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# IntroductiontoControls MEM255

Drexel

GPA 3.65

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This 24 page Class Notes was uploaded by Jada Daniel on Wednesday September 23, 2015. The Class Notes belongs to MEM255 at Drexel University taught by Staff in Fall. Since its upload, it has received 17 views. For similar materials see /class/212406/mem255-drexel-university in Mechanical Engineering at Drexel University.

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Date Created: 09/23/15

MEM 255 Introduction to Control Systems Modeseigenvalues amp eigenvectors Harry G Kwatny Department of Mechanical Engineering amp Mechanics Drexel University Outline Similarity Transformations Eigenvalues amp Eigenvectors 0 Using MATLAB 0 Diagonal Form 0 Modes 0 Complex eigenvalues Similari Transformations 5c Ax bu 5c Ax Bu n m xeR ueR yeRF yCxDu ycxdu Now consider the transformation to new states Z defined by xTzcgt ZT71x 2T 1ATZT IBu T2ATZBu yCTZDu yCTZDu sothat 2AZBquot Aquot T 1ATB T IBC CTD D yCZDu Eigenvalues amp Eigenvectors Consider the square nx n matriXA as a map from R to R ie y Ax Does there exist a nontrivial input vector k such that the output vector y points in the same direction as h ie y 2h 7 where is some scalar Ah 2 Let39s try to solve for h 11 7Ah 0 gt A nontrivial solution exists iff detlI 7A 0 Characteristic polynomial Ml detlI 7A A Diml r 77 an MA 0 has n roots 11 2 possibly complex possibly repeated Choose one say 2 is 1ln and nd the corresponding 11 w 7AM o Example 1 gdistinct roots 73 A 5 32detidiAdet v5 z5l9 3 75 73 15 f10116 2182w 725 73 771 7217Ahl0 0 73 725 772 3771737720 1 gt gt 73m3m0 771 772 h 1772 7 73 0 71 7817Ahl 377 72 1 gt 7 gt i3 f3 fo 771 772 ha J02 Example 2 grepeated roots 5 0 25 0 2 A detM Adet z5 0 5 0 25 22525 55 0 771 0 t39 f df 315 16 Oran 0 55 77 y771 772 2 771 1 0 mm lawH 51 Ah1 0 Diagonal Form 11 12 39 39 A lt Eigenvalues elgensystem of A I h1 h2 39 39 hw Independent elgenvectms rem hz w m 4 h M Am hi m h1 112 hwquotAh1 Ah2 77 Ah h1 h2 77 Wm 1 m M hz WM hz 11wE dfagmr 71quot Example gt De ne A v Compute eigensystem Check similarity trans Use linear solve rather than 111V 03482 08704 03482 gtgt an gtgt ans 90000 0 0 1nvV 90000 700000 700000 VAV 90000 700000 700000 gtgt A3 2 14 gt VVD819 AV 5 A W 0000 W W 0000 amp 61 8581 1907 4767 0 0000 0 0000 0000 0000 0000 W W 4082 8165 4082 Modal Coordinates 1 Consider the system in diagonal form The 2 coordinates are referred to as 39modal coordinates39 2 z1 1x uiln Suppose we solve these equations to obtain 21 t i ln the we can obtain the solution in the original coordinates via 21 I xv mt a hz km 24 m twee r z I Notice that each term hxzx t describes a motion that takes place in the line de ned by 11 Model Coordinates 2 221 I Ix0 h F l zm 9211 Modal Coordinates 3 Suppose the system is unforce u t 0 The initial conditions for the modal coordinates are 2D T 1x The solution is z t el zmil n xt hie1 2 l 7 7 hweyzn The modes are the vector time functions he I is the mode shape The solution is a linear combination of the modes If A is real so is I and the modal response is a simple exponential If there are n linearly independent eigenvectors then the set of solutions hie1quot Jae14 is called a fundamental set of solutions Any solution to the homogeneous equation x Ax is a linear combination of the fundamental solutions R21 1A212R1 A1 6 Exam 1 Example Cont d gtgt A71 12 73 gtgt EVe1gA E 08069 03437 05907 09391 V 02679 0 0 37321 08069 slow mode 31 02679 h1 05907 03437 fast mode 22 37321 h2 0 9391 Example Cont d WW K M 08069 05907I ll IVI 09391 034374 M Vl Complex Modes 1 Ifthe an eigenvalue 1 is complex the so is the eigenvector To make things easier to interpret we can construct real ones Suppose A 039 fa is a complex eigenvalue and xlz Then the corresponding eigenvectors are huh Correspondly the two fundamental solutions are hleltjzfem They are complex We will replace them with real ones Complex Modes 2 De ne x1t 01161 hfe 1 hm jhuemcosatjsinat 2 hLR jhuem cosmt jsinmt hLRem costar hue sinmt 1 x2 I 3 h1821 hle hLRe39 sm at huem cosmt m1k4c12 d x 0 1 x Emth 05v 00559 h 48944 gtgt Ao 174 705 04437 gtgt EVe1gA E 00559 044371 00559 044371 08944 08944 v 02500 198431 0 0 02500 198431 System Transmission Line Turbine Generator Tailwater Differential Equations Q h 2 5 2 4 Qz PZ2X105 90 aZ1021198x105Q 1079 107w 9 425 000 Remaining Elements Splitter Orifice Valve Turbine mass balance Q Q Q Q C3 kshz ow accross orifice P3 kshz Q 1 Q0 Ct B throttle valve charateristic P4 P5 Cth Cma hydro machine characteristic Find QSQt P P4 2 Recall the variables Q hpa 1 are treated as known Sol ns t0 Algebraic Equations P2 909256h2 0001851 00909256Q 00090744160 P4 907441h2 00185118a 00907441Q 00092559a Q 907441h2 00185118Z 0909256Q 0090744160 Q 907441h2 00185118Z 00907441Q 0090744160 df State Eguations h7 722686 0227314 0 70022686 h 7000462795 Q 7 70454628X10394 70104546X10394 0 453721x10 8 Q 92559 9 9 0 0 9 0 60 0943556 943556X10393 7106101 7105157 60 000192485 Modes 7225582 705324408853 705327108853 70150x104 0851 70007 0004 0099 h2 00 00 00 0990 Q 0213 70358 70596 0097 9 70480 0718 00 00 a The Matnx ExponentIal ReVISIted Recall the matrix exponential e Miami111 iAW 2 k Now suppose T is the transformation matrix that diagonalizes the matrix A TquotAT A 2 diagM NJ Construct Tquote quotT Tquotequot T TquotITTquotATI TquotA2th T39Mthk 1 A1 lTquotATTquotATt2 77 iTquotATW 2 k ImlA2t2 iAk1k 2 k diag 21quot Summary Similarity transformations Diagonalization using Eigenvectors 0 Modes amp Modal coordinates Interpreting behavior in terms of modes 0 Converting complex modes into real ones

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