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<h1 id="seo-header">Linear Quadratic Regulators</h1>
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- [理论推导](#理论推导)
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<h2 id="理论推导"><a href="#理论推导" class="headerlink" title="理论推导"></a>理论推导</h2><p>设线性系统</p>
<p>$$<br> \dot{x} = Ax + Bu \<br> y = Cx + Du \tag{1}<br>$$</p>
<p>其中 $x(t) \in R^n$，$u(t) \in R^m$，初始条件是 $x(0)$，且假设这个系统的所有状态变量都是可测量的。<br>设状态反馈控制器</p>
<p>$$ u = -Kx \tag{2}$$</p>
<p>代入线性系统方程得</p>
<p>$$ \dot{x} = (A - BK)x = {A_c}x \tag{3}$$</p>
<p>设能量函数</p>
<p>$$ J = \frac{1}{2} \int_0^\infty (x^TQx + u^TRu)dt \tag{4}$$</p>
<p>其中 $Q$ 是半正定矩阵（$Q = Q^T \geq 0$），$R$ 是正定矩阵（$Q = Q^T > 0$）。</p>
<ul>
<li>方法一</li>
</ul>
<p>将 $(2)$ 式代入 $(4)$ 得</p>
<p>$$ J = \frac{1}{2} \int_0^\infty x^T(Q + K^TRK)xdt \tag{5}$$</p>
<p>假设存在常量矩阵 $P$ 使得</p>
<p>$$ \frac{d}{dt} (x^TPx) = -x^T(Q + K^TRK)x \tag{6}$$</p>
<p>将 $(6)$ 式代入 $(5)$ 式得</p>
<p>$$ J = -\frac{1}{2} \int_0^\infty \frac{d}{dt} (x^TPx)dt = \frac{1}{2} x^T(0)Px(0) \tag{7}$$</p>
<p>将 $(6)$ 式左边微分展开并代入 $(3)$ 式得</p>
<p>$$<br> \dot{x}^TPx + x^TP\dot{x} + x^TQx + x^TK^TRKx = 0 \<br> x^T{A_c}^TPx + x^TPA_cx + x^TQx + x^TK^TRKx = 0 \<br> x^T({A_c}^TP + PA_c + Q + K^TRK)x = 0 \tag{8}<br>$$</p>
<p>要使 $(8)$ 式成立，括号里的项必须恒等于 0，即</p>
<p>$$<br> (A - BK)^TP + P(A - BK) + Q + K^TRK = 0 \<br> A^TP - K^TB^TP + PA - PBK + Q + K^TRK = 0 \<br> A^TP + PA + Q + K^TRK - K^TB^TP - PBK = 0 \tag{9}<br>$$</p>
<p>假设 $K = R^{-1}B^TP$ 代入 $(9)$ 式得</p>
<p>$$<br> A^TP + PA + Q + (R^{-1}B^TP)^TR(R^{-1}B^TP) - (R^{-1}B^TP)^TB^TP - PB(R^{-1}B^TP) = 0 \<br> A^TP + PA + Q - PBR^{-1}B^TP = 0 \tag{10}<br>$$</p>
<p>$(10)$ 式即为代数黎卡提方程（Algebraic Riccati Equation, ARE），根据给定的 $A$，$B$，$Q$，$R$ 解出 $P$ 从而得出 $K$。</p>
<ul>
<li>方法二</li>
</ul>
<p>假设存在 $P = P^T$，使</p>
<p>$$<br> J = x(0)^TPx(0) - x(0)^TPx(0) + \frac{1}{2} \int_0^\infty (x^TQx + u^TRu)dt \<br> J = x(0)^TPx(0) + \frac{1}{2} \int_0^\infty [\frac{d}{dt}(x^TPx) + x^TQx + u^TRu]dt \<br> J = x(0)^TPx(0) + \frac{1}{2} \int_0^\infty (\dot{x}^TPx + x^TP\dot{x} + x^TQx + u^TRu)dt \tag{11}<br>$$</p>
<p>将 $(1)$ 式代入 $(11)$ 式，得</p>
<p>$$<br> J = x(0)^TPx(0) + \frac{1}{2} \int_0^\infty [(Ax + Bu)^TPx + x^TP(Ax + Bu) + x^TQx + u^TRu]dt \<br> J = x(0)^TPx(0) + \frac{1}{2} \int_0^\infty (x^TA^TPx + u^TB^TPx + x^TPAx + x^TPBu + x^TQx + u^TRu)dt \<br> J = x(0)^TPx(0) + \frac{1}{2} \int_0^\infty [x^T(A^TP + PA + Q)x + u^TB^TPx + x^TPBu + u^TRu]dt \tag{12}<br>$$</p>
<p>已知</p>
<p>$$(u + R^{-1}B^TPx)^TR(u + R^{-1}B^TPx) - x^T(PBR^{-1}B^TP)x = u^TB^TPx + x^TPBu + u^TRu \tag{13}$$</p>
<p>将 $(13)$ 式代入 $(12)$ 式，得</p>
<p>$$J = x(0)^TPx(0) + \ \frac{1}{2} \int_0^\infty [x^T(A^TP + PA + Q - PBR^{-1}B^TP)x + (u + R^{-1}B^TPx)^TR(u + R^{-1}B^TPx)]dt \tag{14}$$</p>
<p>使 $(14)$ 式积分中第二部分等于 0，即 $u = -R^{-1}B^TPx$，结合 $(2)$ 式，得 $K = R^{-1}B^TP$；<br>使 $(14)$ 式积分中第一部分等于 0，得</p>
<p>$$A^TP + PA + Q - PBR^{-1}B^TP = 0 \tag{15}$$</p>
<p>$(15)$ 式即为代数黎卡提方程（Algebraic Riccati Equation, ARE），根据给定的 $A$，$B$，$Q$，$R$ 解出 $P$ 从而得出 $K$。</p>
<h2 id="参考文献"><a href="#参考文献" class="headerlink" title="参考文献"></a>参考文献</h2><ul>
<li><a target="_blank" rel="noopener" href="https://underactuated.mit.edu/lqr.html">Linear Quadratic Regulators</a></li>
<li><a target="_blank" rel="noopener" href="https://blog.csdn.net/heyijia0327/article/details/39270597">LQR 的直观推导及简单应用</a></li>
<li><a target="_blank" rel="noopener" href="https://www.youtube.com/watch?v=E_RDCFOlJx4">What Is Linear Quadratic Regulator (LQR) Optimal Control? | State Space, Part 4</a></li>
<li><a target="_blank" rel="noopener" href="https://www.youtube.com/watch?v=ZktL3YjTbB4">Why the Riccati Equation Is important for LQR Control</a></li>
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