Learning the Kalman Filter

My case:

>> clear s
>> s.x = 12;
>> s.A = 1;
>> s.Q = 2^2;
>> s.H = 1;
>> s.R = 2^2;
>> s.B = 0;
>> s.u = 0;
>> s.x = nan;
>> s.P = nan;

>> s.z = [1 2 3 4 5 6 7 8 9 22 3 4 5 6 7 8 88];
>> kalmanf(s);

>> figure
>> hold on
>> grid on
>> hz=plot(s.z,'r.')
>> hk=plot(s,'b-')

Error using plot
Conversion to double from struct is not possible.

Where is the problem ?

Thank u

Actually iam getting the error s not defined what can i do for this and this example gives v.gud idea.

[rectification to the post above]: sry, 's' is not the measurement data. your data goes into 's.z' here.

you are awesome

helpful and clear comments

Nicely documented.
As a practicing engineer I would never use the implementation shown. This is the standard covariance form of the Kalman filter. In operation the statement
s.P = s.P - K*s.H*s.P;
causes significant issues. s.P needs to always be positive definite but with rounding this will tend to violate this assmption making the Kalman filter 'blow up' over time or with poorly conditioned data.

The alternative is to process in the square root domain where the P matrix is expressed as a P=Psr'*Psr. Therefore the resuting matrix must always be positive definite and does not have this issue. An added advantage is the precision is doubled processing in this domain and can be similar to the input data resolution rather than 2x the number of bits to provide comparable precision.

For more information see
'Linear estimation' Kailath, Sayed, Hassibi or
'adaptive filter thoery' Haykin

IMHO one should always to filtering in the SR domain. There are also advantages of processing with separated real/imag data rather then complex if the underling data is complex.

good !

thank you very much!

What is the difference between this and kalman implementation in Control System toolbox ?

Dear Michael Kleder,thank you !

love you so much, your sample is more helpful than those books wrote by some professional guys. easy, straightforward. especially for beginners.

Thanks a lot because of your great explanation.

But, I have a simple question, why results don't change when we give very high measurement noise?
You gave standart deviation as 2 in your example, but when I make it for example 1000, it estimates perfectly. Doesn't supposed to change?

Can anyone explain this to me?
Thanks a lot
-Enver

awesome! makes so much more sense now. thanks.

thank you for sharing!

sehr gut!

[continued from poste above]

I have solved the problem by modifying the line to;

>> s.x = s.x + K' * (s.z-s.H*s.x); %added transpose of K

This was required as my observance vector (z) was a 2X1 matrix(and so is K and hence they couldn't be multiplied). Although the calculation now works I am still wondering why it doesnt work with the original code.

Thanks very much.

Thank you for your hard work
http://learnbyexamples.org/category/matlab

good

Very Helpful, although i would like to see an example with a control vector u.

very good

thanks

Thanks very much

Nice comments. Awful numerics. Why would you consider multiplying by the inverse? That's horribly inaccurate and slow. Replace

x = inv(H)*z
P = inv(H)*R*inv(H')

with, at least

x = H\z
P = (H\R)/H'

likewise

K = P * H' * inv (lots of stuff)

should be

K = P * H' / (lots of stuff)

further more, if H is not changing, it should be factorized just once and the factors kept (LU or CHOL).

Very Useful Indeed

Good for those who want to see what the Kalman Filter is for the first time.
It works and there is a simple example in the m file.

great

This is a very popular file in the File Exchange. The function itself is excelent. However, I just noticed that the example provide with the file is not correct. Somehow, it is misleading to beginers.

In line 131, the process is defined as:
true(end+1) = randn*2 + 12;
i.e. the state is a constant plus a noise. If so, the process in the standard state space form should be:
x(k+1) = 0 * x(k) + 12 + w(k)
i.e. s.A = 0; s.B = 1; s.u = 12;
However, in the file it is wrongly defined as:
s.A = 1; s.B = 0; s.u = 0;
The difference is that the process noise is not dynamically cumulated in the former definition but does in the later.

Be careful. Also we are looking at your downloads records.

how con I find calculation of dry gas filter
for natural gas ? please

this is very good

Actually, I am not a beginner at Kalman filter issue. But, i think this is a very useful tool and i wish i got this m-file when i first started to work with Kalman. Because, it is very important for the beginners to have the simplest form of the problem and to see its solution with a simulation program.

Very easy to use, recommended.

Good demo.
You probably should separete the example, to another m file, named like RunMeDemoKalman.
For the really beginners.

very good tools
thank you very much

discrete kalman filter

It will be better if you separate the comment from the m.file and try to add them as "help" in a pdf format.

Same Error !!!

I get the same error:

kalmanf
??? Input argument "s" is undefined.

Error in ==> kalmanf at 150
if ~isfield(s,'x'); s.x=nan*z; end

I need Kalman Filter program.

Thank You! Great introduction to the Kalman filter, even if you don't use Matlab.

try to replace inv(A)*B by A\B to speed up, although it's not significant when the matrix size is small

Useful Comments in Code.

its great.

VERY GOOD

Simple, pretty, excellent :-))) Thanks a lot!!

Its nice, no doubt.

A good try to explain something. Keep up the good work! There are, however, some mistakes (at the autoinitialization step). Anyway I still don't quite understand the use of Kalman filter. The given example is good, but I'm still confused how to apply this filter to my data.

Nice work. It would be better if there is an example for vector state. Also it would be very cool if someone can put Kalman filter algorithm in simulink so that we can see the estimation of states dynamically. And of course, an extended kalman filter for nonlinear system would be also very useful.

Very fun. Very neat.

very good, I like the idea

This is a very good example and shows how easy it is to apply the Kalman filter. It does, however, require some background knowledge. It would be nice to have a more complicated example with non-zero u and where H and A are not =1. Thanks Michael!

Very helpful. Thanks!!

Excellent!! Thank You.

great!