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    Limit behaviors of extended Kalman filter as a parameter estimator for a sinusoidal signal

    2018-07-31 03:30:18LiXIE
    Control Theory and Technology 2018年3期

    Li XIE

    State Key Laboratory of Alternate Electrical Power System with Renewable Energy Sources,School of Control and Computer Engineering,North China Electric Power University,Beijing 102206,China

    Abstract In this note,the basic limit behaviors of the solution to Riccati equation in the extended Kalman filter as a parameter estimator for a sinusoidal signal are analytically investigated by using lim sup and lim inf in advanced calculus.We show that if the covariance matrix has a limit,then it must be a zero matrix.

    Keywords:Extended Kalman filter,parameter estimator,sinusoidal signal,covariance matrix,limit behavior

    1 Introduction

    Since the standard Kalman filter for linear systems was invented in 1960 by R.E.Kalman,there have been a number of applications both in theory and practice.One of the applications was Apollo navigation system in the 1960s where the actual version of Kalman filter on board was the extended Kalman filter(EKF)adapted for nonlinear systems[1].In EKF,instead of exact state and observation matrices provided by linear models,these matrices are obtained by linearizing nonlinear models around the predicted values or filtered values of state vectors,which introduces model errors and the filter may quickly diverge[2].Hence unlike its linear counterpart,there are stability and convergence issues in EKF for the linearization.

    Concerning the convergence and stability analysis of discrete-time EKF,there are two methods related to the topic of this note.Ljung in[3]developed a differential equation method to analyze the convergence of the estimates for linear systems with unknown parameters.A simple linear system with unknown parameter as a numerical example was used to illustrate the method.Reif et al.in[2]analyzed the stochastic stability of EKF by using Lyapunov function method;see also[4]for the Lyapunov function method.Lower and upper bounds of covariance matrices were required to solve a meansquare boundedness problem.

    In this note,we considera parameterestimation problem for a sinusoidal signal,and EKF is used to estimate an unknown parameter.The measurements of the sinusoidal signal are corrupted by white noise.Since the related observation equation is nonlinear,Ljung’s method cannot be directly adapted for such an estimation problem.Meanwhile,we will see that the covariance matrix in our parameter estimation does not have a strict positive lower bound,thus the stochastic stability cannot be easily established by using Lyapunov function method.Instead of the convergence of parameter estimates and the stochastic stability of covariance matrices,we study the limit of the covariance matrix.The limit behavior of the solution to the Riccatiequation of the covariance matrix is also mainly concerned both in practice and theory.For example,an asymptotic analysis for the covariance matrix was carried out in[5]when an extended complex Kalman filter was used to estimate a sinusoidal signal.The main result is stated in Theorem 1 in Section 2.2.A numerical example is given in Section 3 to show the limit behavior of the EKF.

    2 Two-state extended Kalman filter

    Consider a sinusoidal signal with an additive noise

    We assume that the magnitude a is a known constant and the angular frequency ω as an unknown constant will be estimated.Let φ= ωt.In order to solve the parameter estimation problem of the sinusoidal signal(1)by using discrete-time EKF,we follow a procedure given in[6,Chapter 10]to establish the discrete-time state and measurement equations of(1).By calculating the derivatives of φand ω with respect to the time t,we have

    We then discretize the continuous-time state equation(2)by sampling.After a computation,we have its fundamental matrix as follows:

    in which Tsis the sampling time.We next use the nonnegative integer k to denote the discrete-time kTs.Then the discrete-time state and observation equations are obtained

    where the Gaussian white noise v has zero mean and variance R>0.In this way,we can use a real-valued EKF for the nonlinear system(3)to estimate the unknown parameter ω of the sinusoidal signal(1)with an additive noise.

    2.1 Extended Kalman filter

    Let xk+1denote x(k+1).The extended Kalman filter of the nonlinear system(3)is given as follows:1)State and covariance matrix update

    2)State and covariance matrix predict

    and substituting it into the random Riccati equation(9),after a straightforward calculation,we have

    Here we use cos2to denote cos2()for convenience.By(4)and(7),we also obtain the recursive formulas for the state estimation

    Since Pkis a covariance matrix for the linearized system,it should be a positive semidefinite matrix[10,Page 275]and[9]which can also be seen from the first equality in(9).Hence Pk,11≥0,Pk,22≥0,det(Pk)=Obviously each entry of the matrix Pkis a number since the noise covariance R>0.Using(12),we can rewrite(10)and(11)as

    where bk=

    The next proposition describes the singularity of Pkby its determinant as k increases.

    Proposition 11)If there exists a finite k1≥0 such that det(Pk1)=that is,Pk1is a singular matrix,then

    2)If there exists a finite k1≥0 such thatthat is,Pk1>0,then

    ProofSuppose det(Pk1)straightforward calculation shows that this equality also holds for k+1.Substituting it into(12)–(14),we have

    Therefore,

    Then 1)follows.The statement 2)of this proposition is directly due to the matrix inversion lemma;see(21)in Section 2.4. □

    2.2 The limit of the solution to Riccati equations

    It follows from(12)that

    Hence we have the following lemma.

    Lemma 1The sequence Pk,22monotonically decreases with increasing k.The limit of Pk,22exists and is greater than or equal to zero.Also

    Assumption 1The sequence cos2does not have a limit as k approaches∞.

    Proposition 2If P0/0,12≥0,then Pk1,12≥0 for any k>0;otherwise either there is a finite k1>0 such that for any k≥k1,Pk,12≥0 or under Assumption 1,we have the limits

    ProofSince P1,12∶=P1/0,12=P0/0,12+TsP0/0,22,Pk,12≥0 follows from(14)if P0/0,12≥0.For the other case,once there existsa finite k1>0 such that Pk1,12≥0,then it is easy to see that Pk,12≥ 0,?k> k1from(14).Hence we next assume that Pk,12<0 for any finite k in order to establish(16),then again by(14)that Pk,12≤Pk+1,12<0,thus its limit exists by the monotonic property.Suppose further thatthen by using the equality(14)one more time and the convergence of Pk,22,we have the limit

    If c2≠0,then(15)implies that the limit of cos2equals zero,which contradicts the assumption that cos2does not have a limit.Hence c2=0.Finally by taking the limit of(14),we have

    which also contradicts that c1<0 and the fact that the right-hand side is great than or equal to zero.We conclude that the limits(16)hold. □

    In the sequel we always assume that Pk,12≥0.As we will see subsequently in Theorem 2,if Pkis invertible,then eventually Pk,12>0.We next make use of the inequalities of lim sup and lim inf for two sequences xkand yk,for example,if xk≤yk,then

    and for two non-negative sequences xk,yk

    See[11,Problem 2.4.17]for details.All related inequalities as above make senses provided that both sides are not of the indeterminate forms0×±∞,±∞?∞.For each sequence defined on the extended real line[?∞,∞],the limit superior and the limit inferior always exist.

    Assumption 2

    Remark 1Ifthen∞,that is,Pk,11is unbounded since

    With the help of Assumption 2,we can apply(17)to obtain the main result.

    Theorem 1Under Assumptions 1 and 2,we have

    2)If the limit of Pk,11exists as k→ ∞,we have

    ProofIt follows from(15)that

    from which three possible cases follow

    Due to

    we have

    Further it follows from(14)that Pk+1,12≥TsPk+1,22,then taking lim inf on both sides,

    We now establish the second limit of this part.Since Pk≥0,we haveTaking lim sup on both sides yields

    Hence,

    Corollary 1Under Assumptions 1 and 2,we have

    ProofIt follows from(13)and using(1)in Theorem 1 that

    One can easily see from the last inequality that eitherSince the former leads to the latter,we conclude that

    2.3 The solution to recursive equations

    In this section,we use the transition function of the recursions(13)and(14)to derive their solutions in terms of initial values of the states.

    Denote αk=R/(R+bk)and define the transition function

    Then after a straightforward calculation,we obtain the solution to(14)in terms of the initial value and the transition function

    Notice that in order to obtain(18),we use(8)to calculate P1,12and define α0=1.Introducing a new sequence

    substituting(13)into(19),and using(14),we have

    Then the solution to(13)is given by

    Proposition 3Under Assumption 1,if the initial value of Pk,11is not equal to zero,then it is forgotten in the long run.We also have

    ProofSince Ψ(k+1,0)= αkΨ(k,0),it follows from αk≤ 1 that Ψ(k+1,0)≥ 0 is monotonically decreasing and hence has a limit.We now assume0,then

    Forthe non-trivial case P0,11>0,it follows from(20)thatThanks to the left-hand side inequality of(17),we have

    The second equality is due to the equivalence

    See[12,Page 17]or[13,Page 220,Theorem 4]in which the left-hand side is referred to as a divergent infinite product.The claim of the proposition directly follows from(20)in which the transition function Ψ(k+1,0)is coefficient of P0,11. □

    2.4 The limit of the inverse matrix of Pk

    By the matrix inversion lemma,

    Denote A?T=(AT)?1.Taking the inverse of both sides of(21)gives a Lyapunov equation

    Substituting the partitioned matrixand the inverse of the matrix A

    into the Lyapunov equation(22)yields

    Further we obtain recursive formulas for all entries

    By using(24),the equality(25)can be rewritten as

    Making use of the above equalities,we obtain the following theorem.

    Theorem 2Under Assumptions 1 and 2,suppose that there exists a finite k1≥0 such that det(Pk1)=that is,Pk1>0,then

    The entry Pk,12of Pkeventually is greater than zero.We have also

    ProofBy Proposition 1,the covariance matrix Pk>0 for k>k1which ensures that its inverse exists.Then one can easily obtain its inverse matrix as follows:

    Since we assume that the limit of the sequence cos2φkdoes not exist,we haveotherwise its limit is zero.Then(23)gives

    from which we obtain

    Thanks to(24),we have

    An argument similar to the one used in(27)shows that

    Notice that since

    as k→∞due to Pk,12→0,as k→∞.Therefore we have Pk,11Pk,22→ 0 as k→ ∞,which implies the last claim in this theorem. □

    3 A numerical example

    Consider a sinusoidal signal with an additive noise.The discrete-time state and observation equations are

    where a=0.5 and the Gaussian white noise v has zero mean and variance R=1.We employ the extended Kalman filter to estimate the unknown parameter ω.The true value of ω is 1.5.In simulation we let P0/0be the identity matrix.

    Fig.1 shows a sample path of sinusoidal signal and its measurements with the sampling time Ts=0.1.Indeed,the limit of cos2does not exist.Hence we claim thatby Theorems 1 and 2.This can be seen in Fig.2.In Fig.3,the dependence of the parameter estimatesωkis found on the sampling time Tsfor the same sample path.The parameter estimate is obviously improved by decreasing the sampling time.Notice that the convergence of Pk,22to zero does not necessarily guarantee that the parameter estimates converge to the true value.In Fig.4,we also give the time average for a sample path of the parameter estimateωkand the sample average over 1000 sample paths ofωk.Obviously the EKF as a parameter estimator under consideration is biased.

    Fig.1 The sinusoidal signal and measurements.

    Fig.2 The entries and the determinant of Pk.

    Fig.3 The estimatesωk with different T s s.

    Fig.4 The time and sample averages ofωk.

    4 Conclusions

    In this note,by using lim sup and lim inf,we study the limit behaviors of the extended Kalman filter as a parameter estimator for a sinusoidal signal.The estimation problem of a sinusoidal signal often occurs in power systems.We claim that three entries of the covariance matrix and its determinant have zero limits as the time approaches infinity.Further,if the limit of the covariance matrix exists,we show that it must be a zero matrix.However we also find that it is difficult to obtain the limit behavior of the entry Pk,11of the 2×2 covariance matrix Pk,and the question concerning its existence remains open.Further research could explore the possibility whether or not the differential equation method developed by Ljung in[3]can be adapted to analyze the convergence of the EKF under consideration,and also investigate the ergodic property of the EKF by using operator methods in[14]and references therein.

    Acknowledgements

    The author thanks the reviewers for valuable comments that help to improve the presentation.

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