Part IV · 第 13 章反馈控制 · 13.6Part IV · Ch. 13 Feedback Control · 13.6

操作空间控制Operational-Space Control

想让手爪走直线、在每个方向上都一样"听话"，就要在手爪所在的任务空间里写动力学。操作空间控制用 8.3 节的操作空间惯量 $\Lambda$ 把期望加速度换成力，用 $\boldsymbol\mu$ 抵消速度带来的漂移，再用 $J^\top$ 换成关节力矩。这一节把它和雅可比转置控制放在一起比较，看环境在两种控制下感受到的刚度有何不同，看冗余臂怎样在动力学一致的零空间里做第二任务而不打扰手爪，以及为什么在奇异位形附近必须给等效质量封顶。 To make a gripper move in straight lines and respond the same way in every direction, write the dynamics in the task space where the gripper lives. Operational-space control turns a desired acceleration into a force with the operational-space inertia $\Lambda$ of Lesson 8.3, cancels the velocity-induced drift with $\boldsymbol\mu$, and maps the force to joint torques with $J^\top$. This lesson compares it with Jacobian-transpose control, shows how differently the environment feels the two, shows how a redundant arm runs a second task in the dynamically consistent null space without disturbing the gripper, and explains why the apparent mass must be capped near singularities.

约 40 分钟~40 min 前置：3.4 机器人雅可比 · 7.4 速度运动学 · 8.3 质量矩阵与科氏力 · 13.5 阻抗与导纳控制Prereqs: 3.4 The Robot Jacobian · 7.4 Velocity Kinematics · 8.3 Mass Matrix & Coriolis Terms · 13.5 Impedance & Admittance Control 5 个交互5 interactives

1为什么机器人需要它Why robots need this

很多任务用手爪的坐标来描述最自然：把杯子平移 30 cm，沿直线擦桌子，插孔时轴向推进、侧向保持柔顺。一种做法是先用 7.5 节的逆运动学把手爪路径换成关节轨迹，再交给关节控制器（13.1 节的 PID，或 8.3 节的计算力矩）。位置控制的工业臂通常就这么做。可是对力矩控制的手臂（Franka、KUKA iiwa，以及四足和人形机器人的腿），我们更希望控制器直接"站在手爪处"思考：离目标还差多少，该以多大的加速度靠近，碰到东西时该用多大的力。

最简单的想法来自 3.4 节和 7.6 节的力对偶：在手爪处虚拟一根弹簧加一个阻尼器 $\mathbf F=K_p\mathbf e-K_d\dot{\mathbf x}$，再用 $\boldsymbol\tau=J^\top\mathbf F$ 送到关节。这叫雅可比转置控制。它稳定，而且不需要动力学模型。但 8.3 节已经告诉我们：手爪在不同方向上的等效质量 $\Lambda$ 可以差好几倍甚至几十倍。同一根弹簧，往"轻"的方向拉得快，往"重"的方向拉得慢，于是路径会弯，响应还随姿态改变。

操作空间控制（operational-space control，OSC，Khatib 1987）的答案是：用 8.3 节推导的任务空间动力学 $\Lambda\ddot{\mathbf x}+\boldsymbol\mu+\mathbf p=\mathbf F$ 把手爪看成一个"质点"，像计算力矩控制那样把 $\Lambda$、$\boldsymbol\mu$、$\mathbf p$ 抵消掉，让每个任务方向都变成一个独立的单位质量。这一节回答四个问题：(1) 为什么要乘 $\Lambda$、加 $\boldsymbol\mu$；(2) 和雅可比转置控制相比，得到了什么，又付出了什么；(3) 冗余臂多出来的自由度怎样才能不打扰手爪；(4) 奇异位形附近该怎么办。

全篇沿用 8.3 节的平面两连杆臂（$l_1=1$ m，$l_2=0.8$ m，$m_1=2$ kg，$m_2=1.5$ kg，均匀杆），讲冗余时换成三连杆臂。为了把注意力集中在惯性上，所有手臂都在水平面里运动（俯视图，类似 SCARA），重力不作用在关节上，$\mathbf p=0$；竖直平面里的手臂只要再加上 8.3 节的重力补偿 $\hat{\mathbf p}$ 即可。

Many tasks are most naturally described in the gripper's coordinates: slide a cup 30 cm, wipe a table along a straight line, push along the axis of a peg while staying compliant sideways. One way is to turn the gripper path into a joint trajectory with the inverse kinematics of Lesson 7.5 and hand it to a joint controller (the PID of Lesson 13.1, or the computed torque of Lesson 8.3). Position-controlled industrial arms usually do exactly that. For torque-controlled arms, though (Franka, KUKA iiwa, the legs of quadrupeds and humanoids), we would rather have the controller think "at the gripper": how far it still is from the target, how hard it should accelerate toward it, and how much force to use when it touches something.

The simplest idea is the force duality of Lesson 3.4 and Lesson 7.6: put a virtual spring and damper at the gripper, $\mathbf F=K_p\mathbf e-K_d\dot{\mathbf x}$, and send it to the joints with $\boldsymbol\tau=J^\top\mathbf F$. This is Jacobian-transpose control. It is stable and needs no dynamic model. But Lesson 8.3 showed that the apparent mass $\Lambda$ at the gripper can differ by a factor of several, even tens, between directions. The same spring pulls fast in a "light" direction and slowly in a "heavy" one, so the path bends and the response changes with posture.

The answer of operational-space control (OSC, Khatib 1987) is to use the task-space dynamics $\Lambda\ddot{\mathbf x}+\boldsymbol\mu+\mathbf p=\mathbf F$ derived in Lesson 8.3, treat the gripper as a "point mass", and cancel $\Lambda$, $\boldsymbol\mu$ and $\mathbf p$ the way computed-torque control does, so that every task direction becomes an independent unit mass. This lesson answers four questions: (1) why multiply by $\Lambda$ and add $\boldsymbol\mu$; (2) what that buys, and what it costs, compared with Jacobian-transpose control; (3) how a redundant arm can use its extra freedom without disturbing the gripper; (4) what to do near singular postures.

The page reuses Lesson 8.3's planar two-link arm ($l_1=1$ m, $l_2=0.8$ m, $m_1=2$ kg, $m_2=1.5$ kg, uniform rods) and switches to a three-link arm for redundancy. To keep the focus on inertia, every arm moves in a horizontal plane (a top view, like a SCARA), so gravity does not load the joints and $\mathbf p=0$; an arm in a vertical plane only needs Lesson 8.3's gravity compensation $\hat{\mathbf p}$ on top.

2直觉：力不等于加速度Intuition: force is not acceleration

8.3 节 FIG 3 里我们推过手爪：沿 $\mathbf F$ 推，手爪的加速度是 $\Lambda^{-1}\mathbf F$，一般不沿 $\mathbf F$。做控制时把问题反过来：我们想要一个加速度 $\mathbf a^*$，应该施加什么力？

雅可比转置的想法：把手爪当成质量为 $\bar m$ 的质点，施加 $\mathbf F=\bar m\,\mathbf a^*$。真实的响应是 $\bar m\Lambda^{-1}\mathbf a^*$：大小和方向都被 $\Lambda^{-1}$ 扭曲了。
操作空间的想法：施加 $\mathbf F=\Lambda\mathbf a^*$，用真实的等效质量去"配"想要的加速度，$\Lambda^{-1}\Lambda\mathbf a^*=\mathbf a^*$。

手臂在动的时候还有一层：即使关节力矩为零，手爪也会自己加速。一部分来自关节空间的离心力和科氏力 $C\dot{\mathbf q}$（8.3 节），一部分来自几何本身：关节匀速转动时手爪走的是圆弧，圆弧运动就有向心加速度，这就是 7.4 节末尾预告的 $\dot J\dot{\mathbf q}$。由 $\ddot{\mathbf x}=J\ddot{\mathbf q}+\dot J\dot{\mathbf q}$ 和 $\ddot{\mathbf q}=M^{-1}(\boldsymbol\tau-C\dot{\mathbf q})$（水平面里没有重力）得到

In FIG 3 of Lesson 8.3 we pushed on the gripper: push along $\mathbf F$ and it accelerates by $\Lambda^{-1}\mathbf F$, generally not along $\mathbf F$. Control turns the question around: we want an acceleration $\mathbf a^*$; which force should we apply?

The Jacobian-transpose idea: treat the gripper as a point of mass $\bar m$ and apply $\mathbf F=\bar m\,\mathbf a^*$. The real response is $\bar m\Lambda^{-1}\mathbf a^*$: both its size and its direction are warped by $\Lambda^{-1}$.
The operational-space idea: apply $\mathbf F=\Lambda\mathbf a^*$, matching the wanted acceleration with the real apparent mass, so that $\Lambda^{-1}\Lambda\mathbf a^*=\mathbf a^*$.

When the arm is moving there is one more layer: even with zero joint torque the gripper accelerates by itself. Part of that comes from the joint-space centrifugal and Coriolis terms $C\dot{\mathbf q}$ (Lesson 8.3), part from the geometry itself: with the joints turning at constant rates the gripper travels on arcs, and arcs need centripetal acceleration; that is the $\dot J\dot{\mathbf q}$ previewed at the end of Lesson 7.4. From $\ddot{\mathbf x}=J\ddot{\mathbf q}+\dot J\dot{\mathbf q}$ and $\ddot{\mathbf q}=M^{-1}(\boldsymbol\tau-C\dot{\mathbf q})$ (no gravity in the horizontal plane) we get

$$ \boldsymbol\tau=J^{\top}\mathbf F\quad\Longrightarrow\quad \ddot{\mathbf x}=\hl{Lam}{\Lambda^{-1}}\mathbf F+\hl{mu}{\mathbf d},\qquad \mathbf d=\dot J\dot{\mathbf q}-JM^{-1}C\dot{\mathbf q},\qquad \boldsymbol\mu=-\Lambda\,\mathbf d . $$

$\mathbf d$ 是"漂移"：不施力时手爪自己的加速度。8.3 节的 $\boldsymbol\mu=\Lambda(JM^{-1}C-\dot J)\dot{\mathbf q}$ 恰好是抵消它所需的力（取负号）。所以完整的操作空间指令是 $\mathbf F=\Lambda\hl{a}{\mathbf a^*}+\boldsymbol\mu$，代入得 $\ddot{\mathbf x}=\mathbf a^*$，一点不差。下图左右拖动关节圆圈改变姿态，拖动琥珀色箭头的箭尖改变想要的加速度 $\mathbf a^*$，用滑块给关节加上速度。

$\mathbf d$ is the "drift": the gripper's own acceleration when no force is applied. Lesson 8.3's $\boldsymbol\mu=\Lambda(JM^{-1}C-\dot J)\dot{\mathbf q}$ is exactly the force needed to cancel it (with a minus sign). So the full operational-space command is $\mathbf F=\Lambda\hl{a}{\mathbf a^*}+\boldsymbol\mu$, and substituting gives $\ddot{\mathbf x}=\mathbf a^*$ exactly. In the figure, drag the joint circles to change the posture, drag the head of the amber arrow to change the wanted acceleration $\mathbf a^*$, and give the joints some velocity with the sliders.

FIG 1想要的加速度 → 力 → 关节力矩：三种做法Wanted acceleration → force → joint torque: three ways

俯视 · 拖动关节圆圈与琥珀色箭尖TOP VIEW · drag the joint circles and the amber arrowhead

琥珀色箭头是想要的加速度 $\mathbf a^*$。三种力矩都作用在同一个真实手臂上，箭头是由正动力学算出的手爪实际加速度：紫色是雅可比转置 $\boldsymbol\tau=J^\top(\bar m\mathbf a^*)$（$\bar m=2$ kg），青色是只乘 $\Lambda$、不加 $\boldsymbol\mu$ 的 $\boldsymbol\tau=J^\top\Lambda\mathbf a^*$，绿色圆环标出完整的操作空间控制 $\boldsymbol\tau=J^\top(\Lambda\mathbf a^*+\boldsymbol\mu)$ 的结果。青色虚线箭头是漂移 $\mathbf d$，从 $\mathbf a^*$ 的箭尖指向青色箭尖。蓝色闭合曲线是各方向的等效质量 $m(\mathbf u)$（每 kg 画 0.15 m，超过 6 kg 截断），细的无色箭头是手爪速度 $\dot{\mathbf x}$（1 m/s ↔ 0.15 m）。加速度的比例是 1 m/s² ↔ 0.12 m，长于 1.2 m 的箭头被截短，读数给出完整数值。 The amber arrow is the wanted acceleration $\mathbf a^*$. All three torques act on the same true arm, and each arrow is the actual tip acceleration computed with forward dynamics: purple is Jacobian transpose $\boldsymbol\tau=J^\top(\bar m\mathbf a^*)$ ($\bar m=2$ kg), cyan multiplies by $\Lambda$ but leaves out $\boldsymbol\mu$, $\boldsymbol\tau=J^\top\Lambda\mathbf a^*$, and the green ring marks the result of full operational-space control $\boldsymbol\tau=J^\top(\Lambda\mathbf a^*+\boldsymbol\mu)$. The dashed cyan arrow is the drift $\mathbf d$, from the head of $\mathbf a^*$ to the head of the cyan arrow. The blue closed curve is the apparent mass $m(\mathbf u)$ in each direction (0.15 m per kg, cut at 6 kg), and the thin plain arrow is the tip velocity $\dot{\mathbf x}$ (1 m/s ↔ 0.15 m). Accelerations are drawn at 1 m/s² ↔ 0.12 m; arrows longer than 1.2 m are shortened, and the readout gives the full values.

试一试Try this

关节速度为零时，找一个姿态和 $\mathbf a^*$ 的方向，让雅可比转置的加速度偏离 $\mathbf a^*$ 超过 45°。提示：先看蓝色曲线在哪个姿态下最扁。With zero joint velocity, find a posture and a direction of $\mathbf a^*$ for which the Jacobian-transpose acceleration is more than 45° off $\mathbf a^*$. Hint: first look for the posture where the blue curve is flattest.
给关节加上速度（$\|\dot{\mathbf q}\|\ge1$ rad/s），保持 $\|\mathbf a^*\|\ge2$ m/s²，让"不加 $\boldsymbol\mu$"的加速度偏离 $\mathbf a^*$ 超过 30°，同时确认完整的操作空间控制仍然偏差不到 1°。Give the joints some velocity ($\|\dot{\mathbf q}\|\ge1$ rad/s) and keep $\|\mathbf a^*\|\ge2$ m/s²: make the "without $\boldsymbol\mu$" acceleration more than 30° off $\mathbf a^*$, and check that full operational-space control is still within 1°.

核心洞见Key insight

在手爪处，"力 = 质量 × 加速度"里的质量是一个随姿态变化的矩阵 $\Lambda(\mathbf q)$，而且手臂在动时还有一个与力无关的漂移 $\mathbf d$。操作空间控制就是在手爪处做一次计算力矩：先用 $\boldsymbol\mu$ 抵消漂移，再用 $\Lambda$ 把想要的加速度翻译成力。想要的加速度从哪里来，是下一步的事。

At the gripper, the mass in "force = mass × acceleration" is a posture-dependent matrix $\Lambda(\mathbf q)$, and a moving arm adds a force-independent drift $\mathbf d$. Operational-space control is computed torque done at the gripper: cancel the drift with $\boldsymbol\mu$, then translate the wanted acceleration into a force with $\Lambda$. Where the wanted acceleration comes from is the next step.

3动手玩：雅可比转置 vs 操作空间Play: Jacobian transpose vs. operational space

想要的加速度来自任务空间的 PD：$\mathbf a=\omega_n^2\mathbf e-2\omega_n\dot{\mathbf x}$，$\mathbf e=\mathbf x_d-\mathbf x$。这是一个单位质量的临界阻尼弹簧，带宽 $\omega_n$。下面两个控制器用同一个 $\omega_n$，只差在把加速度换成力的方式：

The wanted acceleration comes from a task-space PD: $\mathbf a=\omega_n^2\mathbf e-2\omega_n\dot{\mathbf x}$, with $\mathbf e=\mathbf x_d-\mathbf x$. That is a critically damped spring on a unit mass with bandwidth $\omega_n$. The two controllers below share the same $\omega_n$ and differ only in how they turn acceleration into force:

$$ \hl{JT}{\boldsymbol\tau_{J^\top}=J^{\top}\,\bar m\big(\omega_n^2\mathbf e-2\omega_n\dot{\mathbf x}\big)},\qquad\qquad \hl{osc}{\boldsymbol\tau_{\mathrm{OSC}}=J^{\top}\Big(\hat\Lambda\big(\omega_n^2\mathbf e-2\omega_n\dot{\mathbf x}\big)+\hat{\boldsymbol\mu}\Big)} . $$

雅可比转置控制的增益按一个"名义质量" $\bar m$ 来定：如果手爪真是一个 $\bar m$ kg 的质点，它就是临界阻尼。操作空间控制用模型 $\hat\Lambda,\hat{\boldsymbol\mu}$（帽子表示控制器心目中的模型）。模型准确时，手爪的误差满足 $\ddot{\mathbf e}+2\omega_n\dot{\mathbf e}+\omega_n^2\mathbf e=0$：两个方向各是一个独立的临界阻尼系统。从静止出发时 $\mathbf e(t)=(1+\omega_nt)e^{-\omega_nt}\,\mathbf e(0)$，误差向量只缩短、不转向，所以手爪走直线。

拖动蓝色目标圈，松手后两条手臂同时从同一个起点出发（起点就是上一个目标）。也可以加一个控制器不知道的末端负载，或者把负载写进模型。

The Jacobian-transpose gains are set for a "nominal mass" $\bar m$: if the gripper really were a point of $\bar m$ kg, it would be critically damped. Operational-space control uses a model $\hat\Lambda,\hat{\boldsymbol\mu}$ (the hats mark the controller's belief). With an exact model the gripper error obeys $\ddot{\mathbf e}+2\omega_n\dot{\mathbf e}+\omega_n^2\mathbf e=0$: two independent critically damped systems, one per direction. Starting from rest, $\mathbf e(t)=(1+\omega_nt)e^{-\omega_nt}\,\mathbf e(0)$: the error vector only shrinks and never turns, so the gripper moves in a straight line.

Drag the blue target ring and release it: both arms set off together from the same start, which is the previous target. You can also add a tip payload the controller does not know about, or write the payload into the model.

FIG 2点到点：雅可比转置与操作空间控制Point to point: Jacobian transpose and operational space

俯视 · 拖动蓝色目标，松手出发TOP VIEW · drag the blue target, release to go

误差分量 (m) vs 时间 (s)ERROR COMPONENTS (m) vs. TIME (s)

琥珀色 = 操作空间控制，紫色 = 雅可比转置；无色虚线是起点到目标的直线。整条路径先淡淡画出，回放到的部分加粗。右图：实线是沿直线还剩的距离 $e_\parallel$，虚线是偏离直线的侧向误差 $e_\perp$。读数里"偏离"是 $|e_\perp|$ 的最大值，"过冲"是手爪沿直线越过目标的最大距离，"到位"是手爪最后一次离目标超过 1 cm 的时刻。仿真用 RK4、1 ms 步长，控制器 1 kHz 零阶保持，没有力矩上限（真实控制器会插值一条参考轨迹，而不是一下子跳到目标）。目标限制在离基座 0.4 到 1.72 m 之间，起点到目标的直线离基座至少 0.3 m，以避开手臂折叠时的奇异位形。 Amber = operational space, purple = Jacobian transpose; the plain dashed line is the straight line from start to target. Each full path is drawn faintly first, and the part played back so far is bold. Right: solid lines are the distance still to go along the line, $e_\parallel$; dashed lines are the sideways error $e_\perp$ off the line. In the readout, "deviation" is the largest $|e_\perp|$, "overshoot" the furthest the tip passes the target along the line, and "settle" the last time the tip is more than 1 cm from the target. Simulation: RK4 at 1 ms, controller at 1 kHz with zero-order hold, no torque limits (a real controller would interpolate a reference trajectory instead of jumping to the target). Targets are kept 0.4 to 1.72 m from the base, and the straight line from start to target at least 0.3 m from the base, to stay clear of the folded-arm singularity.

试一试Try this

拖动目标，找一次运动，让雅可比转置偏离直线超过 15 cm，而操作空间控制偏离不到 2 mm（负载为 0 或已写进模型）。Drag the target and find a move where Jacobian transpose deviates from the straight line by more than 15 cm while operational space stays within 2 mm (no payload, or the payload is in the model).
再找一个目标，让雅可比转置的到位时间比操作空间控制长 1.6 倍以上（同样要求负载为 0 或已建模）。操作空间控制的到位时间只取决于距离，雅可比转置的还取决于什么？Find a target where Jacobian transpose takes more than 1.6 times as long to settle as operational space (again with no payload, or a modelled one). Operational space's settling time depends only on the distance; what else does Jacobian transpose's depend on?
把末端负载调到 1 kg 以上，模型保持"不含负载"，让操作空间控制冲过目标超过 10 cm。为什么雅可比转置几乎不受影响？Set the tip payload to 1 kg or more with the model "without payload", and make operational space overshoot the target by more than 10 cm. Why is Jacobian transpose hardly affected?
负载调到 2 kg 以上，再把模型切换为"含负载"：操作空间控制的路径重新变直（偏离和过冲都小于 2 mm）。With a payload of 2 kg or more, switch the model to "with payload": the operational-space path becomes straight again (deviation and overshoot both below 2 mm).

核心洞见Key insight

雅可比转置控制在手爪处装的是一根"不知道手臂多重"的弹簧，它的响应由 $\Lambda(\mathbf q)^{-1}$ 塑形：方向不同、姿态不同，快慢就不同，路径因此弯曲。操作空间控制先把手爪变成单位质量，再装弹簧，于是每个方向的响应都一样，误差解耦、路径笔直。代价是它依赖模型：负载没写进 $\hat\Lambda$，等效增益就被悄悄缩小，阻尼不足而过冲。

Jacobian-transpose control mounts a spring at the gripper that "does not know how heavy the arm is", so its response is shaped by $\Lambda(\mathbf q)^{-1}$: fast or slow depending on direction and posture, which bends the path. Operational-space control first turns the gripper into a unit mass and then mounts the spring, so every direction responds the same way: the errors are decoupled and the path is straight. The price is its reliance on the model: leave a payload out of $\hat\Lambda$ and the effective gains quietly shrink, the damping falls short, and the arm overshoots.

4第二、第三视角：环境感受到的刚度，与冗余臂的零空间Second and third views: the stiffness the world feels, and the null space of a redundant arm

环境感受到的刚度The stiffness the environment feels

FIG 2 看的是自由运动。现在让手臂停在目标处，用一个外力 $\mathbf F_{\text{ext}}$ 去推手爪。静止时 $\dot{\mathbf x}=0$、$\boldsymbol\mu=0$，两种控制器的手爪力分别是 $\bar m\omega_n^2\mathbf e$ 和 $\Lambda\omega_n^2\mathbf e$，平衡时它们抵消外力，偏移 $\boldsymbol\delta=-\mathbf e$ 满足

FIG 2 was about free motion. Now let the arm hold still at the target and push on the gripper with an external force $\mathbf F_{\text{ext}}$. At rest $\dot{\mathbf x}=0$ and $\boldsymbol\mu=0$, so the two controllers produce gripper forces $\bar m\omega_n^2\mathbf e$ and $\Lambda\omega_n^2\mathbf e$; at equilibrium these cancel the push, and the deflection $\boldsymbol\delta=-\mathbf e$ satisfies

$$ \underbrace{\bar m\,\omega_n^2\,\boldsymbol\delta=\mathbf F_{\text{ext}}}_{J^{\top}}\qquad\qquad \underbrace{\hl{K}{\omega_n^2\,\Lambda(\mathbf q)}\,\boldsymbol\delta=\mathbf F_{\text{ext}}}_{\mathrm{OSC}} . $$

雅可比转置给出一根各向同性的弹簧，刚度 $\bar m\omega_n^2$。操作空间控制给环境的刚度却是 $\omega_n^2\Lambda(\mathbf q)$：轻方向软、重方向硬，并且随姿态变化。对这条手臂，最轻方向的等效质量在各种姿态下都在 0.45 到 0.5 kg 之间，而最重方向在手肘弯成直角附近约 2.2 kg，接近伸直或折叠时超过 20 kg。所以"把手爪变成单位质量"让运动整齐了，环境感受到的刚度却恰恰继承了 $\Lambda$ 的不均匀。下图中拖动琥珀色的推力手柄，两条手臂在同一个推力下各自偏移。

Jacobian transpose gives an isotropic spring of stiffness $\bar m\omega_n^2$. Operational-space control, however, presents the environment with the stiffness $\omega_n^2\Lambda(\mathbf q)$: soft in light directions, stiff in heavy ones, and changing with posture. For this arm the lightest apparent mass stays between 0.45 and 0.5 kg in every posture, while the heaviest is about 2.2 kg with the elbow near a right angle and exceeds 20 kg near the stretched or folded postures. So "turning the gripper into a unit mass" tidies up the motion, but the stiffness felt by the environment inherits exactly the unevenness of $\Lambda$. In the figure, drag the amber push handle: both arms deflect under the same push.

FIG 3推一下手爪：两种控制器的刚度Push the gripper: the stiffness of two controllers

俯视 · 拖动蓝色目标与琥珀色推力手柄TOP VIEW · drag the blue target and the amber push handle

两条手臂（琥珀色 = 操作空间，紫色 = 雅可比转置）实时仿真，都停在蓝色目标处，$\omega_n=8$ rad/s，$\bar m=2$ kg，所以雅可比转置的刚度是 128 N/m。推力从目标指向琥珀色手柄（1 N ↔ 0.04 m，最大 8 N），同样的推力作用在两条手臂的手爪上，画在各自的手爪处。以目标为中心，琥珀色椭圆和紫色圆是按线性刚度预测的偏移范围：推力大小不变、方向转一圈时手爪会偏到的地方（推力小于 2 N 时按 2 N 画）。两种控制器在目标处都没有稳态误差，偏移完全来自外力。 Both arms (amber = operational space, purple = Jacobian transpose) are simulated live and hold the blue target, with $\omega_n=8$ rad/s and $\bar m=2$ kg, so the Jacobian-transpose stiffness is 128 N/m. The push points from the target to the amber handle (1 N ↔ 0.04 m, at most 8 N); the same push acts on both grippers and is drawn at each tip. Centred on the target, the amber ellipse and the purple circle are the deflections predicted by the linear stiffness: where the tip would go if a push of this size swept through all directions (pushes under 2 N are drawn as 2 N). Neither controller has a steady-state error at the target, so the deflection is entirely due to the push.

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用至少 2 N 的推力，找一个方向，让操作空间控制的偏移达到雅可比转置的 3.5 倍以上（等两条手臂停稳）。这是哪个方向？With a push of at least 2 N, find a direction where the operational-space deflection is more than 3.5 times the Jacobian-transpose one (wait for both arms to settle). Which direction is it?
反过来：移动目标改变姿态，找到操作空间控制的偏移不到雅可比转置一半的情况（推力至少 2 N，停稳后判定）。The other way round: move the target to change the posture, and find a case where the operational-space deflection is less than half the Jacobian-transpose one (push at least 2 N, judged once settled).

这就是 13.5 节的阻抗控制要处理的问题：接触任务关心的是环境感受到的刚度和阻尼，应该直接以 N/m 为单位来设计，例如 $\mathbf F=K\mathbf e-D\dot{\mathbf x}+\boldsymbol\mu$（带科氏补偿的雅可比转置），而把 $\Lambda$ 留给自由运动。要同时得到"想要的刚度"和"想要的惯量"，就得测量外力，见 13.5 节。

This is the problem the impedance control of Lesson 13.5 deals with: contact tasks care about the stiffness and damping the environment feels, which should be designed directly in N/m, e.g. $\mathbf F=K\mathbf e-D\dot{\mathbf x}+\boldsymbol\mu$ (Jacobian transpose with Coriolis compensation), leaving $\Lambda$ to free motion. Getting both the wanted stiffness and the wanted inertia requires measuring the external force; see Lesson 13.5.

冗余臂：动力学一致的零空间Redundant arms: the dynamically consistent null space

三连杆平面臂控制手爪的 2 维位置，还多出 1 个自由度：手爪不动时，手肘仍然可以摆（1.6 节的零空间，7.4 节叫它自运动）。我们想用这份自由做第二个任务，比如让第一根连杆朝向角度 $\psi$：关节力矩 $\boldsymbol\tau_0=k_\psi(\psi-\theta_1)\,\mathbf e_1-k_d\dot{\mathbf q}$（$\mathbf e_1$ 是第一个单位向量，只拉关节 1；阻尼作用在所有关节上）。但 $\boldsymbol\tau_0$ 不能直接加上去，它会推动手爪。要先用一个矩阵 $N^\top$ 过滤：$\boldsymbol\tau=J^\top\mathbf F+N^\top\boldsymbol\tau_0$。过滤后的力矩不得改变手爪加速度，即 $JM^{-1}N^\top=0$。

3.4 节和 7.4 节在速度层面用过伪逆投影 $I-J^+J$。把它原样搬到力矩上是一个常见的错误：$JM^{-1}(I-J^+J)$ 一般不为零。正确的选择是用 8.3 节误区 5 提到的动力学一致广义逆 $\bar J=M^{-1}J^\top\Lambda$：

A three-link planar arm controlling the 2-D tip position has one degree of freedom to spare: with the tip still, the elbow can still swing (the null space of Lesson 1.6, called self-motion in Lesson 7.4). We want to use that freedom for a second task, say pointing the first link at an angle $\psi$: joint torques $\boldsymbol\tau_0=k_\psi(\psi-\theta_1)\,\mathbf e_1-k_d\dot{\mathbf q}$ ($\mathbf e_1$ is the first unit vector, so only joint 1 is pulled; the damping acts on every joint). But $\boldsymbol\tau_0$ cannot simply be added: it would push the tip. It must first be filtered by a matrix $N^\top$: $\boldsymbol\tau=J^\top\mathbf F+N^\top\boldsymbol\tau_0$. The filtered torque must not change the tip acceleration, i.e. $JM^{-1}N^\top=0$.

Lessons 3.4 and 7.4 used the pseudo-inverse projector $I-J^+J$ at the velocity level. Carrying it over unchanged to torques is a common mistake: $JM^{-1}(I-J^+J)$ is not zero in general. The right choice uses the dynamically consistent generalized inverse $\bar J=M^{-1}J^\top\Lambda$ mentioned in Pitfall 5 of Lesson 8.3:

$$ \hl{N}{N^{\top}=I-J^{\top}\bar J^{\top}},\qquad \bar J=M^{-1}J^{\top}\Lambda, $$ $$ JM^{-1}N^{\top}=JM^{-1}-\underbrace{JM^{-1}J^{\top}}_{\Lambda^{-1}}\Lambda JM^{-1}=0 . $$

下图用三种过滤器跑同一个第二任务：动力学一致的 $N^\top$，伪逆投影 $I-J^+J$，以及不过滤。按"摆动"让 $\psi$ 以 0.4 Hz、±0.6 rad 来回摆，看手爪误差。

The figure runs the same second task through three filters: the dynamically consistent $N^\top$, the pseudo-inverse projector $I-J^+J$, and no filter at all. Press "Swing" to sweep $\psi$ back and forth at 0.4 Hz by ±0.6 rad, and watch the tip error.

FIG 4三连杆冗余臂：零空间里的第二任务Redundant three-link arm: a second task in the null space

俯视 · 拖动蓝色目标TOP VIEW · drag the blue target

手爪误差 (mm)，对数，最近 6 sTIP ERROR (mm), log, last 6 s

连杆长 0.7、0.6、0.5 m，质量 1.6、1.2、0.8 kg。主任务是手爪位置的操作空间控制（$\omega_n=10$ rad/s），第二任务 $k_\psi=30$ N·m/rad、$k_d=4$ N·m·s/rad。紫色虚线从基座指向想要的第一连杆方向 $\psi$，紫色细线是第一个关节点（连杆 1 的末端）最近走过的轨迹，琥珀色细线是手爪轨迹。右图曲线的颜色表示当时用的过滤器：绿色 = 动力学一致，紫色 = 伪逆，红色 = 不过滤。读数里的"最大误差"与"摆幅"统计最近 3 s（这段时间里没有拖动目标、没有切换过滤器）。 Links of 0.7, 0.6 and 0.5 m with masses 1.6, 1.2 and 0.8 kg. The main task is operational-space control of the tip position ($\omega_n=10$ rad/s); the second task has $k_\psi=30$ N·m/rad and $k_d=4$ N·m·s/rad. The purple dashed line points from the base in the wanted first-link direction $\psi$, the thin purple line is the recent path of the first joint point (the end of link 1), and the thin amber line is the tip's path. On the right, the colour of the curve shows the filter in use at the time: green = dynamically consistent, purple = pseudo-inverse, red = none. The readout's "max error" and "swing" cover the last 3 s (with no target drag and no filter switch in that window).

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用动力学一致的过滤器，让第一连杆在 3 s 内摆动超过 0.6 rad，同时手爪的最大误差保持在 1 mm 以下。With the dynamically consistent filter, make the first link swing by more than 0.6 rad within 3 s while the tip's maximum error stays below 1 mm.
换成伪逆投影 $I-J^+J$，做同样的摆动，让手爪误差超过 2 cm。第二任务的力矩"漏"进了主任务。Switch to the pseudo-inverse projector $I-J^+J$ and swing the same way until the tip error exceeds 2 cm. The second task's torque "leaks" into the main task.

一个概念，三个视角One idea, three views

代数视角：$\bar J$ 是用 $M$ 加权的伪逆，它满足 $J\bar J=I$，在所有满足 $J\dot{\mathbf q}=\dot{\mathbf x}$ 的关节速度里，$\bar J\dot{\mathbf x}$ 的动能最小（8.3 节）。力学视角：速度和力矩是对偶的两类量（3.4 节的 $\dot{\mathbf x}=J\dot{\mathbf q}$ 与 $\boldsymbol\tau=J^\top\mathbf F$）。"不影响手爪"对速度说的是 $J\dot{\mathbf q}=0$，对力矩说的却是"不产生手爪加速度" $JM^{-1}\boldsymbol\tau=0$，中间隔着 $M^{-1}$，所以投影要按 $M$ 加权。只有当 $M$ 与单位阵成比例时，两种投影才一样。层级视角：$N^\top$ 给出一个严格的优先级，主任务永远不受第二任务干扰，第二任务只能在剩下的自由里尽力而为。多层任务的层级就是 13.7 节全身控制的内容。

Algebraic view: $\bar J$ is the pseudo-inverse weighted by $M$; it satisfies $J\bar J=I$, and among all joint velocities with $J\dot{\mathbf q}=\dot{\mathbf x}$ the one $\bar J\dot{\mathbf x}$ has the least kinetic energy (Lesson 8.3). Mechanical view: velocities and torques are dual quantities (Lesson 3.4's $\dot{\mathbf x}=J\dot{\mathbf q}$ and $\boldsymbol\tau=J^\top\mathbf F$). For velocities "leave the tip alone" means $J\dot{\mathbf q}=0$; for torques it means "create no tip acceleration", $JM^{-1}\boldsymbol\tau=0$, with an $M^{-1}$ in between, so the projector has to be weighted by $M$. Only when $M$ is proportional to the identity do the two projectors agree. Hierarchy view: $N^\top$ sets a strict priority: the main task is never disturbed by the second, and the second does its best with the freedom left over. Hierarchies of many tasks are the subject of the whole-body control of Lesson 13.7.

5再看数学The math

提示：把鼠标悬停在公式中带底色的符号上，对应的图元素会被高亮：$\Lambda$ 是 FIG 1 的蓝色等效质量曲线，$\mathbf d$ 是 FIG 1 的青色虚线漂移箭头，$\mathbf a^*$ 是 FIG 1 的琥珀色箭头，两个控制律是 FIG 2 的紫色与琥珀色路径，刚度 $\omega_n^2\Lambda$ 是 FIG 3 的琥珀色椭圆，$N^\top$ 是 FIG 4 的第一关节点轨迹与 $\psi$ 虚线，封顶的 $\Lambda_\varepsilon$ 是 FIG 5 的琥珀色力矩曲线。Tip: hover the highlighted symbols in the formulas and the matching element lights up: $\Lambda$ is FIG 1's blue apparent-mass curve, $\mathbf d$ FIG 1's dashed cyan drift arrow, $\mathbf a^*$ FIG 1's amber arrow, the two control laws FIG 2's purple and amber paths, the stiffness $\omega_n^2\Lambda$ FIG 3's amber ellipse, $N^\top$ FIG 4's first-joint path and $\psi$ line, and the capped $\Lambda_\varepsilon$ FIG 5's amber torque curve.

任务空间动力学（8.3 节的结论）Task-space dynamics (the result of Lesson 8.3)

设任务坐标 $\mathbf x\in\R^m$（这里是手爪位置，$m=2$），$\dot{\mathbf x}=J(\mathbf q)\dot{\mathbf q}$，关节动力学为 8.3 节的 $M(\mathbf q)\ddot{\mathbf q}+\mathbf b(\mathbf q,\dot{\mathbf q})=\boldsymbol\tau$，$\mathbf b=C\dot{\mathbf q}+\mathbf g$。8.3 节把 $\boldsymbol\tau=J^\top\mathbf F$ 代入、消去 $\ddot{\mathbf q}$，得到

Let the task coordinates be $\mathbf x\in\R^m$ (here the tip position, $m=2$), $\dot{\mathbf x}=J(\mathbf q)\dot{\mathbf q}$, and let the joint dynamics be those of Lesson 8.3, $M(\mathbf q)\ddot{\mathbf q}+\mathbf b(\mathbf q,\dot{\mathbf q})=\boldsymbol\tau$ with $\mathbf b=C\dot{\mathbf q}+\mathbf g$. Lesson 8.3 substituted $\boldsymbol\tau=J^\top\mathbf F$ and eliminated $\ddot{\mathbf q}$ to get

$$ \hl{Lam}{\Lambda}\,\ddot{\mathbf x}+\boldsymbol\mu+\mathbf p=\mathbf F,\qquad \Lambda=(JM^{-1}J^{\top})^{-1}, $$ $$ \boldsymbol\mu=\Lambda\big(JM^{-1}C-\dot J\big)\dot{\mathbf q},\qquad \mathbf p=\Lambda JM^{-1}\mathbf g . $$

（$\boldsymbol\mu$ 和 $\mathbf p$ 正是偏置项 $\mathbf b$ 的速度部分和重力部分在任务空间里的样子：$\boldsymbol\mu+\mathbf p=\Lambda(JM^{-1}\mathbf b-\dot J\dot{\mathbf q})$。13.7 节沿用浮动基座文献的习惯，把同一个 $\mathbf b$ 记作 $\mathbf h$。）这里假设 $J$ 行满秩，于是 $JM^{-1}J^\top$ 对称正定、可逆。对非冗余臂（$J$ 方阵）这个式子就是把关节动力学换了坐标；对冗余臂它描述的是"手爪在力 $\mathbf F$ 作用下、关节取动能最小的那种运动时"的动力学，下面的零空间部分会用到。

($\boldsymbol\mu$ and $\mathbf p$ are just the velocity and gravity parts of the bias $\mathbf b$ seen in task space: $\boldsymbol\mu+\mathbf p=\Lambda(JM^{-1}\mathbf b-\dot J\dot{\mathbf q})$. Lesson 13.7 follows the floating-base literature and writes the same $\mathbf b$ as $\mathbf h$.) We assume $J$ has full row rank, so $JM^{-1}J^\top$ is symmetric positive definite and invertible. For a non-redundant arm (square $J$) the equation is just the joint dynamics in new coordinates; for a redundant arm it describes the tip under the force $\mathbf F$ when the joints move in the least-kinetic-energy way, which the null-space part below relies on.

操作空间控制律与解耦The operational-space control law and decoupling

取参考轨迹 $\mathbf x_d(t)$，误差 $\mathbf e=\mathbf x_d-\mathbf x$，令

With a reference trajectory $\mathbf x_d(t)$ and error $\mathbf e=\mathbf x_d-\mathbf x$, choose

$$ \mathbf F=\hat\Lambda\,\hl{a}{\mathbf a}+\hat{\boldsymbol\mu}+\hat{\mathbf p},\qquad \mathbf a=\ddot{\mathbf x}_d+K_d\dot{\mathbf e}+K_p\mathbf e,\qquad \boldsymbol\tau=J^{\top}\mathbf F . $$

模型准确时代入任务空间动力学，$\Lambda\ddot{\mathbf x}=\Lambda\mathbf a$，$\Lambda$ 可逆，于是 $\ddot{\mathbf x}=\mathbf a$，也就是

With an exact model, substituting into the task-space dynamics gives $\Lambda\ddot{\mathbf x}=\Lambda\mathbf a$; since $\Lambda$ is invertible, $\ddot{\mathbf x}=\mathbf a$, i.e.

$$ \ddot{\mathbf e}+K_d\dot{\mathbf e}+K_p\mathbf e=0 . $$

取对角的 $K_p=\omega_n^2I$、$K_d=2\omega_nI$，每个任务方向就是一个独立的临界阻尼二阶系统，与姿态无关：这就是 Khatib 所说的解耦。$\mathbf x_d$ 固定、从静止出发时，解是 $\mathbf e(t)=(1+\omega_nt)e^{-\omega_nt}\mathbf e(0)$，方向不变，手爪走直线；误差降到 $\epsilon$ 所需的时间只由 $\|\mathbf e(0)\|/\epsilon$ 和 $\omega_n$ 决定（FIG 2 的"到位"时间）。

With diagonal $K_p=\omega_n^2I$ and $K_d=2\omega_nI$, every task direction is an independent, critically damped second-order system, regardless of posture: this is what Khatib calls decoupling. For fixed $\mathbf x_d$ and a start from rest the solution is $\mathbf e(t)=(1+\omega_nt)e^{-\omega_nt}\mathbf e(0)$: its direction never changes and the tip moves in a straight line, and the time for the error to fall to $\epsilon$ depends only on $\|\mathbf e(0)\|/\epsilon$ and $\omega_n$ (FIG 2's "settle" time).

与计算力矩的关系。把 $\mathbf F$ 展开，$\boldsymbol\tau=J^\top\Lambda(\mathbf a-\dot J\dot{\mathbf q})+J^\top\Lambda JM^{-1}\mathbf b$。当 $J$ 是可逆方阵时，$J^\top\Lambda J=J^\top J^{-\top}MJ^{-1}J=M$，于是

Relation to computed torque. Expanding $\mathbf F$ gives $\boldsymbol\tau=J^\top\Lambda(\mathbf a-\dot J\dot{\mathbf q})+J^\top\Lambda JM^{-1}\mathbf b$. When $J$ is square and invertible, $J^\top\Lambda J=J^\top J^{-\top}MJ^{-1}J=M$, so

$$ \boldsymbol\tau=M\,J^{-1}\big(\mathbf a-\dot J\dot{\mathbf q}\big)+\mathbf b . $$

这正是 8.3 节的计算力矩控制，只是关节加速度指令取 $\ddot{\mathbf q}^*=J^{-1}(\mathbf a-\dot J\dot{\mathbf q})$，这种做法叫分解加速度控制（resolved-acceleration control，Luh、Walker、Paul 1980）。对非冗余臂，两者在数学上相同；区别在于冗余臂（$J$ 不是方阵，没有 $J^{-1}$）、多任务的组织方式，以及奇异位形附近的处理。

This is Lesson 8.3's computed-torque control with the joint-acceleration command $\ddot{\mathbf q}^*=J^{-1}(\mathbf a-\dot J\dot{\mathbf q})$, known as resolved-acceleration control (Luh, Walker and Paul, 1980). For a non-redundant arm the two are mathematically the same; they differ for redundant arms ($J$ is not square and has no $J^{-1}$), in how several tasks are organized, and in how singular postures are handled.

模型误差。若 $\hat\Lambda\ne\Lambda$，在静止附近 $\ddot{\mathbf x}\approx\Lambda^{-1}\hat\Lambda\mathbf a$：等效增益被矩阵 $\Lambda^{-1}\hat\Lambda$ 缩放。未建模的负载使 $\Lambda$ 变大，$\Lambda^{-1}\hat\Lambda$ 的特征值小于 1；如果把它记作 $s$，等效的 $K_p,K_d$ 都乘以 $s$，阻尼比变成 $\sqrt s$ 倍，系统欠阻尼而过冲（FIG 2 的负载任务）。最轻方向（约 0.5 kg）加上一个 1 kg 的负载，$s$ 约为 1/3。

Model error. If $\hat\Lambda\ne\Lambda$, then near rest $\ddot{\mathbf x}\approx\Lambda^{-1}\hat\Lambda\mathbf a$: the effective gains are scaled by the matrix $\Lambda^{-1}\hat\Lambda$. An unmodelled payload makes $\Lambda$ larger, so the eigenvalues of $\Lambda^{-1}\hat\Lambda$ drop below 1; calling one of them $s$, the effective $K_p$ and $K_d$ are both multiplied by $s$, the damping ratio by $\sqrt s$, and the system becomes underdamped and overshoots (FIG 2's payload task). In the lightest direction (about 0.5 kg), a 1 kg payload makes $s$ about 1/3.

雅可比转置控制：为什么不要模型也稳定Jacobian-transpose control: why it is stable without a model

取固定目标 $\mathbf x_d$、对称正定的 $K_p,K_d$，控制律 $\boldsymbol\tau=J^\top(K_p\mathbf e-K_d\dot{\mathbf x})+\hat{\mathbf g}$（Takegaki、Arimoto 1981），并假设重力补偿准确。取能量函数 $V=\tfrac12\dot{\mathbf q}^\top M\dot{\mathbf q}+\tfrac12\mathbf e^\top K_p\mathbf e$：动能加上虚拟弹簧的势能。利用 8.3 节的功率平衡（$\dot M-2C$ 斜对称）和 $\dot{\mathbf e}=-J\dot{\mathbf q}$：

Take a fixed target $\mathbf x_d$, symmetric positive definite $K_p,K_d$, the law $\boldsymbol\tau=J^\top(K_p\mathbf e-K_d\dot{\mathbf x})+\hat{\mathbf g}$ (Takegaki and Arimoto, 1981), and exact gravity compensation. Use the energy function $V=\tfrac12\dot{\mathbf q}^\top M\dot{\mathbf q}+\tfrac12\mathbf e^\top K_p\mathbf e$: kinetic energy plus the potential energy of the virtual spring. With Lesson 8.3's power balance ($\dot M-2C$ skew-symmetric) and $\dot{\mathbf e}=-J\dot{\mathbf q}$:

$$ \dot V=\dot{\mathbf q}^{\top}(\boldsymbol\tau-\mathbf g)-\mathbf e^{\top}K_pJ\dot{\mathbf q}=\dot{\mathbf q}^{\top}J^{\top}(K_p\mathbf e-K_d\dot{\mathbf x})-\mathbf e^{\top}K_pJ\dot{\mathbf q}=-\dot{\mathbf x}^{\top}K_d\,\dot{\mathbf x}\le0 . $$

能量只减不增，而且没有用到 $M$ 的任何数值，所以对质量误差、未知负载都是鲁棒的。若手臂停在某处不再运动，则 $\boldsymbol\tau-\mathbf g=J^\top K_p\mathbf e=0$；$J$ 行满秩时只能 $\mathbf e=0$，由 LaSalle 不变原理，手爪收敛到目标（在奇异位形处 $J^\top$ 有零空间，手臂可能停在 $\mathbf e\ne0$ 处）。它控制的是终点，而不是过程：在目标附近线性化，$\Lambda\ddot{\mathbf x}\approx K_p\mathbf e-K_d\dot{\mathbf x}$，沿 $\Lambda$ 的特征方向 $\mathbf u$，自然频率和阻尼比分别是

The energy never increases, and no value of $M$ was used, so the result is robust to mass errors and unknown payloads. If the arm comes to rest for good, then $\boldsymbol\tau-\mathbf g=J^\top K_p\mathbf e=0$; when $J$ has full row rank this forces $\mathbf e=0$, and LaSalle's invariance principle gives convergence to the target (at a singular posture $J^\top$ has a null space and the arm may stop with $\mathbf e\ne0$). It controls the end point, not the journey: linearizing near the target, $\Lambda\ddot{\mathbf x}\approx K_p\mathbf e-K_d\dot{\mathbf x}$, and along an eigen-direction $\mathbf u$ of $\Lambda$ the natural frequency and damping ratio are

$$ \omega(\mathbf u)=\sqrt{\frac{k_p}{m(\mathbf u)}},\qquad \zeta(\mathbf u)=\frac{k_d}{2\sqrt{k_p\,m(\mathbf u)}},\qquad k_p=\bar m\,\omega_n^2,\quad k_d=2\bar m\,\omega_n . $$

只有 $m(\mathbf u)=\bar m$ 的方向才得到设计的 $\omega_n$ 和 $\zeta=1$：比 $\bar m$ 轻的方向更快、阻尼更小，比 $\bar m$ 重的方向更慢、阻尼更大。加速度 $\Lambda^{-1}\mathbf F$ 又一般不沿 $\mathbf F$，最大偏角是 $\arctan\frac{\kappa-1}{2\sqrt\kappa}$（$\kappa$ 是 $\Lambda$ 的条件数），所以路径弯曲。这就是 FIG 1 的紫色箭头和 FIG 2 的紫色路径。

Only directions with $m(\mathbf u)=\bar m$ get the designed $\omega_n$ and $\zeta=1$: lighter directions are faster and less damped, heavier ones slower and more damped. And the acceleration $\Lambda^{-1}\mathbf F$ generally does not point along $\mathbf F$; the largest angle between them is $\arctan\frac{\kappa-1}{2\sqrt\kappa}$ ($\kappa$ the condition number of $\Lambda$), so the path bends. These are FIG 1's purple arrow and FIG 2's purple path.

冗余：动力学一致性的推导Redundancy: deriving dynamic consistency

设 $n>m$，力矩 $\boldsymbol\tau=J^\top\mathbf F+N^\top\boldsymbol\tau_0$。手爪加速度是 $\ddot{\mathbf x}=JM^{-1}(\boldsymbol\tau-\mathbf b)+\dot J\dot{\mathbf q}$，所以 $\boldsymbol\tau_0$ 对它的贡献是 $JM^{-1}N^\top\boldsymbol\tau_0$。要求它对任何 $\boldsymbol\tau_0$ 都为零：

Let $n>m$ and $\boldsymbol\tau=J^\top\mathbf F+N^\top\boldsymbol\tau_0$. The tip acceleration is $\ddot{\mathbf x}=JM^{-1}(\boldsymbol\tau-\mathbf b)+\dot J\dot{\mathbf q}$, so $\boldsymbol\tau_0$ contributes $JM^{-1}N^\top\boldsymbol\tau_0$. We require this to vanish for every $\boldsymbol\tau_0$:

$$ JM^{-1}N^{\top}=0 . $$

设 $N^\top=I-J^\top G^\top$，$G$ 是某个满足 $JG=I$ 的广义逆。条件变成 $JM^{-1}=JM^{-1}J^\top G^\top=\Lambda^{-1}G^\top$，即 $G^\top=\Lambda JM^{-1}$，$G=M^{-1}J^\top\Lambda=\bar J$。所以动力学一致的广义逆是唯一的（Khatib 1987）。伪逆对应 $G=J^+=J^\top(JJ^\top)^{-1}$，代入得 $JM^{-1}(I-J^+J)$，一般不为零：在 FIG 4 里它让手爪误差升到厘米量级。另外注意 $N^\top$ 是投影：$(N^\top)^2=N^\top$，它把 $\boldsymbol\tau_0$ 中"会推手爪"的部分剔除，剩下的部分只改变自运动。

Write $N^\top=I-J^\top G^\top$ with some generalized inverse $G$, $JG=I$. The condition becomes $JM^{-1}=JM^{-1}J^\top G^\top=\Lambda^{-1}G^\top$, i.e. $G^\top=\Lambda JM^{-1}$ and $G=M^{-1}J^\top\Lambda=\bar J$. So the dynamically consistent generalized inverse is unique (Khatib, 1987). The pseudo-inverse corresponds to $G=J^+=J^\top(JJ^\top)^{-1}$, which gives $JM^{-1}(I-J^+J)$, generally non-zero: in FIG 4 it drives the tip error to centimetres. Note also that $N^\top$ is a projector, $(N^\top)^2=N^\top$: it strips from $\boldsymbol\tau_0$ the part that would push the tip, and what remains only changes the self-motion.

第二任务本身也可以写成"操作空间"的形式：把手肘位置之类的第二任务坐标 $\mathbf x_2=f_2(\mathbf q)$ 投影到零空间里，再用投影后的等效质量 $\Lambda_{2|1}=(J_2M^{-1}N^\top J_2^\top)^{-1}$ 做解耦（Sentis、Khatib 2005 的层级框架）。更一般的多任务、带不等式约束的版本是 13.7 节的二次规划。

The second task can itself be written in operational-space form: project a second task coordinate $\mathbf x_2=f_2(\mathbf q)$, such as the elbow position, into the null space and decouple it with the projected apparent mass $\Lambda_{2|1}=(J_2M^{-1}N^\top J_2^\top)^{-1}$ (the hierarchical framework of Sentis and Khatib, 2005). The more general version, with many tasks and inequality constraints, is the quadratic program of Lesson 13.7.

奇异位形与封顶的 $\Lambda$Singularities and a capped $\Lambda$

设 $J$ 的最小奇异值是 $\sigma$，相应的手爪方向是 $\mathbf u$。$\Lambda^{-1}=JM^{-1}J^\top$ 在 $\mathbf u$ 方向上大致正比于 $\sigma^2$，于是 $m(\mathbf u)\propto1/\sigma^2$，而 $J^\top$ 在这个方向上只有 $\sigma$ 的放大倍数，所以让手爪沿 $\mathbf u$ 获得加速度 $a$ 所需的力矩

Let $\sigma$ be the smallest singular value of $J$ and $\mathbf u$ the corresponding tip direction. Along $\mathbf u$, $\Lambda^{-1}=JM^{-1}J^\top$ is roughly proportional to $\sigma^2$, so $m(\mathbf u)\propto1/\sigma^2$, while $J^\top$ only has a gain of $\sigma$ in that direction. The torque needed to give the tip an acceleration $a$ along $\mathbf u$ is therefore

$$ \|\boldsymbol\tau\|\;\sim\;\sigma\cdot\frac{1}{\sigma^{2}}\cdot a=\frac{a}{\sigma}\;\xrightarrow{\ \sigma\to0\ }\;\infty,\qquad \hl{cap}{\Lambda_\varepsilon=\big(\Lambda^{-1}+\varepsilon I\big)^{-1}},\qquad m_\varepsilon(\mathbf u)\le\frac1\varepsilon . $$

FIG 5 里目标越过可达边界时，$\mathbf e$ 不会归零，控制器一直要求沿伸直方向加速，于是力矩发散。补救办法有三类：(1) 封顶：用 $\Lambda_\varepsilon$ 代替 $\Lambda$，把等效质量限制在 $m_{\max}=1/\varepsilon$ 以下，这和 3.4 节的阻尼最小二乘是同一个思路，代价是在被封顶的方向上不再精确得到 $\mathbf a$；(2) 降维：对 $\Lambda^{-1}$ 做特征分解，奇异值低于阈值的方向从任务里拿掉，改用雅可比转置或关节空间控制；(3) 躲开：用规划、关节限位和零空间任务让手臂远离奇异位形。实际系统通常三者并用。

In FIG 5, when the target passes the edge of the workspace, $\mathbf e$ cannot vanish, the controller keeps demanding acceleration along the stretched direction, and the torque diverges. There are three families of remedies: (1) cap: use $\Lambda_\varepsilon$ instead of $\Lambda$, limiting the apparent mass to $m_{\max}=1/\varepsilon$, the same idea as the damped least squares of Lesson 3.4, at the cost of no longer getting $\mathbf a$ exactly in the capped direction; (2) drop dimensions: eigen-decompose $\Lambda^{-1}$, remove the directions whose singular values fall below a threshold from the task, and control them with Jacobian transpose or in joint space instead; (3) stay away: use planning, joint limits and null-space tasks to keep the arm clear of singular postures. Real systems usually combine all three.

六维任务与姿态Six-dimensional tasks and orientation

真实手臂的任务通常是手爪的完整位姿，$J$ 是 7.4 节的 $6\times n$ 雅可比。姿态误差不能用欧拉角相减，常用 5.6 节的对数映射，例如 $\mathbf e_\omega=\Log(R_dR^\top)$（以基座系表示的转角向量），并且要与 $J$ 的角速度行用同一个坐标系。6 维的 $\Lambda$ 把 kg 和 kg·m² 混在一起（8.3 节误区 5），所以转动和平移的增益要分开设计；把姿态误差放大一个特征长度再与位置误差比较，也是常见做法。

On a real arm the task is usually the full gripper pose and $J$ is the $6\times n$ Jacobian of Lesson 7.4. Orientation errors cannot be formed by subtracting Euler angles; the usual choice is the logarithm map of Lesson 5.6, e.g. $\mathbf e_\omega=\Log(R_dR^\top)$ (a rotation vector expressed in the base frame), and it must use the same frame as the angular rows of $J$. A 6-D $\Lambda$ mixes kg with kg·m² (Pitfall 5 of Lesson 8.3), so rotational and translational gains are designed separately; scaling the orientation error by a characteristic length before comparing it with the position error is another common practice.

6机器人中的它：奇异位形与真实的力矩控制臂In robotics: singularities and real torque-controlled arms

真实任务里最常见的"事故"之一是让手爪去够一个稍微够不着的点：视觉估计的位置有误差，或者人推着手臂往外拉。下面的目标沿 $y=0.3$ m 的水平线从 $x=0.8$ m 出发，滑到 $x_{\max}$ 再回来，周期 4 s；在这条线上，手爪最远只能到 $x=\sqrt{1.8^2-0.3^2}\approx1.775$ m。控制器是带前馈的操作空间控制（$\omega_n=8$ rad/s），可以选择给等效质量封顶。

One of the most common "accidents" in real tasks is sending the gripper to a point just out of reach: the vision estimate is off, or a person pulls the arm outward. Below, the target slides along the horizontal line $y=0.3$ m from $x=0.8$ m out to $x_{\max}$ and back with a period of 4 s; on that line the gripper can reach at most $x=\sqrt{1.8^2-0.3^2}\approx1.775$ m. The controller is operational-space control with feedforward ($\omega_n=8$ rad/s), with an optional cap on the apparent mass.

FIG 5够不着的目标：奇异位形附近的力矩A target out of reach: torque near the singularity

俯视 · 目标沿虚线往返TOP VIEW · target runs along the dashed line

max |τᵢ| (N·m)，对数 vs 时间 (s)max |τᵢ| (N·m), log vs. TIME (s)

左：无色虚线是目标的往返路径，虚线圆是可达边界（半径 1.8 m），蓝点是此刻的目标，琥珀色细线是手爪走过的路。右：琥珀色是当前设置（封顶或不封顶）的最大关节力矩，紫色虚线是同一条路径上不封顶的结果，红色虚线标出 30 N·m。力矩超过 $10^4$ N·m 时仿真停止，标为"发散"：真实的驱动器早已饱和或报警停机。"可达段 RMS"是目标离基座不超过 1.7 m 的那些时刻的手爪误差均方根。仿真为 RK4、1 ms，没有力矩上限。 Left: the plain dashed line is the target's back-and-forth path, the dashed circle is the reach limit (radius 1.8 m), the blue dot is the current target, and the thin amber line is the path the gripper has taken. Right: amber is the largest joint torque for the current setting (capped or not), the dashed purple line the uncapped result on the same path, and the dashed red line marks 30 N·m. When the torque exceeds $10^4$ N·m the simulation stops and is marked "diverged": a real drive would have saturated or faulted long before. "In-reach RMS" is the root-mean-square tip error over the instants when the target is within 1.7 m of the base. Simulation: RK4 at 1 ms, no torque limits.

试一试Try this

不封顶，把路径终点 $x_{\max}$ 推到 1.8 m 以上（刚刚越过可达边界）：力矩在手臂伸直的那一刻发散。With no cap, push the path end $x_{\max}$ beyond 1.8 m (just past the reach limit): the torque diverges at the moment the arm straightens.
$x_{\max}\ge1.85$ m 时，选一个封顶值，让最大力矩低于 30 N·m，同时可达段的 RMS 误差小于 5 mm。封顶太低、太高各会怎样？With $x_{\max}\ge1.85$ m, choose a cap that keeps the peak torque below 30 N·m while the in-reach RMS error stays below 5 mm. What goes wrong when the cap is too low, or too high?

真实系统里In real systems

力矩控制的手臂：Franka 的 libfranka 以 1 kHz 接受关节力矩指令，franka::Model 给出质量矩阵、科氏力、重力和雅可比，KUKA LBR iiwa 每个关节都有力矩传感器并提供笛卡尔阻抗模式，DLR 的轻型臂（iiwa 的前身）是这类控制的经典平台（Albu-Schäffer、Ott 等）。值得注意的是，franka_ros 自带的笛卡尔阻抗示例控制器用的是 $\boldsymbol\tau=J^\top(-K\tilde{\mathbf x}-D\dot{\mathbf x})+$科氏补偿，零空间用 $J^\top$ 的伪逆投影，也就是本节的"雅可比转置加科氏补偿"，而不是完整的操作空间控制。这和 FIG 3 的结论一致：接触任务里，可预测的刚度往往比笔直的路径更重要。学习里的动作空间：robosuite 给机械臂的默认控制器（例如 Panda 的默认配置）是 OSC_POSE，它让策略输出手爪的位姿增量，由操作空间控制换成力矩；Martín-Martín 等（2019）比较了多种动作空间，发现在接触丰富的任务上，末端空间的可变阻抗控制让强化学习学得更好。人形与全身控制：Khatib 组把操作空间控制推广到带优先级的全身任务（Sentis、Khatib 2005），今天的全身控制器多半把同样的思想写成二次规划（13.7 节）。实践上的经验：Nakanishi 等（2008）在统一的记号下整理了速度、加速度和力三个层面的多种操作空间控制律，并在一条 7 自由度的拟人手臂上做了比较：在模型永远不完美的真机上，一种简化的、基于加速度的做法在任务精度、调参难度和鲁棒性上反而更有优势，而准确、物理上自洽的动力学参数估计是这些实验成功的关键；负载和末端工具的参数必须辨识（8.3 节 FIG 5），Franka 和 iiwa 都要求在软件里设置负载质量与质心。

Torque-controlled arms: Franka's libfranka accepts joint torque commands at 1 kHz and its franka::Model provides the mass matrix, Coriolis terms, gravity and Jacobians; the KUKA LBR iiwa has a torque sensor in every joint and offers a Cartesian impedance mode; the DLR lightweight arm (the iiwa's ancestor) is the classic platform for this kind of control (Albu-Schäffer, Ott and colleagues). Notably, the Cartesian impedance example controller shipped with franka_ros uses $\boldsymbol\tau=J^\top(-K\tilde{\mathbf x}-D\dot{\mathbf x})$ plus Coriolis compensation and a null-space projector built from the pseudo-inverse of $J^\top$: that is this lesson's "Jacobian transpose with Coriolis compensation", not full operational-space control. That fits FIG 3's lesson: in contact tasks a predictable stiffness often matters more than a straight path. Action spaces for learning: robosuite's default arm controller (in the default configuration of the Panda, for example) is OSC_POSE, which lets a policy output gripper pose increments that operational-space control turns into torques; Martín-Martín et al. (2019) compared action spaces and found that variable impedance control in end-effector space helps reinforcement learning on contact-rich tasks. Humanoids and whole-body control: Khatib's group extended operational-space control to prioritized whole-body tasks (Sentis and Khatib, 2005), and today's whole-body controllers mostly write the same ideas as quadratic programs (Lesson 13.7). Practical experience: Nakanishi et al. (2008) put many operational-space control laws, at the velocity, acceleration and force levels, into one notation and compared them on a seven-degree-of-freedom anthropomorphic arm: on hardware, where the model is never perfect, a simplified acceleration-based variant came out ahead in task performance, ease of tuning and robustness, and accurate, physically consistent estimates of the dynamic parameters were crucial to the experiments; payload and tool parameters have to be identified (FIG 5 of Lesson 8.3), and both Franka and iiwa require the payload mass and centre of mass to be set in software.

7常见误区Common pitfalls

误区 1：用速度层面的伪逆投影过滤力矩Pitfall 1: filtering torques with the velocity-level pseudo-inverse projector

$I-J^+J$ 在速度层面是对的（$J(I-J^+J)\dot{\mathbf z}=0$），但力矩要经过 $M^{-1}$ 才变成加速度，$JM^{-1}(I-J^+J)\ne0$。FIG 4 里同样的姿态任务，用伪逆投影时手爪误差达到厘米量级，用 $N^\top=I-J^\top\bar J^\top$ 时只有约 0.01 mm（仿真的离散误差）。如果主任务刚度很高，泄漏表现为看起来"莫名其妙"的小误差和抖动；如果刚度低，就是明显的偏移。直觉在哪里失灵？"零空间"这个词让人以为只有一个，其实"对谁为零"取决于度量：速度用欧氏度量，力矩要用 $M$ 加权。

$I-J^+J$ is right at the velocity level ($J(I-J^+J)\dot{\mathbf z}=0$), but torques pass through $M^{-1}$ before they become accelerations, and $JM^{-1}(I-J^+J)\ne0$. In FIG 4, the same posture task drives the tip error to centimetres with the pseudo-inverse projector and to only about 0.01 mm (the simulation's discretization error) with $N^\top=I-J^\top\bar J^\top$. With a stiff main task the leak shows up as small, "unexplained" errors and jitter; with a soft one it is an obvious offset. Where does intuition fail? The phrase "the null space" suggests there is only one, but "null for what" depends on the metric: velocities use the Euclidean one, torques must be weighted by $M$.

误区 2：直接对 $JM^{-1}J^\top$ 求逆Pitfall 2: inverting $JM^{-1}J^\top$ blindly

在奇异位形附近 $\Lambda$ 沿丢失的方向趋于无穷，力矩按 $1/\sigma$ 增长。FIG 5 里目标只比可达边界远 2.5 cm（$x_{\max}=1.8$ m），不封顶的控制器在 1.8 s 时就发散。目标并不需要"明显"够不着：视觉误差、人拉手臂、轨迹规划的一点超调都可能把手臂推到伸直。代码里要么封顶 $\Lambda$，要么检查 $\sigma_{\min}(J)$ 并降维，并且始终保留力矩限幅。手臂完全折叠（$\theta_2=\pi$）是另一个奇异位形，FIG 2 因此禁止经过基座附近的直线。

Near a singular posture $\Lambda$ tends to infinity along the lost direction and the torque grows like $1/\sigma$. In FIG 5 the target is only 2.5 cm beyond the reach limit ($x_{\max}=1.8$ m) and the uncapped controller diverges at 1.8 s. The target does not need to be "obviously" out of reach: a vision error, a person pulling the arm, or a slight overshoot of a planned trajectory can all stretch the arm straight. In code, either cap $\Lambda$ or check $\sigma_{\min}(J)$ and drop dimensions, and always keep torque limits. The fully folded arm ($\theta_2=\pi$) is another singular posture, which is why FIG 2 forbids straight lines passing near the base.

误区 3：以为操作空间控制让手爪"各向同性地柔顺"Pitfall 3: believing operational-space control makes the gripper "isotropically compliant"

$\ddot{\mathbf e}+K_d\dot{\mathbf e}+K_p\mathbf e=0$ 描述的是没有外力时的运动。有外力时，$\ddot{\mathbf e}+K_d\dot{\mathbf e}+K_p\mathbf e=-\Lambda^{-1}\mathbf F_{\text{ext}}$，环境感受到的刚度是 $\Lambda K_p$，随方向和姿态变化（FIG 3）。要在接触时得到指定的刚度，应该按 13.5 节的阻抗控制，直接以 N/m 设计 $K$；要同时改变表观惯量，则需要力传感器。

$\ddot{\mathbf e}+K_d\dot{\mathbf e}+K_p\mathbf e=0$ describes motion without external forces. With one, $\ddot{\mathbf e}+K_d\dot{\mathbf e}+K_p\mathbf e=-\Lambda^{-1}\mathbf F_{\text{ext}}$, and the stiffness felt by the environment is $\Lambda K_p$, which changes with direction and posture (FIG 3). To get a specified stiffness in contact, design $K$ directly in N/m as in the impedance control of Lesson 13.5; changing the apparent inertia as well requires a force sensor.

误区 4：漏掉 $\boldsymbol\mu$，或者漏掉 $\dot J\dot{\mathbf q}$Pitfall 4: leaving out $\boldsymbol\mu$, or leaving out $\dot J\dot{\mathbf q}$

慢速运动时 $\boldsymbol\mu$ 很小，省掉它看起来没问题；但 $\boldsymbol\mu$ 随速度平方增长，快速运动时漂移 $\mathbf d$ 可以和想要的加速度一样大（FIG 1 第二个任务）。另一个常见的错误是只补偿了关节空间的科氏力 $C\dot{\mathbf q}$，却忘了 $\dot J\dot{\mathbf q}$：即使关节匀速转动（$\ddot{\mathbf q}=0$），手爪也在走圆弧，有向心加速度 $\dot J\dot{\mathbf q}$，而这一项并不包含在 $C\dot{\mathbf q}$ 里。一个诱人的捷径是 $\boldsymbol\tau=J^\top\Lambda\mathbf a+\mathbf b$：代入可得 $\ddot{\mathbf x}=\mathbf a+\dot J\dot{\mathbf q}$，它漏掉的正是 $-J^\top\Lambda\dot J\dot{\mathbf q}$ 这一项。

At low speed $\boldsymbol\mu$ is small and dropping it seems harmless; but $\boldsymbol\mu$ grows with the square of the speed, and in fast motion the drift $\mathbf d$ can be as large as the wanted acceleration (FIG 1's second task). Another common mistake is compensating only the joint-space Coriolis terms $C\dot{\mathbf q}$ and forgetting $\dot J\dot{\mathbf q}$: even with the joints turning at constant rates ($\ddot{\mathbf q}=0$) the gripper travels on an arc and has the centripetal acceleration $\dot J\dot{\mathbf q}$, which is not part of $C\dot{\mathbf q}$. A tempting shortcut is $\boldsymbol\tau=J^\top\Lambda\mathbf a+\mathbf b$: substituting gives $\ddot{\mathbf x}=\mathbf a+\dot J\dot{\mathbf q}$, so what it leaves out is precisely the term $-J^\top\Lambda\dot J\dot{\mathbf q}$.

误区 5：以为操作空间控制总比雅可比转置好Pitfall 5: assuming operational space always beats Jacobian transpose

操作空间控制把模型当成真理：FIG 2 里一个 1 kg 的未建模负载就能让它冲过目标十几厘米，而雅可比转置几乎不受影响，因为它的稳定性证明根本不用 $M$。真实手臂上还有关节摩擦、电机转子惯量、关节弹性、通信延迟，这些都会让 $\hat\Lambda$ 偏离 $\Lambda$。选择取决于任务：自由空间里要走得准、走得直，用操作空间控制并认真辨识模型；接触为主、模型不可靠时，雅可比转置加阻抗设计往往更稳妥。

Operational-space control treats the model as the truth: in FIG 2 a 1 kg unmodelled payload makes it overshoot the target by more than ten centimetres, while Jacobian transpose is hardly affected, because its stability proof never uses $M$. Real arms add joint friction, motor rotor inertia, joint elasticity and communication delay, all of which pull $\hat\Lambda$ away from $\Lambda$. The choice depends on the task: for accurate, straight motion in free space, use operational space and identify the model carefully; when contact dominates and the model is unreliable, Jacobian transpose with an impedance design is often the safer choice.

8小练习Check yourself

1. 模型准确时，操作空间控制 $\mathbf F=\Lambda(K_p\mathbf e-K_d\dot{\mathbf x})+\boldsymbol\mu$（$K_p=\omega_n^2I$，$K_d=2\omega_nI$，固定目标）让手爪从静止出发走直线。根本原因是什么？1. With an exact model, operational-space control $\mathbf F=\Lambda(K_p\mathbf e-K_d\dot{\mathbf x})+\boldsymbol\mu$ ($K_p=\omega_n^2I$, $K_d=2\omega_nI$, fixed target) moves the gripper in a straight line from rest. What is the underlying reason?

$\Lambda$ 沿路径在变（A 错），控制器每一步都用当前的 $\Lambda(\mathbf q)$ 和 $\boldsymbol\mu$ 把它抵消掉，剩下 $\ddot{\mathbf x}=\mathbf a$。各向同性、解耦的线性系统从静止出发，解是 $\mathbf e(t)=(1+\omega_nt)e^{-\omega_nt}\mathbf e(0)$，方向不变。$J^\top$ 本身并不保持方向（B 错，这正是雅可比转置控制路径弯曲的原因），重力只是另一个被 $\hat{\mathbf p}$ 抵消的项（D 错）。$\Lambda$ does change along the path (A is wrong); the controller cancels it at every step with the current $\Lambda(\mathbf q)$ and $\boldsymbol\mu$, leaving $\ddot{\mathbf x}=\mathbf a$. An isotropic, decoupled linear system started from rest has the solution $\mathbf e(t)=(1+\omega_nt)e^{-\omega_nt}\mathbf e(0)$, whose direction never changes. $J^\top$ itself does not preserve direction (B is wrong; that is exactly why Jacobian-transpose paths bend), and gravity is just another term cancelled by $\hat{\mathbf p}$ (D is wrong).

2. 关于雅可比转置控制 $\boldsymbol\tau=J^\top(K_p\mathbf e-K_d\dot{\mathbf x})+\hat{\mathbf g}$，下列哪个说法正确？2. Which statement about Jacobian-transpose control $\boldsymbol\tau=J^\top(K_p\mathbf e-K_d\dot{\mathbf x})+\hat{\mathbf g}$ is correct?

$\dot V=-\dot{\mathbf x}^\top K_d\dot{\mathbf x}\le0$ 的推导只用到斜对称性和 $\dot{\mathbf e}=-J\dot{\mathbf q}$，不用 $M$ 的数值，所以 A 错。在 $\Lambda$ 的特征方向上，自然频率是 $\sqrt{k_p/m(\mathbf u)}$，随方向变化，所以 C 错。在奇异位形处 $J^\top$ 有零空间，$J^\top K_p\mathbf e=0$ 不再推出 $\mathbf e=0$，所以 D 错。The derivation of $\dot V=-\dot{\mathbf x}^\top K_d\dot{\mathbf x}\le0$ uses only skew-symmetry and $\dot{\mathbf e}=-J\dot{\mathbf q}$, never the values of $M$, so A is wrong. Along an eigen-direction of $\Lambda$ the natural frequency is $\sqrt{k_p/m(\mathbf u)}$, which varies with direction, so C is wrong. At a singular posture $J^\top$ has a null space and $J^\top K_p\mathbf e=0$ no longer implies $\mathbf e=0$, so D is wrong.

3. 一条 7 关节手臂用操作空间控制管理手爪位姿，同时想用多余的自由度让手肘远离障碍，手肘的力矩是 $\boldsymbol\tau_0$。怎样加入 $\boldsymbol\tau_0$ 才不会干扰手爪？3. A 7-joint arm controls its gripper pose with operational-space control and wants to use the spare freedom to keep the elbow away from an obstacle, with elbow torque $\boldsymbol\tau_0$. How should $\boldsymbol\tau_0$ be added so that it does not disturb the gripper?

"不干扰手爪"的意思是 $\boldsymbol\tau_0$ 不产生手爪加速度：$JM^{-1}N^\top=0$。只有 $N^\top=I-J^\top\bar J^\top$ 对任何 $\boldsymbol\tau_0$ 都满足它（选项 D）。伪逆投影（选项 B）在速度层面正确，但 $JM^{-1}(I-J^+J)\ne0$，FIG 4 里它把手爪误差推到厘米量级。直接相加（选项 A）误差更大，缩小（选项 C）只是减小泄漏，并没有消除。"Not disturbing the gripper" means $\boldsymbol\tau_0$ creates no tip acceleration: $JM^{-1}N^\top=0$. Only $N^\top=I-J^\top\bar J^\top$ satisfies it for every $\boldsymbol\tau_0$ (option D). The pseudo-inverse projector (option B) is right at the velocity level, but $JM^{-1}(I-J^+J)\ne0$, and in FIG 4 it pushes the tip error to centimetres. Adding directly (option A) is even worse, and scaling down (option C) only shrinks the leak without removing it.

4. 手臂接近伸直时，$J$ 的最小奇异值 $\sigma$ 减小到原来的一半。要让手爪沿这个"难动"的方向得到同样的加速度，操作空间控制需要的关节力矩大约怎样变化？4. As the arm approaches the stretched posture, the smallest singular value $\sigma$ of $J$ halves. To give the gripper the same acceleration along that "hard" direction, roughly how does the joint torque required by operational-space control change?

需要的手爪力是 $m(\mathbf u)a$，$m(\mathbf u)\propto1/\sigma^2$ 变成 4 倍（B 只算了这一半）；这个力经 $J^\top$ 换成力矩时又乘以 $\sigma$（缩小一半），所以 $\|\boldsymbol\tau\|\propto a/\sigma$ 大约变成 2 倍。$M$ 确实没变（C 错），但问题出在 $J$。$\sigma\to0$ 时力矩发散，这就是 FIG 5 里不封顶时发生的事。The required tip force is $m(\mathbf u)a$, and $m(\mathbf u)\propto1/\sigma^2$ quadruples (B stops at this half); turning that force into torque through $J^\top$ multiplies by $\sigma$ (halving it), so $\|\boldsymbol\tau\|\propto a/\sigma$ roughly doubles. $M$ has indeed not changed (C is wrong), but the problem lies in $J$. As $\sigma\to0$ the torque diverges, which is what happens in FIG 5 without a cap.

5. 对一条非冗余、非奇异的手臂（$J$ 是可逆方阵），操作空间控制 $\boldsymbol\tau=J^\top(\Lambda\mathbf a+\boldsymbol\mu+\mathbf p)$ 与下面哪个控制律完全相同？5. For a non-redundant arm away from singularities ($J$ square and invertible), operational-space control $\boldsymbol\tau=J^\top(\Lambda\mathbf a+\boldsymbol\mu+\mathbf p)$ is identical to which law?

$J$ 可逆时 $J^\top\Lambda J=M$，展开得 $\boldsymbol\tau=MJ^{-1}(\mathbf a-\dot J\dot{\mathbf q})+\mathbf b$，即分解加速度控制（Luh、Walker、Paul 1980）。两者的区别只在冗余、多任务和奇异位形处理上显现。When $J$ is invertible, $J^\top\Lambda J=M$, and expanding gives $\boldsymbol\tau=MJ^{-1}(\mathbf a-\dot J\dot{\mathbf q})+\mathbf b$: resolved-acceleration control (Luh, Walker and Paul, 1980). The two differ only once redundancy, multiple tasks or singularity handling come in.

9延伸阅读Further reading

O. Khatib, A Unified Approach for Motion and Force Control of Robot Manipulators: The Operational Space Formulation, IEEE J. Robotics and Automation 3(1), 1987 — 操作空间控制、$\Lambda$、$\boldsymbol\mu$、$\mathbf p$ 与动力学一致广义逆的原始论文The original paper on operational-space control, $\Lambda$, $\boldsymbol\mu$, $\mathbf p$ and the dynamically consistent generalized inverse
M. Takegaki & S. Arimoto, A New Feedback Method for Dynamic Control of Manipulators, ASME J. Dynamic Systems, Measurement, and Control 103(2), 1981 — 雅可比转置加重力补偿的能量稳定性证明The energy-based stability proof of Jacobian transpose with gravity compensation
J. Y. S. Luh, M. W. Walker & R. P. C. Paul, Resolved-Acceleration Control of Mechanical Manipulators, IEEE Trans. Automatic Control 25(3), 1980 — 在关节空间做的同一件事The same idea carried out in joint space
J. Nakanishi, R. Cory, M. Mistry, J. Peters & S. Schaal, Operational Space Control: A Theoretical and Empirical Comparison, Int. J. Robotics Research 27(6), 2008 — 多种操作空间控制律的推导与实验比较Derivations and experimental comparison of many operational-space control laws
L. Sentis & O. Khatib, Synthesis of Whole-Body Behaviors through Hierarchical Control of Behavioral Primitives, Int. J. Humanoid Robotics 2(4), 2005 — 带优先级的多任务操作空间控制Prioritized multi-task operational-space control
C. Ott, Cartesian Impedance Control of Redundant and Flexible-Joint Robots, Springer, 2008 — 力矩控制轻型臂上的笛卡尔阻抗与零空间控制Cartesian impedance and null-space control on torque-controlled lightweight arms
R. Martín-Martín, M. A. Lee, R. Gardner, S. Savarese, J. Bohg & A. Garg, Variable Impedance Control in End-Effector Space: An Action Space for Reinforcement Learning in Contact-Rich Tasks, IROS 2019 — 把末端空间控制当作学习的动作空间End-effector-space control as an action space for learning
K. Lynch & F. Park, Modern Robotics, Ch. 11 — 任务空间运动控制、力控制与混合控制Task-space motion control, force control and hybrid control
B. Siciliano & O. Khatib (eds.), Springer Handbook of Robotics, 2nd ed., Springer, 2016 — 运动控制与力控制的章节The chapters on motion control and force control

1为什么机器人需要它Why robots need this

2直觉：力不等于加速度Intuition: force is not acceleration

3动手玩：雅可比转置 vs 操作空间Play: Jacobian transpose vs. operational space

4第二、第三视角：环境感受到的刚度，与冗余臂的零空间Second and third views: the stiffness the world feels, and the null space of a redundant arm

环境感受到的刚度The stiffness the environment feels

冗余臂：动力学一致的零空间Redundant arms: the dynamically consistent null space

5再看数学The math

任务空间动力学（8.3 节的结论）Task-space dynamics (the result of Lesson 8.3)

操作空间控制律与解耦The operational-space control law and decoupling

雅可比转置控制：为什么不要模型也稳定Jacobian-transpose control: why it is stable without a model

冗余：动力学一致性的推导Redundancy: deriving dynamic consistency

奇异位形与封顶的 \(\Lambda\)Singularities and a capped \(\Lambda\)

六维任务与姿态Six-dimensional tasks and orientation

6机器人中的它：奇异位形与真实的力矩控制臂In robotics: singularities and real torque-controlled arms

7常见误区Common pitfalls

8小练习Check yourself

9延伸阅读Further reading