卡尔曼滤波器(8) -- 一维卡尔曼滤波器(例7例8)

This blog is translated from https://www.kalmanfilter.net/default.aspx.
It’s an excellent tutorial about Kalman Filter written by Alex Becker.

目录
卡尔曼滤波器(1) – 背景知识
卡尔曼滤波器(2) – α−β−γ滤波器(例1)
卡尔曼滤波器(3) – α−β−γ滤波器(例2)
卡尔曼滤波器(4) – α−β−γ滤波器(例3&例4&总结)
卡尔曼滤波器(5) – 一维卡尔曼滤波器(介绍)
卡尔曼滤波器(6) – 一维卡尔曼滤波器(例5&完整模型)
卡尔曼滤波器(7) – 一维卡尔曼滤波器(例6)
卡尔曼滤波器(8) – 一维卡尔曼滤波器(例7&例8)
卡尔曼滤波器(9) – 多维卡尔曼滤波器(前言&预备)
卡尔曼滤波器(10) – 多维卡尔曼滤波器(状态外推方程)
卡尔曼滤波器(11) – 多维卡尔曼滤波器(线性动态系统模型)
卡尔曼滤波器(12) – 多维卡尔曼滤波器(协方差外推方程)
卡尔曼滤波器(13) – 多维卡尔曼滤波器(测量方程)
卡尔曼滤波器(14) – 多维卡尔曼滤波器(简短总结)
卡尔曼滤波器(15) – 多维卡尔曼滤波器(状态更新方程)
卡尔曼滤波器(16) – 多维卡尔曼滤波器(协方差更新方程)
卡尔曼滤波器(17) – 多维卡尔曼滤波器(卡尔曼增益)
卡尔曼滤波器(18) – 多维卡尔曼滤波器(简化的协方差更新方程)
卡尔曼滤波器(19) – 多维卡尔曼滤波器(总结)
卡尔曼滤波器(20) – 多维卡尔曼滤波器(例9)

例7 - 估计罐子里的液体温度Ⅰ

还是估算液体温度的例子。不同的是，液体以每秒 0. 1 ∘ C \color{Red}0.1^\circ\mathrm{C} 0.1∘C 的速度加热。

这里也是静态系统。

卡尔曼滤波器的参数和前面例子的相似：

假设我们的模型是很准确的，过程噪声的方差（ q q q）设为 0.0001 0.0001 0.0001
测量误差（标准差）为 0.1 0.1 0.1 摄氏度
每 5 5 5 秒进行一次测量
系统动态是恒定的
注意，虽然实际上它不是恒定的（静态的），因为液体正在被加热。但是由于温度没有变化，我们仍然把它当成是一个静态系统。
在测量的时候，液体的真实温度为 50.47 9 ∘ C ， 51.02 5 ∘ C ， 51. 5 ∘ C ， 52.00 3 ∘ C ， 52.49 4 ∘ C ， 53.00 2 ∘ C ， 53.49 9 ∘ C ， 54.00 6 ∘ C ， 54.49 8 ∘ C ， 54.99 1 ∘ C 50.479^\circ\mathrm{C}，51.025^\circ\mathrm{C}，51.5^\circ\mathrm{C}，52.003^\circ\mathrm{C}，52.494^\circ\mathrm{C}，53.002^\circ\mathrm{C}，53.499^\circ\mathrm{C}，54.006^\circ\mathrm{C}，54.498^\circ\mathrm{C}，54.991^\circ\mathrm{C} 50.479∘C，51.025∘C，51.5∘C，52.003∘C，52.494∘C，53.002∘C，53.499∘C，54.006∘C，54.498∘C，54.991∘C
测量值为 50.4 5 ∘ C ， 50.96 7 ∘ C ， 51. 6 ∘ C ， 52.10 6 ∘ C ， 52.49 2 ∘ C ， 52.81 9 ∘ C ， 53.43 3 ∘ C ， 54.00 7 ∘ C ， 54.52 3 ∘ C ， 54.9 9 ∘ C 50.45^\circ\mathrm{C}，50.967^\circ\mathrm{C}，51.6^\circ\mathrm{C}，52.106^\circ\mathrm{C}，52.492^\circ\mathrm{C}，52.819^\circ\mathrm{C}，53.433^\circ\mathrm{C}，54.007^\circ\mathrm{C}，54.523^\circ\mathrm{C}，54.99^\circ\mathrm{C} 50.45∘C，50.967∘C，51.6∘C，52.106∘C，52.492∘C，52.819∘C，53.433∘C，54.007∘C，54.523∘C，54.99∘C

测量值和真实值的关系如图所示：

第零次迭代
在第一次迭代之前，需要初始化卡尔曼滤波器，然后预测下一个状态（也就是第一个状态）。

初始化
和之前一样，盲猜温度是 1 0 ∘ C 10^\circ\mathrm{C} 10∘C：
x ^ 0 , 0 = 1 0 ∘ C \hat{x}_{0,0}=10^\circ\mathrm{C} x^0,0=10∘C

由于这个猜测值是非常不精准的，于是把初始估计误差（ σ \sigma σ）设为 100 100 100。初始化的估计不确定性(Estimate Uncertainty)是误差的方差（ σ 2 \sigma^2 σ2）：
p 0 , 0 = 10 0 2 = 10 , 000 p_{0,0}=100^2=10,000 p0,0=1002=10,000

这里方差非常高。如果我们的初始值更合理，卡尔曼滤波器会收敛更快。

预测
基于初始值来预测下一个状态。
由于模型是静态的，预测估计与当前估计相同： x ^ 1 , 0 = 1 0 ∘ C \hat{x}_{1,0}=10^\circ\mathrm{C} x^1,0=10∘C

外推估计不确定性（方差）为：
p 1 , 0 = p 0 , 0 + q = 10000 + 0.0001 = 10000.0001 \begin{aligned} p_{1,0} &= p_{0,0} + q \\ &= 10000 + 0.0001 \\ &= 10000.0001 \end{aligned} p1,0=p0,0+q=10000+0.0001=10000.0001

第1-10次迭代
如下所示：

n n n	z n z_n zn	当前状态估计 ( K n , x ^ n , n , p n , n K_n,\hat{x}_{n,n},p_{n,n} Kn,x^n,n,pn,n)	预测( x ^ n + 1 , n , p n + 1 , n \hat{x}_{n+1,n},p_{n+1,n} x^n+1,n,pn+1,n)
1	50.4 5 ∘ C 50.45^\circ\mathrm{C} 50.45∘C	K 1 = 10000.0001 10000.0001 + 0.01 = 0.999999 x ^ 1 , 1 = 10 + 0.999999 ( 50.45 − 10 ) = 50.4 5 ∘ C p 1 , 1 = ( 1 − 0.999999 ) 10000.0001 = 0.01 K_1=\frac{10000.0001}{10000.0001+0.01}=0.999999 \\ \quad \\ \hat{x}_{1,1}=10+0.999999(50.45−10)=50.45^\circ\mathrm{C} \\ \quad \\ p_{1,1}=(1−0.999999)10000.0001=0.01 K1=10000.0001+0.0110000.0001=0.999999x^1,1=10+0.999999(50.45−10)=50.45∘Cp1,1=(1−0.999999)10000.0001=0.01	x ^ 2 , 1 = x ^ 1 , 1 = 50.4 5 ∘ C p 2 , 1 = 0.01 + 0.0001 = 0.0101 \hat{x}_{2,1}=\hat{x}_{1,1} = 50.45^\circ\mathrm{C} \\ \quad \\ p_{2,1} = 0.01+0.0001=0.0101 x^2,1=x^1,1=50.45∘Cp2,1=0.01+0.0001=0.0101
2	50.96 7 ∘ C 50.967^\circ\mathrm{C} 50.967∘C	K 2 = 0.0101 0.0101 + 0.01 = 0.5025 x ^ 2 , 2 = 50.45 + 0.5025 ( 50.967 − 50.45 ) = 50.7 1 ∘ C p 2 , 2 = ( 1 − 0.5025 ) 0.0101 = 0.005 K_2=\frac{0.0101}{0.0101+0.01}=0.5025 \\ \quad \\ \hat{x}_{2,2}=50.45+0.5025(50.967−50.45)=50.71^\circ\mathrm{C} \\ \quad \\ p_{2,2}=(1−0.5025)0.0101=0.005 K2=0.0101+0.010.0101=0.5025x^2,2=50.45+0.5025(50.967−50.45)=50.71∘Cp2,2=(1−0.5025)0.0101=0.005	x ^ 3 , 2 = x ^ 2 , 2 = 50.7 1 ∘ C p 3 , 2 = 0.005 + 0.0001 = 0.0051 \hat{x}_{3,2}=\hat{x}_{2,2} = 50.71^\circ\mathrm{C} \\ \quad \\ p_{3,2} = 0.005+0.0001=0.0051 x^3,2=x^2,2=50.71∘Cp3,2=0.005+0.0001=0.0051
3	51. 6 ∘ C 51.6^\circ\mathrm{C} 51.6∘C	K 3 = 0.0051 0.0051 + 0.01 = 0.3388 x ^ 3 , 3 = 50.71 + 0.3388 ( 51.6 − 50.71 ) = 51.01 1 ∘ C p 3 , 3 = ( 1 − 0.3388 ) 0.0051 = 0.0034 K_3=\frac{0.0051}{0.0051+0.01}=0.3388 \\ \quad \\ \hat{x}_{3,3}=50.71+0.3388(51.6−50.71)=51.011^\circ\mathrm{C} \\ \quad \\ p_{3,3}=(1−0.3388)0.0051=0.0034 K3=0.0051+0.010.0051=0.3388x^3,3=50.71+0.3388(51.6−50.71)=51.011∘Cp3,3=(1−0.3388)0.0051=0.0034	x ^ 4 , 3 = x ^ 3 , 3 = 51.01 1 ∘ C p 4 , 3 = 0.0034 + 0.0001 = 0.0035 \hat{x}_{4,3}=\hat{x}_{3,3} = 51.011^\circ\mathrm{C} \\ \quad \\ p_{4,3} = 0.0034+0.0001=0.0035 x^4,3=x^3,3=51.011∘Cp4,3=0.0034+0.0001=0.0035
4	52.10 6 ∘ C 52.106^\circ\mathrm{C} 52.106∘C	K 4 = 0.0035 0.0035 + 0.01 = 0.2586 x ^ 4 , 4 = 51.011 + 0.2586 ( 52.106 − 51.011 ) = 51.29 5 ∘ C p 4 , 4 = ( 1 − 0.2586 ) 0.0035 = 0.0026 K_4=\frac{0.0035}{0.0035+0.01}=0.2586 \\ \quad \\ \hat{x}_{4,4}=51.011+0.2586(52.106−51.011)=51.295^\circ\mathrm{C} \\ \quad \\ p_{4,4}=(1−0.2586)0.0035=0.0026 K4=0.0035+0.010.0035=0.2586x^4,4=51.011+0.2586(52.106−51.011)=51.295∘Cp4,4=(1−0.2586)0.0035=0.0026	x ^ 5 , 4 = x ^ 4 , 4 = 51.29 5 ∘ C p 5 , 4 = 0.0026 + 0.0001 = 0.0027 \hat{x}_{5,4}=\hat{x}_{4,4} = 51.295^\circ\mathrm{C} \\ \quad \\ p_{5,4} = 0.0026+0.0001=0.0027 x^5,4=x^4,4=51.295∘Cp5,4=0.0026+0.0001=0.0027
5	52.49 2 ∘ C 52.492^\circ\mathrm{C} 52.492∘C	K 5 = 0.0027 0.0027 + 0.01 = 0.2117 x ^ 5 , 5 = 51.295 + 0.2117 ( 52.492 − 51.295 ) = 51.54 8 ∘ C p 5 , 5 = ( 1 − 0.2117 ) 0.0027 = 0.0021 K_5=\frac{0.0027}{0.0027+0.01}=0.2117 \\ \quad \\ \hat{x}_{5,5}=51.295+0.2117(52.492−51.295)=51.548^\circ\mathrm{C} \\ \quad \\ p_{5,5}=(1−0.2117)0.0027=0.0021 K5=0.0027+0.010.0027=0.2117x^5,5=51.295+0.2117(52.492−51.295)=51.548∘Cp5,5=(1−0.2117)0.0027=0.0021	x ^ 6 , 5 = x ^ 5 , 5 = 51.54 8 ∘ C p 6 , 5 = 0.0021 + 0.0001 = 0.0022 \hat{x}_{6,5}=\hat{x}_{5,5} = 51.548^\circ\mathrm{C} \\ \quad \\ p_{6,5} = 0.0021+0.0001=0.0022 x^6,5=x^5,5=51.548∘Cp6,5=0.0021+0.0001=0.0022
6	52.81 9 ∘ C 52.819^\circ\mathrm{C} 52.819∘C	K 6 = 0.0022 0.0022 + 0.01 = 0.1815 x ^ 6 , 6 = 51.548 + 0.1815 ( 52.819 − 51.548 ) = 51.77 9 ∘ C p 6 , 6 = ( 1 − 0.1815 ) 0.0022 = 0.0018 K_6=\frac{0.0022}{0.0022+0.01}=0.1815 \\ \quad \\ \hat{x}_{6,6}=51.548+0.1815(52.819−51.548)=51.779^\circ\mathrm{C} \\ \quad \\ p_{6,6}=(1−0.1815)0.0022=0.0018 K6=0.0022+0.010.0022=0.1815x^6,6=51.548+0.1815(52.819−51.548)=51.779∘Cp6,6=(1−0.1815)0.0022=0.0018	x ^ 7 , 6 = x ^ 6 , 6 = 51.77 9 ∘ C p 7 , 6 = 0.0018 + 0.0001 = 0.0019 \hat{x}_{7,6}=\hat{x}_{6,6} = 51.779^\circ\mathrm{C} \\ \quad \\ p_{7,6} = 0.0018+0.0001=0.0019 x^7,6=x^6,6=51.779∘Cp7,6=0.0018+0.0001=0.0019
7	53.43 3 ∘ C 53.433^\circ\mathrm{C} 53.433∘C	K 7 = 0.0019 0.0019 + 0.01 = 0.1607 x ^ 7 , 7 = 51.779 + 0.1607 ( 53.433 − 51.779 ) = 52.04 5 ∘ C p 7 , 7 = ( 1 − 0.1607 ) 0.0019 = 0.0016 K_7=\frac{0.0019}{0.0019+0.01}=0.1607 \\ \quad \\ \hat{x}_{7,7}=51.779+0.1607(53.433−51.779)=52.045^\circ\mathrm{C} \\ \quad \\ p_{7,7}=(1−0.1607)0.0019=0.0016 K7=0.0019+0.010.0019=0.1607x^7,7=51.779+0.1607(53.433−51.779)=52.045∘Cp7,7=(1−0.1607)0.0019=0.0016	x ^ 8 , 7 = x ^ 7 , 7 = 52.04 5 ∘ C p 8 , 7 = 0.0016 + 0.0001 = 0.0017 \hat{x}_{8,7}=\hat{x}_{7,7} = 52.045^\circ\mathrm{C} \\ \quad \\ p_{8,7} = 0.0016+0.0001=0.0017 x^8,7=x^7,7=52.045∘Cp8,7=0.0016+0.0001=0.0017
8	54.00 7 ∘ C 54.007^\circ\mathrm{C} 54.007∘C	K 8 = 0.0017 0.0017 + 0.01 = 0.1458 x ^ 8 , 8 = 52.045 + 0.1458 ( 54.007 − 52.045 ) = 52.33 1 ∘ C p 8 , 8 = ( 1 − 0.1458 ) 0.0017 = 0.0015 K_8=\frac{0.0017}{0.0017+0.01}=0.1458 \\ \quad \\ \hat{x}_{8,8}=52.045+0.1458(54.007−52.045)=52.331^\circ\mathrm{C} \\ \quad \\ p_{8,8}=(1−0.1458)0.0017=0.0015 K8=0.0017+0.010.0017=0.1458x^8,8=52.045+0.1458(54.007−52.045)=52.331∘Cp8,8=(1−0.1458)0.0017=0.0015	x ^ 9 , 8 = x ^ 8 , 8 = 52.33 1 ∘ C p 9 , 8 = 0.0015 + 0.0001 = 0.0016 \hat{x}_{9,8}=\hat{x}_{8,8} = 52.331^\circ\mathrm{C} \\ \quad \\ p_{9,8} = 0.0015+0.0001=0.0016 x^9,8=x^8,8=52.331∘Cp9,8=0.0015+0.0001=0.0016
9	54.52 3 ∘ C 54.523^\circ\mathrm{C} 54.523∘C	K 9 = 0.0016 0.0016 + 0.01 = 0.1348 x ^ 9 , 9 = 52.331 + 0.1348 ( 54.523 − 52.331 ) = 52.62 6 ∘ C p 9 , 9 = ( 1 − 0.1348 ) 0.0016 = 0.0014 K_9=\frac{0.0016}{0.0016+0.01}=0.1348 \\ \quad \\ \hat{x}_{9,9}=52.331+0.1348(54.523−52.331)=52.626^\circ\mathrm{C} \\ \quad \\ p_{9,9}=(1−0.1348)0.0016=0.0014 K9=0.0016+0.010.0016=0.1348x^9,9=52.331+0.1348(54.523−52.331)=52.626∘Cp9,9=(1−0.1348)0.0016=0.0014	x ^ 10 , 9 = x ^ 9 , 9 = 52.62 6 ∘ C p 10 , 9 = 0.0014 + 0.0001 = 0.0015 \hat{x}_{10,9}=\hat{x}_{9,9} = 52.626^\circ\mathrm{C} \\ \quad \\ p_{10,9} = 0.0014+0.0001=0.0015 x^10,9=x^9,9=52.626∘Cp10,9=0.0014+0.0001=0.0015
10	54.9 9 ∘ C 54.99^\circ\mathrm{C} 54.99∘C	K 10 = 0.0015 0.0015 + 0.01 = 0.1265 x ^ 10 , 10 = 2.626 + 0.1265 ( 54.99 − 52.626 ) = 52.92 5 ∘ C p 10 , 10 = ( 1 − 0.1265 ) 0.0015 = 0.0013 K_{10}=\frac{0.0015}{0.0015+0.01}=0.1265 \\ \quad \\ \hat{x}_{10,10}=2.626+0.1265(54.99−52.626)=52.925^\circ\mathrm{C} \\ \quad \\ p_{10,10}=(1−0.1265)0.0015=0.0013 K10=0.0015+0.010.0015=0.1265x^10,10=2.626+0.1265(54.99−52.626)=52.925∘Cp10,10=(1−0.1265)0.0015=0.0013	x ^ 11 , 10 = x ^ 10 , 10 = 52.92 5 ∘ C p 11 , 10 = 0.0013 + 0.0001 = 0.0014 \hat{x}_{11,10}=\hat{x}_{10,10} = 52.925^\circ\mathrm{C} \\ \quad \\ p_{11,10} = 0.0013+0.0001=0.0014 x^11,10=x^10,10=52.925∘Cp11,10=0.0013+0.0001=0.0014

下面的表画出了真实值，测量值，估计值：

看来这次的卡尔曼滤波器并没有给出可靠的估计。这种现象说明在本例的估计中存在滞后误差(lag error)。我们之前在例3中遇到过滞后误差，在该示例中，我们假定飞机速度是恒定的，用 α − β \alpha-\beta α−β 滤波器估算了加速飞机的位置。在例4中把 α − β \alpha-\beta α−β 滤波器换成了带有加速度的 α − β − γ \alpha-\beta-\gamma α−β−γ 滤波器，解决了滞后误差。

有 2 个原因导致这次卡尔曼滤波器的滞后误差：

系统模型不适合这种情况。
过程模型的可靠性。我们选择了很低的过程噪声（ q = 0.0001 q=0.0001 q=0.0001），然而实际温度的波动要大得多。

Note: 注意：滞后误差应为常数，因此估计曲线的斜率应该和真实值曲线的斜率相同。上面的图仅显示了10个初始测量值，不足以进行收敛。下图显示了前100次测量的恒定滞后误差。

有两种办法来解决滞后误差：

如果我们知道液体温度是线性变化的，则可以定义一个新模型，该模型考虑液体温度可能发生的线性变化。在例4中就是这么做的，这是首选方法。如果无法对温度变化进行建模，则此方法将不会改善卡尔曼滤波器的性能。
如果找不到一个合适的模型，可以通过增加过程噪声 q q q 来调整过程模型的可靠性。详细做法看下一个例子。

例8 - 估计罐子里的液体温度Ⅱ

和上面的例子很相似，只做了1处修改，把过程不确定性（ q q q）从 0.0001 0.0001 0.0001 升到 0.15 0.15 0.15。

第零次迭代
在第一次迭代之前，需要初始化卡尔曼滤波器，然后预测下一个状态（也就是第一个状态）。

初始化
和之前一样，盲猜温度是 1 0 ∘ C 10^\circ\mathrm{C} 10∘C：
x ^ 0 , 0 = 1 0 ∘ C \hat{x}_{0,0}=10^\circ\mathrm{C} x^0,0=10∘C

这里方差非常高。如果我们的初始值更合理，卡尔曼滤波器会收敛更快。

预测
基于初始值来预测下一个状态。
由于模型是静态的，预测估计与当前估计相同： x ^ 1 , 0 = 1 0 ∘ C \hat{x}_{1,0}=10^\circ\mathrm{C} x^1,0=10∘C

外推估计不确定性（方差）为：
p 1 , 0 = p 0 , 0 + q = 10000 + 0.15 = 10000.15 \begin{aligned} p_{1,0} &= p_{0,0} + q \\ &= 10000 + 0.15 \\ &= 10000.15 \end{aligned} p1,0=p0,0+q=10000+0.15=10000.15

第1-10次迭代
如下所示：

n n n	z n z_n zn	当前状态估计 ( K n , x ^ n , n , p n , n K_n,\hat{x}_{n,n},p_{n,n} Kn,x^n,n,pn,n)	预测( x ^ n + 1 , n , p n + 1 , n \hat{x}_{n+1,n},p_{n+1,n} x^n+1,n,pn+1,n)
1	50.4 5 ∘ C 50.45^\circ\mathrm{C} 50.45∘C	K 1 = 10000.15 10000.15 + 0.01 = 0.999999 x ^ 1 , 1 = 10 + 0.999999 ( 50.45 − 10 ) = 50.4 5 ∘ C p 1 , 1 = ( 1 − 0.999999 ) 10000.15 = 0.01 K_1=\frac{10000.15}{10000.15+0.01}=0.999999 \\ \quad \\ \hat{x}_{1,1}=10+0.999999(50.45−10)=50.45^\circ\mathrm{C} \\ \quad \\ p_{1,1}=(1−0.999999)10000.15=0.01 K1=10000.15+0.0110000.15=0.999999x^1,1=10+0.999999(50.45−10)=50.45∘Cp1,1=(1−0.999999)10000.15=0.01	x ^ 2 , 1 = x ^ 1 , 1 = 50.4 5 ∘ C p 2 , 1 = 0.01 + 0.15 = 0.016 \hat{x}_{2,1}=\hat{x}_{1,1} = 50.45^\circ\mathrm{C} \\ \quad \\ p_{2,1} = 0.01+0.15=0.016 x^2,1=x^1,1=50.45∘Cp2,1=0.01+0.15=0.016
2	50.96 7 ∘ C 50.967^\circ\mathrm{C} 50.967∘C	K 2 = 0.16 0.16 + 0.01 = 0.9412 x ^ 2 , 2 = 50.45 + 0.9412 ( 50.967 − 50.45 ) = 50.9 4 ∘ C p 2 , 2 = ( 1 − 0.9412 ) 0.16 = 0.0094 K_2=\frac{0.16}{0.16+0.01}=0.9412 \\ \quad \\ \hat{x}_{2,2}=50.45+0.9412(50.967−50.45)=50.94^\circ\mathrm{C} \\ \quad \\ p_{2,2}=(1−0.9412)0.16=0.0094 K2=0.16+0.010.16=0.9412x^2,2=50.45+0.9412(50.967−50.45)=50.94∘Cp2,2=(1−0.9412)0.16=0.0094	x ^ 3 , 2 = x ^ 2 , 2 = 50.9 4 ∘ C p 3 , 2 = 0.0094 + 0.15 = 0.1594 \hat{x}_{3,2}=\hat{x}_{2,2} = 50.94^\circ\mathrm{C} \\ \quad \\ p_{3,2} = 0.0094+0.15=0.1594 x^3,2=x^2,2=50.94∘Cp3,2=0.0094+0.15=0.1594
3	51. 6 ∘ C 51.6^\circ\mathrm{C} 51.6∘C	K 3 = 0.1594 0.1594 + 0.01 = 0.941 x ^ 3 , 3 = 50.94 + 0.941 ( 51.6 − 50.94 ) = 51.5 6 ∘ C p 3 , 3 = ( 1 − 0.941 ) 0.1594 = 0.0094 K_3=\frac{0.1594}{0.1594+0.01}=0.941 \\ \quad \\ \hat{x}_{3,3}=50.94+0.941(51.6−50.94)=51.56^\circ\mathrm{C} \\ \quad \\ p_{3,3}=(1−0.941)0.1594=0.0094 K3=0.1594+0.010.1594=0.941x^3,3=50.94+0.941(51.6−50.94)=51.56∘Cp3,3=(1−0.941)0.1594=0.0094	x ^ 4 , 3 = x ^ 3 , 3 = 51.5 6 ∘ C p 4 , 3 = 0.0094 + 0.15 = 0.1594 \hat{x}_{4,3}=\hat{x}_{3,3} = 51.56^\circ\mathrm{C} \\ \quad \\ p_{4,3} = 0.0094+0.15=0.1594 x^4,3=x^3,3=51.56∘Cp4,3=0.0094+0.15=0.1594
4	52.10 6 ∘ C 52.106^\circ\mathrm{C} 52.106∘C	K 4 = 0.1594 0.1594 + 0.01 = 0.941 x ^ 4 , 4 = 51.56 + 0.941 ( 52.106 − 51.56 ) = 52.0 7 ∘ C p 4 , 4 = ( 1 − 0.941 ) 0.1594 = 0.0094 K_4=\frac{0.1594}{0.1594+0.01}=0.941 \\ \quad \\ \hat{x}_{4,4}=51.56+0.941(52.106−51.56)=52.07^\circ\mathrm{C} \\ \quad \\ p_{4,4}=(1−0.941)0.1594=0.0094 K4=0.1594+0.010.1594=0.941x^4,4=51.56+0.941(52.106−51.56)=52.07∘Cp4,4=(1−0.941)0.1594=0.0094	x ^ 5 , 4 = x ^ 4 , 4 = 52.4 7 ∘ C p 5 , 4 = 0.0094 + 0.15 = 0.1594 \hat{x}_{5,4}=\hat{x}_{4,4} = 52.47^\circ\mathrm{C} \\ \quad \\ p_{5,4} = 0.0094+0.15=0.1594 x^5,4=x^4,4=52.47∘Cp5,4=0.0094+0.15=0.1594
5	52.49 2 ∘ C 52.492^\circ\mathrm{C} 52.492∘C	K 5 = 0.1594 0.1594 + 0.01 = 0.941 x ^ 5 , 5 = 52.07 + 0.941 ( 52.492 − 52.07 ) = 52.4 7 ∘ C p 5 , 5 = ( 1 − 0.941 ) 0.1594 = 0.0094 K_5=\frac{0.1594}{0.1594+0.01}=0.941 \\ \quad \\ \hat{x}_{5,5}=52.07+0.941(52.492−52.07)=52.47^\circ\mathrm{C} \\ \quad \\ p_{5,5}=(1−0.941)0.1594=0.0094 K5=0.1594+0.010.1594=0.941x^5,5=52.07+0.941(52.492−52.07)=52.47∘Cp5,5=(1−0.941)0.1594=0.0094	x ^ 6 , 5 = x ^ 5 , 5 = 52.4 7 ∘ C p 6 , 5 = 0.0094 + 0.15 = 0.1594 \hat{x}_{6,5}=\hat{x}_{5,5} = 52.47^\circ\mathrm{C} \\ \quad \\ p_{6,5} = 0.0094+0.15=0.1594 x^6,5=x^5,5=52.47∘Cp6,5=0.0094+0.15=0.1594
6	52.81 9 ∘ C 52.819^\circ\mathrm{C} 52.819∘C	K 6 = 0.1594 0.1594 + 0.01 = 0.941 x ^ 6 , 6 = 52.47 + 0.941 ( 52.819 − 52.47 ) = 52. 8 ∘ C p 6 , 6 = ( 1 − 0.941 ) 0.1594 = 0.0094 K_6=\frac{0.1594}{0.1594+0.01}=0.941 \\ \quad \\ \hat{x}_{6,6}=52.47+0.941(52.819−52.47)=52.8^\circ\mathrm{C} \\ \quad \\ p_{6,6}=(1−0.941)0.1594=0.0094 K6=0.1594+0.010.1594=0.941x^6,6=52.47+0.941(52.819−52.47)=52.8∘Cp6,6=(1−0.941)0.1594=0.0094	x ^ 7 , 6 = x ^ 6 , 6 = 52. 8 ∘ C p 7 , 6 = 0.0094 + 0.15 = 0.1594 \hat{x}_{7,6}=\hat{x}_{6,6} = 52.8^\circ\mathrm{C} \\ \quad \\ p_{7,6} = 0.0094+0.15=0.1594 x^7,6=x^6,6=52.8∘Cp7,6=0.0094+0.15=0.1594
7	53.43 3 ∘ C 53.433^\circ\mathrm{C} 53.433∘C	K 7 = 0.1594 0.1594 + 0.01 = 0.941 x ^ 7 , 7 = 52.8 + 0.941 ( 53.433 − 52.8 ) = 53. 4 ∘ C p 7 , 7 = ( 1 − 0.941 ) 0.1594 = 0.0094 K_7=\frac{0.1594}{0.1594+0.01}=0.941 \\ \quad \\ \hat{x}_{7,7}=52.8+0.941(53.433−52.8)=53.4^\circ\mathrm{C} \\ \quad \\ p_{7,7}=(1−0.941)0.1594=0.0094 K7=0.1594+0.010.1594=0.941x^7,7=52.8+0.941(53.433−52.8)=53.4∘Cp7,7=(1−0.941)0.1594=0.0094	x ^ 8 , 7 = x ^ 7 , 7 = 53. 4 ∘ C p 8 , 7 = 0.0094 + 0.15 = 0.1594 \hat{x}_{8,7}=\hat{x}_{7,7} = 53.4^\circ\mathrm{C} \\ \quad \\ p_{8,7} = 0.0094+0.15=0.1594 x^8,7=x^7,7=53.4∘Cp8,7=0.0094+0.15=0.1594
8	54.00 7 ∘ C 54.007^\circ\mathrm{C} 54.007∘C	K 8 = 0.1594 0.1594 + 0.01 = 0.941 x ^ 8 , 8 = 53.4 + 0.941 ( 54.007 − 53.4 ) = 53.9 7 ∘ C p 8 , 8 = ( 1 − 0.941 ) 0.1594 = 0.0094 K_8=\frac{0.1594}{0.1594+0.01}=0.941 \\ \quad \\ \hat{x}_{8,8}=53.4+0.941(54.007−53.4)=53.97^\circ\mathrm{C} \\ \quad \\ p_{8,8}=(1−0.941)0.1594=0.0094 K8=0.1594+0.010.1594=0.941x^8,8=53.4+0.941(54.007−53.4)=53.97∘Cp8,8=(1−0.941)0.1594=0.0094	x ^ 9 , 8 = x ^ 8 , 8 = 53.9 7 ∘ C p 9 , 8 = 0.0094 + 0.15 = 0.1594 \hat{x}_{9,8}=\hat{x}_{8,8} = 53.97^\circ\mathrm{C} \\ \quad \\ p_{9,8} = 0.0094+0.15=0.1594 x^9,8=x^8,8=53.97∘Cp9,8=0.0094+0.15=0.1594
9	54.52 3 ∘ C 54.523^\circ\mathrm{C} 54.523∘C	K 9 = 0.1594 0.1594 + 0.01 = 0.941 x ^ 9 , 9 = 53.97 + 0.941 ( 54.523 − 53.97 ) = 54.4 9 ∘ C p 9 , 9 = ( 1 − 0.941 ) 0.1594 = 0.0094 K_9=\frac{0.1594}{0.1594+0.01}=0.941 \\ \quad \\ \hat{x}_{9,9}=53.97+0.941(54.523−53.97)=54.49^\circ\mathrm{C} \\ \quad \\ p_{9,9}=(1−0.941)0.1594=0.0094 K9=0.1594+0.010.1594=0.941x^9,9=53.97+0.941(54.523−53.97)=54.49∘Cp9,9=(1−0.941)0.1594=0.0094	x ^ 10 , 9 = x ^ 9 , 9 = 54.4 9 ∘ C p 10 , 9 = 0.0094 + 0.15 = 0.1594 \hat{x}_{10,9}=\hat{x}_{9,9} = 54.49^\circ\mathrm{C} \\ \quad \\ p_{10,9} = 0.0094+0.15=0.1594 x^10,9=x^9,9=54.49∘Cp10,9=0.0094+0.15=0.1594
10	54.9 9 ∘ C 54.99^\circ\mathrm{C} 54.99∘C	K 10 = 0.1594 0.1594 + 0.01 = 0.941 x ^ 10 , 10 = 54.49 + 0.941 ( 54.99 − 54.49 ) = 54.9 6 ∘ C p 10 , 10 = ( 1 − 0.941 ) 0.1594 = 0.0094 K_{10}=\frac{0.1594}{0.1594+0.01}=0.941 \\ \quad \\ \hat{x}_{10,10}=54.49+0.941(54.99−54.49)=54.96^\circ\mathrm{C} \\ \quad \\ p_{10,10}=(1−0.941)0.1594=0.0094 K10=0.1594+0.010.1594=0.941x^10,10=54.49+0.941(54.99−54.49)=54.96∘Cp10,10=(1−0.941)0.1594=0.0094	x ^ 11 , 10 = x ^ 10 , 10 = 54.9 6 ∘ C p 11 , 10 = 0.0094 + 0.15 = 0.1594 \hat{x}_{11,10}=\hat{x}_{10,10} = 54.96^\circ\mathrm{C} \\ \quad \\ p_{11,10} = 0.0094+0.15=0.1594 x^11,10=x^10,10=54.96∘Cp11,10=0.0094+0.15=0.1594

下面的表是真实值，测量值，估计值：

可以看到估计值紧跟着测量值，没有滞后误差。

下面的表是卡尔曼增益：

由于过程不确定性较高，测量权重远高于估计权重，所以卡尔曼增益很高，最后去到 0.94 0.94 0.94。

小结

我们可以把过程不确定性设高一点，以此解决滞后误差。但由于采用的模型并不是很符合这个例子，得到的带噪声的估计几乎等于测量值，这并不是卡尔曼滤波器的目的。

最佳的方案是采用非常接近实际的模型，然后仅留一点点空间给过程噪声。
不过，精准模型也不是一直可用的，例如飞行员可以突然改变方向，改变预测的飞机轨迹。这种情况下就应该增加过程噪声。