Kalman Filter Velocity data looking too similar to position data
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Wyatt
on 8 Oct 2024 at 18:23
Edited: James Tursa
on 8 Oct 2024 at 21:21
I'm creating a Kalman filter that intakes Accelerometer and Angular Velocity measurements from Matlab mobil IMU sensors, and filters it, then calculates the position and velocity. My question is on if I am going about calculating the position and velocity wrong, of if this is a good result but it looks strange for some reason?
Here is an abridged look at the sections of code I'm concerned with
dt = 0.1;
% Initial R and V calculation
aRaw = [Accx, Accy, Accz];
vi = aRaw .* dt;
ri = .5 * aRaw .* (dt^2);
% ~~Data is filtered here~~
% Filtered R and V calculation
af = [Accxf, Accyf, Acczf];
vf = af .* dt;
rf = .5 * af .* (dt^2);
af is the filtered acceleration data in x,y,z. I just followed basic kinematics equations, is that the wrong way to calculate it? Initial velocity and positioned are calculated the same way but from the raw data, charts are provided.
If this isn't the issue, where should I be looking to figure this out? My only other guess is maybe its an issue with the state variables. Thank you.
Tried changing sensor data, adjusting kinematic formulas, and debugging. Expected different looking results for position and velocity.
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Accepted Answer
James Tursa
on 8 Oct 2024 at 20:51
Edited: James Tursa
on 8 Oct 2024 at 21:21
Your velocity vf and position rf calculations as shown represent "change" over dt, not current value. Think about it. If you had constant acceleration the entire time you would expect the overall velocity to be linear and position to be parabolic, right? But what you would have for the calculations you show are constants for both. I.e., the "changes" are constant, but the current values would not be constant. You need to be using some type of integration scheme (i.e., summing up the "changes") with these acceleration measurements, like cumtrapz( ), to get current velocity and position.
Example with constant acceleration to drive home this point:
dt = 0.1;
n = 50;
aRaw = zeros(n,3) + 0.1; % constant acceleration
t = (0:(n-1)) * dt;
vi = aRaw .* dt;
vt = cumtrapz(aRaw) * dt;
plot(t,vi(:,1)); title('vi = aRaw*dt'); xlabel('t'); ylabel('x velocity');
plot(t,vt(:,1)); title('vt = cumtrapz(aRaw)*dt'); xlabel('t'); ylabel('x velocity');
More Answers (1)
Les Beckham
on 8 Oct 2024 at 19:33
It appears that you aren't differentiating the position to get the velocity, you are just multiplying it by dt. Thus the two plots look the same with only a scale factor difference. Similarly for the acceleration, you should be differentiating the velocity, not just applying a scale factor.
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