With sampling time $T$, and a continuous measuring model: $$ \begin{align} y(t) &= Cx(t)+v(t) \\ v(t) & \sim \text{N}(0,R_c) \end{align} $$ we can change it into a practical discrete one, which is $$ \begin{align} y_k &= Cx_k+v_k \\ v_k & \sim \text{N}(0,R) \end{align} $$ and $$ \mathbf{R=\frac{R_c}{T}} \tag{$*$}\label{a} $$ how to get the $\eqref{a}$ ?

I don't quit understand this because for a sampling process, we say $$ v_k=v(kT)=v(t) $$ so $v_k$ is as same as $v(t)$ in a particular time $t$. so $v_k$ and $v(t)$ have the same distribution, leading $$\text{Cov}(v_k)=\text{Cov}(v(t))$$

that means $R=R_c$.

Or using equation $\eqref{a}$ to promise the discrte Kalman filter to have the same conclusion with the continuous one?

EDIT: equation $\eqref{a}$ is used in the MIT lecture.

  • $\begingroup$ I agree with you, replacing R by R/T when one samples at time intervals of length T seems absurd. The source you quote does not help much either. $\endgroup$ – Did Sep 15 '15 at 7:15
  • $\begingroup$ @Did this source just uses the equation and gives a citation(Brown and Hwang, S) . But I don't have access to it. $\endgroup$ – SandyX Sep 15 '15 at 7:26

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