I understand that in Kalman filtering, the minimum variance estimator can be found by orthogonal projection of X(k) on the space spanned by linear combinations of observations Y(0), Y(1),...Y(k). However I went to a seminar and one of the speaker ws saying that Kalman filter implicitly weights the more recent value greater than the past values i.e. the weights associating to an observation decays exponentially as time goes by. May I know which is correct? My intuition is that the Kalman filter treats has no preference for recent values. Please enlighten. Cheers!

# Question about Kalman Filters

Started by ●February 5, 2007

Reply by ●February 5, 20072007-02-05

jionglong wrote:> I understand that in Kalman filtering, the minimum variance estimator > can be found by > orthogonal projection of X(k) on the space spanned by linear > combinations of observations Y(0), Y(1),...Y(k). > > However I went to a seminar and one of the speaker ws saying that > Kalman filter implicitly weights the more recent value greater than > the past values i.e. the weights associating to an observation decays > exponentially as time goes by. > > May I know which is correct? My intuition is that the Kalman filter > treats has no preference for recent values. > > Please enlighten. Cheers! >This is an engineering question, with the engineering answer: It Depends. Most Kalman filters will put more weight on recent values (they'll often not put more weight on _the most_ recent value, but there will often be some peak in the weighting, then a drop off). This is because most processes have states that are more sensitive to recent values of the inputs, and the Kalman filter will reflect that. If the process that you're observing is modeled as unchanging and the noise as white, then you're just using the Kalman filter to get rid of noise and the weighting will, indeed, be constant over time for any given reading. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html

Reply by ●February 6, 20072007-02-06

The Kalman filter gives a linear combination of the current measurement and privious state, i.e. x_k=xhat_k+k(zk-zhat_k) for scalar case, replacing zhat_k by zhat_k=H*xhat_k, it would be x_k=xhat_k+k(zk-zhat_k)=a1*xhat_k+a2*zk where a1+a2=1. The past measurements are carried by xhat_k and weighted by a1 at every step. Therefore, its weight decreases exponentially fast. On Feb 5, 2:08 pm, Tim Wescott <t...@seemywebsite.com> wrote:> jionglong wrote: > > I understand that in Kalman filtering, the minimum variance estimator > > can be found by > > orthogonal projection of X(k) on the space spanned by linear > > combinations of observations Y(0), Y(1),...Y(k). > > > However I went to a seminar and one of the speaker ws saying that > > Kalman filter implicitly weights the more recent value greater than > > the past values i.e. the weights associating to an observation decays > > exponentially as time goes by. > > > May I know which is correct? My intuition is that the Kalman filter > > treats has no preference for recent values. > > > Please enlighten. Cheers! > > This is an engineering question, with the engineering answer: It Depends. > > Most Kalman filters will put more weight on recent values (they'll often > not put more weight on _the most_ recent value, but there will often be > some peak in the weighting, then a drop off). This is because most > processes have states that are more sensitive to recent values of the > inputs, and the Kalman filter will reflect that. > > If the process that you're observing is modeled as unchanging and the > noise as white, then you're just using the Kalman filter to get rid of > noise and the weighting will, indeed, be constant over time for any > given reading. > > -- > > Tim Wescott > Wescott Design Serviceshttp://www.wescottdesign.com > > Posting from Google? Seehttp://cfaj.freeshell.org/google/ > > "Applied Control Theory for Embedded Systems" came out in April. > See details athttp://www.wescottdesign.com/actfes/actfes.html- Hide quoted text - > > - Show quoted text -

Reply by ●February 6, 20072007-02-06

zhanluezhao@gmail.com wrote: (top posting fixed)> > On Feb 5, 2:08 pm, Tim Wescott <t...@seemywebsite.com> wrote: > >>jionglong wrote: >> >>>I understand that in Kalman filtering, the minimum variance estimator >>>can be found by >>>orthogonal projection of X(k) on the space spanned by linear >>>combinations of observations Y(0), Y(1),...Y(k). >> >>>However I went to a seminar and one of the speaker ws saying that >>>Kalman filter implicitly weights the more recent value greater than >>>the past values i.e. the weights associating to an observation decays >>>exponentially as time goes by. >> >>>May I know which is correct? My intuition is that the Kalman filter >>>treats has no preference for recent values. >> >>>Please enlighten. Cheers! >> >>This is an engineering question, with the engineering answer: It Depends. >> >>Most Kalman filters will put more weight on recent values (they'll often >>not put more weight on _the most_ recent value, but there will often be >>some peak in the weighting, then a drop off). This is because most >>processes have states that are more sensitive to recent values of the >>inputs, and the Kalman filter will reflect that. >> >>If the process that you're observing is modeled as unchanging and the >>noise as white, then you're just using the Kalman filter to get rid of >>noise and the weighting will, indeed, be constant over time for any >>given reading. >> The Kalman filter gives a linear combination of the current > measurement and privious state, i.e. > x_k=xhat_k+k(zk-zhat_k) > for scalar case, replacing zhat_k by zhat_k=H*xhat_k, it would be > x_k=xhat_k+k(zk-zhat_k)=a1*xhat_k+a2*zk > where a1+a2=1. > The past measurements are carried by xhat_k and weighted by a1 at > every step. Therefore, its weight decreases exponentially fast. >Once it reaches steady state a Kalman exhibits exponential decay. While the state update matrix is evolving the decay is not, in general, exponential. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html