The paper Convergence Analysis of the Unscented Kalman Filter for Filtering Noisy Chaotic Signals

presents the convergence analysis of Unscented Kalman Filter

download http://www.eie.polyu.edu.hk/~cktse/pdf-paper/ISCAS07-Feng1.pdf . The Measurement update equations from Eq16 -- Eq19 are clear to me but I don't understand how Eq(20) is derived which is $P_n = K_nR_n$ where $P_n$ is the covariance of the filtering error and $R_n$ is the covariance matrix of the measurement noise. Can somebody please show the steps how Eq(20) is reached ? I tried something like this starting from Eq(18) but I am stuck --

$$ P_n = P_n^- - K_n P_{y_n y_n}K_n^T,$$ $$ = P_{y_n y_n} - R_n - K_n P_{y_n y_n}K_n^T $$ (substituting Eq 19) $$ = P_{x_ny_n} - K_n P_{y_n y_n}K_n^T $$


1 Answer 1


In order to derive equation (20) you can use the following steps:

  • From substitution of equation (16) into equation (18)

$$ P_n = P_{n|n-1} - K_nP^t_{x_ny_n}$$

  • Now plugging right side equation of (19) into the last equation

$$ P_n = P_{n|n-1} - K_n(P_{y_ny_n} - R_n)=$$ $$ =P_{n|n-1} - K_nP_{y_ny_n} + K_nR_n $$

  • Now Pluging (16) again we know that $$ K_nP_{y_ny_n} = P_{x_ny_n}$$
  • putting this back we get

$$ P_n=P_{n|n-1} - P_{x_ny_n} + K_nR_n $$

  • Using left side equation of (19) we get the desired result $$P_n = K_nR_n$$
  • $\begingroup$ Works for me! :-) $\endgroup$
    – Peter K.
    Nov 2, 2016 at 13:57
  • $\begingroup$ @DoronPor: Sorrry to sound noisy, but I noticed that there is a transpose in the term $P^T_{xy}$ that is present in the first step of your solution, so plugging right side of eq(19) should result in transpose terms. These are missing and so the result is quite different. Could you please check again? $\endgroup$
    – SKM
    Nov 4, 2016 at 20:46
  • $\begingroup$ So, the step is $P_n = P_{n|n-1} - K_n{( P_{y_n y_n} -R_n)}^T$ instead of the one that you have. $\endgroup$
    – SKM
    Nov 4, 2016 at 20:53
  • 1
    $\begingroup$ I just skipped a stage. Transpose of matrix addition is equal to the addition of matrix transpose. Both of the matrices are covariant matrix so they are symmetric. That is why I removed the transpose operation. $\endgroup$
    – DoronPor
    Nov 4, 2016 at 21:10
  • $\begingroup$ @DoronPor: Thank you very much for the clarification. I have another favor to ask, if you may kindly see the Question posted dsp.stackexchange.com/questions/35222/… This question is related to Kalman filters. I am a beginner in kalman filters, due to which I am having difficulty in solving the problem which requires sound knowledge. Was wondering if you could help. The question is its unclear what the functional form of covariance matrix is so that it equals to the inverse of fisher information matrix. $\endgroup$
    – SKM
    Nov 7, 2016 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.