Derivation of (13)
The true position of a target under constant acceleration is expressed as
$$\begin{array}{@{}rcl@{}} x_{\mathrm{t}k} = x_{\mathrm{t}k-1}+Tv_{\mathrm{t}k-1}+\left(T^{2}/2\right)a_{\mathrm{t}k-1}, \end{array} $$
((42))
where v
t and a
t are the true velocity and acceleration. With (1) and (42), the variance of the predicted position errors is
$$ {\fontsize{8.6pt}{9.6pt}\selectfont{\begin{aligned} \sigma_{\mathrm{p}}^{2} &= E\left[\left(x_{\mathrm{p}k}-x_{\mathrm{t}k}\right)^{2}\right]\\ &=E\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)^{2}\right]+T^{2} E\left[\left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)^{2}\right]\\ &\quad+\left(T^{4}/4\right)E\left[\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)^{2}\right]\\ &\quad+2TE\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)\left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)\right]\\ &\quad+T^{2}E\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)\right]\\ &\quad+T^{3}E\left[\left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)\right]. \end{aligned}}} $$
((43))
Because we assume a steady state, the variances and covariances in (43) do not depend on k. Consequently, we can define these variances and covariances as:
$${} \begin{aligned} \sigma_{\text{sx}}^{2} &= E\left[\left(x_{\mathrm{s}k}-x_{\mathrm{t}k}\right)^{2}\right] = E\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)^{2}\right], \end{aligned} $$
((44))
$${} \begin{aligned} \sigma_{\text{sv}}^{2} &=& E\left[\left(v_{\mathrm{s}k}-v_{\mathrm{t}k}\right)^{2}\right] = E\left[\left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)^{2}\right], \end{aligned} $$
((45))
$${} \begin{aligned} \sigma_{\text{sa}}^{2} &=& E\left[\left(a_{\mathrm{s}k}-a_{\mathrm{t}k}\right)^{2}\right] = E\left[\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)^{2}\right], \end{aligned} $$
((46))
$${} \begin{aligned} \sigma_{\text{sxv}}^{2} &= E\left[\left(x_{\mathrm{s}k}-x_{\mathrm{t}k}\right)\left(v_{\mathrm{s}k}-v_{\mathrm{t}k}\right)\right] \\&= E\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)\left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)\right], \end{aligned} $$
((47))
$${} \begin{aligned} \sigma_{\text{sxa}}^{2} &= E\left[\left(x_{\mathrm{s}k}-x_{\mathrm{t}k}\right)\left(a_{\mathrm{s}k}-a_{\mathrm{t}k}\right)\right] \\&= E\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)\right], \end{aligned} $$
((48))
$${} \begin{aligned} \sigma_{\text{sva}}^{2} &= E\left[\left(v_{\mathrm{s}k}-v_{\mathrm{t}k}\right)\left(a_{\mathrm{s}k}-a_{\mathrm{t}k}\right)\right] \\&= E\left[\left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)\right]. \end{aligned} $$
((49))
Substituting (44) to (49) into (43), we have
$${} {\fontsize{9.2pt}{9.6pt}\selectfont{\begin{aligned} \sigma_{\mathrm{p}}^{2} = \sigma_{\text{sx}}^{2}+T^{2} \sigma_{\text{sv}}^{2}+\left(T^{4}/4\right) \sigma_{\text{sa}}^{2}+2T\sigma_{\text{sxv}}^{2}+T^{2} \sigma_{\text{sxa}}^{2}+T^{3}\sigma_{\text{sva}}^{2}. \end{aligned}}} $$
((50))
The variances and covariances in this equation are derived as functions of the filter gains and the variances of measurement noise. With (1) and (10), we have
$$ x_{\mathrm{s}k} = (1-\alpha)\left(x_{\mathrm{s}k-1}+Tv_{\mathrm{s}k-1}+\left(T^{2}/2\right)a_{\mathrm{s}k-1}\right)+\alpha x_{\mathrm{o}k}. $$
((51))
We can rewrite (42) as
$$ x_{\mathrm{t}k} = (1-\alpha)\left(x_{\mathrm{t}k-1}+Tv_{\mathrm{t}k-1}+ \left(T^{2}/2\right)a_{\mathrm{t}k-1}\right)+\alpha x_{\mathrm{t}k}. $$
((52))
Using (51) and (52), the smoothing error is expressed as
$$\begin{array}{@{}rcl@{}} x_{\mathrm{s}k}-x_{\mathrm{t}k} &=& (1-\alpha) \left\{ (x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1})+T(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1})\right.\\&& +\left.\left(T^{2}/2\right) (a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1})\right\}\\ && +\alpha (x_{\mathrm{o}k}-x_{\mathrm{t}k}). \end{array} $$
((53))
Thus, the variance of this error is calculated as
$${} {\fontsize{8.8pt}{9.6pt}\selectfont{\begin{aligned} \sigma_{\text{sx}}^{2} &= E\left[\left(x_{\mathrm{s}k}-x_{\mathrm{t}k}\right)^{2}\right] \\ &= \left(1-\alpha\right)^{2} \left\{ E\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)^{2}\right] + T^{2}E\left[\left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)^{2}\right]\right. \\ &\quad \left. +\left(T^{4}/4\right)E\left[\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)^{2}\right]\right.\\ &\quad \left. + 2TE\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right) \left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)\right] \right. \\ &\quad \left. +T^{2}E\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)\right] \right. \\ &\quad\left.+T^{3}E\left[\left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)\right] {\vphantom{E\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)^{2}\right]}}\right\} \\&\quad+\alpha^{2} E\left[\left(x_{\mathrm{o}k}-x_{\mathrm{t}k}\right)^{2}\right] \\ &\quad +2\alpha(1-\alpha) \left\{ E\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)\left(x_{\mathrm{o}k}-x_{\mathrm{t}k}\right)\right]\right. \\&\quad \left.+ TE\left[\left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)\left(x_{\mathrm{o}k}-x_{\mathrm{t}k}\right)\right] \right.\\ &\quad\left.+\left(T^{2}/2\right)E\left[\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)\left(x_{\mathrm{o}k}-x_{\mathrm{t}k}\right)\right] \right\}. \end{aligned}}} $$
((54))
Here,
$$\begin{array}{@{}rcl@{}} E\left[\left(x_{\mathrm{o}k}-x_{\mathrm{t}k}\right)^{2}\right]=B_{\mathrm{x}}. \end{array} $$
((55))
The following relations are satisfied because of the steady-state assumption and because the smoothed parameters are a linear combination of the measured parameters:
$$ E\left[\left(x_{\mathrm{s}k-1}-x_{\mathrm{t}k-1}\right)\left(x_{\mathrm{o}k}-x_{\mathrm{t}k}\right)\right]=0, $$
((56))
$$ E\left[\left(v_{\mathrm{s}k-1}-v_{\mathrm{t}k-1}\right)\left(x_{\mathrm{o}k}-x_{\mathrm{t}k}\right)\right]=0, $$
((57))
$$ E\left[\left(a_{\mathrm{s}k-1}-a_{\mathrm{t}k-1}\right)\left(x_{\mathrm{o}k}-x_{\mathrm{t}k}\right)\right]=0. $$
((58))
Substituting (44) to (49) and (55) to (58) into (54), we obtain
$$\begin{array}{@{}rcl@{}} \sigma_{\text{sx}}^{2} &=& (1-\alpha)^{2}\left(\sigma_{\text{sx}}^{2}+T^{2}\sigma_{\text{sv}}^{2}+\left(T^{4}/4\right) \sigma_{\text{sa}}^{2}+2T\sigma_{\text{sxv}}^{2}\right.\\&&\left.+T^{2}\sigma_{\text{sxa}}^{2}+T^{3}\sigma_{\text{sva}}^{2}\right) +\alpha^{2}B_{\mathrm{x}}. \end{array} $$
((59))
This can be simplified to
$$\begin{array}{@{}rcl@{}} &\alpha&\!\!(2-\alpha)\sigma_{\text{sx}}^{2}-(1-\alpha)^{2}\left(T^{2}\sigma_{\text{sv}}^{2}+\left(T^{4}/4\right) \sigma_{\text{sa}}^{2}+2T\sigma_{\text{sxv}}^{2}\right.\\&&\left.+T^{2}\sigma_{\text{sxa}}^{2}+T^{3}\sigma_{\text{sva}}^{2}\right)=\alpha^{2}B_{\mathrm{x}}. \end{array} $$
((60))
In the same way, other variances and covariances are calculated using (1) to (3) and (10) to (12) as follows:
$$ \begin{aligned} \sigma_{\text{sv}}^{2} &= E\left[\left(v_{\mathrm{s}k}-v_{\mathrm{t}k}\right)^{2}\right] \\ &=(1-\beta)^{2}\left(\sigma_{\text{sv}}^{2}+T^{2}\sigma_{\text{sa}}^{2}+2T\sigma_{\text{sva}}^{2}\right)+\beta^{2} B_{\mathrm{v}}, \end{aligned} $$
((61))
$${} \begin{aligned} \sigma_{\text{sa}}^{2} &= E\left[\left(a_{\mathrm{s}k}-a_{\mathrm{t}k}\right)^{2}\right]\\ &= \sigma_{\text{sa}}^{2}+(\gamma^{2}/T^{2})B_{\mathrm{v}}+ \left(\gamma^{2}/T^{2}\right)\!\left(\sigma_{\text{sv}}^{2}+T^{2}\sigma_{\text{sa}}^{2} +2T\sigma_{\text{sva}}^{2}\right)\\&\quad-(2\gamma/T)\left(\sigma_{\text{sva}}^{2}+\sigma_{\text{sa}}^{2}\right) \end{aligned} $$
((62))
$$ \begin{aligned} \sigma_{\text{sxv}}^{2} &=E\left[\left(x_{\mathrm{s}k}-x_{\mathrm{t}k}\right)\left(v_{\mathrm{s}k}-v_{\mathrm{t}k}\right)\right]\\ &= (1-\alpha)(1-\beta)\left(\sigma_{\text{sxv}}^{2}+T\sigma_{\text{sxa}}^{2}+T\sigma_{\text{sv}}^{2} +T^{2}\sigma_{\text{sva}}^{2}\right.\\&\quad\left.+(T^{2}/2)\sigma_{\text{sva}}^{2}+(T^{3}/2)\sigma_{\text{sa}}^{2}\right) \end{aligned} $$
((63))
$$ \begin{aligned} \sigma_{\text{sxa}}^{2} &= E\left[\left(x_{\mathrm{s}k}-x_{\mathrm{t}k}\right)\left(a_{\mathrm{s}k}-a_{\mathrm{t}k}\right)\right]\\ &=(1-\alpha)\left(\sigma_{\text{sxa}}^{2}+T\sigma_{\text{sva}}^{2}+\left(T^{2}/2\right)\sigma_{\text{sa}}^{2}\right)\\&\quad-(g(1-a)/T) \left(\sigma_{\text{sxv}}^{2}+T\sigma_{\text{sxa}}^{2}+T\sigma_{\text{sv}}^{2}\right. \\ &\quad\left. +T^{2}\sigma_{\text{sva}}^{2}+\left(T^{2}/2\right)\sigma_{\text{sva}}^{2}+\left(T^{3}/2\right) \sigma_{\text{sa}}^{2}\right), \end{aligned} $$
((64))
$$ \begin{aligned} \sigma_{\text{sva}}^{2} &= E\left[\left(v_{\mathrm{s}k}-v_{\mathrm{t}k}\right)\left(a_{\mathrm{s}k} -a_{\mathrm{t}k}\right)\right]\\ &= (1-\beta)(1-\gamma)\left(\sigma_{\text{sva}}^{2}+T\sigma_{\text{sa}}^{2}\right)- (\gamma(1-\beta)/T)\\&\quad\times\left(\sigma_{\text{sv}}^{2}+T\sigma_{\text{sva}}^{2}\right)+(\beta \gamma /T) B_{\mathrm{v}}, \end{aligned} $$
((65))
where
$$\begin{array}{@{}rcl@{}} E\left[\left(v_{\mathrm{o}k}-v_{\mathrm{t}k}\right)^{2}\right] = B_{\mathrm{v}}, \end{array} $$
((66))
and the following is satisfied because we assume that the measurement position and velocity noise are uncorrelated:
$$\begin{array}{@{}rcl@{}} E\left[\left(x_{\mathrm{o}k}-x_{\mathrm{t}k})(v_{\mathrm{o}k}-v_{\mathrm{t}k}\right)\right] = 0. \end{array} $$
((67))
Equations (61) to (65) can be simplified to:
$${} \begin{aligned} \beta(2-\beta)\sigma_{\text{sv}}^{2}-(1-\beta)^{2}\left(T^{2}\sigma_{\text{sa}}^{2}+2T\sigma_{\text{sva}}^{2}\right) = \beta^{2} B_{\mathrm{v}}, \end{aligned} $$
((68))
$$ \begin{aligned} &\gamma(2-\gamma)\sigma_{\text{sa}}^{2}-\left(\gamma^{2}/T^{2}\right) \sigma_{\text{sv}}^{2}-\left(2\gamma(1-\gamma)/T\right)\\&\sigma_{\text{sva}}^{2} = \left(\gamma^{2}/T^{2}\right)B_{\mathrm{v}}, \end{aligned} $$
((69))
$${} \begin{aligned} &\left(\alpha+\beta-\alpha \beta\right) \sigma_{\text{sxv}}^{2}-(1-\alpha)(1-\beta)\\&\left(T\sigma_{\text{sv}}^{2}+\left(T^{3}/2\right)\sigma_{\text{sa}}^{2}+ T\sigma_{\text{sxa}}^{2}+\left(3T^{2}/2\right)\sigma_{\text{sva}}^{2}\right) = 0, \end{aligned} $$
((70))
$$ \begin{aligned} &\left(\alpha+\gamma-\alpha \gamma\right) \sigma_{\text{sxa}}^{2} +(\gamma(1-\alpha)/T)\left(T\sigma_{\text{sv}}^{2}+\sigma_{\text{sxv}}^{2}\right)\\&- \left(\left(\alpha-1\right)\left(\gamma-1\right)T^{2}/2\right)\sigma_{\text{sa}}^{2}\\ &-\left(\left(\alpha-1\right)\left(3\gamma-2\right)T/2\right)\sigma_{\text{sva}}^{2} = 0, \end{aligned} $$
((71))
$$ \begin{aligned} &\left(\beta+2\gamma-2 \beta \gamma\right) \sigma_{\text{sva}}^{2}+\gamma(1-\beta)\sigma_{\text{sv}}^{2}\\&-(1-\beta)(1-\gamma)T\sigma_{\text{sa}}^{2}=(\beta \gamma/T)B_{\mathrm{v}}. \end{aligned} $$
((72))
Solving the linear system involving (59) and (68) to (72), we obtain:
$$ \begin{aligned} \sigma_{\text{sx}}^{2} = \frac{\alpha}{2-\alpha}B_{\mathrm{x}}+ \frac{f_{1}(\alpha,\beta,\gamma)(1-\alpha)^{2}}{f_{2}(\alpha, \beta, \gamma)}T^{2} B_{\mathrm{v}}, \end{aligned} $$
((73))
$$ \begin{aligned} \sigma_{\text{sv}}^{2} &= \frac{2\beta^{2}+2\gamma-3\beta \gamma}{\beta(4-2\beta-\gamma)} B_{\mathrm{v}},\\ \end{aligned} $$
((74))
$$ \begin{aligned} \sigma_{\text{sa}}^{2} &= \frac{2\gamma^{2}}{\beta(4-2\beta - \gamma)} \frac{B_{\mathrm{v}}}{T^{2}}, \\ \end{aligned} $$
((75))
$${} {\fontsize{8.6pt}{9.6pt}\selectfont{\begin{aligned} \sigma_{\text{sxv}}^{2} &= \frac{(1-\alpha)(1-\beta)\left(4\alpha \beta^{2}-\alpha \gamma^{2} -4\alpha \beta \gamma+4\alpha \gamma+4\beta \gamma\right)}{f_{2}(\alpha,\beta,\gamma)/\alpha} TB_{\mathrm{v}}, \end{aligned}}} $$
((76))
$$ \begin{aligned} \sigma_{\text{sxa}}^{2} &= \frac{\gamma(\alpha-1)\left(4\alpha \beta^{2}-4\alpha \beta-\gamma^{2}+4\gamma\right)}{f_{2}(\alpha,\beta,\gamma)/\alpha} B_{\mathrm{v}}, \\ \end{aligned} $$
((77))
$$ \begin{aligned} \sigma_{\text{sva}}^{2} &= \frac{\gamma(2\beta-\gamma)}{\beta(4-2\beta - \gamma)} \frac{B_{\mathrm{v}}}{T}, \end{aligned} $$
((78))
where f
1(α,β,γ) and f
2(α,β,γ) are expressed as (14) and (15). Substituting (73) to (78) into (50), we arrive at (13).
Derivation of (16)
We first derive the relationship between the measured signals (x
ok
and v
ok
) and the predicted position x
pk
in the z-domain and then obtain the tracking performance index using the final value theorem. Applying a z-transform to (1) to (3) and (10) to (12), we obtain:
$$ X_{\mathrm{p}}(z) = X_{\mathrm{s}}(z)/z+TV_{\mathrm{s}}(z)/z+\left(T^{2}/2\right)A_{\mathrm{s}}(z)/z, $$
((79))
$$ V_{\mathrm{p}}(z) = V_{\mathrm{s}}(z)/z+TA_{\mathrm{s}}(z)/z, $$
((80))
$$ A_{\mathrm{p}}(z) = A_{\mathrm{s}}(z)/z, $$
((81))
$$ X_{\mathrm{s}}(z) = X_{\mathrm{p}}(z)+\alpha(X_{\mathrm{o}}(z)-X_{\mathrm{p}}(z)), $$
((82))
$$ V_{\mathrm{s}}(z) = V_{\mathrm{p}}(z)+\beta(V_{\mathrm{o}}(z)-V_{\mathrm{p}}(z)), $$
((83))
$$ A_{\mathrm{s}}(z) = A_{\mathrm{p}}(z)+(\gamma/T)(V_{\mathrm{o}}(z)-V_{\mathrm{p}}(z)). $$
((84))
Substituting (84) into (81), we have
$$\begin{array}{@{}rcl@{}} A_{\mathrm{p}}(z) &=& \frac{\gamma}{z-1} \cdot \frac{V_{\mathrm{o}}(z)-V_{\mathrm{p}}(z)}{T}. \end{array} $$
((85))
Substituting (83) into (80) gives
$$\begin{array}{@{}rcl@{}} (z+\beta-1)V_{\mathrm{p}}(z) &=& \beta V_{\mathrm{o}}(z) + zTA_{\mathrm{p}}(z). \end{array} $$
((86))
Substituting (85) into (86), the relationship between the predicted and measured velocities is calculated as
$$\begin{array}{@{}rcl@{}} V_{\mathrm{p}}(z) &=& \frac{(\beta+\gamma)z-\beta}{z^{2}+(\beta+\gamma-2)z-\beta+1}V_{\mathrm{o}}(z). \end{array} $$
((87))
Substituting (87) into (85), the relationship between the predicted acceleration and the measured velocities is calculated as
$$\begin{array}{@{}rcl@{}} A_{\mathrm{p}}(z) &=& \frac{\gamma(z-1)}{z^{2}+(\beta+\gamma-2)z-\beta+1} \frac{V_{\mathrm{o}}(z)}{T}. \end{array} $$
((88))
Substituting (82) to (84), (87), and (88) into (79), the relationship between the predicted position and the measured position and velocity is written as
$${} {\fontsize{9.2pt}{9.6pt}\selectfont{\begin{aligned} X_{\mathrm{p}}(z) &= \frac{\alpha}{z+\alpha-1}X_{\mathrm{o}}(z)\\&\quad+\frac{z((2\beta+\gamma)z-2\beta+\gamma)} {2(z+\alpha-1)(z^{2}+(\beta+\gamma-2)z-\beta+1)} TV_{\mathrm{o}}(z). \end{aligned}}} $$
((89))
Thus, the z-transform of the error x
ok
−x
pk
is expressed as
$${} {\fontsize{9.2pt}{9.6pt}\selectfont{\begin{aligned} E_{\mathrm{p}}(z) &= \frac{z-1}{z+\alpha-1}X_{\mathrm{o}}(z)\\&\quad- \frac{z((2\beta+\gamma)z-2\beta+\gamma)}{2(z+\alpha-1)(z^{2}+(\beta+\gamma-2)z-\beta+1)} TV_{\mathrm{o}}(z). \end{aligned}}} $$
((90))
Here, the measured position and velocity of a target with constant jerk J are:
$$ x_{\mathrm{o}k} = J(kT)^{3}/6, $$
((91))
$$ v_{\mathrm{o}k} = J(kT)^{2}/2, $$
((92))
and their z-transforms are:
$$ X_{\mathrm{o}}(z) = \frac{z\left(z^{2}+4z+1\right)}{6(z-1)^{4}}JT^{3}, $$
((93))
$$ V_{\mathrm{o}}(z) = \frac{z(z+1)}{2(z-1)^{3}}JT^{2}. $$
((94))
Substituting (93) and (94) into (90), we have
$${} {\fontsize{8.8pt}{9.6pt}\selectfont{\begin{aligned} E_{\mathrm{p}}(z) = \frac{z\left(2z^{2}+(8-4\beta-\gamma)\,z+2-2\beta\right)}{12(z-1)(z+\alpha-1) \left(z^{2}+(\beta+\gamma-2)\,z-\beta+1\right)} JT^{3}. \end{aligned}}} $$
((95))
With the final value theorem \({\lim }_{z \to 1}(z-1)E_{\mathrm {p}}(z)\), we have (16).
Derivation of (20)
Using the same procedure as for the A-V filter, the linear system with respect to the variances and covariances of the smoothing parameters is calculated using (1) to (3) and (17) to (19) as:
$${} \begin{aligned} &\alpha(2-\alpha)\sigma_{\text{sx}}^{2}-(1-\alpha)^{2}\left(T^{2}\sigma_{\text{sv}}^{2} +\left(T^{4}/4\right)\sigma_{\text{sa}}^{2}+2T\sigma_{\text{sxv}}^{2}\right.\\&\left.+T^{2}\sigma_{\text{sxa}}^{2} +T^{3}\sigma_{\text{sva}}^{2}\right)=\alpha^{2}B_{\mathrm{x}}, \end{aligned} $$
((96))
$$ \begin{aligned} & \beta(2-\beta)\sigma_{\text{sv}}^{2}-(1-\beta)^{2}\left(T^{2}\sigma_{\text{sa}}^{2} +2T\sigma_{\text{sva}}^{2}\right) = \beta^{2} B_{\mathrm{v}}, \end{aligned} $$
((97))
$${} \begin{aligned} & \left(\gamma(4-\gamma)/4\right)\sigma_{\text{sa}}^{2} - \left(\gamma-2/T^{4}\right)\left(\sigma_{\text{sx}}^{2}+T^{2}\sigma_{\text{sv}}^{2} +2T\sigma_{\text{sxv}}^{2}\right)\\&+\gamma(2-\gamma)\left(\sigma_{\text{sxa}}^{2}/T^{2}+\sigma_{\text{sva}}^{2}/T\right)= \left(\gamma^{2}/T^{4}\right)B_{\mathrm{x}}, \end{aligned} $$
((98))
$$ \begin{aligned} & \left(\alpha+\beta-\alpha \beta\right) \sigma_{\text{sxv}}^{2}-(1-\alpha)(1-\beta)\\& \left(T\sigma_{\text{sv}}^{2}+\left(T^{3}/2\right)\sigma_{\text{sa}}^{2} +T\sigma_{\text{sxa}}^{2}+(3T^{2}/2)\sigma_{\text{sva}}^{2}\right) \sigma_{\text{sxv}}^{2} = 0, \end{aligned} $$
((99))
$$ \begin{aligned} &\left(\alpha+\gamma-\alpha\gamma\right)\sigma_{\text{sxa}}^{2}-(\alpha-1)(\gamma-1)T\sigma_{\text{sva}}^{2}\\&- \left((\alpha-1)(\gamma-2)/4\right)T^{2}\sigma_{\text{sa}}^{2}\\ & +\gamma(1-\alpha)(\sigma_{\text{sx}}^{2}/T^{2}+\sigma_{\text{sv}}^{2}+2\sigma_{\text{sxv}}^{2}/T) = \left(\alpha \gamma/T^{2}\right)B_{\mathrm{x}}, \\ \end{aligned} $$
((100))
$$ \begin{aligned} &\left(2\beta+3\gamma-3\beta\gamma\right) \sigma_{\text{sva}}^{2}/2+\gamma(1-\beta)\\&\times\left(\sigma_{\text{sxv}}^{2}/T^{2}+\sigma_{\text{sv}}^{2}/T+\sigma_{\text{sxa}}^{2}/T^{2}\right)\\ & -(\beta-1)(\gamma-2)T\sigma_{\text{sa}}^{2}/2=0. \end{aligned} $$
((101))
Solving the linear system involving (96) to (101), we obtain:
$$ \begin{aligned} \sigma_{\text{sx}}^{2} = \frac{g_{\text{1x}}(\alpha, \beta, \gamma)B_{\mathrm{x}}+(1-\alpha)^{2}g_{2}(\alpha,\beta)T^{2}B_{\mathrm{v}}}{g_{3}(\alpha, \beta, \gamma)}, \end{aligned} $$
((102))
$$ \begin{aligned} \sigma_{\text{sv}}^{2} = \frac{g_{\text{1v}}(\alpha, \beta, \gamma)B_{\mathrm{x}}/T^{2}+g_{\text{2v}}(\alpha,\beta,\gamma)B_{\mathrm{v}}}{(2-\beta)g_{3}(\alpha, \beta, \gamma)}, \end{aligned} $$
((103))
$$ \begin{aligned} \sigma_{\text{sa}}^{2} = \frac{g_{\text{1a}}(\alpha, \beta, \gamma)B_{\mathrm{x}}/T^{4}+g_{\text{2a}}(\alpha,\beta,\gamma)B_{\mathrm{v}}/T^{2}}{(2-\beta)g_{3}(\alpha, \beta, \gamma)}, \end{aligned} $$
((104))
$$ \begin{aligned} \sigma_{\text{sxv}}^{2} = \frac{g_{\text{1xv}}(\alpha, \beta, \gamma)B_{\mathrm{x}}/T+g_{\text{2xv}}(\alpha,\beta,\gamma)TB_{\mathrm{v}}}{g_{3}(\alpha, \beta, \gamma)}, \end{aligned} $$
((105))
$${} {\fontsize{9.2pt}{9.6pt}\selectfont{\begin{aligned} \sigma_{\text{sxa}}^{2} &= \frac{g_{\text{1xa}}(\alpha, \beta, \gamma)B_{\mathrm{x}}/T^{2}+2(1-\alpha)(2-\beta)\gamma^{2} g_{2}(\alpha,\beta)B_{\mathrm{v}}}{\gamma(2-\beta)g_{3}(\alpha, \beta, \gamma)}, \end{aligned}}} $$
((106))
$$ \begin{aligned} \sigma_{\text{sva}}^{2} &= \frac{g_{\text{1va}}(\alpha, \beta, \gamma)B_{\mathrm{x}}/T^{3}+g_{\text{2va}}(\alpha,\beta,\gamma)B_{\mathrm{v}}/T}{(2-\beta)g_{3}(\alpha, \beta, \gamma)}, \end{aligned} $$
((107))
where g
2(α,β) and g
3(α,β,γ) are expressed as (22) and (23), and
$${} \begin{aligned} g_{\text{1x}}(\alpha, \beta, \gamma)&=8\alpha^{3}\beta(2-\beta)(\beta-1)-2\alpha^{2}\left(3\beta^{3}\gamma-4\beta^{3}\right.\\&\left.\quad-9\beta^{2}\gamma+8\beta^{2}+2\beta \gamma+4\gamma\right)\\ &\quad+\alpha \gamma\left(10\beta^{3}\,+\,\beta^{2}\gamma\,-\,28\beta^{2}-\beta \gamma +8\beta+16\right)\\&\quad-4\beta\gamma(2-\beta)^{2}, \end{aligned} $$
((108))
$${} \begin{aligned} g_{\text{1v}}(\alpha, \beta, \gamma)&=& 8\gamma^{2}(1-\beta)^{2}(2-\beta)(\alpha+\beta-\alpha \beta -2), \end{aligned} $$
((109))
$${} \begin{aligned} g_{\text{2v}}(\alpha, \beta, \gamma)&= 8\alpha^{3}\beta^{2}(2-\beta)(1-\beta)+2\alpha^{2}\beta^{2}(\beta-2)\\&\quad\times(3\beta \gamma-12\beta-3\gamma+8)\\ &\quad -\alpha \beta^{2}\left(22\beta^{2}\gamma-16\beta^{2}+\beta \gamma^{2}-64\beta \gamma \right.\\&\quad\left.+32\beta-\gamma^{2}+40\gamma\right)\\ &\quad +2\beta^{2} \gamma(\beta-1)(8\beta+\gamma-16), \end{aligned} $$
((110))
$${} \begin{aligned} g_{\text{1a}}(\alpha, \beta, \gamma)&= 4\beta \gamma^{2}(\beta-2)\left(2\alpha \beta^{2}-6\alpha \beta-2\beta^{2}-\beta \gamma\right.\\&\quad\left.+4\alpha+4\beta+\gamma\right), \end{aligned} $$
((111))
$${} \begin{aligned} g_{\text{2a}}(\alpha, \beta, \gamma)&= 4\beta^{2} \gamma(\alpha-2)\left(2\alpha \beta^{2}-6\alpha \beta-2\beta^{2}-\beta \gamma\right.\\&\quad\left.+4\alpha+4\beta+\gamma\right), \end{aligned} $$
((112))
$${} \begin{aligned} g_{\text{1xv}}(\alpha, \beta, \gamma)&= 2\beta \gamma(\alpha-1)(\beta-2)(\beta-1)(4\alpha+\gamma), \end{aligned} $$
((113))
$$ \begin{aligned} g_{\text{2xv}}(\alpha, \beta, \gamma)&= 2\beta^{2}(\alpha-1)(\alpha-2)(\beta-1)(4\alpha+\gamma), \end{aligned} $$
((114))
$${} \begin{aligned} g_{\text{1xa}}(\alpha, \beta, \gamma)&= \gamma^{2}(2-\beta)\left(8\alpha^{2}\beta^{3}-24\alpha^{2}\beta^{2}+16\alpha^{2}\beta\right.\\&\quad\left.+2\alpha \beta^{3} \gamma-8\alpha \beta^{3}-6\alpha \beta^{2} \gamma +16\alpha \beta^{2}\right.\\ &\quad\left.-4\alpha \beta \gamma+8\alpha \gamma -2\beta^{3} \gamma-\beta^{2} \gamma^{2}+\beta \gamma^{2}\right.\\&\quad\left.+16 \beta \gamma -16\gamma\!\vphantom{\left(8\alpha^{2}\beta^{3}-24\alpha^{2}\beta^{2}+16\alpha^{2}\beta\right.}\right), \end{aligned} $$
((115))
$${} {\fontsize{8.9pt}{9.6pt}\selectfont{\begin{aligned} g_{\text{1va}}(\alpha, \beta, \gamma)= 2\beta \gamma^{2}(2-\beta)(\beta-1)(4\alpha \beta-8\alpha-4\beta-\gamma+8), \end{aligned}}} $$
((116))
$${} {\fontsize{8.9pt}{9.6pt}\selectfont{\begin{aligned} g_{\text{2va}}(\alpha, \beta, \gamma)= 2\beta^{2} \gamma(2-\alpha)(\beta-1)(4\alpha \beta-8\alpha-4\beta-\gamma+8). \end{aligned}}} $$
((117))
Substituting (102) to (107) into (50), we arrive at (20).
Derivation of (24)
The derivation process is the same as for the A-V filter. By applying a z-transform to (1) to (3) and (17) to (19) and their simplified forms, the predicted parameters in the z-domain are derived as:
$$ \begin{aligned} A_{\mathrm{p}}(z) = \frac{\gamma z}{z-1} \cdot \frac{X_{\mathrm{o}}(z)-X_{\mathrm{p}}(z)}{T^{2}}, \end{aligned} $$
((118))
$$ \begin{aligned} V_{\mathrm{p}}(z) &= \frac{\gamma z^{2}}{(z-1)(z+\beta-1)}\cdot \frac{X_{\mathrm{o}}(z)-X_{\mathrm{p}}(z)}{T}\\&\quad+\frac{\beta}{z+\beta-1}V_{\mathrm{o}}(z), \end{aligned} $$
((119))
$${} {\fontsize{8.4pt}{9.6pt}\selectfont{\begin{aligned} X_{\mathrm{p}}(z) = \frac{(2\alpha+\gamma)X_{\mathrm{o}}(z)+2\beta TV_{\mathrm{o}}(z)+2(1-\beta)TV_{\mathrm{p}}(z)+T^{2}A_{\mathrm{p}}(z)}{2z+2\alpha+\gamma-2}. \end{aligned}}} $$
((120))
Substituting (118) and (119) into (120), the relationship between the predicted position and the measured position and velocity is expressed as
$$\begin{array}{@{}rcl@{}} X_{\mathrm{p}}(z) = \frac{h_{1}(z)}{h_{2}(z)}X_{\mathrm{o}}(z)+\frac{2\beta z(z-1)}{h_{2}(z)}TV_{\mathrm{o}}(z), \end{array} $$
((121))
where
$${} \begin{aligned} h_{1}(z) &= \gamma z^{3}+(2\alpha-\beta \gamma+2\gamma)z^{2}\\&\quad+\!\left(2\alpha \beta-4\alpha+\beta \gamma-2\gamma\!\right)z-2\alpha \beta -\beta \gamma+2\alpha+\gamma\!, \end{aligned} $$
((122))
$$ \begin{aligned} h_{2}(z) &= (\gamma+2)z^{3}+(2\alpha+2\beta+2\gamma-\beta \gamma-6)z^{2} \\ &\quad+(2\alpha \beta+\beta \gamma-4\alpha-4\beta-2\gamma+6)z \\ &\quad-2\alpha \beta -\beta \gamma+2\alpha+2\beta+\gamma-2. \end{aligned} $$
((123))
From (121), the z-transform of the predicted error is
$${} \begin{aligned} E_{\mathrm{p}}(z) &= X_{\mathrm{o}}(z)-X_{\mathrm{p}}(z)\\&=\frac{2(z-1)^{2}(z+\beta-1)} {h_{2}(z)}X_{\mathrm{o}}(z)-\frac{2\beta z(z-1)}{h_{2}(z)}TV_{\mathrm{o}}(z). \end{aligned} $$
((124))
Substituting (93) and (94) into (124), the error for a target with constant jerk is given by
$$\begin{array}{@{}rcl@{}} E_{\mathrm{p}}(z) &=& \frac{z\left(z^{2}-2\beta z+4z-\beta+1\right)}{3(z-1)h_{2}(z)}JT^{3}. \end{array} $$
((125))
Applying the final value theorem to (125), we have (24).