Question concerning the extended kalman filter

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The discussion centers on the Extended Kalman Filter (EKF) and the calculation of the Jacobian matrix H for the measurement function h. The user presents a vehicle state vector represented as x=\begin{pmatrix} x \\ y \\ \varphi \\ v \end{pmatrix} and expresses confusion regarding the Jacobian matrix derived from the measurement function. The user incorrectly assumes that the Jacobian H becomes a zero matrix, which would invalidate the correction step of the EKF. Clarification is provided that h is a vector-valued function, not a constant matrix, and that the Jacobian should be calculated by differentiating the function h(x,y,\varphi,v).

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Anonymous123
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Good day,

i read a lot about the kalman filter and the extended kalman filter, but some things are still not clear to me. E.g. I have one question concerning the jacobian matrix of the measurement matrix h. I want to point out my problem with a concrete example:

A vehicle is represented by the following state vector: x=\begin{pmatrix} x \\ y \\ \varphi \\ v \end{pmatrix} (position, rotation and speed).

The equations of the motion model are the following ones:
<br /> x_x = x_x + x_v \cdot sin(x_\varphi);<br />
<br /> x_y = x_y + x_v \cdot cos(x_\varphi);<br />

x y phi and v can every second be measured with a failure.

To remind Extended Kalman Filter: http://www.embedded-world.eu/fileadmin/user_upload/pdf/batterie2011/Armbruster.pdf (Slide 9)


Question: As visible on the slide 9, I have to calculate the jacobian matrix H for my measurement-function h. The slide points that very well out, in the correction step is H the jacobian matrix and h is my measurement function.

So if I want to consider all elements of the measurement vektor, h would be in my opinion the following matrix:

h = \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \end{pmatrix}

because

z_k = h*x_t = \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \end{pmatrix} * \begin{pmatrix} x_{t,x} \\ x_{t,y} \\ x_{t,\varphi} \\ x_{t,v} \end{pmatrix}

(x_t is the current measurement vector)
Therefore z_k = x_t

But in that case the jacobian matrix J(h)=H becomes

H = \begin{pmatrix} 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \end{pmatrix}

And this means, that the whole correction step of the EKF does not work, because the Kalman gain-matrix also becomes a zero matrix (See again slide 9, there are some matrix multiplications where H is used. When H contains just zeros the gain K_K also becomes zero).

So to conclude: I think I haven't understood the meaning of h and how to calculate H. I appreciate any help and apologize for my english, because I am not a native speaker :)
 
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Anonymous123 said:
So if I want to consider all elements of the measurement vektor, h would be in my opinion the following matrix:

h = \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \end{pmatrix}

In the treatments of the extended Kalman filter that I can read (such as http://www.google.com/url?sa=t&rct=...sg=AFQjCNFjl1YZdkoZZiaJ7-NMEi3vDKnU5Q&cad=rja
), h is a vector valued function, not a matrix. In you example it would be the function h(x,y,\varphi,v) = (x,y,\varphi,v). So the Jacobian of this function is not found by differentiating constants.
 

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