LaTex与数学公式

w(t) longrightarrow igg[frac{sqrt{2sigma ^2eta}}{s+eta}igg]  longrightarrow igg[frac{1}{s}igg] longrightarrow y

$w(t) longrightarrow igg[frac{sqrt{2sigma ^2eta}}{s+eta}igg]  longrightarrow igg[frac{1}{s}igg] longrightarrow y$

usepackage{amsmath}  %可以使用oldsymbol加粗罗马字符;mathbf对罗马字符不起作用。

mathbf{x}_{k+1} = oldsymbol{phi}_k mathbf{x}_k + mathbf{w}_k

$mathbf{x}_{k+1} = oldsymbol{phi}_k mathbf{x}_k + mathbf{w}_k$

%注意{和}是特殊字符,使用{和}

mathbf{Q}_k=E[mathbf{w}_kmathbf{w}_k^T]

=Eig{   ig[ int_{t_k}^{t_{k+1}} oldsymbol{phi}(t_{k+1}, u) mathbf{G}(u) mathbf{w}(u)du ig]  ig[ int_{t_k}^{t_{k+1}}oldsymbol{phi}(t_{k+1},v) mathbf{G}(v) mathbf{w}(v)dv ig]^T   ig}

=int_{t_k}^{t_{k+1}} int_{t_k}^{t_{k+1}} oldsymbol{phi}(t_{k+1}, u)mathbf{G}(u)E[mathbf{w}(u)mathbf{w}^T(v)]mathbf{G}^T(v)oldsymbol{phi}^T(t_{k+1},v)dudv

$mathbf{Q}_k=E[mathbf{w}_kmathbf{w}_k^T]$

$=Eig{   ig[ int_{t_k}^{t_{k+1}} oldsymbol{phi}(t_{k+1}, u) mathbf{G}(u) mathbf{w}(u)du ig]  ig[ int_{t_k}^{t_{k+1}}oldsymbol{phi}(t_{k+1},v) mathbf{G}(v) mathbf{w}(v)dv ig]^T   ig}$

$=int_{t_k}^{t_{k+1}} int_{t_k}^{t_{k+1}} oldsymbol{phi}(t_{k+1}, u)mathbf{G}(u)E[mathbf{w}(u)mathbf{w}^T(v)]mathbf{G}^T(v)oldsymbol{phi}^T(t_{k+1},v)dudv$

left[egin{matrix}

dot{x_1}\dot{x_2}

end{matrix} ight]

= left[

egin{matrix}

0&1\0&-eta

end{matrix}

ight]

left[egin{matrix}

x_1\x_2

end{matrix} ight] +

left[egin{matrix}

0\sqrt{2sigma^2eta}

end{matrix} ight]w(t)

$left[egin{matrix}dot{x_1}\dot{x_2}end{matrix} ight] = left[egin{matrix}0&1\0&-etaend{matrix} ight] left[egin{matrix}x_1\x_2end{matrix} ight] + left[egin{matrix}0\sqrt{2sigma^2eta}end{matrix} ight]w(t)$

y=left[egin{matrix}

1&0

end{matrix} ight]

left[egin{matrix}

x_1\x_2

end{matrix} ight]

$y=left[egin{matrix}1&0end{matrix} ight]left[egin{matrix}x_1\x_2end{matrix} ight]$

三角形帽子表示估计

mathbf{hat{x}}_k^-=oldsymbol{Phi}_kmathbf{hat{x}}_{k-1}+mathbf{G}_kmathbf{u}_k

$mathbf{hat{x}}_k^-=oldsymbol{Phi}_kmathbf{hat{x}}_{k-1}+mathbf{G}_kmathbf{u}_k$

原文地址:https://www.cnblogs.com/byeyear/p/6682053.html