# Question concerning the extended kalman filter

1. Apr 19, 2013

### Anonymous123

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:
$$x_x = x_x + x_v \cdot sin(x_\varphi);$$
$$x_y = x_y + x_v \cdot cos(x_\varphi);$$

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

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:

$$h = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

because

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

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

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

$$H = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

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 appriciate any help and apologize for my english, because Im not a native speaker :)

2. Apr 20, 2013

### Stephen Tashi

), $h$ 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.