s + 1)(s + 10).s(s2 + s + 25)1. Sketch straightline approximation for magnitude Bode plot.2. Find errors at corner frequencies, and make corrections with dashed line.3. Repeat 1
LSU - EE - 4580
Solution to Midterm IProblem 2: For departure angle, we have 2 = dep ,1 = 135o ,3 = 90o ,1 = 225o ,and 2 = 90o + tan1 (3/7) = 113.2o . Hencedep = 1 + 2 1 3 180o = 113.2o 180o = 66.8o .For arrival angle, we have 1 = tan1 (5), 2 = 45o ,3 = tan1 (7/3
LSU - EE - 4580
LSU - EE - 4580
LSU - EE - 4580
LSU - EE - 4580
LSU - EE - 4580
Jason McAllisterNovember 29, 2004UNCOMPENSATED SYSTEMStep Response1.41.2System: Closed Loop: r to yI/O: r to ySettling Time (sec): 9.961System: Closed Loop: r to yI/O: r to yPeak amplitude: 1Overshoot (%): 0.314At time (sec): 10.9Amplitude
LSU - EE - 4580
The take home part of the second midterm is:Deasign dynamic compensator for (s + 5)G(s) = - s(s^2 + s + 10)such that the closed loop system has overshot no more than 15%, settling time no more than5s, and ess \leq 0.1 for the unit ramp inpiut.The
LSU - EE - 4580
Analysis of the Continuous Linear SystemThrough the design in project 1, the plant with the servo system was calculated as12.88. Figure 1 shows the frequency response of this system.s + 6.5s 3 + 20.7 s 24Figure 1Due to the required response time of
LSU - EE - 4580
A1Appendix ADesign of Continuous-Time SystemTransfer Function and Kp/Kv Derivation:KpAmAmKpAmKpTm= Tm * s + AmKv + 1 ==2AmKpAmKv + 1 AmKpdTm * s + ( AmKv + 1) * s + AmKp1+s 2 + s()+Tm * s + AmKv + 1TmTmn =2AmKpTmAND 2 n = 4.8 ~ 8
LSU - EE - 4580
Page: 1Project 1EE 4002Alicia CorbettMichael DantinScott HammattPatrick QuebedeauxIntroductionThe first logical step in designing a feedback controller for the ball and beamsystem is to analyze the DC motor, the part of the system that causes the
LSU - EE - 4580
Page: 1Project 1EE 4002Alicia CorbettMichael DantinScott HammattPatrick QuebedeauxIntroductionThe first logical step in designing a feedback controller for the ball and beamsystem is to analyze the DC motor, the part of the system that causes the
LSU - EE - 4580
J ohnnyLaraEE4002DesignProject2:DigitalControlofthe BallBeamSystemM ay3,2004IntroductionThe purpose of this project is to gain experience and knowledge of theperformance of an uncompensated system and how it reflects our design if the system isun
LSU - EE - 4580
Page: 1Karrie HugghinsPatrick QuebedeauxMatthew BullEE 4580Report 1 of Design Project 1September 12, 2003Executive SummaryThe first logical step in designing a feedback controller for the ball and beam systemis to analyze the DC motor, the part o
LSU - EE - 4580
Page: 1Project 1EE 4002Alicia CorbettMichael DantinScott HammattPatrick QuebedeauxIntroductionThe first logical step in designing a feedback controller for the ball and beamsystem is to analyze the DC motor, the part of the system that causes the
LSU - EE - 4580
J ohnnyLaraEE4002DesignProject2:DigitalControlofthe BallBeamSystemM ay3,2004IntroductionThe purpose of this project is to gain experience and knowledge of theperformance of an uncompensated system and how it reflects our design if the system isun
LSU - EE - 4580
#Jason McAllister#J#a#s#o#n##M#c#A#l#l#i#s#t#e#r#0###
LSU - EE - 4580
Page: 1EE-4580 CONTROL DESIGNPROJECT 2- BALL AND BEAM SYSTEMJASON MCALLISTER438-49-472611-17-04Page: 2Project SummaryOur goal is to design two compensators: one for the actuator which is a D.C. Motor(inner-loop), and the other is for the ball and
LSU - EE - 4580
LSU - EE - 4580
LSU - EE - 4580
Design Project 1 for EE4580As in the distributed notes, the linearized model for inverted pendulum consists of twotransfer functions. We aim to design Gyu (s) and Gu (s) such that the dominant dynamics forthe closed-loop system have damping ratio 0.83,
LSU - EE - 4580
EE-4580 CONTROL DESIGNPROJECT 1- INVERTED PENDULUMJASON MCALLISTER438-49-472610-18-04Introduction:In this project we have analyzed an inverted pendulum. This is a system which has bothan angle and distance considerations. The control of such a syst
LSU - EE - 4580
LSU - EE - 4580
Design Project 1 of EE4580: Root Locus DesignThe plant model is a linearized pitch axis missile that has a transfer function = 2 + 8 7+ 25 = 2The design speci cations are given as follows::GsTransient response:P:O:Steady-state response:s 12,ess
LSU - EE - 4580
LSU - EE - 4002
,. meclrJo.V1iCQ\&lt;';~E&gt;ievYJbQ) lineaof' motioY)fo.~~f~ ~Hr~l~)Mmotion.!9UJc~'0.fdrtdV&quot;'.tM'Mot). o.Md~clt.tfr 0. Dot.(=?u= ~&quot;-~o.s'M+DFI.tdc.TIon It&gt;)&quot;rfI~'16 b ;nertiai V\plAt pv-Je,('Sef'fdc~d&quot; LAt M~(-I-t)~1\V) &quot; tdr \J.b) n
LSU - EE - 4002