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Solution Manual For Optimal State Estimation: Kalman, H Infinity, and Nonlinear Approaches, 1st Edition

Solution Manual For Optimal State Estimation: Kalman, H Infinity, and Nonlinear Approaches, 1st Edition is your textbook problem-solver, offering clear and concise solutions to difficult questions.

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    | | CONTENTS -— Introduction v 1 Linear systems theory 1 2 Probability theory 15 3 Least squares estimation 29 4 Propagation of states and covariances 41 5 The discrete-time Kalman filter 51 6 Altemate Kalman filter formulations 65 7 Kalman filter generalizations 79 8 The continuous-time Kalman filter 93 9 Optimal smoothing 103 iii | =n

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    | | iv CONTENTS 10 Additional topics in Kalman filtering 113 11 The Hy filter 127 12 Additional topics in Ho, filtering 141 13 Nonlinear Kalman filtering 149 14 The unscented Kalman filter 169 15 The particle filter 179 Source code for computer problems 193 ll

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    -_— OOOO O_o This solution manual is & companion to the text Optimal State Estimation; Kalman, Hoo, and Nonlinear Approaches, by Dan Simon (John Wiley & Sons, 2006). The MATLAB! source code for the computer exercise solutions is given at the end of this solution manual. The references in this solution manual refer to the references section in the text Optimal State Estimation. The equation numbers in this solution manual refer to the equations in the book Optimal State Estimation. Although the MATLAB code for the solutions is not available og the Internet, MATLAB-based source code for the examples in the text is available at the author's Web site? The author's e-mail address is also available on the Web site, and I eagerly invite feedback, comments, suggestions for improvements, and corrections. A note on notation Three dots between delimiters (parenthesis, brackets, or braces) means that the duantity between the delimiters is the same as the quantity between the previous set of identical delimiters in the same equation. For example, U+BEDY + (97 = (A+ BCD) + (A+ BoD)™ A+IBC-DIEL] = A+B DI E[B(C + D)] MATLAB is a registered trademark of The MathWorks, Inc, http:/ facademic.csuohio, edu/simond /estimation — if the Web site address changes, it shouid be easy to find with an Internet search. v

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    CHAPTER 1 ee ———————————————————————— Linear systems theory J Problems Written exercises ; foo 1.1 Find she rank of the matrix | o |- Solution The rank of a matrix A can be defined as the dimension of the largest submarrix consisting of rows and columns of A whose determinant is nonzero. With this definition we see that the rank of the zero matrix is zero. 1.2 Find two 2 x 2 matrices A and B such that A # B, neither A nor B are diagonal, 4 # cB for any scalar ¢, and AB = BA. Find the eigenvectors of 4 and B. Note that they share an eigenvector, Interestingly, every pair of commuting matrices shares ut least one eigenvector [Hor83, p. 51]. Optimal State Estimation. Solution Manual, First Edition. By Dan J. Simon 1 (©2006 John Wiley & Sons. Inc. — | — + StudyXY

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    2 LINEAR SYSTEMS THECRY Solution Suppose 4 and B are given as _ Ja a rela] _ hob B= [ bz by Then we see that ap — | @bitamh ashy + ass - azby + aghy axby + ashy BA = arb + ashy ashy + ashy . - @1by + ashy azby + ashy We see that 48 = BA if a;by +azbs = aby + aghy. This will be true, for example, Hay=10y=2a3=1,b=1,by=3 and by = 1. This gives 12 ao [st] [13 = 0s? For these matrices A has the eigenvalues —1 and 3, B has the eigenvalues —2 and 4, and both 4 and B hove the eigenvectors [ ~1 17 and [1 1". 1.3 Prove the three identities of Equation (1.26). Solution a). Suppose A is ann x m matrix, and B is an m x p matrix. Then Ano im 1] Bao By 1\T “py” = Co Poe An or An Bri + Bp T TAB; CAB, LAB oo YC AyBy | YABy oo YAuBa | AGB 3 AwByy Bu + Bm Ann Am BTAT = FE : JE : Bip + Bmp Aim Agm | YBuAy co YL Bud | Bp; oY Bipdy,

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    PROBLEMS 3 QED b). Suppose that (AB)! = C. Then CAB = I. Postmultiplying both sides of this equation by B~! gives C4 = B~!. Postmultiplying both sides of this equation by A=! gives C= B~'A~1. Hence we see that (AB)~> = B~1A~1, QED c). Suppose A is ann x m matrix, and B is an m x n matrix. Then An oc Aim Bix +o Bin (4B) = || : AE Au Aum Bry + Bon SAB; cr Yo AyBie = Tr : ., : YC AnBy co TAnBin nom = 33 AB == ( Bui +r Big An oo Aim] Tr(BA} = Tr : : FE : { Bry +++ Bun An Aum J Bud co LU Bidim = Tr : iS : 2 BrjAjr cr 2 Bmidjm mn - Ys 11 QED 1.4 Find the partial derivative of the trace of AB with respect to A. Solution Suppose 4 is an 7 x 7 matrix, and B is an m x n matrix. Then Tr(AB) 8 “ar T Alun i=15m1 | Shr Tim Dim Aub wh Li Ei Au | oh Si Sh AB wh Ui 0k ABs | By +++ Bm Bin + Bm = BT

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    4 LINEAR SYSTEMS THEORY 1.5 Consider the matrix a_[a® “lb oe Recall that the eigenvalues of A are found by find the roots of the polynomial P(3) = |\ = A]. Show that P(A) = 0. (This is an illustration of the Cayley— Hamilton theorem [Bay99, Che99, Kai00).) Solution Py = A-4 _lA-a Tl = oa-e = M_(a+tcr+ac~b? P(A) = A*—(a+c)A+ (ac—b)I a? +6 ab+be a b ac— 0 = [G18 a IP hl EE _ [oo loo 1.6 Suppose that A is invertible and 4 A174] Jo Bal ciT1 Find B and Cin terms of A [Lie67]. Solution Multiplying out the matrix equation gives the following two equations. ALAC = 0 BA+AC = 1 Solving for B and C in terms of 4 gives B = A+a! C = -A L.7 Show that AB may not be symmetric even though both A and B are sym- metric. Solution Suppose the symmetric matrices A and B are given as 4 = a ey ag az J EE

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    PROBLEMS 5 Then we see that Ap — | Gbitashy aby + ashy ashy +aghy azbs + ashy AB is not symmetric if aybz + a2bs # azby + aba. 1.8 Consider the matrix ace e Tle oe where a, b, and ¢ are real, and a and ¢ are nonnegative. a) Compute the solutions of the characteristic polynomial of A to prove that the eigenvalues of A are real. b) For what values of b is A positive semidefinite? Solution a). The characteristic polynomial of A is PQ) = I-A] _r-a —b - | —b A-c = M-(a+r+ac—b Finding Lhe roots of this gives 1 5 A= 3 [aes vias c)? +47 The discriminant is non-negative so A is real. QED b). In order for A to be positive semidefinite, its eigenvalues must be positive. The eigenvalues are H [aver Nicer A= 3 fase Nicer The first eigenvalue is always non-negative. The second eigenvalue js non- negative if a + ¢ > \/(e — cj? + 46%. Solving this equdtion gives |b] < v/ac as the condition of positive semidefiniteness. 1.9 Derive the properties of the state transition matrix given in Equation {1.72) Solution d = (At) EEO i tf | Vv | TT StudyXY

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    6 LINEAR SYSTEMS THEORY _d (41)? (4? = 5 [roar 2 Ft Ag? = Aran ACL 2 At)? = Alva BEL] 2! = Ae™ This proves the first equality. After writing the third expression of the above sequence of equations, we can bring the common factor A out to the right to obtain d at (At)? “4, = At 2207 Fr I+ At+ 5 + A = ety QED 1.10 Suppose that the matrix A has eigenvalues A; and eigenvectors v; (i = 1,.-+,n). What are the eigenvalues and eigenvectors of — A? Solution Av; = Ay, therefore — Av; = — Mw; From this we sce that —4 has eigenvalues —A; and eigenvectors v;. 1.11 Show that |e**| = eM for any square matrix A Solution: A? ett = Toate =0 0 . A%? led = rj AE A? = in paps APE = elk QED 1.12 Show that if A = BA. then 414] —— = Tr(B)| 4 = Tima Solution: The equation A = BA can be solved as A = €8*.4(0). Taking the determinant of this equation gives [Al = [Aw

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    PROBLEMS 7 = [e™lA@)] = FA) From this we see that 44 84 «|B 11) 4¢0) Fr [Blel?F|A(0)] = TrB)A| QED 1.13 The linear position p of an object under constant acceleration is Ll, P= pot ptt Git where po is the initiel position of the object. a) Define a state vectorasz = [ pp p |" and write the state space equa- tion & = Az for this system. b) Use all three expressions in Equation (1.71) to find the state transition matrix eA? for the system. ©) Prove for the state transition matrix found above that ¢4% = I. Solution a). ar 010 p lp =joo01 p B 000s b). From the first expression in Equation (1.72) we obtain A _ esa Pa =o (ant ar (An? (an =o thtm tram to ot 0 00 #2 = 1+|00 t +00 0 | +064 000 00 0 1 t 2/2 = lo: ¢ 00 1 From the second expression in Equation {1.72) we obtain eM o= L7(sI- A)7Y s -1 017" = cto 5s 1 000 os

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    8 LINEAR SYSTEMS THEORY 1s 1/s* 1/8? = £70 0 ls 1s? 00 1s 1 22 = [01 ¢ 00 From the third expression in Equation (1.72) we obtain At Qettgt The eigendata of 4 are found to be A = {00,0} 1 0 [i v= of, lt],l0 0 0 1 Actually we can note that A is already in Jordan form, which means that its eigenvalues are on the disgonal, and its eigenvectors form the identity matrix when augmented together. Recall for a third order Jordan block that Soph Lon eo) 0 et fed 0 0 ert In our case A = 0 so 1 ot $82 ef=10 1 00 1 ©). From the above expression for 4, if we substitute ¢ = 0 we see that ct — I. QED 1.14 Consider the following system matrix. 10 r=[o 5] Show that the matrix sme] 0 To 2 satisfies the relation $2) = A(t), but S(¢) is not the state transition matrix of the system. Solution sip = [dO EE 0 2

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    PROBLEMS. 9 As — | 100 e& 0 @ = 0 21] ]o ze _ [et 0 “lo -2t We see that $(¢) = AS(t). However, the state transition matrix is found to be et 0 t= [ 0 et | QED 1.15 Give an example of a discrete-time system that is marginally stable but not asymptotically stable. Solution The system 7341 = Tj is marginally stable, because the state is bounded for any initial bounded state, but it is not asymptotically stable, because it is not true that the state approaches zero for all initial states. 1.16 Show (H, Fis an observable matrix pair if and only if (H, F~*) is observable (assuming that F is nonsingular). Solution If (H. F) is observable, then Qx # 0 for all nonzero x, where o Hr Q= . HE! Since F is nonsingular F~®=1z spans the entire n-dimensional space. (That is, any n-element. vector can be written as F~"~9z for some n-element vector z.) So the observability of (H, F) is equivalent to QF "=z 2 0 for all nonzero z. This is equivalent to Q'z # 0 for all nonzero x, where HF=n=U HF=(n=2) Q = . H which is the observability matrix of (H, 1). QED Computer exercises 1.17 The dynamics of a DC motor can be described as J§+F§=T

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    10 LINEAR SYSTEMS THEORY. where 6 is the angular position of the motor, J is the moment of inertia, F is the coefficient. of viscous friction, ard T is the torque applied to the motor a) Generate a two-state linear system equation for this motor in the form i=Az+ Bu b) Simulate the system for 5 s and plot the angular position and velocity. Use J = 10 kg w?, F = 100 kg m*/s, 2(@) = {0 0% and T= 10 N m. Use rectangular integration with a step size of 0.05 s. Do the output plots look correct? What happens when you change the step size At to 0.287 What happens when you change the step size to 0.5 s? What are the eigenvalues of the A matrix, and how can you relate their magnitudes to the step size that is required for a correct simulation? Solution a). Let 2; = 0 and 74 = 6. Then [0 1 0 [5 Laer b). Output plots for various simulation step sizes are shown in Figures 1.1-1.3, With At = 0.05 the simulation works fine. With At = 0.2 the simulation results are obviously incorrect, although the simulation is sill stable. With At = 0.5 the simulation blows up. The eigenvalues of A are 0 and —10. The simuistion step size should be appreciably smaller than 1/(Alugs, which implies that the step size should be smaller than 1/10. which is consistent with our experimental results. dt=0.05 0.5 position 0.45 velocity 0.4] 0.35 0.3] 0.25) 0.2 0.15 c1 0.05 % 1 2 3 4 5 Seconds. Figure 1.1 Problem 1,17 simulation with Af — 0.05. Good simulation 1.18 The dynamic equations for a series RIC circuit can be written as w = IR+LI+V,

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    PROBLEMS 11 ai=02 0.5 —— position c4 025] 03 025 0.2 0.15) 0.1 005 0 o 1 2 3 4 5 Seconds Figure 1.2 Problem 1.17 simulation with Af = 0.2. Poor simulation (x10 &=05 —— positon 2 | velocity Of 2] By | 6h | _8- | ~10r ji al 2% 1 2 3 4 5 Seconds Figure 1.3 Problem 1.17 simulation with At = 0.5. Unstable simulation I = ov, where u is the applied voltage, 1 is the current through the circuit, and V, is the voltage across the capacitor. a) Write a state-space equation in matrix form for this system with ry as the capacitor voltage and 25 as the current, b) Suppose that R =3, I = 1, and C= 0.5. Pind an analytical expression for the capacitor voltage for 2 0, assuming that the initial state is zero, and the input voltage is u(t) = ¢~ 2, c) Simulate the system using rectangular, trapezoidal, and fourth-order Runge— Kutta integration to obtain a numerical solution for the capacitor voltage.

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