Kalman Assumption
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From: Tim Wescott on 27 Apr 2010 16:04 HardySpicer wrote: > On Apr 28, 6:43 am, Tim Wescott <t...(a)seemywebsite.now> wrote: >> HardySpicer wrote: >>> On Apr 27, 2:53 pm, Tim Wescott <t...(a)seemywebsite.now> wrote: >>>> HardySpicer wrote: >>>>> On Apr 27, 4:40 am, Tim Wescott <t...(a)seemywebsite.now> wrote: >>>>>> Cagdas Ozgenc wrote: >>>>>>> Hello, >>>>>>> In Kalman filtering does the process noise have to be Gaussian or >>>>>>> would any uncorrelated covariance stationary noise satisfy the >>>>>>> requirements? >>>>>>> When I follow the derivations of the filter I haven't encountered any >>>>>>> requirements on Gaussian distribution, but in many sources Gaussian >>>>>>> tag seems to go together. >>>>>> The Kalman filter is only guaranteed to be optimal when: >>>>>> * The modeled system is linear. >>>>>> * Any time-varying behavior of the system is known. >>>>>> * The noise (process and measurement) is Gaussian. >>>>>> * The noise's time-dependent behavior is known >>>>>> (note that this means the noise doesn't have to be stationary -- >>>>>> just that it's time-dependent behavior is known). >>>>>> * The model exactly matches reality. >>>>>> None of these requirements can be met in reality, but the math is at its >>>>>> most tractable when you assume them. Often the Gaussian noise >>>>>> assumption comes the closest to being true -- but not always. >>>>>> If your system matches all of the above assumptions _except_ the >>>>>> Gaussian noise assumption, then the Kalman filter that you design will >>>>>> have the lowest error variance of any possible _linear_ filter, but >>>>>> there may be nonlinear filters with better (perhaps significantly >>>>>> better) performance. >>>>> Don't think so. You can design an H infinity linear Kalman filter >>>>> which is only a slight modification and you don't even need to know >>>>> what the covariance matrices are at all. >>>>> H infinity will give you the minimum of the maximum error. >>>> But strictly speaking the H-infinity filter isn't a Kalman filter. It's >>>> certainly not what Rudi Kalman cooked up. It is a state-space state >>>> estimator, and is one of the broader family of "Kalmanesque" filters, >>>> however. >>>> And the H-infinity filter won't minimize the error variance -- it >>>> minimizes the min-max error, by definition. >>>> -- >>>> Tim Wescott >>>> Control system and signal processing consultingwww.wescottdesign.com >>> Who says that minimum mean-square error is the best? That's just one >>> convenient criterion. >> Not me! I made the point in another branch of this thread -- my >> "optimum" may well not be your "optimum". Indeed, my "optimum" may be a >> horrendous failure to fall inside the bounds of your "good enough". >> >> Minimum mean-square error certainly makes the math easy, though. >> >>> For example, the optimal control problem with a Kalman filter is >>> pretty bad. It doesn't even have integral action. >>> Simple PID gives better results for many occasions. >> OTOH, if you model the plant as having an uncontrolled integrator and >> you track that integrator with your Kalman, you suddenly have an 'I' term. >> >> -- >> Tim Wescott >> Control system and signal processing consultingwww.wescottdesign.com > > That's right and what people did, but it doesn't come out naturally, > whereas it does in H infinity control. > Kalman filters are not robust to changes in the plant either. No, and H-infinity filters are. The biggest drawback from the perspective of my current project is that H-infinity filters require a lot of computation at design time, and I'm working on an extended Kalman filter (it's actually morphed into a hybrid extended-unscented filter), for which the filter must compute the gains -- essentially doing a design cycle -- at each iteration. The gain computation is easy with a Kalman-Kalman, but extracting all the eigenvalues for an H-infinity-Kalman is _expensive_. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
From: HardySpicer on 28 Apr 2010 15:17 On Apr 28, 8:04 am, Tim Wescott <t...(a)seemywebsite.now> wrote: > HardySpicer wrote: > > On Apr 28, 6:43 am, Tim Wescott <t...(a)seemywebsite.now> wrote: > >> HardySpicer wrote: > >>> On Apr 27, 2:53 pm, Tim Wescott <t...(a)seemywebsite.now> wrote: > >>>> HardySpicer wrote: > >>>>> On Apr 27, 4:40 am, Tim Wescott <t...(a)seemywebsite.now> wrote: > >>>>>> Cagdas Ozgenc wrote: > >>>>>>> Hello, > >>>>>>> In Kalman filtering does the process noise have to be Gaussian or > >>>>>>> would any uncorrelated covariance stationary noise satisfy the > >>>>>>> requirements? > >>>>>>> When I follow the derivations of the filter I haven't encountered any > >>>>>>> requirements on Gaussian distribution, but in many sources Gaussian > >>>>>>> tag seems to go together. > >>>>>> The Kalman filter is only guaranteed to be optimal when: > >>>>>> * The modeled system is linear. > >>>>>> * Any time-varying behavior of the system is known. > >>>>>> * The noise (process and measurement) is Gaussian. > >>>>>> * The noise's time-dependent behavior is known > >>>>>> (note that this means the noise doesn't have to be stationary -- > >>>>>> just that it's time-dependent behavior is known). > >>>>>> * The model exactly matches reality. > >>>>>> None of these requirements can be met in reality, but the math is at its > >>>>>> most tractable when you assume them. Often the Gaussian noise > >>>>>> assumption comes the closest to being true -- but not always. > >>>>>> If your system matches all of the above assumptions _except_ the > >>>>>> Gaussian noise assumption, then the Kalman filter that you design will > >>>>>> have the lowest error variance of any possible _linear_ filter, but > >>>>>> there may be nonlinear filters with better (perhaps significantly > >>>>>> better) performance. > >>>>> Don't think so. You can design an H infinity linear Kalman filter > >>>>> which is only a slight modification and you don't even need to know > >>>>> what the covariance matrices are at all. > >>>>> H infinity will give you the minimum of the maximum error. > >>>> But strictly speaking the H-infinity filter isn't a Kalman filter. It's > >>>> certainly not what Rudi Kalman cooked up. It is a state-space state > >>>> estimator, and is one of the broader family of "Kalmanesque" filters, > >>>> however. > >>>> And the H-infinity filter won't minimize the error variance -- it > >>>> minimizes the min-max error, by definition. > >>>> -- > >>>> Tim Wescott > >>>> Control system and signal processing consultingwww.wescottdesign.com > >>> Who says that minimum mean-square error is the best? That's just one > >>> convenient criterion. > >> Not me! I made the point in another branch of this thread -- my > >> "optimum" may well not be your "optimum". Indeed, my "optimum" may be a > >> horrendous failure to fall inside the bounds of your "good enough". > > >> Minimum mean-square error certainly makes the math easy, though. > > >>> For example, the optimal control problem with a Kalman filter is > >>> pretty bad. It doesn't even have integral action. > >>> Simple PID gives better results for many occasions. > >> OTOH, if you model the plant as having an uncontrolled integrator and > >> you track that integrator with your Kalman, you suddenly have an 'I' term. > > >> -- > >> Tim Wescott > >> Control system and signal processing consultingwww.wescottdesign.com > > > That's right and what people did, but it doesn't come out naturally, > > whereas it does in H infinity control. > > Kalman filters are not robust to changes in the plant either. > > No, and H-infinity filters are. The biggest drawback from the > perspective of my current project is that H-infinity filters require a > lot of computation at design time, and I'm working on an extended Kalman > filter (it's actually morphed into a hybrid extended-unscented filter), > for which the filter must compute the gains -- essentially doing a > design cycle -- at each iteration. The gain computation is easy with a > Kalman-Kalman, but extracting all the eigenvalues for an > H-infinity-Kalman is _expensive_. > > -- > Tim Wescott > Control system and signal processing consultingwww.wescottdesign.com Always suspicious about extended Kalman filters since they are not guaranteed to converge. I would do a separate estimation of the plant with say a Volterra type LMS estimator and use that in some way to feed an estimator of the states. Hardy
From: Tim Wescott on 28 Apr 2010 16:10
HardySpicer wrote: > On Apr 28, 8:04 am, Tim Wescott <t...(a)seemywebsite.now> wrote: >> HardySpicer wrote: >>> On Apr 28, 6:43 am, Tim Wescott <t...(a)seemywebsite.now> wrote: >>>> HardySpicer wrote: >>>>> On Apr 27, 2:53 pm, Tim Wescott <t...(a)seemywebsite.now> wrote: >>>>>> HardySpicer wrote: >>>>>>> On Apr 27, 4:40 am, Tim Wescott <t...(a)seemywebsite.now> wrote: >>>>>>>> Cagdas Ozgenc wrote: >>>>>>>>> Hello, >>>>>>>>> In Kalman filtering does the process noise have to be Gaussian or >>>>>>>>> would any uncorrelated covariance stationary noise satisfy the >>>>>>>>> requirements? >>>>>>>>> When I follow the derivations of the filter I haven't encountered any >>>>>>>>> requirements on Gaussian distribution, but in many sources Gaussian >>>>>>>>> tag seems to go together. >>>>>>>> The Kalman filter is only guaranteed to be optimal when: >>>>>>>> * The modeled system is linear. >>>>>>>> * Any time-varying behavior of the system is known. >>>>>>>> * The noise (process and measurement) is Gaussian. >>>>>>>> * The noise's time-dependent behavior is known >>>>>>>> (note that this means the noise doesn't have to be stationary -- >>>>>>>> just that it's time-dependent behavior is known). >>>>>>>> * The model exactly matches reality. >>>>>>>> None of these requirements can be met in reality, but the math is at its >>>>>>>> most tractable when you assume them. Often the Gaussian noise >>>>>>>> assumption comes the closest to being true -- but not always. >>>>>>>> If your system matches all of the above assumptions _except_ the >>>>>>>> Gaussian noise assumption, then the Kalman filter that you design will >>>>>>>> have the lowest error variance of any possible _linear_ filter, but >>>>>>>> there may be nonlinear filters with better (perhaps significantly >>>>>>>> better) performance. >>>>>>> Don't think so. You can design an H infinity linear Kalman filter >>>>>>> which is only a slight modification and you don't even need to know >>>>>>> what the covariance matrices are at all. >>>>>>> H infinity will give you the minimum of the maximum error. >>>>>> But strictly speaking the H-infinity filter isn't a Kalman filter. It's >>>>>> certainly not what Rudi Kalman cooked up. It is a state-space state >>>>>> estimator, and is one of the broader family of "Kalmanesque" filters, >>>>>> however. >>>>>> And the H-infinity filter won't minimize the error variance -- it >>>>>> minimizes the min-max error, by definition. >>>>>> -- >>>>>> Tim Wescott >>>>>> Control system and signal processing consultingwww.wescottdesign.com >>>>> Who says that minimum mean-square error is the best? That's just one >>>>> convenient criterion. >>>> Not me! I made the point in another branch of this thread -- my >>>> "optimum" may well not be your "optimum". Indeed, my "optimum" may be a >>>> horrendous failure to fall inside the bounds of your "good enough". >>>> Minimum mean-square error certainly makes the math easy, though. >>>>> For example, the optimal control problem with a Kalman filter is >>>>> pretty bad. It doesn't even have integral action. >>>>> Simple PID gives better results for many occasions. >>>> OTOH, if you model the plant as having an uncontrolled integrator and >>>> you track that integrator with your Kalman, you suddenly have an 'I' term. >>>> -- >>>> Tim Wescott >>>> Control system and signal processing consultingwww.wescottdesign.com >>> That's right and what people did, but it doesn't come out naturally, >>> whereas it does in H infinity control. >>> Kalman filters are not robust to changes in the plant either. >> No, and H-infinity filters are. The biggest drawback from the >> perspective of my current project is that H-infinity filters require a >> lot of computation at design time, and I'm working on an extended Kalman >> filter (it's actually morphed into a hybrid extended-unscented filter), >> for which the filter must compute the gains -- essentially doing a >> design cycle -- at each iteration. The gain computation is easy with a >> Kalman-Kalman, but extracting all the eigenvalues for an >> H-infinity-Kalman is _expensive_. >> >> -- >> Tim Wescott >> Control system and signal processing consultingwww.wescottdesign.com > > Always suspicious about extended Kalman filters since they are not > guaranteed to converge. > I would do a separate estimation of the plant with say a Volterra type > LMS estimator and use that in some way to feed an estimator of the > states. This particular Kalman was pretty strongly dependent on 3-D angles; it worked OK as an extended Kalman, but really started to shine when it got turned into an unscented Kalman. I didn't consider doing the Volterra series, because what's a few more terms in a really severely nonlinear transform like 3-D angles? But the unscented version is working like dynamite -- and not in the sense that it's blowing up in my face. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com |