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We cannot apply the commutative law to subtraction problems and expect valid results Let s look at these two subtractions: 5 10 = 5 but 10 5 = 5

Figure 57: Feedforward complementary lter implementation of the algorithm in Table 56 the feedback algorithm summarized in Table 58, the error state is absent; however, the equations for Kk , P+ and P are exactly the same These k k+1 issues often cause confusion and therefore deserve further comment The algorithm in Table 56 is illustrated in Figure 57 Integration of the vehicle kinematics provides the reference trajectory x which allows calculation of the output y The EKF estimates and propagates the error state which is added to the navigation system output to produce the total navigation state estimate The algorithm in Table 58 is illustrated in Figure 58 Again, integration of the vehicle kinematics provides the EKF with the nominal trajec tory and output At initialization x0 = E x(0) ; therefore, x+ (0) = 0 and P+ = var( x+ (0)) = var(x(0)) Assuming that the rst measurement 0 occurs for k = 1, the error state and its covariance at time t1 are computed as x 1 P 1 = = 0 x+ (0) = 0 0 P+ 0 + Qd0 0

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Question 4-9

047 05 056 068 1 2 22 33 47 56 1 2

Because of the fact that x = 0, the measurement update 1 x1 x1 + = + K1 z1 H1 x1 reduces to + = K1 z1 x1

Answer 4-9

We can t apply the associative law to subtraction done twice and expect valid results Here s an example showing its failure: (5 10) 15 = 5 15 = 20 but 5 (10 15) = 5 ( 5) =5+5 = 10

(5103)

Part One Question 4-10

1 1 2 3 4 8 7 6 5

1 with K1 and P+ computed as shown in Table 58 Given x and + , the x1 1 best estimate of the total state after the measurement is 1 1 1 x1 x+ = x + + = x + K1 z1 (5104)

Suppose a, b, and c are integers and we see the expression a b + c without any parentheses in it Can we use parentheses for grouping, apply the associative law, and expect valid results in this situation

which is equal to the expected value of x(t1 ) after accounting for the measurement information at time t1 , which is denoted as x+ Because the state 1 correction of eqn (5104) results in x+ (t1 ) = E x+ then it is now true 1 that 1 (5105) + = x+ E x+ = 0 x1 1 The software implementation of the algorithm must make the assignment + = 0 immediately after implementing eqn (5104) Failure to reset x1 + 1 to zero after correcting the total state estimate as in eqn (5104) x would result in double accounting for + in the subsequent measurement x1 correction The result of eqn (5104) serves as the initial condition for the next period of integration Because + = 0, we will have that x = 0 x1 2 and the measurement correction at t2 will still reduce as in eqn (5103) The above procedure can be replicate at each measurement instant Note that it is not proper to state that the error state is not propagated through time It is correct to state that the error state time propagation is trivial and is therefore not implemented A detailed example of the use of the error state to correct the total state is presented in Section 126

Answer 4-10

14 14 13 12 11 10 9 8

No! Here s an example: (5 10) + 15 = 5 + 15 = 10 but 5 (10 + 15) = 5 25 = 20

There are many excellent books on Kalman lter theory and design The main references for the presentation of this chapter were [31, 34, 58, 60, 93] The complementary lter and its relation to the Kalman lter is described well in [30, 31] Alternative nonlinear estimation approaches, including the unscented and particle lters, are presented in a uni ed notation similar to that used herein in [121]

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