
#1
Apr1913, 07:03 AM

P: 1

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: [tex]x=\begin{pmatrix} x \\ y \\ \varphi \\ v \end{pmatrix}[/tex] (position, rotation and speed). The equations of the motion model are the following ones: [tex] x_x = x_x + x_v \cdot sin(x_\varphi); [/tex] [tex] x_y = x_y + x_v \cdot cos(x_\varphi); [/tex] x y phi and v can every second be measured with a failure. To remind Extended Kalman Filter: http://www.embeddedworld.eu/fileadm...Armbruster.pdf (Slide 9) Question: As visible on the slide 9, I have to calculate the jacobian matrix H for my measurementfunction 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: [tex]h = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} [/tex] because [tex]z_k = h*x_t = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} * \begin{pmatrix} x_{t,x} \\ x_{t,y} \\ x_{t,\varphi} \\ x_{t,v} \end{pmatrix} [/tex] (x_t is the current measurement vector) Therefore [tex]z_k = x_t[/tex] But in that case the jacobian matrix [tex]J(h)=H[/tex] becomes [tex]H = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} [/tex] And this means, that the whole correction step of the EKF does not work, because the Kalman gainmatrix 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 [tex]K_K[/tex] also becomes zero). So to conclude: I think I haven't understood the meaning of h and how to calculate H. I appriciate any help and apologize for my english, because Im not a native speaker :) 



#2
Apr2013, 12:52 AM

Sci Advisor
P: 3,170

http://www.google.com/url?sa=t&rct=j...DKnU5Q&cad=rja
), [itex] h [/itex] is a vector valued function, not a matrix. In you example it would be the function [itex] h(x,y,\varphi,v) = (x,y,\varphi,v) [/itex]. So the Jacobian of this function is not found by differentiating constants. 


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