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Home/ Questions/Q 3692930
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Editorial Team
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Editorial Team
Asked: May 19, 20262026-05-19T04:19:08+00:00 2026-05-19T04:19:08+00:00

I am following a probably well-known tutorial about Kalman filter. From these lines of

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I am following a probably well-known tutorial about Kalman filter.

From these lines of code:

figure;
plot(t,pos, t,posmeas, t,poshat);
grid;
xlabel('Time (sec)');
ylabel('Position (feet)');
title('Figure 1 - Vehicle Position (True, Measured, and Estimated)')

I understand that x is the true position, y is measured position, xhat is estimated position. Then, if we can compute x (this code: x = a * x + b * u + ProcessNoise;), why do we need to estimated x anymore?

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  1. Editorial Team
    Editorial Team
    2026-05-19T04:19:08+00:00Added an answer on May 19, 2026 at 4:19 am

    …

    OK, after a second look at the referenced article, I think I see the confusion. Apparently, the program in the article is a simulation of a linear system (and thus, it repeatedly generates new x‘s as new states in the simulated system). Then it also simulates a “noisy” measurement of x, and from that (simulated) noisy measurement, then demonstrates using a Kalman Filter on the noisy data to try to estimate the actual (simulated) x‘s.

    So, the exact x calculations that you ask about are just part of the Simulation, and not part of the Kalman Filter itself or the data that is available to the Kalman Filter algorithm.

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