VECTOR RANDOM PROCESSES

Transcription

1 ECE5530: Multivariable Control Systems II. 1 VECTOR RANDOM PROCESSES.1: Scalar random variables We will be working with systems whose state is affected by noise and whose output measurements are also corrupted by noise. Disturbance w.t/ System x.t/ Sensor noise v.t/ By definition, noise is not deterministic it is random in some sense. So, to discuss the impact of noise on the system dynamics, we must y.t/ review the concept of a random variable, (RV) X. Cannot predict exactly what we will get each time we measure or sample the random variable, but We can characterize the probability of each sample value by the probability density function (pdf). Probability density functions (pdf) We denote probability density function (pdf) of RV X as fx.x/. f X.x/ dx fx.x 0 / dx is the probability that x 0 x random variable X is between Œx 0 ;x 0 C dx.

2 ECE5530, VECTOR RANDOM PROCESSES Properties that are true of all pdfs: 1. f X.x/ 0 8 x.. Z 1 f X.x/ dx D Pr.X x 0 / D Z x0 distribution function (cdf). 1 f X.x/ dx 4 D F X.x 0 /,whichistherv scumulative Problem: Apart from simple examples it is often difficult to determine f X.x/ accurately. Use approximations to capture the key behavior. Need to define key characteristics of fx.x/. EXPECTATION: Describes the expected outcome of a random trial. Nx D EŒX D Z 1 xf X.x/ dx. Expectation is a linear operator (very important for working withit). So, for example, the first moment about the mean: EŒX Nx D 0. STATISTICAL AVERAGE: Different from expectation. Consider a (discrete) RV X that can assume n values x1, x,...x n. Define the average by making many measurements N!1.Then, m i is the number of times the value of the measurement is i. Nx D 1 N.m 1x 1 C m x CCm n x n / D m 1 N x 1 C m N x CC m n N x n. In the limit, m1 =N! Pr.X D x 1 / and so forth (assuming ergodicity), NX Nx D x i Pr.X D x i /. id1 1 Some call this a probability distribution function, with acronym PDF (uppercase), as different from a probability density function, which has acronym pdf (lowercase).

3 ECE5530, VECTOR RANDOM PROCESSES 3 So, statistical means can converge to expectation in the limit. (Can show similar property for continuous RVs... but harder to do.) VARIANCE: Second moment about the mean. var.x/ D EŒ.X Nx/ D D EŒX.EŒX /, Z 1.x Nx/ f X.x/ dx or is equal to the mean-square minus the square-mean. STANDARD DEVIATION: Measure of dispersion about the mean of the samples of X: X D p var.x/. The expectation and variance capture key features of the actual pdf. Higher-order moments are available, but we won t need them! KEY POINT FOR UNDERSTANDING VARIANCE: Chebychev s inequality Chebychev s inequality states (for positive ") Pr.jX Nxj "/ X ", which implies that probability is concentrated around the mean. It may be proven as follows: Pr.jX Nxj "/ D Z Nx " f X.x/ dx C Z 1 NxC" f X.x/ dx. For the two regions of integration jx Nxj=" 1 or.x Nx/ =" 1. So, Pr.jX Nxj "/ Z Nx ".x Nx/ " f X.x/ dx C Since fx.x/ is positive, then we also have Pr.jX Nxj "/ Z 1.x Nx/ Z 1 NxC".x Nx/ " f X.x/ dx D X ". " f X.x/ dx.

4 ECE5530, VECTOR RANDOM PROCESSES 4 This inequality shows that probability is clustered around the mean, and that the variance is an indication of the dispersion of the pdf. That is, variance (later on, covariance too) informs us of how uncertain we are about the value of a random variable. Low variance means that we are very certain of its value; High variance means that we are very uncertain of its value. The mean and variance give us an estimate of the value of a random variable, and how certain we are of that estimate. The most important distribution for this course The Gaussian (normal) distribution is of key importance to us. (We will explain why this is true later see main point #7 on pg. 14.) Its pdf is defined as: 1.x Nx/ f X.x/ D p exp. X Symmetric about Nx. X Peak proportional to 1 X at Nx. Notation: X N. Nx; X / Probability that X within X of Nx is 68%; probabilitythatx within X of Nx is 96%; probabilitythatx within 3 X of Nx is 99:7%. A 3 X range almost certainly covers observed samples. Narrow distribution Sharp peak. High confidence in predicting X. Wide distribution Poor knowledge in what to expect for X.

5 ECE5530, VECTOR RANDOM PROCESSES 5.: Vector random variables With very little change in the preceding, we can also handle vectors of random variables. X D 6 4 X 1 X : X n I let x 0 D 6 4 x 1 x : x n X described by (scalar function) joint pdf fx.x/ of vector X. fx.x 0 / means f X.X 1 D x 1 ;X D x X n D x n /. That is, fx.x 0 / dx 1 dx dx n is the probability that X is between x 0 and x 0 C dx. Properties of joint pdf fx.x/: 1. f X.x/ 0 8 x. Sameasbefore.. Z 1 Z 1 3. Nx D EŒX D Z 1 f X.x/ dx 1 dx dx n D 1. Basicallythesame. Z 1 Z 1 Z 1 4. Correlation matrix: Different. X D EŒXX T (outer product) D Z 1 Z 1 ::: xf X.x/ dx 1 dx dx n.basicallysame. Z 1 xx T f X.x/ dx 1 dx dx n. 5. Covariance matrix: Different. Define e X D X Nx. Then, X D e X D EŒ.X Nx/.X Nx/ T D Z 1 Z 1 Z 1.x Nx/.x Nx/ T f X.x/ dx 1 dx dx n.

6 ECE5530, VECTOR RANDOM PROCESSES 6 X is symmetric and positive-semi-definite (psd). This means y T X y 0 8 y. PROOF: For all y 0, 0 EŒ.y T.X Nx// D y T EŒ.X Nx/.X Nx/ T y D y T X y. Notice that correlation and covariance are the same for zero-mean random vectors. The covariance entries have specific meaning:. X / ii D X i. X / ij D ij Xi Xj D. X / ji. The diagonal entries are the variances of each vector component; The correlation coefficient ij is a measure of linear dependence between X i and X j. j ij j1. The most important multivariable distribution for this course The multivariable Gaussian is of key importance for Kalman filtering. Nx 1 Nx

7 ECE5530, VECTOR RANDOM PROCESSES 7 f X.x/ D 1./ n= j X j 1= exp 1.x Nx/T X.x Nx/ : j X jddet. X /; X requires positive-definite X. Notation: X N. Nx; X /. Contours of constant fx.x/ are hyper-ellipsoids, centered at Nx, directions governed by X.Principleaxesdecouple X (eigenvectors). Two-dimensional zero-mean case: (Let 1 D X1 and D X ) " # X D 1 1 j X jd1.1 1 /: 0 1 x1 x 1 1 x B 1 f X1 ;X.x 1 ;x / D 1 q1 1 1 C x C / A.

8 ECE5530, VECTOR RANDOM PROCESSES 8.3: Uncorrelated versus independent INDEPENDENCE: Iff jointly-distributed RVs are independent, then f X.x 1 ;x ;:::;x n / D f X1.x 1 /f X.x / f Xn.x n /. Joint distribution can be split up into the product of individual distributions for each RV Equivalent condition: For all functions f./ and g./, EŒf.X 1 /g.x / D EŒf.X 1 / EŒg.X /. The particular value of the random variable X1 has no impact on what value we would obtain for the random variable X. UNCORRELATED: Two jointly-distributed RVs X 1 and X are uncorrelated if their second moments are finite and which implies 1 D 0. cov.x 1 ;X / D EŒ.X 1 Nx 1 /.X Nx / D 0 Uncorrelated means that there is no linear relationship between X1 and X. MAIN POINT #1: If jointly-distributed RV X 1 and X are independent then they are uncorrelated. Independence implies uncorrelation. To see this, notice that independence means EŒX1 X D EŒX 1 EŒX. Therefore, therefore uncorrelated. cov.x 1 ;X / D EŒX 1 X EŒX 1 EŒX D 0,

9 ECE5530, VECTOR RANDOM PROCESSES 9 Does uncorrelation imply independence? Consider the example Y 1 D sin.x/ and Y D cos.x/ with X uniformly distributed on Œ0; 1. We can show that EŒY1 D EŒY D EŒY 1 Y D 0. So, cov.y1 ;Y / D 0 and Y 1 and Y are uncorrelated. But, Y 1 C Y D 1 and therefore Y 1 and Y are clearly not independent. Note: cov.x1 ;X / D 0 (uncorrelated) implies that but independence requires for all f./ and g./. EŒX 1 X D EŒX 1 EŒX EŒf.X 1 /g.x / D EŒf.X 1 / EŒg.X / Therefore, independence is much stronger than uncorrelated. COROLLARY: Consider a RV X with uncorrelated components. The covariance matrix X D EŒ.X Nx/.X Nx/ T is diagonal. PROOF: Notice that: The diagonal elements are. X / ii D EŒ.X i Nx i / D i : The off-diagonal elements are. X / ij D EŒ.X i Nx i /.X j Nx j / D EŒX i X j X i Nx j Nx i X j CNx i Nx j D EŒX i EŒX j EŒX i EŒX j C EŒX i EŒX j D 0. MAIN POINT #: If jointly normally distributed RVs are uncorrelated, then they are independent. (This is a special case.)

10 ECE5530, VECTOR RANDOM PROCESSES 10.4: Functions of random variables MAIN POINT #3: Suppose we are given two random variables Y and X with Y D g.x/. Alsoassumethatg exists, g and g are continuously differentiable, then f Y.y/ D f ˇ, where kjjk means to take the absolute value of the determinant. So what? Say we know the pdf of X1 and X and ) Y 1 D X 1 C X Y D AX Y D X we can now form the pdf of Y handy! EXAMPLE: Y D kx and X N.0; X /. 1 X D k Y and so g.y/ D 1.y/ k y.then,@g D k. y ˇˇˇˇ 1 f Y.y/ D f X k kˇ D 1 1.y=k/ p exp jkj X X 1 y D p.x k/ exp.. X k/ Therefore, multiplication by gain k results in a normally-distributed RV with scale change in standard deviation. Y N.0; k X /. EXAMPLE: Y D AX C B where A is a constant (non-singular) matrix, B is a constant vector, and X N. Nx; X /. X D A Y A B so g.y/ D A y D A.

11 ECE5530, VECTOR RANDOM PROCESSES 11 fx.x/ D Therefore, f Y.y/ D D D 1 exp./ n= j X j1= ja j exp./ n= j X j1= 1./ n=.jajj X jja T j/ 1 exp./ n= j Y j1= 1.x Nx/T X.x Nx/. 1.A.y B/ Nx/ T 1= exp X.A.y B/ Nx/ 1.y Ny/T.A / T X A.y Ny/ 1.y Ny/ Y.y Ny/, if Y D A X A T and Ny D A Nx C B. Thatis,Y N.A Nx C B;A X A T /. CONCLUSION: Sum of Gaussians is Gaussian very special case. RANDOM NOTE: How to use randn.m to simulate non-zero mean Gaussian noise with covariance Y? Y N. Ny; Y / but randn.m returns X N.0; I /. Try y DNy C A T x where A is square with the same dimension as Y ; A T A D Y.(A is the Cholesky decomposition of positive-definite symmetric matrix Y ). ybar = [1; ]; covar = [1, 0.5; 0.5, 1]; A = chol(covar); x = randn([, 1]); y = ybar + A'*x; When Y is non-positive definite (but also non-negative definite) y coordinate samples with mean [1;] and covar [1,0.5; 0.5,1] [L,D] = ldl(covar); x = randn([,5000]); y = ybar(:,ones([1 5000])) +(L*sqrt(D))*x; x coordinate

12 ECE5530, VECTOR RANDOM PROCESSES 1 MAIN POINT #4: Normality is preserved in a linear transformation (extension of above example). Let X and W be independent vector random variables, a is a constant (vector): X N. Nx; X /, W N. Nw; W /. Let Z D AX C BW C a EŒZ DŃDANx C B Nw C a Z D A X A T C B W B T. Z is also a normal RV and Z N.Ń; Z /.

13 ECE5530, VECTOR RANDOM PROCESSES 13.5: Conditioning MAIN POINT #5: Conditional probabilities. Given jointly-distributed RVs, it is often of extreme interest to find the pdf of some of the RVs given known values for the rest. For example, given the joint pdf fx;y.x; y/ for RVs X and Y,wewant the conditional pdf of f XjY.x j y/ which is the pdf of X for a known value Y D y. Can also think of it as fxjy.x D x j Y D y/. DEFINE: Conditional pdf f XjY.x j y/ D f X;Y.x; y/ f Y.y/ is the probability that X D x given that Y D y has happened. NOTE I: The marginal probability f Y.y/ may be calculated as f Y.y/ D Z 1 For each y, integrateouttheeffectofx. f X;Y.x; y/ dx. NOTE II: If X; Y independent, f X;Y.x; y/ D f X.x/f Y.y/. Therefore f XjY.x j y/ D f X;Y.x; y/ f Y.y/ D f X.x/f Y.y/ f Y.y/ D f X.x/. Knowing that Y D y has occurred provides no information about X. (Is this what you would expect?) DIRECT EXTENSION: f X;Y.x; y/ D f XjY.x j y/f Y.y/ D f Y jx.y j x/f X.x/,

14 ECE5530, VECTOR RANDOM PROCESSES 14 Therefore, f XjY.x j y/ D f Y jx.y j x/f X.x/. f Y.y/ This is known as Bayes rule. It relates the aposterioriprobability to the aprioriprobability. It forms a key step in the Kalman filter derivation. MAIN POINT #6: Conditional expectation. Now that we have a way of expressing a conditional probability, we can also express a conditional mean. What do we expect the value of X to be given that Y D y has happened? EŒX D x j Y D y D EŒX j Y D Z 1 xf XjY.x j Y/dx. Conditional expectation is not a constant (as expectation is) but a random variable. It is a function of the conditioning random variable (i.e., Y ). Note: Conditional expectation is critical. TheKalmanfilterisan algorithm to compute EŒx k j Z k,wherewedefinez k later. MAIN POINT #7: Central limit theorem. X If Y D X i and the X i are independent and identically distributed i IID), and the X i have finite mean and variance, then Y will be approximately normally distributed. The approximation improves as the number of summed RVs gets large.

15 ECE5530, VECTOR RANDOM PROCESSES 15 Since the state of our dynamic system adds up the effects of lots of independent random inputs, it is reasonable to assume that the distribution of the state tends to the normal distribution. This leads to the key assumptions for the derivation of the Kalman filter, as we will see: We will assume that state x k is a normally-distributed RV; We will assume that process noise w k is a normally-distributed RV; We will assume that sensor noise v k is a normally-distributed RV; We will assume that w k and v k are uncorrelated with each other. Even when these assumptions are broken in practice, the Kalman filter often works quite well. Exceptions to this rule tend to be with very highly nonlinear systems, for which particle filters must sometimes be employed to get good estimates.

16 ECE5530, VECTOR RANDOM PROCESSES 16.6: Vector random (stochastic) processes Astochasticrandomprocessisafamilyofrandomvectorsindexed by aparameterset( time inourcase). For example, the random process X.t/. The value of the random process at any time t 0 is RV X.t 0 /. Usually assume stationarity. The pdf of the random variables are time-shift invariant. Therefore, EŒX.t/ DNx for all t and EŒX.t 1 /X T.t / D R X.t 1 t /. Properties 1. Autocorrelation: R X.t 1 ;t / D EŒX.t 1 /X T.t /. Ifstationary, R X./ D EŒX.t/X T.t C /. Provides a measure of correlation between elements of the process having time displacement. RX.0/ D X for zero-mean X. RX.0/ is always the maximum value of R X./.. Autocovariance: C X.t 1 ;t / D EŒ.X.t 1 / EŒX.t 1 / /.X.t / EŒX.t / / T.If stationary, C X./ D EŒ.X.t/ Nx/.X.t C / Nx/ T. White noise Correlation from one time instant to the next is a key property ofa random process. R X./ and C X./ very important. Some processes have a unique autocorrelation:

17 ECE5530, VECTOR RANDOM PROCESSES Zero mean,. R X./ D EŒX.t/X.t C / T D S X ı./ where ı./ is the Dirac delta. Therefore, the process is uncorrelated in time. Clearly an abstraction, butprovestobeavery useful one. 4 White noise 0. Correlated noise Value 0 Value Time Power spectral density (PSD) Time Consider stationary random processes with autocorrelation defined as R X./ D EŒX.t/X.t C / T. If a process varies slowly, time-adjacent samples are very correlated, and R X./ will drop off slowly Mostly low-frequency content. If a process varies quickly, time-adjacent samples aren t very correlated, and R X./ drops off quickly More high-freq. content. Therefore, the autocorrelation function tells us about the frequency content of the random signal. Consider scalar case. We define the power spectral density (PSD) as the Fourier transform of the autocorrelation function:

18 ECE5530, VECTOR RANDOM PROCESSES 18 Z 1 S X.!/ D R X./ D 1 Z 1 R X./e j! d S X.!/e j! d!. SX.!/ gives a frequency-domain interpretation of the power (energy) in the random process X.t/, andisreal,symmetricand positive definite for real random processes. EXAMPLE: R X./ D e jj for >0.Then, S X.!/ D Z 1 R X./ Peak of e jj e j! d D! C. S X.!/Peak of Autocorrelation is a bandwidth measure. PSD! INTERPRETATION: 1. Area under PSD S X.!/ for! 1!! provides a measure of energy in the signal in that frequency range.. White noise can be thought of as a limiting process as!1. From our prior example, let!1. As!1, RX./! ı./; thereforewhite. Bandwidth of spectral content is approximately, sogoestoinfinity. Abstraction since implies that signal has infinite energy!

19 ECE5530, VECTOR RANDOM PROCESSES 19.7: Discrete-time dynamic systems with random inputs What are the statistics of the state of a system driven by a random process? That is, how do the system dynamics influence the time-propagation of x.t/? Do not know x.t/ exactly, but can develop a pdf for x.t/. Primary example: A linear system driven by white noise w.t/, possibly in addition to deterministic input u.t/. u.t/ w.t/ A; B; C; D x.t/ y.t/ x.t/ Contrast to deterministic simulation, which can be easily simulated Px.t/ D Ax.t/ C Bu.t/; x.0/ D x 0 ; x.0/; u.t/known. System response is completely specified. Therefore, no uncertainty. The stochastic propagation is different since there are inputs that are not known exactly. Best we can do is to say how uncertainty in the state changes with time. x Pr.x/ x.t f / x x.t f / x.t o / Deterministic x.t x o / 1 x 1 Stochastic Deterministic: x.t0 / and inputs known State trajectory deterministic.

20 ECE5530, VECTOR RANDOM PROCESSES 0 Stochastic: pdf for x.t0 / and inputs known Can find only pdf for x.t/. We will work with Gaussian noises to a large extent, which are uniquely defined by the first- and second central moments of the statistics Gaussian assumption not essential. We will only ever track the first two moments. NOTATION: Until now, we have always used capital letters for random variables. The state of a system driven by a random process is a random vector, so we could now call it X.t/ or X k.however,itismore common to retain the standard notation x.t/ or x k and understand from the context that we are now discussing an RV Discrete-time systems We will start with a discrete system and then look at how to get an equivalent answer for continuous systems. Model: x k D A d;kx k C B d;ku k C w k, where x k W State vector; arandomprocess: u k W Deterministic control inputs: w k W Noise that drives the process.process noise/; arandomprocess. Ad and B d assumed known. Can be time-varying. For now, assume fixed. Generalize later if necessary. Key assumptions about driving noise:

21 ECE5530, VECTOR RANDOM PROCESSES 1 1. Zero mean: EŒw k D 0 8 k.. White: EŒw k1 wk T D w.k 1 k /. w is called the spectral density of the noise signal w k. Uncertainty at startup: Statistics of initial condition. EŒx 0 DNx 0 I EŒ.x 0 Nx 0 /wk T D 0 8 k x;0 D EŒ.x 0 Nx 0 /.x 0 Nx 0 / T. Key question: How do we propagate the statistics of this system? Mean value: The mean value is EŒx k DNx k D EŒA d x k C B d u k C w k D A d Nx k C B d u k. Therefore, the mean propagation is Nx 0 W Given Nx k D A d Nx k C B d u k. Deterministic simulation and mean values treated the same way. Variations about mean To study the random variations about the mean, we need to form the second central moment of the statistics. x;k D EŒ.x k Nx k /.x k Nx k / T.

22 ECE5530, VECTOR RANDOM PROCESSES Easiest to study if we note that x k Nx k D A d x k C B d u k C w k Thus, A d Nx k B d u k D A d.x k Nx k / C w k. x;k D EŒ.A d.x k Nx k / C w k /.A d.x k Nx k / C w k / T. Three terms: 1. EŒA d.x k Nx k /.x k Nx k / T A T d D A d x;k A T d.. EŒw k w T k D w. 3. EŒA d.x k Nx k /wk T D? The third term is a cross-correlation term. But, x k depends only on x 0 and inputs w m for m D 0:::k. w k is white noise uncorrelated with x 0. Therefore, third term is zero. Therefore, the covariance propagation is x;0 W Given x;k D A d x;k A T d C w. In this equation, A d x;k A T d is the homogeneous part; w is the driving term. Note: If Ad and w are constant and A d stable, there is a steady-state solution. As k!1, x;k D x;kc1 D x.

23 ECE5530, VECTOR RANDOM PROCESSES 3 Then, x D A d x A T d C w. This form is called a discrete Lyapunov equation. In MATLAB, dlyap.m EXAMPLE: Can solve by hand in the scalar case. Consider: xk D x k C w k. EŒw k D 0; w D 1; Nx 0 D 0. Then, x.1 / D 1 x D x T C 1 x D 1 1. Valid for j j <1;otherwiseunstable. In the example below, we plot 100 random trajectories for D 0:75 and x;0 D 50. Compare propagation of x;k with steady-state dlyap.m solution. 50 x D :9 30 Propagation of state, with error bounds 40 0 x;k 30 0 Value Time Time (100 simulations plotted) The error bounds plotted are 3 bounds (i.e., 3 p x;k ).

24 ECE5530, VECTOR RANDOM PROCESSES 4.8: Continuous-time dynamic systems with random inputs For continuous-time systems we have where Px.t/ D A.t/x.t/ C B u.t/u.t/ C B w.t/w.t/, x.t/ W State vector; arandomprocess: u.t/ W Deterministic control inputs: w.t/ W Noise driving the process.process noise/; arandomprocess. Further: EŒw.t/ D 0I EŒw.t/w./ T D S w ı.t / EŒx.0/ DNx.0/I EŒ.x.0/ Nx.0//.x.0/ Nx.0// T D x.0/. A, Bu and B w assumed known. Can be time-varying. For now, assume fixed. Generalize later if necessary. Sw is the spectral density of w.t/. Easiest analysis is to discretize model; use results obtained earlier; let t! 0. Drop deterministic inputs u.t/ for simplicity. (They affect only the mean, and in known ways.) Continuous-time propagation of statistics Mean value Starting with the discrete case, we have Nx k D A d Nx k C B d u k A d D e A t I C A t C O. t /.

25 ECE5530, VECTOR RANDOM PROCESSES 5 Let t! 0 and consider case when uk D 0 for all k. Nx k D.A t C I/Nx k As t! 0 Nx k Nx k t D A Nx k. PNx.t/ D A Nx.t/ as expected. Therefore, the mean propagation is Nx.0/ W Given PNx.t/ D A Nx.t/ C Bu.t/. Deterministic simulation and mean values treated the same way. Variations about mean The result for discrete-time systems was: x;k D A d x;k A T d C w. But, we need a way to relate discrete w to a continuous spectral density S w before we can proceed. Recall the discrete system response in terms of continuous system matrices: x k D e A t x k C Z k t.k/ t e A.k t / B w w./ d D e A t x k C w k. Integral explicitly accounts for variations in the noise during t. RECALL: Discrete white noise of the form 8 < EŒw k wl T D w ; k D li : 0; k l:

26 ECE5530, VECTOR RANDOM PROCESSES 6 But noise input term to our converted state dynamics is defined asan integral w k D Z k t.k/ t e A.k t / B w w./ d. Form outer product to get equivalent w Z! k t Z k t w D E 4 e A.k t / B w w./ d D E " Z k t.k/ t Z k t.k/ t.k/ t.k/ t! 3 T e A.k t / B w w./ d 5 # e A.k t / B w w./w./ T Bw T eat.k t / d d. Big ugly mess but has two saving graces 1. Expectation can go inside integrals.. EŒw./w./ T D S w ı. / One of the integrals drops out! So, we have w D Z k t.k/ t e A.k t / B w S w B T w eat.k t / d. KEY POINT: While S w may have simple form, w will be full matrix in general. To solve integral, we can approximate: As t! 0, thenk t! 0 and e A.k t / I C A.k t / C That is, e A.k t / I.Then, w B w S w Bw T t. We will see a better method to evaluate w when t 0, butfornow we continue with this result to determine the continuous-time system covariance propagation.

27 ECE5530, VECTOR RANDOM PROCESSES 7 Start with x;k D A d x;k A T d C w,andsubstitute w B w S w B T w t. Also, use A d.i C A t/ x;k x;k t As t! 0 x;k.i C A t/ x;k.i C A t/ T C B w S w B T w t initialized with x.0/. D x;k C t.a x;k C x;k A T C B w S w B T w / C O. t / D A x;k C x;k A T C B w S w B T w C O. t/. P x.t/ D A x.t/ C x.t/a T C B w S w B T w, This is a matrix differential equation. Symmetric, so don t need to solve for every element. Two effects A x.t/ C x.t/a T :Homogeneouspart.ContractiveforstableA. Reduces covariance. B w S w B T w :Impactofprocessnoise.Tendstoincreasecovariance. Steady-state solution: Effects balance for systems with constant dynamics (A, B w, S w )andstablea. A x;ss C x;ss A T C B w S w B T w D 0. This is a continuous-time Lyapunov equation. In MATLAB, lyap.m The covariance propagation is x.0/ W Given P x.t/ D A x.t/ C x.t/a T C B w S w B T w. In this equation, A x.t/ C x.t/a T is the homogeneous part; B w S w B T w is the driving term.

28 ECE5530, VECTOR RANDOM PROCESSES 8 EXAMPLE: Px.t/ D Ax.t/ C B w w.t/, wherea and B w are scalars. Then, P x D A x C B w S w.thiscanbesolvedinclosedformtofind x.t/ D B w S w A.eAt 1/ C x.0/e At. If A<0(stable) then the initial condition contribution goes to zero. independent of x.0/. Example: Let A D, Bw D 1, S w D, and x.0/ D 5. Find x.t/. x;ss D B w S w A 5 4 x;ss D 1 Solution: Increased x;ss as more noise added: Bw S w term. Decreased x;ss as A becomes more stable. x.t/ Time

29 ECE5530, VECTOR RANDOM PROCESSES 9.9: Relating w to S w precisely: A little trick Want to find w precisely, regardless of t. AssumeweknowS w and plant dynamics. Define Z D w D Z t 0 4 A B ws w B T w 0 A T e A. t / B w S w B T w eat. t / d. 3 5 ; C D e Z t D 4 c 11 c 1 0 c We can compute C D L Œ.sI Z/ ˇˇtD t and find that c 11 D L Œ.sI 11 / ˇˇtD t D L Œ.sI C A/ ˇˇtD t D e A t 3 5 c 1 D L Œ.sI 11 / 1.sI / ˇˇtD t D L Œ.sI C A/ B w S w Bw T.sI AT / ˇˇtD t c D L Œ.sI / D L ˇˇtD t Œ.sI A T / D e ˇˇtD t AT t D A T d So, c T D A d.alsoc T c 1 D w.withoneexpm.m command, can quickly switch from continuous-time to discrete-time. To show the second identity, recognize that c1 is a convolution of two impulse responses: one due to the system.si C A/ B w and the other due to the system S w Bw T.sI AT /. The first of these has impulse response Ie At B w and the second has impulse response S w B T w eat t.convolvingthemweget Z t 0 Ie A B w S w B T w eat. t / d.

30 ECE5530, VECTOR RANDOM PROCESSES 30 Finally, consider c T c 1 D EXAMPLE: Let Z t 0 with S w D 0:06 and t D =16. e A. t / B w S w B T w eat. t / d D w. " # " # Px.t/ D x.t/ C w.t/ 1 Z D :06 I C D 3 0:91 0:47 0:006 0:0046 0:47 1:38 0:0046 0:03 : :93 0: :3 0:6 So, A d D " 0:93 0:3 0:3 0:6 # I w D " 0:0009 0:0030 0:0030 0:0156 # I B w S w B T w D " :06 #. Exact w and approximate w very different due to large t. Compare predictions of steady-state performance. Continuous: " # A x;ss C x;ss A T C B w S w Bw T D 0I 0:03 0 x;ss D : 0 0:03 " # Discrete: x;ss D A d x;ss A T d C 0:03 0 wi x;ss D : 0 0:03 Because we used discrete white noise scaled to be equivalent to the continuous noise, the steady-state predictions are the same. Conversion now complete:

31 ECE5530, VECTOR RANDOM PROCESSES 31 DISCRETE: A d, w, B d Given x;0 W x;k D A d x;k A T d C w Given Nx 0 W Nx k D A d Nx k C B d u k. CONTINUOUS: A, S w, B u, B w Given x.0/ W P x.t/ D A x.t/ C x.t/a T C B w S w B T w Given Nx.0/ W PNx.t/ D A Nx.t/ C B u u.t/. All matrices may be time varying. Equivalent w for given S w available.

32 ECE5530, VECTOR RANDOM PROCESSES 3.10: Shaping filters We have assumed that the inputs to the linear dynamic systems are white (Gaussian) noises. Therefore, the PSD is flat The input has content at all frequencies. Pretty limiting assumption, but one that can be easily fixed Can use second linear system to shape the noise and modify the PSD as desired. White noise w.t/ Previous Picture G.s/ y.t/ Shaped noise w 1.t/ New Picture G.s/ y.t/ White noise w.t/ H.s/ w 1.t/ Shaped G.s/ y.t/ Therefore, we can drive our linear system with noise that has a desired PSD by introducing a shaping filter H.s/ that itself is driven by white noise. The combined system GH.s/ looks exactly the same as before, but the system G.s/ is not driven by pure white noise any more. Analysis quite simple: Augment original model with filter states. Original system has Px.t/ D Ax.t/ C B w w 1.t/ y.t/ D Cx.t/ New shaping filter with white input and desired PSD output has

33 ECE5530, VECTOR RANDOM PROCESSES 33 Px s.t/ D A s x s.t/ C B s w.t/ w 1.t/ D C s x s.t/. Combine into one system Px.t/ 5 D 4 A B wc s 5 4 x.t/ 5 C w.t/ Px s.t/ 0 A s x s.t/ B s h i 3 y.t/ D C 0 4 x.t/ 5 x s.t/ Augmented system just a larger-order linear system driven by white noise. KEY QUESTION: Given a PSD for w 1,howdowefindH.s/ (or, A s, B s, C s ). Fact: If w is unit-variance white noise, then PSD for w 1 is S w1.!/ D H. j!/h T.j!/. Find H.j!/ by spectral factorization of Sw1.!/. Takethe minimum-phase part (keep the poles and zeros that have negative real part). EXAMPLE: Let Then S w1.!/ D D! C p p C j! j!.

34 ECE5530, VECTOR RANDOM PROCESSES 34 EXAMPLE: Wind model. H.s/ D p s C ; Px s.t/ D x s.t/ C p w.t/ w 1.t/ D x s.t/. Wind gusts have some high-frequency content, but typically are dominated by low-frequency disturbances White noise a good model? We would typically think of a PSD for the wind that puts more emphasis on the low-frequency component of the signal. Think of this as the output of a shaping filter H.s/ D 1 w s C 1. NOTE: If the bandwidth of the wind filter.1= w />bandwidth of plant dynamics, then white noise is a good approximation to system input. Otherwise, must use shaping filters in plant model and drive with white noise. The size of the A matrix becomes a limitation. Use the simplest noise model possible.