Started by February 5, 2007
```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.

```
```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.
>
>
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

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html
```
```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.
>
>
> 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
>
> --
>
> Tim Wescott
> Wescott Design Serviceshttp://www.wescottdesign.com
>
>
> "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 -

```
```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.
>>
>>
>>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
>> 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