Thanks to everyone who posted comments/answers to my query yesterday (Implementing a Kalman filter for position, velocity, acceleration ). I've been looking at what was recommended, and in particular at both (a) the wikipedia example on one dimensional position and velocity and also another website that considers a similar thing.

Update 26-Apr-2013: the original question here contained some errors, related to the fact that I hadn't properly understood the the wikipedia example on one dimensional position and velocity. With my improved understanding of what's going on, I've now redrafted the question and focused it more tightly.

Both examples that I refer to in the introductory paragraph above assume that it's only position that's measured. However, neither example has any kind of calculation $(x_k-x_{k-1})/dt$ for speed. For example, the Wikipedia example specifies the ${\bf H}$ matrix as ${\bf H} = [1\ \ \ 0]$, which means that only position is input. Focussing on the Wikipedia example, the state vector ${\bf x}_k$ of the Kalman filter contains position $x_k$and speed $\dot{x}_{k}$, i.e.

$$ \begin{align*} \mathbf{x}_{k} & =\left(\begin{array}[c]{c}x_{k}\\ \dot{x}_{k}\end{array} \right) \end{align*} $$

Suppose the measurement of position at time $k$ is $\hat{x}_k$. Then if the position and speed at time $k-1$ were $x_{k-1}$ and $\dot{x}_{k-1}$, and if $a$ is a constant acceleration that applies in the time interval $k-1$ to $k$, from the measurement of $\hat{x}$ it's possible to deduce a value for $a$ using the formula

$$ \hat{x}_k = x_{k-1} + \dot{x}_{k-1} dt + \frac{1}{2} a dt^2 $$

This implies that at time $k$, a measurement $\hat{\dot{x}}_k$ of the speed is given by

$$ \hat{\dot{x}}_k = \dot{x}_{k-1} + a dt = 2 \frac{\hat{x}_k - {x}_{k-1}}{dt} - \dot{x}_{k-1} $$

All the quantities on the right hand side of that equation (i.e. $\hat{x}_k$, $x_{k-1}$ and $\dot{x}_{k-1}$) are normally distributed random variables with known means and standard deviations, so the $\bf R$ matrix for the measurement vector

$$ \begin{align*} \mathbf{\hat{x}}_{k} & =\left(\begin{array}[c]{c}\hat{x}_{k}\\ \hat{\dot{x}}_{k}\end{array} \right) \end{align*} $$

can be calculated. Is this a valid way of introducing speed estimates into the process?

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    $\begingroup$ I didn't look through all of your calculations. However, speaking of the Wikipedia example, you seem to be a bit confused on its structure. You're correct in that only position is measured. However, a so-called "constant-velocity" model is used. This means that the velocity is considered to be constant in the state transition matrix. $\endgroup$
    – Jason R
    Commented Apr 25, 2013 at 18:43
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    $\begingroup$ Changes in the velocity are modeled using the process noise matrix. Therefore, you're inherently assuming that the velocity is going to change randomly with some specified covariance. Surprisingly enough, this often works well. It is common to use process noise one derivative above your highest state-variable derivative in this way. For instance, if you included acceleration in your model, then you might have a random jerk component included in your process noise. $\endgroup$
    – Jason R
    Commented Apr 25, 2013 at 18:44
  • $\begingroup$ @JasonR with the wikipedia model (assuming zero initial covariance between position and speed), the speed estimate is always it's initial value (as you say, a "constant-velocity" model). However, the variance of the speed grows monotonically via the process noise, and there are no measurements which can reduce it. What is the advantage of this over a model that only models position and assumes a constant speed? $\endgroup$ Commented Apr 25, 2013 at 19:02
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    $\begingroup$ The variance of the velocity estimate shouldn't increase monotonically. Process noise simply introduces a stochastic component to the state transition equation, allowing you to express some uncertainty at exactly how the system state will evolve from time step to time step. If you don't include process noise, then your filter would truly output a constant velocity. That probably isn't what you want. $\endgroup$
    – Jason R
    Commented Apr 25, 2013 at 20:02
  • $\begingroup$ Well @JasonR, if you look at the wikipedia model, you'll see that monotinically increasing variance of velocity is what it gives you! $\endgroup$ Commented Apr 25, 2013 at 20:05

1 Answer 1


Is this a valid way of introducing speed estimates into the process?

If you choose your state appropriately, then the speed estimates come "for free". See the derivation of the signal model below (for the simple 1-D case we've been looking at).

Signal Model, Take 2

So, we really need to agree on a signal model before we can move this forward. From your edit, it looks like your model of the position, $x_k$, is:

$$ \begin{array} xx_{k+1} &=& x_{k} + \dot{x}_{k} \Delta t + \frac{1}{2} a (\Delta t)^2\\ \dot{x}_{k+1} &=& \dot{x}_{k} + a \Delta t \end{array} $$

If our state is as before: $$ \begin{align*} \mathbf{x}_{k} & =\left(\begin{array}[c]{c}x_{k}\\ \dot{x}_{k}\end{array} \right) \end{align*} $$ then the state update equation is just: $$ \mathbf{x}_{k+1} = \left(\begin{array}[c]{c} 1\ \ \Delta t\\ 0\ \ 1\end{array} \right) \mathbf{x}_{k} + \left(\begin{array}[c]{c} \frac{(\Delta t)^2}{2} \\ \Delta t \end{array} \right) a_k $$ where now our $a_k$ is the normally distributed acceleration.

That gives different $\mathbf{G}$ matrix from the previous version, but the $\mathbf{F}$ and $\mathbf{H}$ matrices should be the same.

If I implement this in scilab (sorry, no access to matlab), it looks like:

// Signal Model
DeltaT = 0.1;
F = [1 DeltaT; 0 1];
G = [DeltaT^2/2; DeltaT];
H = [1 0];

x0 = [0;0];
sigma_a = 0.1;

Q = sigma_a^2;
R = 0.1;

N = 1000;

a = rand(1,N,"normal")*sigma_a;

x_truth(:,1) = x0;
for t=1:N,
    x_truth(:,t+1) = F*x_truth(:,t) + G*a(t);
    y(t) = H*x_truth(:,t) + rand(1,1,"normal")*sqrt(R);

Then, I can apply the Kalman filter equations to this $y$ (the noisy measurements).

// Kalman Filter
p0 = 100*eye(2,2);

xx(:,1) = x0;
pp = p0;
pp_norm(1) = norm(pp);
for t=1:N,
    [x1,p1,x,p] = kalm(y(t),xx(:,t),pp,F,G,H,Q,R);
    xx(:,t+1) = x1;
    pp = p1;
    pp_norm(t+1) = norm(pp);

So we have our noisy measurements $y$, and we've applied the Kalman filter to them and used the same signal model to generate $y$ as we do to apply the Kalman filter (a pretty big assumption, sometimes!).

Then following plots show the result.

Plot 1: $y$ and $x_k$ versus time.

enter image description here

Plot 2: A zoomed view of the first few samples:

enter image description here

Plot 3: Something you never get in real life, the true position vs the state estimate of the position.

enter image description here

Plot 4: Something you also never get in real life, the true velocity vs the state estimate of the velocity.

enter image description here

Plot 5: The norm of the state covariance matrix (something you should always monitor in real life!). Note that it very quickly goes from its initial very large value to something very small, so I've only shown the first few samples.

enter image description here

Plot 6: Plots of the error between the true position and velocity and their estimates.

enter image description here

If you study the case where the position measurements are exact, then you find that the Kalman udpate equations produce exact results for BOTH position and speed. Mathematically it's straightforward to see why. Using the same notation as the wikipedia article, exact measurements mean that $\mathbf{z}_{k+1}=x_{k+1}$. If you assume that the initial position and speed are known so that $\mathbf{P}_k=0$, then $\mathbf{P}_{k+1}^{-}=\mathbf{Q}$ and the Kalman gain matrix $\mathbf{K}_{k+1}$ is given by

$$ \mathbf{K}_{k+1} = \left(\begin{array}[c]{c}1\\ 2/dt\end{array} \right) $$

This means that the Kalman update procedure produces

$$ \begin{align*} \mathbf{\hat{x}}_{k+1} & = \mathbf{F}_{k+1}\mathbf{x}_k + \mathbf{K}_{k+1}\left(\mathbf{z}_{k+1} - \mathbf{H}_{k+1} \mathbf{F}_{k+1}\mathbf{x}_k\right)\\ & = \left(\begin{array}[c]{c}x_k + \dot{x}_k dt\\ \dot{x}_k\end{array} \right) + \left(\begin{array}[c]{c}1\\ 2/dt\end{array} \right) \left(x_{k+1} - \left( x_k + \dot{x}_k dt\right) \right)\\ & = \left(\begin{array}[c]{c}x_{k+1}\\ 2 \left(x_{k+1} - x_k \right) /dt - \dot{x}_k\end{array} \right) \end{align*} $$

As you can see, the value for the speed is given by exactly the formula you were proposing to use for the speed estimate. So although you couldn't see any kind of calculation $(x_k-x_{k-1})/dt$ for speed, in fact it is hidden in there after all.

  • $\begingroup$ Thanks for all your help so far. My original question did contain some misunderstandings, so I've tried to refocus it and with an attempt to answer your question about z_k. $\endgroup$ Commented Apr 26, 2013 at 14:57
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    $\begingroup$ thanks vm for all your efforts :-). Your work spurred me on to do a bit of maths which I've tacked on to your answer, hope you don't mind. Anyway, I'm now 100% convinced that the standard method is good, because I can see formulas in there for speed after all. Thanks again $\endgroup$ Commented Apr 27, 2013 at 13:34
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    $\begingroup$ Glad to be able to help! No problem with adding to the answer. $\endgroup$
    – Peter K.
    Commented Apr 27, 2013 at 15:16

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