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Midterm Exam with Solution | Advanced Nonlinear Control Strategy | E C E 874, Exams of Electrical and Electronics Engineering

Material Type: Exam; Professor: Dawson; Class: ADV NONLINEAR CNTRL; Subject: ELECTRICAL AND COMPUTER ENGINEERING; University: Clemson University; Term: Spring 2009;

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Uploaded on 07/28/2009

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Download Midterm Exam with Solution | Advanced Nonlinear Control Strategy | E C E 874 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! Midterm Examination  Name: ___________________________________________ Class: ECE 874 - Nonlinear Control Spring Semester - 2009 Instructor: Darren M. Dawson 1. Equilibrium Points - For the system     1 2 1 2 1 1 2 2 tan x x x x x x          Find all of the equilibrium points. Find the linear approximation about each equilibrium point, find the eigenvalues of the resulting A matrix, and classify the stability of each point. Solution: Midterm Examination  2. Autonomous System - Consider the system defined by the following equations: 3 1 1 2 1 2 1 3 x x x x x x              Describe the stability of the equilibrium point xe = (0, 0), in the following cases:  i 0 (ii) 0 (iii) 0       3. Nonautonomous System - Study the stability of the equilibrium point xe = (0, 0) for the system:    2 2 1 1 1 2 2 2 2 2 1 2 cos sin x x x x t x x x x t         4. Backstepping Control Design - Given the following system: 3 1 1 2 2 4 2 1 2 1 x x x x x x x u        Using the backstepping approach outlined in class in which intermediate tracking variables are explicitly defined, design a state feedback control law to stabilize the equilibrium point at the origin. Discuss stability and show all signals are bounded. Midterm Examination gen | — | Lind 4K) to ote be/ Y eae ™ Ke co Stet [ree X, =k + aK, oa top subsystena = Mike —Xs t= Find 410i, ,he) fo sdebilize 4% iil Ob eet f galt, we) = -! v= Ne a : \ ok, re a. aes, € + dex,) | 4.x > Ke eR K i Pas eee ee 54 4,Cx,,4.) = — x, [fea #90) ¥0] an, A { Example, 55 ~ kan d(x) 1 ~ Gtono} Idx) - 2k, => 24, aye Ox, : mi fet eam 4) (Ke) = “tee ahaa eae iS See 2 7 PAK ort = 24 P 4K) 2K, Xq - 3K GK, t3X, +k) th ; IK ty, tt Ns. 2 = 2k," —BX, +/0K, ~Xike + thy z z Y= Vt Alye-¢ey| > K+ 20x =H hesy, 1” Midterm Examination il Cad we Cong dei Dhele. 24 ote rt oa Xs to as — 7 fatstietnt fet rg be 4 six) CG, neni tintas XXa | = XX. x,t KK Ke ft. Jalna, Ks) | | 4 (estate) > Zien ; 2 Est) Bs, 0) a 2 go -kKLksy- $:(kj ke) ~ Gls) 4 2 bil rte) = [ 2 lx, Xo) 249k, rm) OK ak, Oke * GK, 16K, +12 Ky -Xx, “tf 2b pb No < SNES GOs (One te \(-1c 24k) tk 4K Xs) = -bxi epee box? ke the 32%, - both (0X, + 20K, +1oks, AD - DMs “Df +4 krk, = 6X14 26K7 - 43.4" 4 20%,-4 4K, +O Ke HONK. al be) sian) = E 2M) DWC te nia tt | x ’ 27 MAG). Cy, ¥, on ‘) 2 = Xa-K, +3%) Midterm Examination Marron a ont 8K? 454," + 20K, ~NM, 110K, 1 ObP% Kt = BK = 3K, + Pint = Ke lky = Bx ~ EX * 10K a eH) | iI lst ke! F dali ghayXq) = 5K." + $0X,? -SOK 97K “SKKe + 13K HN I ~ Xs I A = (x, x. Se) Vez Vat 2 PK, — dfx, x> 1 (swen Wis 15 a Cuslempdic es Iq approach we Knots 2 ig) nts CAS # X, 72 pNe= ee (x, %) %))7e X, 20 Pix) 20 = X, PO X,,kp—o =e kyo Masi at lke iolanc meats kip keykg 20 f. PO dis bboy) =0 - since %, Xp Xo, Po XnXa ike jbo Kg-o all signals are aaa
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