Appendix A: Determining the region estimated by the Kalman filtering

 

 

Table: Definitions of symbols for the Kalman filtering

Symbol	Estimation of Ck,0(t)	Estimation of Dk,0(t)
Observed signal ${\bf {y}}_m$	Xk(tm)	$\exp(j\phi_k(t_m))$
State variable ${\bf {x}}_m$	Ck(tm)	$\exp(j D_{k}(t_m))$
Observed noise ${\bf {v}}_m$	X2,k(tm)	X2,k(tm)/Sk(tm)
System noise ${\bf {w}}_m$	wm	wm
State transition matrix ${\bf {F}}_m$	$\Delta C_k(t_m)$	$\Delta D_k(t_m)$
Observation matrix ${\bf {H}}_m$	$\exp(j\omega_k t_m)$	$\hat{C}_{k}(t_m)/S_k(t_m)$
Driving matrix ${\bf {G}}_m$	1	1
Initial value $\hat{\bf {x}}_{0\vert-1}$	0	1
Initial value $\hat{\bf {\Sigma}}_{0\vert-1}$	Sk(t0)	$\exp(j\phi_k(t_0))$

In this paper, we consider how to estimate C_k,0 and D_k,0 from the observed component X_k(t) using the Kalman filter. The estimation duration is [T_h-1-T_h]. It is then decomposed into discrete time $t_m=m\cdot \Delta t$, $m=0, 1, 2, \cdots, M$, where the sampling period is $\Delta t=1/f_s$ and f_s is the sampling frequency. First, let C_k(t) and D_k(t) be $C_k(t)=\int C_{k,0}(t)dt$ and $D_k(t)=\int D_{k,0}(t)dt$, respectively. Here, let the temporal variations of C_k(t) and D_k(t) at discrete time t_m be

  

$$C_k(t_m)\Delta C_k(t_m)+w_m,$$ C_k(t_m+1) = (37)

$$\Delta C_k(t_m) 1+\frac{C_k(t_m)-C_k(t_{m-1})}{C_k(t_m)},$$ = (38)

$$D_k(t_m)\Delta D_k(t_m)+w_m,$$ D_k(t_m+1) = (39)

$$\Delta D_k(t_m) 1+\frac{D_k(t_m)-D_k(t_{m-1})}{D_k(t_m)},$$ = (40)

where t₀=T_h-1 and t_M=T_h. It is assumed that the variation error is represented by white noise with mean 0 and variance $\sigma_m^2$.

Next, for the system of the Kalman filtering problem:

  

$${\bf {x}}_{m+1} {\bf {F}}_m{\bf {x}}_m+{\bf {G}}_m{\bf {w}}_m \qquad \mbox{(state)},$$ = (41)

$${\bf {y}}_{m} {\bf {H}}_m{\bf {x}}_m+{\bf {v}}_m \qquad \mbox{(observation)},$$ = (42)

in order to estimate C_k,0(t), we apply Eq. () to Eq. () and apply Eq. () to Eq. (). Using the same steps as for the system of the Kalman filtering problem, in order to estimate D_k,0(t), we apply Eq. () to Eq. () and apply the normalized Eq. () to Eq. (). The parameters in Eqs. () and () are shown in Table .

In these systems, the mean and variance of the terms, ${\bf {x}}_0$, ${\bf {w}}_m$, and ${\bf {v}}_m$, are known. And ${\bf {F}}_m$, ${\bf {G}}_m$, ${\bf {H}}_m$, and ${\bf {v}}_m$ are known matrices. The Kalman filtering problem is to determine the minimum variance requirement $\hat{\bf {x}}_{m\vert m}$ from the observed ${\bf {y}}_m$, $m=0, 1, 2, \cdots, M$ as follows.

$$\hat{\bf {x}}_{m\vert m}=E({\bf {x}}_m+{\bf {y}}_0,\cdots, {\bf {y}}_m)$$ (43)

It is calculated by sequentially solving the following equations:

1.

Filtering equation

$$\hat{{\bf {x}}}_{m\vert m}=\hat{{\bf {x}}}_{m\vert m-1}+{\bf {K}}_m({\bf {y}}_m-{\bf {H}}_m\hat{{\bf {x}}}_{m\vert m-1})$$ (44)

$$\hat{{\bf {x}}}_{m+1\vert m}={\bf {F}}_m\hat{{\bf {x}}}_{m\vert m}$$ (45)

2.

Kalman gain

$${\bf {K}}_m=\frac{\hat{{\bf {\Sigma}}}_{m\vert m-1}{\bf {H}}_... ... {\Sigma}}}_{m\vert m-1}{\bf {H}}_m^{*T}+{\bf {\Sigma}}_{v_m}}$$ (46)

3.

Covariance equation for the estimated-error

$$\hat{{\bf {\Sigma}}}_{m\vert m} \hat{\bf {\Sigma}}_{m\vert m-1}-{\bf {K}}_m {\bf {H}}_m \hat{{\bf {\Sigma}}}_{m\vert m-1}$$ = (47)

$$\hat{{\bf {\Sigma}}}_{m+1\vert m} \hat{\bf {F}}_m{\bf {\Sigma}}_{m\vert m}{\bf {F}}_m^{*T}+{\bf {G}}_m {\bf {\Sigma}}_{w_m}{\bf {G}}_m^{*T}$$ = (48)

4.

Initial state

$$\hat{\bf {x}}_{0\vert-1}=\bar{\bf {x}}_0, \qquad \hat{\bf {\Sigma}}_{0\vert-1}={\bf {\Sigma}}_{x_0},$$ (49)

The symbols $\bar{\qquad}$ and ${\bf {\Sigma}}$ are the mean and variance of a random variable, respectively.

Finally, performing the Kalman filtering according to Eqs. () and (), we obtain the minimal-variance estimated value $\hat{\bf {x}}(t_m)=\hat{\bf {x}}_{m\vert m}$ and the covariance matrix $\hat{\bf {e}}(t_m)=\hat{\bf {\Sigma}}_{m\vert m}$ at discrete time t_m. As a result, the estimated $\hat{C}_{k,0}(t)$ and $\hat{D}_{k,0}(t)$, and the estimated errors P_k(t) and Q_k(t) are determined by $\hat{C}_{k,0}(t)=\vert d\hat{\bf {x}}(t)/dt\vert$ and $P_k(t)=\vert d\hat{\bf {e}}(t)/dt\vert$, $\hat{D}_{k,0}(t)=\arg (d\hat{\bf {x}}(t)/dt)$, and $Q_k(t)=\arg (d\hat{\bf {e}}(t)/dt)$, respectively.