ECE210a_lecture1_small

ECE210a_lecture1_small - Introduction Matrix Analysis and...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Introduction Matrix Analysis and Computation ECE 210a, CMPSC 211a, ME210a, Math206a, ChE 211a Roy Smith 5117 Harold Frank Hall (805) 893-2967 roy@ece . ucsb . edu Roy Smith, ECE 210a, 1: 1 Introduction Course information: Course administration: Gauchospace: https : / /gauchospace . ucsb . edu (look for ECE 210a) Lecture notes: http: / /www. ece . ucsb . edu/~roy (also linked from Gauchospace) Roy Smith, ECE 210a, 1: 2 Simultaneous equations Simultaneous equations (Mam; pr 1) “Three sheafs of a good crop, two sheafs of a mediocre crop, and one sheaf of a bad crop are sold for 39 dou. Two sheafs of good, three mediocre, and one bad are sold for 34 dou; and one good, two mediocre, and three bad are sold for 26 dou. What is the price received for each sheaf of a good crop, each sheaf of a mediocre crop, and each sheaf of a bad crop?” Chapter VIII, Chin-Chang Suan-shu (Nine chapters on Arithmetic). circa 200 BC. Solution: 9:1 : good crop; 3:2 : mediocre crop; 3:3 : bad crop Simultaneous equations Matrix form 3% + 2962 + x3 : 39, 3 2 1 x1 39 2x1+3x2+x3=34, 2 3 1 932 i 34 1E1+ 2172 + 3113: 1 2 3 £173 Roy Smith, ECE 210a, 1: 3 Simultaneous equations Solution: Simultaneous equations Matrix form 3271 W 21132 W 233 2111 31132 ID; = [\D 00 x1+2x2+3mg = 26 1 2 3 x3 3$1+2$2—-.’E3 I 1 0 0 5$2——x3 = 24 (372—271) —2 3 0 49024—8373 = 39 (3r3—r1) —1 0 3 3:51 ‘1‘ 2$2 -- {E3 _ 1 0 0 5:172 -- (133 = 24 0 1 0 O 5 36.233 * 4T2) 3 2 1 x1 39 0 5 1 $2 = 24 ‘. 0 0 36 $3 99 So an : 9%, m2 : 4i and $3 : 2‘ Mo. Roy Smith, ECE 210a, 1: 4 Simultaneous equations Interpretation: The solution equations are: 1% x = Z3 3 2 1 39 0 5 1 a: = 44 0 0 36 99 1 0 0 1 0 0 1 0 0 1 0 0 0 1 0 —2 3 0 A :c = 0 1 0 —2 3 0 b, 0 —4 5 —1 0 3 0 —4 5 —1 0 3 TQTlA $ = T2T1b. So R : T 2 T1 A7 or equivalently, A = Tfng'lR = QR. In general we can always find a unitary Q and upper triangular R such that, A = QR, (Q*Q = QQ* = I) then Rx = Q*b is a triangular system. Roy Smith, ECE 210a, 1: 5 Planetary orbit determination Planetary orbit determination: Ceres 1801, 1st January: Giuseppe Piazzi finds and makes 22 observations of a new “planet’7 over 41 days. 1801, February: Ceres is obscured by the light of the sun. The observations correspond to a 9 degree are of its orbit. 1801, June: Zach publishes the observational data and many astronomers and mathematicians attempt to calcu— late its orbit and predict its reappearance. Hubble image of Ceres. 930 km dia. (I 'n a > C U) Vi * u \i 5! q ... .a m U! U1 1801, September: Carl Friedrich Gauss uses his method w t l— of Least Squares to calculate the orbit and Zach publishes 5 , 8 his ephemeris (along with those of other astronomers). m m ’ ‘ ' ' l‘ d": aft ‘ W Gauss s differ Signlficantly. 3 t . LI-l 5.1.: 1‘ Q o z 1801, 7th December: Zach locates Ceres within 1/2 of a , a degree of Gauss’s prediction. Gauss refused to disclose his method and is accused by some of using sorcery. Roy Smith, ECE 210a, 1: 6 Planetary orbit determination Planetary orbit determination 1802 Gauss also calculates the orbit of Pallas. 1805 Legendre publishes a paper on the method of least squares. 1807 Gauss becomes director of Gottingen observatory. Olbers discovers Vesta and Gauss takes only 10 hours to calculate its orbit. Ceres’s orbit took him 100 hours to calculate. 1809 Gauss finally publishes the details of his method and claims priority over Legendre. Gauss’s approach Basic assumptions: Kepler’s laws of planetary motion: — Planetary orbits are ellipses with the sun at one of the foci. — A line drawn between the Sun and the planets sweeps out equal areas in equal times. — The ratio of the square of the periods of the orbits of two planets is equal to ratio of the cubes of their major semi-axes. Roy Smith, ECE 210a, 1: 7 Planetary orbit determination Gauss’s approach The orbit of a planet is defined by five variables (orbital elements) — Two angles that orient the plane of the orbit with respect to the ecliptic plane. 7 The size and eccentricity of the orbit. Equivalently the length of the major and minor semi—axes. — The tilt of the main axis of the orbit with respect to that of Earth’s orbit. Ceres — Knowing the location on the orbit at a particular time completely specifies the location in space. Ecliptic plane Roy Smith, ECE 210a, 1: 8 Planetary orbit determination Steps in Gauss’s approach — Select three observations: 1st January, 21st January and 11th February. Each observation is two angles (Ceres with respect to the background stars) and a time. — Calculate a nominal orbit matching the observations. — Taylor series approximations to the nonlinear differential equations. — Iterative refinement. — Adjust linearized versions of the orbit to minimize the sum of squares of all 22 observations. Problem formulation: H is the nonlinear orbital prediction model. RA of obs. #1 dec. of obs. #1 yniodel = = H where :1: = orbital elements. RA. of obs. #22 dec. of obs. #22 Roy Smith, ECE 210a, 1: 9 Planetary orbit determination Problem formulation: gobb are the actual (noisy) observations. 6 = yobs — H (x. t) e is the residual. r 71 The cost function is J = R is a weighting matrix. Objective: Ioptimal : mTin . . 8] Nominal orbit w/ 4' R i ) Nominal orbit determination Least squares orbit fit 6 equations in 6 unknown parameters 44 equations in 6 unknown parameters Roy Smith, ECE 210a, 1: 10 Planetary orbit determination Linearize the problem: The nominal orbit is specified by 3:0. Close to this, HUD) N H($0) + H.7(55 — 530% Where Hi] = (Jacobian). Define 6x = as — :60 and 6y 2 yobs — H(:c0). Then Jhnmwx) : g “930 + 5x). 2 The linearized cost function is close to the true cost function. And it’s quadratic. The gradient of Jhnearwm) is7 8Jlinczu’ 6x : —H}R’l (6y — Hm). Setting this to zero is equivalent to solving: HfR’lHJéx = HfR’ldy Note the form: Ax = b. It the nonlinear case we would apply this iteratively. Roy Smith, ECE 210a, 1: 11 Planetary orbit determination Gauss’s sketch ' Piazzi‘s last I l- ' u I h. ' observation £9.33: ’E PMZZI a iiibt ._ ' - J, —...j “A,” observation of. I Ceres found ’ “4"- r—F h_ - ’- -r-- by zaCh l/fl References: Wfiwm-r ‘ I m-En‘lfil- "fl"? Skizze der Bahnen der Kleinplaneten f a' / Cores, Pallas 11nd Vesta. “Astronomic- . / . ' elm Untersuchungen and Rechmmgen 5'31; 3N nornelmzlieh ber‘ die Ceres Ferdin(1,71,de(1,, ” J} _|1 \- 1802. SUB Gottingen: Codi Ms Gaufi .r. ' 36m? Ilandbuch 4 B1. 1. I‘ allu- . x ; 5. " a? ‘ “\ sun I 9,. \\ Ia‘h'h-lar- - I ' ' ' R. Bannister, Am J. Phys.) 71(12), pp. "I \ b “)4” \_ ' 125871275. DOC. 2003. '. " - or 110 ‘ as " “”' ' ' - | '- orbit of" Earth ' “fig; . - I L. Weiss. Gauss and Ceres. , (www. ‘\ \ciuf _ ' . ' ' - math.rutgers.edu/Eherlin/History/ N‘K- " PaperslS99/weiss.html). Roy Smith, ECE 210a, 1: 12 Vibrations of polyatomic molecules Vibrations of polyatomic molecules The potential energy, V(:E), is a function of the positions of the atoms in molecule. Expanding about an equilibrium (denoted by 0)7 av 1 82V V = V 0 L- — o + 2 (W + W 0 0 ) + ... 0 So for small displacements, 1 82V . V — 5 [Warmly-7 km- — ( 8237-89“) 0 . (generalized force constants) In the movement of N atoms there are 3N — 6 vibrational modes plus 3 translations and 3 rotations. (3N — 5 vibrational modes for linear molecules). What are the normal vibrational modes? Reference: Atkins 81, Fricdnmn. Alrilmzulm' Quantum Almzhunlris. 3rd Ed, Oxford, 1997. Roy Smith, ECE 210a, 1: 13 Vibrations of polyatomic molecules Newton’s second law: d2$i m,- dt2 = — ,7' Key feature: ICU = Introduce mass—weighted coordinates: (17: = «mi, 15,. 612% kij 2 = ‘2 . ‘17' = EXAM- dt j 1/mn’ri j ,7' Matrix form: d2 ‘1‘ W ' K1 1 ' ' ' K1 91 (11 = 3 ' ' . 5 3 = K : (1N K N1 ' ' ‘ KNN (IN (IN This is a multi—modal oscillator with modal frequency given by the square root of the eigenvalues of K. The “shapes” of the modes are given by the eigenvectors. Roy Smith, ECE 210a, 1: 14 Vibrations of polyatomic molecules Diagonalizing transform A1 0 Find a unitary U (UTU = UUT = I) and diagonal A: ' d2 (11 (11 K:UAUT, s0 3 :UAUT ; . E (IN QN - . :| such that7 0 AN Define new coordinates (normal coordinates), inl iqll I = U T I . QN (1N Then, d2 Q1 (11 Q1 E g : UTUAUT ; = —A 5 N independent oscillators. QN (1N QN Roy Smith, ECE 210a, 1: 15 Vibrations of polyatomic molecules Carbon Dioxide Example (1 dimensional) k .:C:. m0 mC mo I | | l > I x1 x2 X; x k i k V — 5(271 — 1‘2)2 + 5(275 —x2)2 (92V kin: , ', ‘=1,2,3. J axixj Z .7 k) —k‘ 0 16/7710 —k‘/‘/’ITL()TTL(7 0 [kw] = —k 2k —k , and K = —k/‘/m0mc Qk/mc —k/./m0mc . 0 —k k 0 —k/‘/m0mc k/mo Roy Smith, ECE 210a, 1: 16 Vibrations of polyatomic molecules Normal modes: k/mo —k/ 1 [mama 0 The eigenvalues of K : —k/‘ Mummy 2k/mp —k/‘ Mining are: 0 —k/ 1 flinging k/mo Eigenvalues Eigenvectors Eigenvectors /\1- (normal coord.) (physical coord.) V mO 1 no restoring force 0 1/m 1/7710 1/771 1 . (translat1onal mode) a /m0 1 1 1 independent of m , /—y ’ C k/mO l/fi Pl 1/ 2mg 31 (carbon doesn’t move) 4mg \/77l(jr/TIZ() . . kin/mama 1/\/ 2m —2‘ mm 1/\/ 2m *2 mo/mo carbon osmllatmg between oxygens \/””C' V mn/‘mo (m is the total mass of the molecule: in : mo + 2mg). Roy Smith, ECE 210a, 1: 17 Non-diugonulizable dynamics Tank mixing problem (Meyer example 7197) We have three tanks (of volume V gallons) containing polluted water with the ith tank initially containing 0,- pounds of pollutant. To remove the polluted water all valves are opened and fresh water flows in at 1" gallons/ second. The flow between tanks is also 7" gallons / second. Fresh ' *—""Wflter Eh r gal/sec i] r gal/ sec Assume that mixing in the tanks is instantaneous and continuous g r gal/sec r gal/sec ii Question: What are the pollutant levels in each tank as a function of time? Roy Smith, ECE 210a, 1: 18 Non-diagonulizable dynamics Matrix problem formulation The concentration of pollutant in tank i at time t > 0 is, ci(t)/ V pounds / gallon. The rate of pollutant flowing into tank 2' is T071105) / V pounds / second (except for tank 3), and the rate (pollutant flowing out is rci(t) / V. The differential equations can be written in matrix form: d clgt; T —1 1 0 clgt; (:18 ( ) — Cgt 2— 0 —1 1 0215 =A Cgt =Act. dt egg) V 0 0 —1 03a) 03(75) Solution in terms of a matrix exponential The solution is, C(t)=e"1tc(0), where, e‘“ = er'r‘f/V Roy Smith, ECE 210a, 1: 19 Markov processes game (Meyer: example 7.108) A “pea” is hidden under one of four “shells” which are randomly shuffled according to the probability rules: XVI/Kr“: @@@@ VVV 1/2 1/2 1 The pea must be moved with each shuflie. Questions 1. If we know where the pea starts, can we say anything about the probability of where it will end up after N moves? 2. Is there a limiting value to the probability and does it depend on the pea’s starting shell? Roy Smith, ECE 210a, 1: 20 Markov processes Markov process: The probability of the pea being under a particular shell depends only on the shell it was previously under. If 10,; is the probability that the pea is under shell 1', then, in going from shuffle k to k + 17 “(19771) 0 1/2 0 0 191(k) 2(k—1) 1 0 1/2 0 Que) plk+1)_ gags—1) Z 0 1/2 0 1 $30?) 2 AM)‘ 4(k——1) 0 0 1/2 0 4(k) 4 Note that AU 2 0 and 2 AU 2 11:1 Question 1: Applying the above iteratively gives, p(N) = ANp(0). Question 2: The solution would be given by, lim Al" Istfiwx but unfortunately this limit does not exist in this case. What can be said about the long term probabilities? Roy Smith, ECE 210a, 1: 21 ...
View Full Document

Page1 / 11

ECE210a_lecture1_small - Introduction Matrix Analysis and...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online