07 卡尔曼滤波器的数学原理.ipynb

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   "source": [
    "# 7. 卡尔曼滤波器的数学原理"
   ]
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   "source": [
    "%matplotlib inline"
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    "#format the book\n",
    "import book_format\n",
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   "cell_type": "markdown",
   "id": "af9c3dec",
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   "source": [
    "如果你已经读到这里，我希望你觉得卡尔曼滤波器的可怕名声有些不太应该。当然，我有忽略了一些方程，但我希望你对实现过程还是比较理解的。其基本概念相当简单——取两个测量值，或者一次测量和一次预测，然后选择输出值介于两者之间。如果你更相信测量值，你的估计会更接近测量值；如果你更相信预测更准确，你的估计会更接近它。这不是什么高深的科学（开个小玩笑——正是这种数学让阿波罗号登月并安全返回了地球！）。\n",
    "\n",
    "老实说，我一直精心选择问题。对于任意问题，设计卡尔曼滤波器的矩阵可能会非常困难。不过，我并没有搞得太复杂。像牛顿运动方程这样的方程在卡尔曼滤波器的应用中可以轻松计算，而且它们构成了我们想要解决的问题的大部分。\n",
    "\n",
    "我用代码和推理阐述了概念，而不是数学。但有些主题需要比我目前使用的数学更多。这一章介绍了你在本书后续内容中会用到的数学知识。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8adc98b0",
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   "source": [
    "# 动态系统建模\n",
    "\n",
    "动态系统是指其状态（位置、温度等）随时间演变的物理系统。微积分是研究变化值的数学，因此我们使用微分方程来建模动态系统。有些系统无法用微分方程建模，但在本书中我们不会遇到这些系统。\n",
    "\n",
    "对动态系统建模通常是几门大学课程的主题。在某种程度上，几个学期的普通微分方程和偏微分方程，再加上一门控制系统理论的研究生课程，是无可替代的。如果你是一个业余爱好者，或者在工作中尝试解决一个非常具体的滤波问题，你可能没有时间和/或倾向去花费一年或更长时间来接受这种教育。\n",
    "\n",
    "幸运的是，我可以介绍足够的理论知识，让我们能够为许多不同的卡尔曼滤波器创建系统方程。我的目标是让你能够阅读一篇论文，并理解得足够透彻，以便实现这些算法。背景数学很深奥，但在实践中，我们最终使用了一些简单的技术。\n",
    "\n",
    "这是本书中纯数学内容最长的一部分。你需要掌握本节中的所有内容才能理解扩展卡尔曼滤波器（EKF），这是最常见的非线性滤波器。我确实介绍了一些不需要太多数学知识的更现代的滤波器。你可以选择现在快速浏览一遍，如果决定学习EKF再回头来复习。\n",
    "\n",
    "我们需要首先了解卡尔曼滤波器使用的基本方程和假设。我们试图对现实世界的现象进行建模，那么我们需要考虑什么？\n",
    "\n",
    "每个物理系统都有一个过程。例如，以特定速度行驶的汽车在固定的时间内行驶一定距离，其速度随加速度变化。我们用高中学过的著名的牛顿方程描述了这种行为。\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "v&=at\\\\\n",
    "x &= \\frac{1}{2}at^2 + v_0t + x_0\n",
    "\\end{aligned}\n",
    "$$\n",
    "\n",
    "一旦我们学了微积分，就能以这种形式看到它们：\n",
    "\n",
    "$$ \\mathbf v = \\frac{d \\mathbf x}{d t}, \n",
    "\\quad \\mathbf a = \\frac{d \\mathbf v}{d t} = \\frac{d^2 \\mathbf x}{d t^2}\n",
    "$$\n",
    "\n",
    "典型的汽车跟踪问题会让你计算在恒定速度或加速度下行驶的距离，就像我们在之前的章节中所做的那样。但是，当然我们知道实际情况并不仅限于此。没有汽车会在完美的道路上行驶。路上会有颠簸、风阻，还有山丘会使速度上下波动。\n",
    "\n",
    "排除最琐碎的问题之外完美地对系统进行建模是不可能的。我们被迫进行简化。在任何时刻$t$，我们说真实状态（比如我们的汽车位置）是来自不完美模型的预测值再加上一些未知的*过程噪声*：\n",
    "\n",
    "$$\n",
    "x(t) = x_{pred}(t) + noise(t)\n",
    "$$\n",
    "\n",
    "这并不意味着$noise(t)$是我们可以通过分析推导出的函数。这只是一个事实陈述——我们总是可以将真实值描述为预测值加上过程噪声。“噪声”并不意味着随机事件。如果我们在大气中追踪一个抛出的球，并且我们的模型假设球在真空中，那么空气阻力的影响就是在这种情况下的过程噪声。\n",
    "\n",
    "在接下来的部分，我们将学习将一组高阶微分方程转换为一组一阶微分方程的技巧。在转换后，没有噪声的系统模型是：\n",
    "\n",
    "$$ \\dot{\\mathbf x} = \\mathbf{Ax}$$\n",
    "\n",
    "矩阵 $\\mathbf A$ 被称为 *系统动态矩阵*，因为它描述了系统的动态。现在我们需要对噪声建模。我们将其称为 $\\mathbf w$，并将其添加到方程中。\n",
    "\n",
    "$$ \\dot{\\mathbf x} = \\mathbf{Ax} + \\mathbf w$$\n",
    "\n",
    "$\\mathbf w$ 这个命名可能让你觉得不太合适，但你很快就会看到，卡尔曼滤波器假设是*白噪声*。\n",
    "\n",
    "最后，我们需要考虑系统中的任何输入。我们假设有一个输入 $\\mathbf u$，并且存在一个线性模型来定义该输入如何改变系统。例如，踩汽车油门会加速汽车，重力会使球下落。这两者都是控制输入。我们需要一个矩阵 $\\mathbf B$ 将 $u$ 转换为对系统的影响，并将其添加到我们的方程中：\n",
    "\n",
    "$$ \\dot{\\mathbf x} = \\mathbf{Ax} + \\mathbf{Bu} + \\mathbf{w}$$\n",
    "\n",
    "这就是其中一个方程。卡尔曼博士尝试解决的方程之一，他找到了一个最优的估计器，如果我们假设了 $\\mathbf w$ 的某些特性。"
   ]
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   "source": [
    "# 动态系统状态空间的表达\n",
    "\n",
    "我们推导出了这个方程：\n",
    "\n",
    "$$ \\dot{\\mathbf x} = \\mathbf{Ax}+ \\mathbf{Bu} + \\mathbf{w}$$\n",
    "\n",
    "然而，我们对 $\\mathbf x$ 本身感兴趣，而不是它的导数。暂时忽略噪声，我们希望找到一个递归地以 $t_{k-1}$ 时刻的 $\\mathbf x$ 表达 $t_k$ 时刻 $\\mathbf x$ 值的方程：\n",
    "\n",
    "$$\\mathbf x(t_k) = \\mathbf F(\\Delta t)\\mathbf x(t_{k-1}) + \\mathbf B(t_k)\\mathbf u (t_k)$$\n",
    "\n",
    "惯例允许我们将 $\\mathbf x(t_k)$ 写成 $\\mathbf x_k$，这表示 $t$ 的第 $k$ 个值时 $\\mathbf x$ 的值。\n",
    "\n",
    "$$\\mathbf x_k = \\mathbf{Fx}_{k-1} + \\mathbf B_k\\mathbf u_k$$\n",
    "\n",
    "矩阵 $\\mathbf F$ 是我们熟悉的 *状态转移矩阵*，因为它能够在离散时间步之间转换状态的值。它与系统动态矩阵 $\\mathbf A$ 非常相似。不同之处在于 $\\mathbf A$ 模拟一组线性微分方程，是连续的。而 $\\mathbf F$ 是离散的，表示一组线性方程（不是微分方程），它在离散的时间步长 $\\Delta t$ 内将 $\\mathbf x_{k-1}$ 转换为 $\\mathbf x_k$。\n",
    "\n",
    "找到这个矩阵通常是相当困难的。方程 $\\dot x = v$ 是最简单的微分方程，我们可以轻松地对其进行积分：\n",
    "\n",
    "$$ \\int\\limits_{x_{k-1}}^{x_k}  \\mathrm{d}x = \\int\\limits_{0}^{\\Delta t} v\\, \\mathrm{d}t $$\n",
    "$$x_k-x_{k-1} = v \\Delta t$$\n",
    "$$x_k = v \\Delta t + x_{k-1}$$\n",
    "\n",
    "这个方程是*递归*的：我们根据其在 $k-1$ 时刻的值来计算 $k$ 时刻的 $x$ 值。这种递归形式使我们能够将系统（过程模型）表示为卡尔曼滤波器所需的形式：\n",
    "\n",
    "$$\\begin{aligned}\n",
    "\\mathbf x_k &= \\mathbf{Fx}_{k-1}  \\\\\n",
    "&= \\begin{bmatrix} 1 & \\Delta t \\\\ 0 & 1\\end{bmatrix}\n",
    "\\begin{bmatrix}x_{k-1} \\\\ \\dot x_{k-1}\\end{bmatrix}\n",
    "\\end{aligned}$$\n",
    "\n",
    "我们之所以能这样做，只是因为 $\\dot x = v$ 是可能的最简单的微分方程。几乎所有其他物理系统中的微分方程都更复杂，无法采用这种方法。\n",
    "\n",
    "*状态空间* 方法在阿波罗任务时期变得流行，这在很大程度上要归功于卡尔曼博士的工作。其思想很简单。用一组 $n$ 阶微分方程对系统进行建模。将它们转换为等价的一阶微分方程集合。将它们放入前面章节中使用的向量-矩阵形式：$\\dot{\\mathbf x} = \\mathbf{Ax} + \\mathbf{Bu}$。一旦达到这种形式，我们就会使用几种技术将这些线性微分方程转换为递归方程：\n",
    "\n",
    "$$ \\mathbf x_k = \\mathbf{Fx}_{k-1} + \\mathbf B_k\\mathbf u_k$$\n",
    "\n",
    "一些书籍将状态转移矩阵称为*基本矩阵*。许多书籍使用 $\\mathbf \\Phi$ 而不是 $\\mathbf F$。在很大程度上基于控制理论的资料倾向于使用这些形式。\n",
    "\n",
    "这些被称为*状态空间*方法，因为我们是用系统状态来表达微分方程的解。"
   ]
  },
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   "cell_type": "markdown",
   "id": "b2b06f3d",
   "metadata": {},
   "source": [
    "# 将高阶微分方程转化为一阶微分方程\n",
    "\n",
    "许多物理系统的模型需要带有控制输入 $u$ 的二阶或更高阶微分方程：\n",
    "\n",
    "$$a_n \\frac{d^ny}{dt^n} + a_{n-1} \\frac{d^{n-1}y}{dt^{n-1}} +  \\dots + a_2 \\frac{d^2y}{dt^2} + a_1 \\frac{dy}{dt} + a_0 = u$$\n",
    "\n",
    "状态空间方法需要一阶方程。任何高阶方程组都可以通过为导数定义额外的变量，然后进行求解来化简为一阶方程。\n",
    "\n",
    "我们来做一个例子。给定系统 $\\ddot{x} - 6\\dot x + 9x = u$，找到等价的一阶方程。我用点符号表示时间导数以便更清楚。\n",
    "\n",
    "第一步是将最高阶的项隔离到方程的一侧。\n",
    "\n",
    "$$\\ddot{x} = 6\\dot x - 9x + u$$\n",
    "\n",
    "我们定义两个新变量:\n",
    "\n",
    "$$\\begin{aligned} x_1(t) &= x \\\\\n",
    "x_2(t) &= \\dot x\n",
    "\\end{aligned}$$\n",
    "\n",
    "现在我们将这些代入原方程并求解。解得到一组关于这些新变量的一阶方程。通常为了方便起见，我们在符号表示中省略 $(t)$。\n",
    "\n",
    "我们知道 $\\dot x_1 = x_2$，以及 $\\dot x_2 = \\ddot{x}$。因此\n",
    "\n",
    "$$\\begin{aligned}\n",
    "\\dot x_2 &= \\ddot{x} \\\\\n",
    "         &= 6\\dot x - 9x + u\\\\\n",
    "         &= 6x_2-9x_1 + u\n",
    "\\end{aligned}$$\n",
    "\n",
    "因此，我们得到的一阶方程组是：\n",
    "\n",
    "$$\\begin{aligned}\\dot x_1 &= x_2 \\\\\n",
    "\\dot x_2 &= 6x_2-9x_1 + u\\end{aligned}$$\n",
    "\n",
    "练习一段时间你就能熟练掌握了。将最高阶项隔离出来，定义一个新变量及其导数，然后进行代换。"
   ]
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   "id": "25e8d1b8",
   "metadata": {},
   "source": [
    "# 一阶微分方程的状态空间形式\n",
    "\n",
    "从前一节定义的新变量代入：\n",
    "\n",
    "$$\\frac{dx_1}{dt} = x_2,\\,  \n",
    "\\frac{dx_2}{dt} = x_3, \\, ..., \\, \n",
    "\\frac{dx_{n-1}}{dt} = x_n$$\n",
    "\n",
    "代入一阶方程得到：\n",
    "\n",
    "$$\\frac{dx_n}{dt} = \\frac{1}{a_n}\\sum\\limits_{i=0}^{n-1}a_ix_{i+1} + \\frac{1}{a_n}u\n",
    "$$\n",
    "\n",
    "使用向量-矩阵表示法，我们有：\n",
    "\n",
    "$$\\begin{bmatrix}\\frac{dx_1}{dt} \\\\ \\frac{dx_2}{dt} \\\\ \\vdots \\\\ \\frac{dx_n}{dt}\\end{bmatrix} = \n",
    "\\begin{bmatrix}\\dot x_1 \\\\ \\dot x_2 \\\\ \\vdots \\\\ \\dot x_n\\end{bmatrix}=\n",
    "\\begin{bmatrix}0 & 1 & 0 &\\cdots & 0 \\\\\n",
    "0 & 0 & 1 & \\cdots & 0 \\\\\n",
    "\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n",
    "-\\frac{a_0}{a_n} & -\\frac{a_1}{a_n} & -\\frac{a_2}{a_n} & \\cdots & -\\frac{a_{n-1}}{a_n}\\end{bmatrix}\n",
    "\\begin{bmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{bmatrix} + \n",
    "\\begin{bmatrix}0 \\\\ 0 \\\\ \\vdots \\\\ \\frac{1}{a_n}\\end{bmatrix}u$$\n",
    "\n",
    "然后我们将其写作 $\\dot{\\mathbf x} = \\mathbf{Ax} + \\mathbf{B}u$。"
   ]
  },
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   "id": "dea8a098",
   "metadata": {},
   "source": [
    "# 寻找时不变系统的基本矩阵\n",
    "\n",
    "我们用状态空间形式表示系统方程，其中\n",
    "\n",
    "$$ \\dot{\\mathbf x} = \\mathbf{Ax}$$\n",
    "\n",
    "其中 $\\mathbf A$ 是系统动态矩阵，我们想要找到*基本矩阵* $\\mathbf F$，它用以下方程在时间间隔 $\\Delta t$ 内传播状态 $\\mathbf x$：\n",
    "\n",
    "$$\\begin{aligned}\n",
    "\\mathbf x(t_k) = \\mathbf F(\\Delta t)\\mathbf x(t_{k-1})\\end{aligned}$$\n",
    "\n",
    "换句话说，$\\mathbf A$ 是一组连续微分方程，而我们需要 $\\mathbf F$ 是一组离散线性方程，用于计算 $\\mathbf A$ 在离散时间步长内的变化。\n",
    "\n",
    "通常会省略 $t_k$ 和 $(\\Delta t)$，使用记号\n",
    "\n",
    "$$\\mathbf x_k = \\mathbf {Fx}_{k-1}$$\n",
    "\n",
    "总的来说，有三种常用的方法来为卡尔曼滤波器找到这个矩阵。最常用的技术是矩阵指数。线性时不变理论，也称为LTI系统理论，是第二种技术。最后，还有数值技术。你可能知道其他方法，但在卡尔曼滤波器的文献和实践中，最常遇到的就是这三种。"
   ]
  },
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   "id": "f211d50e",
   "metadata": {},
   "source": [
    "# 矩阵指数\n",
    "\n",
    "求解方程 $\\frac{dx}{dt} = kx$ 可以通过以下步骤找到：\n",
    "\n",
    "$$\\begin{gathered}\\frac{dx}{dt} = kx \\\\\n",
    "\\frac{dx}{x} = k\\, dt \\\\\n",
    "\\int \\frac{1}{x}\\, dx = \\int k\\, dt \\\\\n",
    "\\log x = kt + c \\\\\n",
    "x = e^{kt+c} \\\\\n",
    "x = e^ce^{kt} \\\\\n",
    "x = c_0e^{kt}\\end{gathered}$$\n",
    "\n",
    "当 $t=0$ 时，$x=x_0$。将这些代入上述方程。\n",
    "\n",
    "$$\\begin{gathered}x_0 = c_0e^{k(0)} \\\\\n",
    "x_0 = c_01 \\\\\n",
    "x_0 = c_0 \\\\\n",
    "x = x_0e^{kt}\\end{gathered}$$\n",
    "\n",
    "用类似的数学方法，一阶方程的解是：\n",
    "\n",
    "$$\\dot{\\mathbf x} = \\mathbf{Ax} ,\\, \\, \\, \\mathbf x(0) = \\mathbf x_0$$\n",
    "\n",
    "其中 $\\mathbf A$ 是一个常数矩阵，是\n",
    "\n",
    "$$\\mathbf x = e^{\\mathbf At}\\mathbf x_0$$\n",
    "\n",
    "将 $F = e^{\\mathbf At}$ 代入，我们可以写成\n",
    "\n",
    "$$\\mathbf x_k = \\mathbf F\\mathbf x_{k-1}$$\n",
    "\n",
    "这就是我们要找的形式！我们将寻找基本\n",
    "\n",
    "矩阵的问题简化为寻找 $e^{\\mathbf At}$ 的值。\n",
    "\n",
    "$e^{\\mathbf At}$ 被称为[矩阵指数](https://en.wikipedia.org/wiki/Matrix_exponential)。它可以用以下幂级数计算：\n",
    "\n",
    "$$e^{\\mathbf At} = \\mathbf{I} + \\mathbf{A}t  + \\frac{(\\mathbf{A}t)^2}{2!} + \\frac{(\\mathbf{A}t)^3}{3!} + ... $$\n",
    "\n",
    "这个级数是通过对 $e^{\\mathbf At}$ 进行泰勒级数展开得到的，我在这里不会详细介绍。\n",
    "\n",
    "让我们利用这个来找到牛顿方程的解。假设速度为常数，用 $v$ 代替 $\\dot x$，我们得到线性矩阵-向量形式\n",
    "\n",
    "$$\\begin{bmatrix}\\dot x \\\\ \\dot v\\end{bmatrix} =\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix} \\begin{bmatrix}x \\\\ v\\end{bmatrix}$$\n",
    "\n",
    "这是一个一阶微分方程，所以我们可以设 $\\mathbf{A}=\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix}$，并解下面的方程。我将间隔 $\\Delta t$ 替换为 $t$，以强调基本矩阵是离散的：\n",
    "\n",
    "$$\\mathbf F = e^{\\mathbf A\\Delta t} = \\mathbf{I} + \\mathbf A\\Delta t  + \\frac{(\\mathbf A\\Delta t)^2}{2!} + \\frac{(\\mathbf A\\Delta t)^3}{3!} + ... $$\n",
    "\n",
    "如果你进行乘法运算，你会发现 $\\mathbf{A}^2=\\begin{bmatrix}0&0\\\\0&0\\end{bmatrix}$，这意味着所有更高幂次的 $\\mathbf{A}$ 也都是 $\\mathbf{0}$。因此，我们得到了一个精确的答案，而不需要无穷多个项：\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "\\mathbf F &=\\mathbf{I} + \\mathbf A \\Delta t + \\mathbf{0} \\\\\n",
    "&= \\begin{bmatrix}1&0\\\\0&1\\end{bmatrix} + \\begin{bmatrix}0&1\\\\0&0\\end{bmatrix}\\Delta t\\\\\n",
    "&= \\begin{bmatrix}1&\\Delta t\\\\0&1\\end{bmatrix}\n",
    "\\end{aligned}$$\n",
    "\n",
    "我们将这个代入 $\\mathbf x_k= \\mathbf{Fx}_{k-1}$ 得到\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "x_k &=\\begin{bmatrix}1&\\Delta t\\\\0&1\\end{bmatrix}x_{k-1}\n",
    "\\end{aligned}$$\n",
    "\n",
    "这个矩阵正是我们在**多元卡尔曼滤波器**章节中对恒定速度卡尔曼滤波器进行解析推导得到的矩阵。\n",
    "\n",
    "SciPy 的 linalg 模块包括一个用于计算矩阵指数的例程 `expm()`。它并不使用泰勒级数方法，而是使用[佩德算法近似](https://en.wikipedia.org/wiki/Pad%C3%A9_approximant)。有许多（至少19种）计算矩阵指数的方法，所有这些方法都存在数值困难[1]。你应该注意到这些问题，特别是当 $\\mathbf A$ 很大时。如果你搜索“佩德算法矩阵指数”，你会找到许多致力于解决这个问题的论文。\n",
    "\n",
    "在实践中，对于卡尔曼滤波器，我们通常只考虑泰勒级数的前两项，因此这可能不会成为你关注的问题。但不要假设我对这个问题的处理就是完整的，然后盲目尝试将这种技术用于其他问题，而不对这种技术的性能进行数值分析。有趣的是，解决 $e^{\\mathbf At}$ 的一个受欢迎的方法是使用广义的常微分方程求解器。换句话说，他们做的与我们相反——将 $\\mathbf A$ 转换成一组微分方程，然后使用数值技术求解这组方程！\n",
    "\n",
    "这里是使用 `expm()` 解决 $e^{\\mathbf At}$ 的一个例子。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "84cfffa1",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[1. , 0.1],\n",
       "       [0. , 1. ]])"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "import numpy as np\n",
    "from scipy.linalg import expm\n",
    "\n",
    "dt = 0.1\n",
    "A = np.array([[0, 1], \n",
    "              [0, 0]])\n",
    "expm(A*dt)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6c51d55d",
   "metadata": {},
   "source": [
    "# 时不变系统\n",
    "\n",
    "如果系统的行为取决于时间，我们可以说一个动态系统由一阶微分方程描述：\n",
    "\n",
    "$$ g(t) = \\dot x$$\n",
    "\n",
    "然而，如果系统是*时不变*的，方程的形式则为：\n",
    "\n",
    "$$ f(x) = \\dot x$$\n",
    "\n",
    "什么是*时不变*呢？如果你在时间 $t$ 输入信号 $x$，它将输出一些信号 $f(x)$。如果你在时间 $t + \\Delta t$ 进行输入，输出信号将是相同的 $f(x)$，只是在时间上有所偏移。\n",
    "\n",
    "一个反例是 $x(t) = \\sin(t)$，其系统为 $f(x) = t\\,  x(t) = t \\sin(t)$。这不是时不变的；由于乘以 t，不同时间的值会有所不同。飞机也不是时不变的。如果你在稍后的时间对飞机进行控制输入，它的行为将会不同，因为它将消耗燃料，因此失去重量。较轻的重量会导致不同的行为。\n",
    "\n",
    "我们可以通过对每一边进行积分来解这些方程。我之前展示了对时不变系统 $v = \\dot x$ 进行积分。然而，对时不变方程 $\\dot x = f(x)$ 进行积分并不那么直接。使用*分离变量*技术，我们除以 $f(x)$ 并将 $dt$ 项移到右边，以便对每一边进行积分：\n",
    "\n",
    "\n",
    "$$\\begin{gathered}\n",
    "\\frac{dx}{dt} = f(x) \\\\\n",
    "\\int^x_{x_0} \\frac{1}{f(x)} dx = \\int^t_{t_0} dt\n",
    "\\end{gathered}$$\n",
    "\n",
    "如果我们令 $F(x) = \\int \\frac{1}{f(x)} dx$，我们会得到\n",
    "\n",
    "$$F(x) - F(x_0) = t-t_0$$\n",
    "\n",
    "然后我们解出 x，得到\n",
    "\n",
    "$$\\begin{gathered}\n",
    "F(x) = t - t_0 + F(x_0) \\\\\n",
    "x = F^{-1}[t-t_0 + F(x_0)]\n",
    "\\end{gathered}$$\n",
    "\n",
    "换句话说，我们需要找到 $F$ 的反函数。这并不是简单的问题，在STEM教育中有很多课程专门致力于找到这个棘手的解析解。\n",
    "\n",
    "然而，它们都是一些技巧，许多简单形式的 $f(x)$ 要么没有闭合形式的解，要么具有极大的困难。因此，实际工程师转向状态空间方法来寻找近似解。\n",
    "\n",
    "矩阵指数的优势在于我们可以用它来处理任意一组*时不变*的微分方程。然而，即使方程并非时不变，我们通常也会使用这种技术。例如，随着飞机飞行，它会燃烧燃料并减轻重量。然而，在一秒钟内失重很小，因此系统在这个时间步长上几乎是线性的。只要时间步长很短，我们的答案仍然会相当准确。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c073caea",
   "metadata": {},
   "source": [
    "# 例子：质量-弹簧-阻尼器模型\n",
    "\n",
    "假设我们想要追踪弹簧上的一个质量，并连接到阻尼器，比如汽车的悬挂系统。其运动方程是 $m$ 为质量，$k$ 为弹簧常数，$c$ 为阻尼力，在某个输入 $u$ 下的运动方程为：\n",
    "\n",
    "$$m\\frac{d^2x}{dt^2} + c\\frac{dx}{dt} +kx = u$$\n",
    "\n",
    "为了符号的简便，我将其写成：\n",
    "\n",
    "$$m\\ddot x + c\\dot x + kx = u$$\n",
    "\n",
    "为了方便记号，我省略了 $(t)$。我可以通过设定 $x_1(t)=x(t)$ 来将其转化为一组一阶方程，并进行如下代换：\n",
    "\n",
    "$$\\begin{aligned}\n",
    "x_1 &= x \\\\\n",
    "x_2 &= \\dot x_1 \\\\\n",
    "\\dot x_2 &= \\ddot x_1 = \\ddot x\n",
    "\\end{aligned}$$\n",
    "\n",
    "这里常见的做法是省略了 $(t)$ 以方便记号。这给出了方程：\n",
    "\n",
    "$$m\\dot x_2 + c x_2 +kx_1 = u$$\n",
    "\n",
    "解出 $\\dot x_2$，我们得到一个一阶方程：\n",
    "\n",
    "$$\\dot x_2 = -\\frac{c}{m}x_2 - \\frac{k}{m}x_1 + \\frac{1}{m}u$$\n",
    "\n",
    "我们将其写成矩阵形式：\n",
    "\n",
    "$$\\begin{bmatrix} \\dot x_1 \\\\ \\dot x_2 \\end{bmatrix} = \n",
    "\\begin{bmatrix}0 & 1 \\\\ -\\frac{k}{m} & -\\frac{c}{m} \\end{bmatrix}\n",
    "\\begin{bmatrix} x_1 \\\\ x_2 \\end{bmatrix} + \n",
    "\\begin{bmatrix} 0 \\\\ \\frac{1}{m} \\end{bmatrix}u$$\n",
    "\n",
    "现在我们使用矩阵指数来找到状态转移矩阵：\n",
    "\n",
    "$$\\Phi(t) = e^{\\mathbf At} = \\mathbf{I} + \\mathbf At  + \\frac{(\\mathbf At)^2}{2!} + \\frac{(\\mathbf At)^3}{3!} + ... $$\n",
    "\n",
    "前两项给出：\n",
    "\n",
    "$$\\mathbf F = \\begin{bmatrix}1 & t \\\\ -\\frac{k}{m} t & 1-\\frac{c}{m} t \\end{bmatrix}$$\n",
    "\n",
    "这可能给出足够的精度，但也可能不够。你可以通过计算对应常数的 $\\frac{(\\mathbf At)^2}{2!}$ 并观察这个矩阵对结果的贡献来检验这一点。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8f4d8e7b",
   "metadata": {},
   "source": [
    "# 线性时不变理论\n",
    "\n",
    "[*线性时不变理论*](https://en.wikipedia.org/wiki/LTI_system_theory)，也称为LTI系统理论，为我们提供了一种使用逆拉普拉斯变换来找到 $\\Phi$ 的方法。我在这本书中不会使用拉普拉斯变换。LTI系统理论告诉我们：\n",
    "\n",
    "$$ \\Phi(t) = \\mathcal{L}^{-1}[(s\\mathbf{I} - \\mathbf{A})^{-1}]$$\n",
    "\n",
    "我没有打算深入讨论这个问题，只是说一下拉普拉斯变换 $\\mathcal{L}$ 将信号转换为一个不包括时间的空间 $s$，但找到上述方程的解并不简单。如果你感兴趣，可以参考维基百科上关于LTI系统理论的文章进行了解。我提到LTI是因为你会发现一些文献使用它来为复杂的问题设计卡尔曼滤波器的矩阵。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a5d0ee8d",
   "metadata": {},
   "source": [
    "# 数值解\n",
    "\n",
    "最后，还有数值技术来找到 $\\mathbf F$。随着滤波器变得更大，找到解析解变得非常繁琐（尽管像SymPy这样的包使其更容易）。C. F. van Loan [2] 发展了一种技术来数值计算 $\\Phi$ 和 $\\mathbf Q$。给定连续模型\n",
    "\n",
    "$$ \\dot x = Ax + Gw$$\n",
    "\n",
    "其中 $w$ 是单位白噪声，van Loan 的方法计算出 $\\mathbf F_k$ 和 $\\mathbf Q_k$。\n",
    "\n",
    "我在 `FilterPy` 中实现了 van Loan 的方法。你可以这样使用：\n",
    "\n",
    "```python\n",
    "from filterpy.common import van_loan_discretization\n",
    "\n",
    "A = np.array([[0., 1.], [-1., 0.]])\n",
    "G = np.array([[0.], [2.]]) # white noise scaling\n",
    "F, Q = van_loan_discretization(A, G, dt=0.1)\n",
    "```\n",
    "\n",
    "在 *数值积分的差分方程* 部分，我介绍了在卡尔曼滤波中非常常用的另一种替代方法。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7038726f",
   "metadata": {},
   "source": [
    "# 过程噪声矩阵的设计\n",
    "\n",
    "通常来说，设计 $\\mathbf Q$ 矩阵是卡尔曼滤波器设计中最困难的部分之一。这是由于几个因素造成的。首先，需要对信号理论有良好的基础。其次，我们试图对我们知之甚少的某些事物中的噪声进行建模。想象一下试图为抛出的棒球建模过程噪声。我们可以将其建模为在空气中运动的一个球体，但这留下了许多未知因素——球的旋转和旋转衰减、带缝球的阻力系数、风力和空气密度的影响等等。我们为给定的过程模型开发了精确的数学解的方程，但由于过程模型不完整，$\\mathbf Q$ 的结果也将不完整。这对卡尔曼滤波器的行为有很多影响。如果 $\\mathbf Q$ 太小，滤波器将对其预测模型过于自信，并偏离实际解。如果 $\\mathbf Q$ 太大，滤波器将受到测量中的噪声的不当影响，并表现不佳。在实践中，我们花费大量时间运行模拟并评估收集的数据，以尝试选择适当的 $\\mathbf Q$ 值。但让我们首先看看数学原理。\n",
    "\n",
    "假设一个运动系统——可以使用牛顿运动方程建模的某个系统。我们可以对这个过程作出几种不同的假设。\n",
    "\n",
    "我们一直在使用一个过程模型：\n",
    "\n",
    "$$ \\dot{\\mathbf x} = \\mathbf{Ax} + \\mathbf{Bu} + \\mathbf{w}$$\n",
    "\n",
    "其中 $\\mathbf{w}$ 是过程噪声。运动系统是*连续*的——它们的输入和输出可以在任意时间点变化。然而，我们的卡尔曼滤波器是*离散*的（卡尔曼滤波器有连续形式，但我们在本书中不涉及）。我们以固定间隔对系统进行采样。因此，我们必须找到上述方程中噪声项的离散表示。这取决于我们对噪声行为的假设。我们将考虑两种不同的噪声模型。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "350f1b6d",
   "metadata": {},
   "source": [
    "# 连续白噪声模型\n",
    "\n",
    "我们使用牛顿方程来建模运动系统。我们要么使用位置和速度，要么使用位置、速度和加速度作为系统的模型。我们可以进一步推广，比如可以模拟加加速度、跃度、猛度等。通常我们不这样做，因为超出实际系统动力学范围的项会降低估计的准确性。\n",
    "\n",
    "假设我们需要对位置、速度和加速度进行建模。然后，我们可以假设每个离散时间步长内加速度是恒定的。当然，系统中存在过程噪声，因此实际上加速度并不是恒定的。由于未建模的外部力会随时间改变被跟踪对象的加速度。在这一部分，我们将假设加速度通过连续时间的零均值白噪声 $w(t)$ 发生变化。换句话说，我们假设速度的微小变化随时间平均为0（零均值）。\n",
    "\n",
    "由于噪声在不断变化，我们需要进行积分，以获得我们选择的离散化间隔内的离散噪声。我们不会在这里证明，但离散噪声的方程是：\n",
    "\n",
    "$$\\mathbf Q = \\int_0^{\\Delta t} \\mathbf F(t)\\mathbf{Q_c}\\mathbf F^\\mathsf{T}(t) dt$$\n",
    "\n",
    "其中 $\\mathbf{Q_c}$ 是连续噪声。总体思路应该很清楚。$\\mathbf F(t)\\mathbf{Q_c}\\mathbf F^\\mathsf{T}(t)$ 是基于我们的过程模型 $\\mathbf F(t)$ 在时刻 $t$ 的连续噪声的投影。我们想知道在离散时间间隔 $\\Delta t$ 内系统添加了多少噪声，因此我们要对该表达式在区间 $[0, \\Delta t]$ 上进行积分。\n",
    "\n",
    "我们知道牛顿系统的基本矩阵是：\n",
    "\n",
    "$$F = \\begin{bmatrix}1 & \\Delta t & {\\Delta t}^2/2 \\\\ 0 & 1 & \\Delta t\\\\ 0& 0& 1\\end{bmatrix}$$\n",
    "\n",
    "我们将连续噪声定义为:\n",
    "\n",
    "$$\\mathbf{Q_c} = \\begin{bmatrix}0&0&0\\\\0&0&0\\\\0&0&1\\end{bmatrix} \\Phi_s$$\n",
    "\n",
    "其中 $\\Phi_s$ 是白噪声的谱密度。这个公式是可以推导出来的，但超出了本书的范围。详细信息请参考任何关于随机过程的标准文本。实际上，我们经常不知道噪声的谱密度，因此这变成了一个“工程”因素——一个我们通过实验调整的数字，直到我们的滤波器按预期运行为止。你可以看到 $\\Phi_s$ 乘以的矩阵实际上将功率谱密度分配给了加速度项。这是有道理的；我们假设系统具有恒定的加速度，除了噪声引起的变化。噪声改变了加速度。\n",
    "\n",
    "我们可以自己进行这些计算，但我更喜欢使用 SymPy 来解方程。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "d52ff35f",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\left[\\begin{matrix}\\frac{\\Delta{t}^{5}}{20} & \\frac{\\Delta{t}^{4}}{8} & \\frac{\\Delta{t}^{3}}{6}\\\\\\frac{\\Delta{t}^{4}}{8} & \\frac{\\Delta{t}^{3}}{3} & \\frac{\\Delta{t}^{2}}{2}\\\\\\frac{\\Delta{t}^{3}}{6} & \\frac{\\Delta{t}^{2}}{2} & \\Delta{t}\\end{matrix}\\right] \\Phi_{s}$"
      ],
      "text/plain": [
       "⎡         5           4           3⎤      \n",
       "⎢\\Delta{t}   \\Delta{t}   \\Delta{t} ⎥      \n",
       "⎢──────────  ──────────  ──────────⎥      \n",
       "⎢    20          8           6     ⎥      \n",
       "⎢                                  ⎥      \n",
       "⎢         4           3           2⎥      \n",
       "⎢\\Delta{t}   \\Delta{t}   \\Delta{t} ⎥      \n",
       "⎢──────────  ──────────  ──────────⎥⋅\\Phiₛ\n",
       "⎢    8           3           2     ⎥      \n",
       "⎢                                  ⎥      \n",
       "⎢         3           2            ⎥      \n",
       "⎢\\Delta{t}   \\Delta{t}             ⎥      \n",
       "⎢──────────  ──────────  \\Delta{t} ⎥      \n",
       "⎣    6           2                 ⎦      "
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "import sympy\n",
    "from sympy import (init_printing, Matrix, MatMul, \n",
    "                   integrate, symbols)\n",
    "\n",
    "init_printing(use_latex='mathjax')\n",
    "dt, phi = symbols('\\Delta{t} \\Phi_s')\n",
    "F_k = Matrix([[1, dt, dt**2/2],\n",
    "              [0,  1,      dt],\n",
    "              [0,  0,       1]])\n",
    "Q_c = Matrix([[0, 0, 0],\n",
    "              [0, 0, 0],\n",
    "              [0, 0, 1]])*phi\n",
    "\n",
    "Q = integrate(F_k * Q_c * F_k.T, (dt, 0, dt))\n",
    "\n",
    "# factor phi out of the matrix to make it more readable\n",
    "Q = Q / phi\n",
    "MatMul(Q, phi)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6c6aee22",
   "metadata": {},
   "source": [
    "为了完整起见，让我们计算零阶和一阶方程的公式。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "40b96867",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0th order discrete process noise\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\left[\\begin{matrix}\\Delta{t} \\Phi_{s}\\end{matrix}\\right]$"
      ],
      "text/plain": [
       "[\\Delta{t}⋅\\Phiₛ]"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F_k = Matrix([[1]])\n",
    "Q_c = Matrix([[phi]])\n",
    "\n",
    "print('0th order discrete process noise')\n",
    "integrate(F_k*Q_c*F_k.T,(dt, 0, dt))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "8e8bb9af",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1st order discrete process noise\n"
     ]
    },
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\left[\\begin{matrix}\\frac{\\Delta{t}^{3}}{3} & \\frac{\\Delta{t}^{2}}{2}\\\\\\frac{\\Delta{t}^{2}}{2} & \\Delta{t}\\end{matrix}\\right] \\Phi_{s}$"
      ],
      "text/plain": [
       "⎡         3           2⎤      \n",
       "⎢\\Delta{t}   \\Delta{t} ⎥      \n",
       "⎢──────────  ──────────⎥      \n",
       "⎢    3           2     ⎥      \n",
       "⎢                      ⎥⋅\\Phiₛ\n",
       "⎢         2            ⎥      \n",
       "⎢\\Delta{t}             ⎥      \n",
       "⎢──────────  \\Delta{t} ⎥      \n",
       "⎣    2                 ⎦      "
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F_k = Matrix([[1, dt],\n",
    "              [0, 1]])\n",
    "Q_c = Matrix([[0, 0],\n",
    "              [0, 1]]) * phi\n",
    "\n",
    "Q = integrate(F_k * Q_c * F_k.T, (dt, 0, dt))\n",
    "\n",
    "print('1st order discrete process noise')\n",
    "# factor phi out of the matrix to make it more readable\n",
    "Q = Q / phi\n",
    "MatMul(Q, phi)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1ed3c92d",
   "metadata": {},
   "source": [
    "# 分段白噪声模型\n",
    "\n",
    "另一个噪声模型假设最高阶项（比如加速度）在每个时间段内是恒定的，但在每个时间段内不同，并且这些项在时间段之间是不相关的。换句话说，在每个时间步长上加速度会出现不连续的跃变。这与上述模型略有不同，上述模型假设最后一项受到连续变化的噪声信号的影响。\n",
    "\n",
    "我们将其建模为：\n",
    "\n",
    "$$f(x)=Fx+\\Gamma w$$\n",
    "\n",
    "其中 $\\Gamma$ 是系统的*噪声增益*，$w$ 是常数分段加速度（或速度、或者加加速度等）。\n",
    "\n",
    "让我们先看一个一阶系统。在这种情况下，我们有状态转移函数\n",
    "\n",
    "$$\\mathbf{F} = \\begin{bmatrix}1&\\Delta t \\\\ 0& 1\\end{bmatrix}$$\n",
    "\n",
    "在一个时间段内，速度的变化将是 $w(t)\\Delta t$，而位置的变化将是 $w(t)\\Delta t^2/2$，这给出了\n",
    "\n",
    "$$\\Gamma = \\begin{bmatrix}\\frac{1}{2}\\Delta t^2 \\\\ \\Delta t\\end{bmatrix}$$\n",
    "\n",
    "因此，过程噪声的协方差是\n",
    "\n",
    "$$Q = \\mathbb E[\\Gamma w(t) w(t) \\Gamma^\\mathsf{T}] = \\Gamma\\sigma^2_v\\Gamma^\\mathsf{T}$$\n",
    "\n",
    "我们可以用 SymPy 来计算它："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "id": "3fe61941",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\left[\\begin{matrix}\\frac{\\Delta{t}^{4}}{4} & \\frac{\\Delta{t}^{3}}{2}\\\\\\frac{\\Delta{t}^{3}}{2} & \\Delta{t}^{2}\\end{matrix}\\right] \\sigma^{2}_{v}$"
      ],
      "text/plain": [
       "⎡         4           3⎤    \n",
       "⎢\\Delta{t}   \\Delta{t} ⎥    \n",
       "⎢──────────  ──────────⎥    \n",
       "⎢    4           2     ⎥    \n",
       "⎢                      ⎥⋅σ²ᵥ\n",
       "⎢         3            ⎥    \n",
       "⎢\\Delta{t}            2⎥    \n",
       "⎢──────────  \\Delta{t} ⎥    \n",
       "⎣    2                 ⎦    "
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "var = symbols('sigma^2_v')\n",
    "v = Matrix([[dt**2 / 2], [dt]])\n",
    "\n",
    "Q = v * var * v.T\n",
    "\n",
    "# factor variance out of the matrix to make it more readable\n",
    "Q = Q / var\n",
    "MatMul(Q, var)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "12b147f9",
   "metadata": {},
   "source": [
    "第二阶系统遵循相同的数学原理。\n",
    "\n",
    "$$\\mathbf{F} = \\begin{bmatrix}1 & \\Delta t & {\\Delta t}^2/2 \\\\ 0 & 1 & \\Delta t\\\\ 0& 0& 1\\end{bmatrix}$$\n",
    "\n",
    "在这里，我们假设白噪声是离散时间的维纳过程。这给了我们\n",
    "\n",
    "$$\\Gamma = \\begin{bmatrix}\\frac{1}{2}\\Delta t^2 \\\\ \\Delta t\\\\ 1\\end{bmatrix}$$\n",
    "\n",
    "这个模型并没有所谓的“真实性”，它只是方便并且提供了良好的结果。例如，我们可以假设噪声作用于加加速度，尽管这会导致更复杂的方程。\n",
    "\n",
    "因此，过程噪声的协方差是\n",
    "\n",
    "$$Q = \\mathbb E[\\Gamma w(t) w(t) \\Gamma^\\mathsf{T}] = \\Gamma\\sigma^2_v\\Gamma^\\mathsf{T}$$\n",
    "\n",
    "我们可以用 SymPy 来计算它："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "id": "500ee8cc",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\left[\\begin{matrix}\\frac{\\Delta{t}^{4}}{4} & \\frac{\\Delta{t}^{3}}{2} & \\frac{\\Delta{t}^{2}}{2}\\\\\\frac{\\Delta{t}^{3}}{2} & \\Delta{t}^{2} & \\Delta{t}\\\\\\frac{\\Delta{t}^{2}}{2} & \\Delta{t} & 1\\end{matrix}\\right] \\sigma^{2}_{v}$"
      ],
      "text/plain": [
       "⎡         4           3           2⎤    \n",
       "⎢\\Delta{t}   \\Delta{t}   \\Delta{t} ⎥    \n",
       "⎢──────────  ──────────  ──────────⎥    \n",
       "⎢    4           2           2     ⎥    \n",
       "⎢                                  ⎥    \n",
       "⎢         3                        ⎥    \n",
       "⎢\\Delta{t}            2            ⎥    \n",
       "⎢──────────  \\Delta{t}   \\Delta{t} ⎥⋅σ²ᵥ\n",
       "⎢    2                             ⎥    \n",
       "⎢                                  ⎥    \n",
       "⎢         2                        ⎥    \n",
       "⎢\\Delta{t}                         ⎥    \n",
       "⎢──────────  \\Delta{t}       1     ⎥    \n",
       "⎣    2                             ⎦    "
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "var = symbols('sigma^2_v')\n",
    "v = Matrix([[dt**2 / 2], [dt], [1]])\n",
    "\n",
    "Q = v * var * v.T\n",
    "\n",
    "# factor variance out of the matrix to make it more readable\n",
    "Q = Q / var\n",
    "MatMul(Q, var)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e7734d81",
   "metadata": {},
   "source": [
    "这个模型并不能说比连续模型更正确或更错误——它们都是对实际对象发生情况的近似。只有经验和实验才能指导你选择合适的模型。实际上，你通常会发现这两种模型都能提供合理的结果，但其中一种通常比另一种表现更好。\n",
    "\n",
    "第二个模型的优势在于我们可以用 $\\sigma^2$ 来描述噪声，这可以根据运动和我们预期的误差量来描述。而第一个模型则需要我们指定谱密度，这并不是很直观，但它更容易处理时间取样的变化，因为噪声是在时间段内被积累的。然而，这些并不是固定的规则——根据滤波器的性能测试或者你对物理模型行为的了解，可以选择任何一个模型（或自己设计的模型）。\n",
    "\n",
    "一个经验法则是将 $\\sigma$ 设置在 $\\frac{1}{2}\\Delta a$ 到 $\\Delta a$ 之间，其中 $\\Delta a$ 是加速度在采样周期之间最大变化的量。在实践中，我们选择一个数值，对数据进行模拟，并选择效果良好的值。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1e2b7506",
   "metadata": {},
   "source": [
    "# 使用FilterPy库计算Q\n",
    "\n",
    "FilterPy提供了几种计算 $\\mathbf Q$ 矩阵的例程。`Q_continuous_white_noise()` 函数根据给定的 $\\Delta t$ 值和谱密度计算 $\\mathbf Q$。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "id": "3901a17a",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[0.333 0.5  ]\n",
      " [0.5   1.   ]]\n"
     ]
    }
   ],
   "source": [
    "from filterpy.common import Q_continuous_white_noise\n",
    "from filterpy.common import Q_discrete_white_noise\n",
    "\n",
    "Q = Q_continuous_white_noise(dim=2, dt=1, spectral_density=1)\n",
    "print(Q)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "id": "18972f37",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[0.05  0.125 0.167]\n",
      " [0.125 0.333 0.5  ]\n",
      " [0.167 0.5   1.   ]]\n"
     ]
    }
   ],
   "source": [
    "Q = Q_continuous_white_noise(dim=3, dt=1, spectral_density=1)\n",
    "print(Q)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "08d05577",
   "metadata": {},
   "source": [
    "函数 `Q_discrete_white_noise()` 假设噪声采用分段模型来计算 $\\mathbf Q$。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "id": "0e175b02",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[0.25 0.5 ]\n",
      " [0.5  1.  ]]\n"
     ]
    }
   ],
   "source": [
    "Q = Q_discrete_white_noise(2, var=1.)\n",
    "print(Q)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "id": "86e1402e",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[0.25 0.5  0.5 ]\n",
      " [0.5  1.   1.  ]\n",
      " [0.5  1.   1.  ]]\n"
     ]
    }
   ],
   "source": [
    "Q = Q_discrete_white_noise(3, var=1.)\n",
    "print(Q)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0320f880",
   "metadata": {},
   "source": [
    "# Q的简化\n",
    "\n",
    "许多处理方法对 $\\mathbf Q$ 使用了更简单的形式，将其设置为除了右下角的噪声项之外全为零。这样做是否合理呢？嗯，考虑一下在 $\\Delta t$ 很小的情况下 $\\mathbf Q$ 的值。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "id": "d9d7c4e2",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[0.00000002 0.00000078 0.00002083]\n",
      " [0.00000078 0.00004167 0.00125   ]\n",
      " [0.00002083 0.00125    0.05      ]]\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "\n",
    "np.set_printoptions(precision=8)\n",
    "Q = Q_continuous_white_noise(\n",
    "    dim=3, dt=0.05, spectral_density=1)\n",
    "print(Q)\n",
    "np.set_printoptions(precision=3)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b8c037b8",
   "metadata": {},
   "source": [
    "我们可以看到大多数项都非常小。回想一下，唯一使用这个矩阵的方程是：\n",
    "\n",
    "$$ \\mathbf P=\\mathbf{FPF}^\\mathsf{T} + \\mathbf Q$$\n",
    "\n",
    "如果相对于 $\\mathbf P$，$\\mathbf Q$ 的值很小，那么它对 $\\mathbf P$ 的计算几乎没有贡献。将 $\\mathbf Q$ 设置为零矩阵，除了右下角的项\n",
    "\n",
    "$$\\mathbf Q=\\begin{bmatrix}0&0&0\\\\0&0&0\\\\0&0&\\sigma^2\\end{bmatrix}$$\n",
    "\n",
    "虽然这样做并不正确，但通常是一个有用的近似。如果你在重要应用中这样做，你将不得不进行大量研究，以确保你的滤波器在各种情况下都能正常工作。\n",
    "\n",
    "如果这样做，‘右下角的项’意味着每个变量中变化最快的项。如果状态是 $x=\\begin{bmatrix}x & \\dot x & \\ddot{x} & y & \\dot{y} & \\ddot{y}\\end{bmatrix}^\\mathsf{T}$，那么 $\\mathbf Q$ 将是 6x6 的；$\\mathbf Q$ 中 $\\ddot{x}$ 和 $\\ddot{y}$ 的元素都必须设为非零。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "33af87e7",
   "metadata": {},
   "source": [
    "# 后验协方差的稳定计算\n",
    "\n",
    "我提出了计算后验协方差的方程为\n",
    "\n",
    "$$\\mathbf P = (\\mathbf I - \\mathbf{KH})\\mathbf{\\bar P}$$\n",
    "\n",
    "严格来说，这是正确的，但这并不是我在`FilterPy`中计算的方法，我使用了*Joseph*方程\n",
    "\n",
    "$$\\mathbf P = (\\mathbf I-\\mathbf {KH})\\mathbf{\\bar P}(\\mathbf I-\\mathbf{KH})^\\mathsf T + \\mathbf{KRK}^\\mathsf T$$\n",
    "\n",
    "我经常收到邮件和GitHub的问题，声称这种实现是一个bug。这不是bug，我使用它有几个原因。首先，减法 $(\\mathbf I - \\mathbf{KH})$ 可能由于浮点误差导致矩阵结果不对称。协方差必须是对称的，所以变成不对称通常会导致卡尔曼滤波器发散，甚至因为`NumPy`内置的检查而引发异常。\n",
    "\n",
    "保持对称性的传统方法是以下公式：\n",
    "\n",
    "$$\\mathbf P = (\\mathbf P + \\mathbf P^\\mathsf T) / 2$$\n",
    "\n",
    "这是安全的，因为矩阵中所有协方差 $\\sigma_{ij} = \\sigma_{ji}$。因此，如果它们由于浮点误差发散，此操作会平均两个值的差异。\n",
    "\n",
    "如果你看一下上面的Joseph形式的方程，你会发现两个项中都有类似 $\\mathbf{ABA}^\\mathsf T$ 的模式。因此，它们都保持对称性。但这个方程从何而来？为什么我使用它而不是\n",
    "\n",
    "$$\\mathbf P = (\\mathbf I - \\mathbf{KH})\\mathbf{\\bar P} \\\\\n",
    "\\mathbf P = (\\mathbf P + \\mathbf P^\\mathsf T) / 2$$\n",
    "\n",
    "让我们从第一原理推导出这个方程。这并不太糟糕，你需要理解推导来理解方程的目的，更重要的是，如果你的滤波器因数值不稳定性而发散，就需要诊断问题。这个推导来自Brown[4]。\n",
    "\n",
    "首先，一些符号。$\\mathbf x$ 是我们系统的真实状态。$\\mathbf{\\hat x}$ 是我们系统的估计状态 - 后验状态。而 $\\mathbf{\\bar x}$ 是系统的先验估计。\n",
    "\n",
    "鉴于此，我们可以定义我们的模型为\n",
    "\n",
    "$$\\mathbf x_{k+1} = \\mathbf F_k \\mathbf x_k + \\mathbf w_k \\\\\n",
    "\\mathbf z_k = \\mathbf H_k \\mathbf x_k + \\mathbf v_k$$\n",
    "\n",
    "换句话说，系统的下一个状态 $\\mathbf x_{k+1}$ 是当前状态 $k$ 乘以某个过程 $\\mathbf F_k$ 加上一些噪声 $\\mathbf w_k$。\n",
    "\n",
    "请注意，这些只是定义。没有系统完全遵循数学模型，所以我们用噪声项 $\\mathbf w_k$ 来模拟。而由于传感器误差，没有测量是完美的，所以我们用 $\\mathbf v_k$ 来模拟。\n",
    "\n",
    "我将在剩下的推导中省略下标 $k$，因为在接下来的推导中，我们只考虑步骤 $k$ 的值，从不考虑步骤 $k+1$。\n",
    "\n",
    "现在我们定义估计误差为真实状态和估计状态之间的差异\n",
    "\n",
    "$$ \\mathbf e = \\mathbf x - \\mathbf{\\hat x}$$\n",
    "\n",
    "同样，这是一个定义；我们不知道如何计算 $\\mathbf e$，它只是真实状态和估计状态之间定义的差异。\n",
    "\n",
    "这使我们能够定义我们估计的协方差，这被定义为 $\\mathbf{ee}^\\mathsf T$ 的期望值：\n",
    "\n",
    "$$\\begin{aligned}\n",
    "P &= E[\\mathbf{ee}^\\mathsf T] \\\\\n",
    "&= E[(\\mathbf x - \\mathbf{\\hat x})(\\mathbf x - \\mathbf{\\hat x})^\\mathsf T]\n",
    "\\end{aligned}$$\n",
    "\n",
    "接下来，我们定义后验估计为\n",
    "\n",
    "$$\\mathbf {\\hat x} = \\mathbf{\\bar x} + \\mathbf K(\\mathbf z - \\mathbf{H \\bar x})$$\n",
    "\n",
    "看起来像卡尔曼滤波器中的方程，原因很简单。但就像到目前为止的数学一样，这是一个**定义**。特别地，我们还没有定义 $\\mathbf K$，你不应该将其视为卡尔曼增益，因为我们为*任何*问题解决此问题，而不仅仅是线性卡尔曼滤波器。这里，$\\mathbf K$ 只是0到1之间的一些未指定的混合值。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3b9b1f6d",
   "metadata": {},
   "source": [
    "现在我们有了定义，让我们进行一些替换和代数运算。\n",
    "\n",
    "表达式 $(\\mathbf x - \\mathbf{\\hat x})$ 可以通过用上面的定义替换 $\\mathbf{\\hat x}$ 来展开：\n",
    "\n",
    "$$(\\mathbf x - \\mathbf{\\hat x}) = \\mathbf x - (\\mathbf{\\bar x} + \\mathbf K(\\mathbf z - \\mathbf{H \\bar x}))$$\n",
    "\n",
    "现在我们用 $\\mathbf z$ 替换为 $\\mathbf H \\mathbf x + \\mathbf v$：\n",
    "\n",
    "$$\\begin{aligned}\n",
    "(\\mathbf x - \\mathbf{\\hat x})\n",
    "&= \\mathbf x - (\\mathbf{\\bar x} + \\mathbf K(\\mathbf z - \\mathbf{H \\bar x})) \\\\\n",
    "&= \\mathbf x - (\\mathbf{\\bar x} + \\mathbf K(\\mathbf H \\mathbf x + \\mathbf v - \\mathbf{H \\bar x})) \\\\\n",
    "&= (\\mathbf x - \\mathbf{\\bar x}) - \\mathbf K(\\mathbf H \\mathbf x + \\mathbf v - \\mathbf{H \\bar x}) \\\\\n",
    "&= (\\mathbf x - \\mathbf{\\bar x}) - \\mathbf{KH}(\\mathbf x - \\mathbf{ \\bar x}) - \\mathbf{Kv} \\\\\n",
    "&=  (\\mathbf I - \\mathbf{KH})(\\mathbf x - \\mathbf{\\bar x}) - \\mathbf{Kv}\n",
    "\\end{aligned}$$\n",
    "\n",
    "现在我们可以解出 $\\mathbf P$，如果注意到 $(\\mathbf x - \\mathbf{\\bar x})$ 的期望值是先验协方差 $\\mathbf{\\bar P}$，$\\mathbf v$ 的期望值是 $E[\\mathbf{vv}^\\mathbf T] = \\mathbf R$：\n",
    "\n",
    "$$\\begin{aligned}\n",
    "\\mathbf P &= \n",
    "   E\\big[[(\\mathbf I - \\mathbf{KH})(\\mathbf x - \\mathbf{\\bar x}) - \\mathbf{Kv})]\n",
    "  [(\\mathbf I - \\mathbf{KH})(\\mathbf x - \\mathbf{\\bar x}) - \\mathbf{Kv}]^\\mathsf T\\big ] \\\\\n",
    "  &= (\\mathbf I - \\mathbf{KH})\\mathbf{\\bar P}(\\mathbf I - \\mathbf{KH})^\\mathsf T + \\mathbf{KRK}^\\mathsf T\n",
    "\\end{aligned}$$\n",
    "\n",
    "这就是我们要证明的内容。\n",
    "\n",
    "请注意，这个方程对*任何* $\\mathbf K$ 都有效，不仅仅是由卡尔曼滤波器计算得到的最佳 $\\mathbf K$。这就是我使用这个方程的原因。实际上，由滤波器计算得到的卡尔曼增益*并不是*最优值，因为现实世界永远不是真正的线性和高斯的，计算中的浮点误差也会导致不确定性。相对于真实世界条件，这个方程大大减少了卡尔曼滤波器发散的可能性。\n",
    "\n",
    "那么 $\\mathbf P = (\\mathbf I - \\mathbf{KH})\\mathbf{\\bar P}$ 是从何而来的呢？让我们完成推导，这很简单。回想一下，卡尔曼滤波器（最优）增益由以下公式给出：\n",
    "\n",
    "$$\\mathbf K = \\mathbf{\\bar P H^\\mathsf T}(\\mathbf{H \\bar P H}^\\mathsf T + \\mathbf R)^{-1}$$\n",
    "\n",
    "现在我们将其代入我们刚刚推导出的方程：\n",
    "\n",
    "$$\\begin{aligned}\n",
    "&= (\\mathbf I - \\mathbf{KH})\\mathbf{\\bar P}(\\mathbf I - \\mathbf{KH})^\\mathsf T + \\mathbf{KRK}^\\mathsf T\\\\\n",
    "&= \\mathbf{\\bar P} - \\mathbf{KH}\\mathbf{\\bar P} - \\mathbf{\\bar PH}^\\mathsf T\\mathbf{K}^\\mathsf T + \\mathbf K(\\mathbf{H \\bar P H}^\\mathsf T + \\mathbf R)\\mathbf K^\\mathsf T \\\\\n",
    "&= \\mathbf{\\bar P} - \\mathbf{KH}\\mathbf{\\bar P} - \\mathbf{\\bar PH}^\\mathsf T\\mathbf{K}^\\mathsf T + \\mathbf{\\bar P H^\\mathsf T}(\\mathbf{H \\bar P H}^\\mathsf T + \\mathbf R)^{-1}(\\mathbf{H \\bar P H}^\\mathsf T + \\mathbf R)\\mathbf K^\\mathsf T\\\\\n",
    "&= \\mathbf{\\bar P} - \\mathbf{KH}\\mathbf{\\bar P} - \\mathbf{\\bar PH}^\\mathsf T\\mathbf{K}^\\mathsf T + \\mathbf{\\bar P H^\\mathsf T}\\mathbf K^\\mathsf T\\\\\n",
    "&= \\mathbf{\\bar P} - \\mathbf{KH}\\mathbf{\\bar P}\\\\\n",
    "&= (\\mathbf I - \\mathbf{KH})\\mathbf{\\bar P}\n",
    "\\end{aligned}$$\n",
    "\n",
    "因此，$\\mathbf P = (\\mathbf I - \\mathbf{KH})\\mathbf{\\bar P}$ 在增益最优时是数学上正确的，但 $(\\mathbf I - \\mathbf{KH})\\mathbf{\\bar P}(\\mathbf I - \\mathbf{KH})^\\mathsf T + \\mathbf{KRK}^\\mathsf T$ 也是正确的，当增益不是最优时，它更加数值稳定。因此在 FilterPy 中我使用这种计算方法。\n",
    "\n",
    "你的滤波器仍然可能会发散，特别是如果它运行了数百或数千个周期。你需要检查这些方程。文献提供了这种计算的其他形式，可能更适用于你的问题。一如既往，如果你正在解决真正的工程问题，其中失败可能意味着设备或生命的损失，你需要查阅工程文献。如果你处理的是‘玩具’问题，失败并不会有太大损失，如果你检测到发散，你可以将 $\\mathbf P$ 的值重置为一些‘合理’的值，然后继续进行。例如，你可以将非对角线元素清零，使矩阵只包含方差，然后可能乘以一个比一大一点的常数，以反映刚刚注入到滤波器中的信息损失。发挥你的想象力，并进行测试。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "13434da5",
   "metadata": {},
   "source": [
    "# 推导卡尔曼增益方程\n",
    "\n",
    "这个推导是关于卡尔曼滤波器方程的。请注意，这个推导*不*使用贝叶斯方程。我见过至少四种不同的推导卡尔曼滤波器方程的方式；这个推导是文献中典型的方式，是从上一节推导出来的。源自 Brown [4]。\n",
    "\n",
    "在上一节中，我们使用了一个未指定的缩放因子 $\\mathbf K$ 来推导协方差方程的 Joseph 形式。如果我们要得到一个最优的滤波器，我们需要使用微积分来最小化方程中的误差。你应该熟悉这个想法。如果你想找到一个函数 $f(x)$ 的最小值，你会对它求导并令导数等于零：$\\frac{x}{dx}f(x) = 0$。\n",
    "\n",
    "在我们的问题中，误差由协方差矩阵 $\\mathbf P$ 表示。特别是，对角线表示状态向量中每个元素的误差（方差）。因此，为了找到最优增益，我们希望对对角线的和（迹）的导数进行求导。\n",
    "\n",
    "Brown 提醒了我们两个涉及迹的导数的公式：\n",
    "\n",
    "$$\\frac{d\\, trace(\\mathbf{AB})}{d\\mathbf A} = \\mathbf B^\\mathsf T$$\n",
    "\n",
    "$$\\frac{d\\, trace(\\mathbf{ACA}^\\mathsf T)}{d\\mathbf A} = 2\\mathbf{AC}$$\n",
    "\n",
    "其中 $\\mathbf{AB}$ 是方阵，$\\mathbf C$ 是对称矩阵。\n",
    "\n",
    "我们展开 Joseph 方程：\n",
    "\n",
    "$$\\mathbf P = \\mathbf{\\bar P} - \\mathbf{KH}\\mathbf{\\bar P} - \\mathbf{\\bar P}\\mathbf H^\\mathsf T \\mathbf K^\\mathsf T + \\mathbf K(\\mathbf H \\mathbf{\\bar P}\\mathbf H^\\mathsf T + \\mathbf R)\\mathbf K^\\mathsf T$$\n",
    "\n",
    "现在我们需要对 $\\mathbf K$ 求 $\\mathbf P$ 的迹的导数：$\\frac{d\\, trace(\\mathbf P)}{d\\mathbf K}$。\n",
    "\n",
    "第一项对 $\\mathbf K$ 的迹的导数为 $0$，因为它的表达式中没有 $\\mathbf K$。\n",
    "\n",
    "第二项的迹的导数是 $(\\mathbf H\\mathbf{\\bar P})^\\mathsf T$。\n",
    "\n",
    "我们可以通过注意到 $\\mathbf{\\bar P}\\mathbf H^\\mathsf T \\mathbf K^\\mathsf T$ 是 $\\mathbf{KH}\\mathbf{\\bar P}$ 的转置来找到第三项的迹的导数。矩阵的迹等于它的转置的迹，因此它的导数将与第二项相同。\n",
    "\n",
    "最后，第四项的迹的导数是 $2\\mathbf K(\\mathbf H \\mathbf{\\bar P}\\mathbf H^\\mathsf T + \\mathbf R)$。\n",
    "\n",
    "这给了我们最终的值\n",
    "\n",
    "$$\\frac{d\\, trace(\\mathbf P)}{d\\mathbf K} = -2(\\mathbf H\\mathbf{\\bar P})^\\mathsf T + 2\\mathbf K(\\mathbf H \\mathbf{\\bar P}\\mathbf H^\\mathsf T + \\mathbf R)$$\n",
    "\n",
    "我们将这个值设为零并解出方程得到最小化误差的 $\\mathbf K$ 方程：\n",
    "\n",
    "$$-2(\\mathbf H\\mathbf{\\bar P})^\\mathsf T + 2\\mathbf K(\\mathbf H \\mathbf{\\bar P}\\mathbf H^\\mathsf T + \\mathbf R) = 0 \\\\\n",
    "\\mathbf K(\\mathbf H \\mathbf{\\bar P}\\mathbf H^\\mathsf T + \\mathbf R) = (\\mathbf H\\mathbf{\\bar P})^\\mathsf T \\\\\n",
    "\\mathbf K(\\mathbf H \\mathbf{\\bar P}\\mathbf H^\\mathsf T + \\mathbf R) = \\mathbf{\\bar P}\\mathbf H^\\mathsf T \\\\\n",
    "\\mathbf K= \\mathbf{\\bar P}\\mathbf H^\\mathsf T (\\mathbf H \\mathbf{\\bar P}\\mathbf H^\\mathsf T + \\mathbf R)^{-1}\n",
    "$$\n",
    "\n",
    "这个推导并不是完全严格的，因为我省略了关于为什么最小化迹可以最小化总误差的论证，但我认为这对这本书来说已经足够了。任何标准文献都会对这个问题有更深入的讨论。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "66ac685f",
   "metadata": {},
   "source": [
    "# 微分方程的数值积分\n",
    "\n",
    "我们已经接触了几种解线性微分方程的数值技术。这些方法包括状态空间方法、拉普拉斯变换和范·洛恩方法。\n",
    "\n",
    "这些方法对于线性常微分方程（ODEs）效果很好，但对于非线性方程效果不佳。例如，考虑尝试预测一辆快速转弯的汽车的位置。汽车通过转动前轮来操控。这使它们围绕后轴作为中心点进行转弯。因此，路径会持续变化，线性预测必然会产生错误的值。如果系统的变化相对于 $\\Delta t$ 足够小，这通常可以产生足够的结果，但对于我们接下来将研究的非线性卡尔曼滤波器来说，这种情况很少会发生。\n",
    "\n",
    "因此，出于这些原因，我们需要了解如何对常微分方程进行数值积分。这可能是一个涉及多本书的广阔话题。然而，我将介绍一些简单的技术，它们适用于你遇到的大多数问题。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "248b4f0e",
   "metadata": {},
   "source": [
    "# 欧拉法\n",
    "\n",
    "我们假设有初值问题：\n",
    "\n",
    "$$\\begin{gathered}\n",
    "y' = y, \\\\ y(0) = 1\n",
    "\\end{gathered}$$\n",
    "\n",
    "我们碰巧知道精确解是 $y=e^t$，因为我们之前解过它，但对于任意的常微分方程，我们不会知道精确解。一般来说，我们只知道方程的导数，即斜率。我们还知道初始值：在 $t=0$ 时，$y=1$。如果我们知道这两个信息，我们就可以使用 $t=0$ 处的斜率和 $y(0)$ 的值来预测 $y(t=1)$ 的值。我在下面绘制了这个图像。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "id": "a63fd361",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 900x400 with 1 Axes>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "import matplotlib.pyplot as plt\n",
    "\n",
    "t = np.linspace(-1, 1, 10)\n",
    "plt.plot(t, np.exp(t))\n",
    "t = np.linspace(-1, 1, 2)\n",
    "plt.plot(t,t+1, ls='--', c='k');"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1230f407",
   "metadata": {},
   "source": [
    "你可以看到在 $t=0.1$ 处，斜率非常接近曲线，但在 $t=1$ 处却相距很远。但让我们继续以步长为1进行一会儿。我们可以看到在 $t=1$ 处，$y$ 的估计值为2。现在，我们可以通过计算 $t=1$ 处曲线的斜率，并将其加到我们的初始估计中来计算 $t=2$ 处的值。根据 $y'=y$，在 $t=1$ 处的斜率是2。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "id": "f2b09a06",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 900x400 with 1 Axes>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "import kf_book.book_plots as book_plots\n",
    "\n",
    "t = np.linspace(-1, 2, 20)\n",
    "plt.plot(t, np.exp(t))\n",
    "t = np.linspace(0, 1, 2)\n",
    "plt.plot([1, 2, 4], ls='--', c='k')\n",
    "book_plots.set_labels(x='x', y='y');"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "64029872",
   "metadata": {},
   "source": [
    "这里我们看到下一个 $y$ 的估计值是4。误差迅速变大，可能让你印象不佳。但1是一个非常大的步长。让我们将这个算法编写成代码，并通过使用一个小的步长来验证它是否有效。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "id": "85cf8c48",
   "metadata": {},
   "outputs": [],
   "source": [
    "def euler(t, tmax, y, dx, step=1.):\n",
    "    ys = []\n",
    "    while t < tmax:\n",
    "        y = y + step*dx(t, y)\n",
    "        ys.append(y)\n",
    "        t +=step        \n",
    "    return ys"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "id": "652d420c",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2.0\n",
      "4.0\n"
     ]
    }
   ],
   "source": [
    "def dx(t, y): return y\n",
    "\n",
    "print(euler(0, 1, 1, dx, step=1.)[-1])\n",
    "print(euler(0, 2, 1, dx, step=1.)[-1])"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0123b983",
   "metadata": {},
   "source": [
    "这看起来是正确的。现在让我们绘制一个更小步长的结果。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "id": "e023664d",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 900x400 with 2 Axes>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "ys = euler(0, 4, 1, dx, step=0.00001)\n",
    "plt.subplot(1,2,1)\n",
    "plt.title('Computed')\n",
    "plt.plot(np.linspace(0, 4, len(ys)),ys)\n",
    "plt.subplot(1,2,2)\n",
    "t = np.linspace(0, 4, 20)\n",
    "plt.title('Exact')\n",
    "plt.plot(t, np.exp(t));"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "id": "69c87a74",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "exact answer= 54.598150033144236\n",
      "euler answer= 54.59705808834125\n",
      "difference = 0.0010919448029866885\n",
      "iterations = 400000\n"
     ]
    }
   ],
   "source": [
    "print('exact answer=', np.exp(4))\n",
    "print('euler answer=', ys[-1])\n",
    "print('difference =', np.exp(4) - ys[-1])\n",
    "print('iterations =', len(ys))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a2b35b36",
   "metadata": {},
   "source": [
    "在这里，我们可以看到误差相当小，但要达到三位有效数字却需要大量迭代次数。实际上，对于大多数问题，欧拉方法速度太慢了，我们会使用更复杂的方法。\n",
    "\n",
    "在继续之前，让我们正式推导一下欧拉方法，因为它是下一节中使用的更高级的龙格-库塔方法的基础。事实上，欧拉方法是龙格-库塔方法的最简单形式。\n",
    "\n",
    "这里是 $y$ 泰勒展开的前三项。无限展开将给出精确答案，所以 $O(h^4)$ 表示由于有限展开而产生的误差。\n",
    "\n",
    "$$y(t_0 + h) = y(t_0) + h y'(t_0) + \\frac{1}{2!}h^2 y''(t_0) + \\frac{1}{3!}h^3 y'''(t_0) +  O(h^4)$$\n",
    "\n",
    "在这里，我们可以看到欧拉方法使用了泰勒展开的前两项。随着项数增加，每个后续项都比前面的项小，所以我们可以确信估计值不会偏离正确值太远。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ae7b99df",
   "metadata": {},
   "source": [
    "# 龙格库塔法\n",
    "\n",
    "龙格-库塔方法是数值积分的主力军。文献中有大量的方法。在实践中，使用我在这里介绍的龙格-库塔算法将能解决你面临的大多数问题。它提供了很好的速度、精度和稳定性平衡，是“常用”的数值积分方法，除非你有非常充分的理由选择其他方法。\n",
    "\n",
    "让我们深入探讨。我们从某个微分方程开始\n",
    "\n",
    "$$\\ddot{y} = \\frac{d}{dt}\\dot{y}$$\n",
    "\n",
    "我们可以用函数 f 替换 y 的导数，就像这样\n",
    "\n",
    "$$\\ddot{y} = \\frac{d}{dt}f(y,t)$$.\n",
    "\n",
    "推导这些方程不在本书的范围之内，但是龙格-库塔 RK4 方法是根据这些方程定义的。\n",
    "\n",
    "$$y(t+\\Delta t) = y(t) + \\frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) + O(\\Delta t^4)$$\n",
    "\n",
    "$$\\begin{aligned}\n",
    "k_1 &= f(y,t)\\Delta t \\\\\n",
    "k_2 &= f(y+\\frac{1}{2}k_1, t+\\frac{1}{2}\\Delta t)\\Delta t \\\\\n",
    "k_3 &= f(y+\\frac{1}{2}k_2, t+\\frac{1}{2}\\Delta t)\\Delta t \\\\\n",
    "k_4 &= f(y+k_3, t+\\Delta t)\\Delta t\n",
    "\\end{aligned}\n",
    "$$\n",
    "\n",
    "这是相应的代码："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "id": "28c4a0c3",
   "metadata": {},
   "outputs": [],
   "source": [
    "def runge_kutta4(y, x, dx, f):\n",
    "    \"\"\"computes 4th order Runge-Kutta for dy/dx.\n",
    "    y is the initial value for y\n",
    "    x is the initial value for x\n",
    "    dx is the difference in x (e.g. the time step)\n",
    "    f is a callable function (y, x) that you supply \n",
    "    to compute dy/dx for the specified values.\n",
    "    \"\"\"\n",
    "    \n",
    "    k1 = dx * f(y, x)\n",
    "    k2 = dx * f(y + 0.5*k1, x + 0.5*dx)\n",
    "    k3 = dx * f(y + 0.5*k2, x + 0.5*dx)\n",
    "    k4 = dx * f(y + k3, x + dx)\n",
    "    \n",
    "    return y + (k1 + 2*k2 + 2*k3 + k4) / 6."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5f1947e3",
   "metadata": {},
   "source": [
    "让我们用一个简单的例子来演示。\n",
    "\n",
    "$$\\dot{y} = t\\sqrt{y(t)}$$\n",
    "\n",
    "初始值为\n",
    "\n",
    "$$\\begin{aligned}t_0 &= 0\\\\y_0 &= y(t_0) = 1\\end{aligned}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "id": "abd0963c",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "max error 0.00005\n"
     ]
    },
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 900x400 with 1 Axes>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "import math\n",
    "import numpy as np\n",
    "t = 0.\n",
    "y = 1.\n",
    "dt = .1\n",
    "\n",
    "ys, ts = [], []\n",
    "\n",
    "def func(y,t):\n",
    "    return t*math.sqrt(y)\n",
    "\n",
    "while t <= 10:\n",
    "    y = runge_kutta4(y, t, dt, func)\n",
    "    t += dt\n",
    "    ys.append(y)\n",
    "    ts.append(t)\n",
    "\n",
    "exact = [(t**2 + 4)**2 / 16. for t in ts]\n",
    "plt.plot(ts, ys)\n",
    "plt.plot(ts, exact)\n",
    "\n",
    "error = np.array(exact) - np.array(ys)\n",
    "print(f\"max error {max(error):.5f}\")"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1cd49961",
   "metadata": {},
   "source": [
    "# 贝叶斯滤波\n",
    "\n",
    "首先，我们讨论贝叶斯滤波。假设我们正在追踪一个物体。我们将其在特定时间的*状态*定义为其位置、速度等。例如，我们可能将时刻 $t$ 的状态写为 $\\mathbf x_t = \\begin{bmatrix}x_t &\\dot x_t \\end{bmatrix}^\\mathsf T$。\n",
    "\n",
    "当我们对物体进行测量时，我们实际上是在测量状态或其部分。传感器存在噪声，因此测量结果会受到干扰。但显然，测量是由状态决定的。也就是说，状态的改变可能会导致测量的变化，但测量的改变不会影响状态。\n",
    "\n",
    "在滤波中，我们的目标是从时间 0 到时间 $t$ 计算一组状态 $\\mathbf x_{0:t}$ 的最佳估计。如果我们知道 $\\mathbf x_{0:t}$，那么计算与这些状态对应的一组测量 $\\mathbf z_{0:t}$ 就会非常容易。但是，我们接收到一组测量 $\\mathbf z_{0:t}$，并希望计算相应的状态 $\\mathbf x_{0:t}$。这被称为*统计反演*，因为我们试图根据输出来计算输入。\n",
    "\n",
    "反演是一个困难的问题，因为通常情况下并不存在唯一的解。对于给定的状态 $\\mathbf x_{0:t}$，只有一种可能的测量集（加上噪声），但对于给定的测量集，可能有很多不同的状态集可以导致这些测量。\n",
    "\n",
    "回顾一下贝叶斯定理：\n",
    "\n",
    "$$P(x \\mid z) = \\frac{P(z \\mid x)P(x)}{P(z)}$$\n",
    "\n",
    "其中 $P(z \\mid x)$ 是测量 $z$ 的*似然*，$P(x)$ 是基于我们的过程模型的*先验*，$P(z)$ 是归一化常数。$P(x \\mid z)$ 是*后验*，即在融合测量 $z$ 后的分布，也称为*证据*。\n",
    "\n",
    "这是一个*统计反演*，因为它从 $P(z \\mid x)$ 推导出 $P(x \\mid z)$。我们滤波问题的解可以表示为：\n",
    "\n",
    "$$P(\\mathbf x_{0:t} \\mid \\mathbf z_{0:t}) = \\frac{P(\\mathbf z_{0:t} \\mid \\mathbf x_{0:t})P(\\mathbf x_{0:t})}{P(\\mathbf z_{0:t})}$$\n",
    "\n",
    "这一切看起来都很好，但是当下一个测量 $\\mathbf z_{t+1}$ 来临时，我们需要重新计算整个范围 $0:t+1$ 的表达式。\n",
    "\n",
    "在实践中，这是不可行的，因为我们试图计算状态 $\\mathbf x_{0:t} \\mid \\mathbf z_{0:t}$ 在所有时间步长上的后验分布。但是我们真的关心第三步（比如说）的概率分布吗，当我们刚收到第十个测量时？通常不关心。因此，我们放宽了要求，只计算当前时间步长的分布。\n",
    "\n",
    "第一个简化是我们将我们的过程（例如移动物体的运动模型）描述为*马尔可夫链*。也就是说，我们说当前状态仅依赖于上一个状态和一个转移概率 $P(\\mathbf x_k \\mid \\mathbf x_{k-1})$，这只是从上一个状态到当前状态的概率。我们写成：\n",
    "\n",
    "$$\\mathbf x_k \\sim P(\\mathbf x_k \\mid \\mathbf x_{k-1})$$\n",
    "\n",
    "在实践中，这是非常合理的，因为许多事物都具有*马尔可夫性质*。如果你在停车场开车，你在下一秒的位置是否取\n",
    "\n",
    "决于你一分钟前是从高速公路上下来还是在土路上缓慢行驶？不会。你在下一秒的位置仅取决于你当前的位置、速度和控制输入，而不取决于一分钟前发生了什么。因此，汽车具有马尔可夫性质，我们可以进行这种简化而不会损失精度或泛化性。\n",
    "\n",
    "我们接下来的简化是将*测量模型*定义为取决于当前状态 $\\mathbf x_k$ 的条件测量概率：$P(\\mathbf z_k \\mid \\mathbf x_k)$。我们写成：\n",
    "\n",
    "$$\\mathbf z_k \\sim P(\\mathbf z_k \\mid \\mathbf x_k)$$\n",
    "\n",
    "我们现在有了一个递归，因此我们需要一个初始条件来终止它。因此，我们说初始分布是状态 $\\mathbf x_0$ 的概率：\n",
    "\n",
    "$$\\mathbf x_0 \\sim P(\\mathbf x_0)$$\n",
    "\n",
    "这些术语被用于贝叶斯公式。如果我们有状态 $\\mathbf x_0$ 和第一个测量，我们可以估计 $P(\\mathbf x_1 | \\mathbf z_1)$。运动模型创建了先验 $P(\\mathbf x_2 \\mid \\mathbf x_1)$。我们将其反馈到贝叶斯定理中，以计算 $P(\\mathbf x_2 | \\mathbf z_2)$。我们继续这个预测-校正算法，递归地根据时间 $t$ 的状态和分布仅依赖于时间 $t-1$ 的状态和分布以及时间 $t$ 的测量来计算状态和分布。\n",
    "\n",
    "这种计算的数学细节因问题而异。**离散贝叶斯**和**单变量卡尔曼滤波器**章节提供了两种不同的公式，你应该可以推理出来。单变量卡尔曼滤波器假设对于标量状态，噪声和过程模型都是线性模型，受零均值、不相关高斯噪声的影响。\n",
    "\n",
    "多变量卡尔曼滤波器做出了相同的假设，但针对状态和测量是矢量而不是标量。卡尔曼博士能够证明，如果这些假设成立，那么卡尔曼滤波器在最小二乘意义上是*最优*的。俗称的意思是无法从嘈杂的测量中获得更多信息。在本书的其余部分，我将介绍放宽线性和高斯噪声约束的滤波器。\n",
    "\n",
    "在继续之前，关于统计反演还有更多要说。正如 Calvetti 和 Somersalo 在《贝叶斯科学计算导论》中所写的那样，“我们采取了贝叶斯观点：*随机性仅意味着信息的缺乏*”。我们的状态参数化了我们原则上可以测量或计算的物理现象：速度、空气阻力等。我们缺乏足够的信息来计算或测量它们的值，因此我们选择将其视为随机变量。严格来说，它们并不是随机的，因此这是一种主观的观点。\n",
    "\n",
    "他们在书中专门讨论了这个主题。我可以少说一点。贝叶斯滤波器之所以可能，是因为我们将未知参数的统计性质归因于随机性。在卡尔曼滤波器的情况下，我们有闭式解来找到最优估计。其他滤波器，比如我们在后面一章中介绍的离散贝叶斯滤波器或粒子滤波器，以更加临时、非最优的方式建模概率。我们的技术之所以强大，在于将缺乏信息视为随机变量，并将该随机变量描述为概率分布，然后使用贝叶斯定理来解决统计推断问题。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f6c80cdf",
   "metadata": {},
   "source": [
    "# 把卡尔曼滤波器转换为 g-h 滤波器\n",
    "\n",
    "我指出了卡尔曼滤波器是 g-h 滤波器的一种形式。只需要进行一些代数运算就可以证明这一点。在一维情况下更容易证明，所以我将从那里开始。回想一下，\n",
    "\n",
    "$$\n",
    "\\mu_{x}=\\frac{\\sigma_1^2 \\mu_2 + \\sigma_2^2 \\mu_1} {\\sigma_1^2 + \\sigma_2^2}\n",
    "$$\n",
    "\n",
    "我将这个公式写成更容易理解的形式：\n",
    "\n",
    "$$\n",
    "\\mu_{x}=\\frac{ya + xb} {a+b}\n",
    "$$\n",
    "\n",
    "我们可以通过以下代数将其转换为 g-h 形式：\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "\\mu_{x}&=(x-x) + \\frac{ya + xb} {a+b} \\\\\n",
    "\\mu_{x}&=x-\\frac{a+b}{a+b}x  + \\frac{ya + xb} {a+b} \\\\ \n",
    "\\mu_{x}&=x +\\frac{-x(a+b) + xb+ya}{a+b} \\\\\n",
    "\\mu_{x}&=x+ \\frac{-xa+ya}{a+b}  \\\\\n",
    "\\mu_{x}&=x+ \\frac{a}{a+b}(y-x)\\\\\n",
    "\\end{aligned}\n",
    "$$\n",
    "\n",
    "我们几乎完成了，但回忆一下估计值的方差由以下公式给出：\n",
    "\n",
    "$$\\begin{aligned}\n",
    "\\sigma_{x}^2 &= \\frac{1}{\\frac{1}{\\sigma_1^2} +  \\frac{1}{\\sigma_2^2}} \\\\\n",
    "&= \\frac{1}{\\frac{1}{a} +  \\frac{1}{b}}\n",
    "\\end{aligned}$$\n",
    "\n",
    "我们可以通过观察到的事实将这个项纳入我们上面的方程中：\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "\\frac{a}{a+b} &= \\frac{a/a}{(a+b)/a} = \\frac{1}{(a+b)/a}  \\\\\n",
    " &= \\frac{1}{1 + \\frac{b}{a}} = \\frac{1}{\\frac{b}{b} + \\frac{b}{a}}  \\\\\n",
    " &= \\frac{1}{b}\\frac{1}{\\frac{1}{b} + \\frac{1}{a}} \\\\\n",
    " &= \\frac{\\sigma^2_{x}}{b}\n",
    " \\end{aligned}\n",
    "$$\n",
    "\n",
    "我们可以用以下形式将所有这些联系起来：\n",
    "\n",
    "$$\n",
    "\\begin{aligned}\n",
    "\\mu_{x}&=x+ \\frac{a}{a+b}(y-x) \\\\\n",
    "&= x + \\frac{\\sigma^2_{x}}{b}(y-x) \\\\\n",
    "&= x + g_n(y-x)\n",
    "\\end{aligned}\n",
    "$$\n",
    "\n",
    "其中\n",
    "\n",
    "$$\n",
    "g_n = \\frac{\\sigma^2_{x}}{\\sigma^2_{y}}\n",
    "$$\n",
    "\n",
    "最终结果是将两个测量值的残差乘以一个常数，并加到先前的值上，这是 g-h 滤波器的 $g$ 方程。$g$ 是新估计的方差与测量值的方差之比。当然，在这种情况下，$g$ 不是一个常数，因为它随着方差的变化而变化。我们也可以以同样的方式推导出 $h$ 的公式。这不是一个特别启发式的推导，我会跳过。最终结果是\n",
    "\n",
    "$$\n",
    "h_n = \\frac{COV (x,\\dot x)}{\\sigma^2_{y}}\n",
    "$$\n",
    "\n",
    "要点是 $g$ 和 $h$ 完全由时间 $n$ 的测量和预测的方差和协方差指定。换句话说，我们通过两个输入的质量来确定测量和预测之间的某个点。"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2b3634e9",
   "metadata": {},
   "source": [
    "# 参考\n",
    "\n",
    " * [1] C.B. Molwer and C.F. Van Loan \"Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later,\", *SIAM Review 45, 3-49*. 2003.\n",
    "\n",
    "\n",
    " * [2] C.F. van Loan, \"Computing Integrals Involving the Matrix Exponential,\" IEEE *Transactions Automatic Control*, June 1978.\n",
    " \n",
    " \n",
    " * [3] Calvetti, D and Somersalo E, \"Introduction to Bayesian Scientific Computing: Ten Lectures on Subjective Computing,\", *Springer*, 2007.\n",
    " \n",
    " * [4] Brown, R. G. and Hwang, P. Y.C., \"Introduction to Random Signals and Applied Kalman Filtering\", *Wiley and Sons*, Fourth Edition, p.143-147, 2012. "
   ]
  }
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