Electrical – Is it possible for the Kalman gain to become 1 or 0

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Is it possible for the Kalman gain to ever become exactly 0 or 1? Because as far as I understand, if that would happen (no matter if it were 0 or 1) the Kalman gain would stay 0 in all the following cycles?

Best Answer

There is an additive Gaussian noise in the state transition model of a Kalman filter. For example \$ \epsilon_t \$ in the equation below:

\$ x_t = A_tx_{t-1} + B_tu_t + \epsilon_t \$

\$ \epsilon_t \$ has a zero mean and a covariance (let's call it \$ R_t \$). The purpose of \$\epsilon_t\$ is to model the uncertainty associated with the state transition process. Without this term (or more precisely if \$R_t\$ is a vector of zeros), we have no uncertainty in the state transition process. This essentially means that we have found the perfect model that can predict the future with 100% certainty and therefore there would be no need for any measurement.

Based on the above model the covariance matrix would be estimated as follows in the prediction phase of the Kalman filter:

\$ \overline{\Sigma}_t = A_t \Sigma_{t-1} A_t^T + R_t\$

\$\Sigma\$ is the covariance matrix for the state estimate, and the predicted state covariance is denoted by \$\overline{\Sigma} \$. As you can see in the above equation the predicted covariance can never be zero unless \$ R_t \$ is taken to be zero (which makes no sense as already explained).

Assume the measurement data is formulated as:

\$ z_t = C_tx_t + \delta_t \$

\$cov(\delta_t) = Q_t \$

Now we can formulate the Kalman gain which would be:

\$K_t = \overline{\Sigma}_tC_t^T(C_t\overline{\Sigma}_tC_t^T + Q_t)^{-1} \$

Looking at the above equation, it is clear that it would not lock on zero even if the previous gain somehow ended up being zero.

The case of a Kalman gain equal to 1 only happens when the measurement has an uncertainty of zero (again not really possible). Even if that were to be the case, Kalman gain only depends on the covariance of the current measurement, so it would not get locked on 1 unless the measurement covariance was consistently showing up as zero (which would not be a flaw of the filter, but the measurement system).