The tranditional Kalman filter equation is as follows: x(n)=F x(n-1)+w(n) y(n)=H x(n)+v(n) And x(n) is called "state". If there are two states, x1(n) and x2(n), x1(n)=F1 x1(n-1)+w1(n) x2(n)=F2 x2(n-1)+w2(n) y(n)=H1 x1(n)+H2 x2(n)+v(n) Is it still the Kalman filter? How to make estimation? Or can you refer me some papers? Thanks.

# Kalman filter

Started by ●May 30, 2007

Reply by ●May 30, 20072007-05-30

On May 30, 6:19 pm, heng <likemurs...@gmail.com> wrote:> The tranditional Kalman filter equation is as follows: > x(n)=F x(n-1)+w(n) > y(n)=H x(n)+v(n) > And x(n) is called "state". > If there are two states, x1(n) and x2(n), > x1(n)=F1 x1(n-1)+w1(n) > x2(n)=F2 x2(n-1)+w2(n) > y(n)=H1 x1(n)+H2 x2(n)+v(n) > Is it still the Kalman filter? How to make estimation? Or can you > refer me some papers? > Thanks.The Kalman filter equation applies to the multi-dimensional case. This is what you have: x, y (using the traditional form) are two-dimensional signals. Let x_vec = [x1 x2]^T, H_vec = [H1 H2], and so on and so forth. Hope that helps, Julius

Reply by ●May 2, 20082008-05-02

>On May 30, 6:19 pm, heng <likemurs...@gmail.com> wrote: >> The tranditional Kalman filter equation is as follows: >> x(n)=F x(n-1)+w(n) >> y(n)=H x(n)+v(n) >> And x(n) is called "state". >> If there are two states, x1(n) and x2(n), >> x1(n)=F1 x1(n-1)+w1(n) >> x2(n)=F2 x2(n-1)+w2(n) >> y(n)=H1 x1(n)+H2 x2(n)+v(n) >> Is it still the Kalman filter? How to make estimation? Or can you >> refer me some papers? >> Thanks. > >The Kalman filter equation applies to the multi-dimensional >case. This is what you have: x, y (using the traditional >form) are two-dimensional signals. > >Let x_vec = [x1 x2]^T, H_vec = [H1 H2], and so on and >so forth. > >Hope that helps, >Julius > >your new system now would be of the form x(n+1)=Fx(n)+w(n) withx(n)=[x1,x2]^T and F=diag[F1,F2] the same with H the covariance matrices must have the appropriate dimensions so as the multiplications can be done.