Robot joint control and force control

Why it matters

A trajectory is only a wish list until each joint servo makes the arm follow it despite gravity, payload changes, friction and coupling between joints. And when the robot must touch things — polishing, insertion, assembly, grinding — controlling position alone either loses contact or crushes the part, so the force at the tool must be controlled too.

Key ideas

Joint drive. Each joint has a motor (usually a brushless servo), a gear reduction of ratio N (often 50–160 in industrial arms), an encoder for position and velocity feedback, and a servo amplifier with current (torque) control in the innermost loop. The cascade is: current loop (fastest) → velocity loop → position loop.

Independent joint control. The simplest and most common scheme treats each joint as a single-input single-output system and ignores coupling as a disturbance. This works well because a high gear ratio divides the link-side inertia variations seen by the motor by N².

PD and PID control. τ = K_p·e + K_d·ė (+ K_i·∫e dt), with e = θ_d − θ.

• For a joint of effective inertia J, the closed-loop PD dynamics are J·θ̈ + K_d·θ̇ + K_p·θ = K_p·θ_d, a second-order system with ω_n = √(K_p/J) and ζ = K_d / (2·√(K_p·J)). Robots are tuned near critical damping (ζ ≈ 1) so there is no overshoot that could hit a part.
• A constant load torque τ_L (gravity) gives a steady-state error e_ss = τ_L / K_p under PD. Fix it with an integral term, or better, add gravity compensation g(q) as feed-forward.

Model-based control. Computed-torque (inverse dynamics) control: τ = M(q)·(q̈_d + K_d·ė + K_p·e) + C(q, q̇)·q̇ + g(q). If the model is exact, every joint behaves as an independent linear double integrator with chosen gains. Feed-forward of the planned torque greatly reduces tracking error at high speed.

Force control. Needed when the tool is in contact with a stiff environment.

• Stiffness control — make the tool behave like a spring: F = K·Δx. Contact force is set by commanding a small "penetration" into the surface.
• Impedance control — make the tool behave like a mass-spring-damper: F = M_d·Δẍ + B_d·Δẋ + K_d·Δx. Uses a force/torque sensor or motor currents; robust and widely used (e.g. collaborative robots).
• Admittance control — the inverse: measure force, compute a motion correction, send it to a stiff position controller; suits industrial arms with stiff position loops.
• Hybrid position/force control (Raibert-Craig) — a selection matrix S chooses, per task direction, whether to control force (constrained directions, e.g. normal to a surface) or position (free directions, e.g. along the surface). You cannot control both force and position in the same direction.
• Joint torques for a desired tool force come from τ = Jᵀ·F.

Sensors. Wrist six-axis force/torque sensors (strain-gauge), joint torque sensors (cobots) or motor current estimates.

Formulas

τ = K_p·e + K_d·ė + K_i·∫e dt, e = θ_d − θ

• K_p — N·m/rad; K_d — N·m·s/rad; K_i — N·m/(rad·s).

ω_n = √(K_p / J), ζ = K_d / (2·√(K_p·J)); critical damping: K_d = 2·√(K_p·J)

• J — effective inertia (kg·m²).

e_ss = τ_L / K_p (PD with constant load torque τ_L)

J_eff (at motor) = J_m + J_L / N²; θ_m = N·θ_L, τ_L = N·τ_m (ideal gearbox)

• J_m — motor rotor inertia; J_L — link inertia about the joint; N — gear ratio.

Computed torque: τ = M(q)·(q̈_d + K_d·ė + K_p·e) + C(q, q̇)·q̇ + g(q)

Stiffness/impedance: F = K·Δx (+ B·Δẋ + M·Δẍ); τ = Jᵀ·F

Worked examples

Example 1 (standard). A joint has effective inertia J = 0.5 kg·m² and a PD controller with K_p = 200 N·m/rad. (a) Find ω_n and the K_d for critical damping. (b) A constant gravity torque of 10 N·m acts; find the steady-state position error.

1. ω_n = √(200 / 0.5) = √400 = 20 rad/s.
2. K_d = 2·√(K_p·J) = 2·√(200·0.5) = 2·10 = 20 N·m·s/rad.
3. e_ss = τ_L / K_p = 10 / 200 = 0.05 rad = 2.86°.

Answer: ω_n = 20 rad/s, K_d = 20 N·m·s/rad, e_ss = 0.05 rad (2.86°). Gravity feed-forward or an integral term removes this error.

Example 2 (GATE level). A motor with rotor inertia J_m = 2 × 10⁻⁴ kg·m² drives a link of inertia J_L = 0.5 kg·m² through a gearbox of ratio N = 50. Find (a) the inertia seen by the motor, (b) the motor torque needed to accelerate the link at 10 rad/s² (friction neglected).

1. J_eff = J_m + J_L / N² = 2 × 10⁻⁴ + 0.5 / 2500 = 2 × 10⁻⁴ + 2 × 10⁻⁴ = 4 × 10⁻⁴ kg·m².
2. Motor acceleration = N × 10 = 500 rad/s².
3. τ_m = J_eff × 500 = 4 × 10⁻⁴ × 500 = 0.20 N·m.
4. Check: rotor needs 2 × 10⁻⁴ × 500 = 0.10 N·m; link needs 0.5 × 10 = 5 N·m at the link, i.e. 5/50 = 0.10 N·m at the motor; total 0.20 N·m ✓. Here N = √(J_L/J_m) = 50, the inertia-matched ratio.

Answer: (a) 4 × 10⁻⁴ kg·m², (b) 0.20 N·m.

Example 3 (stiffness control). The tool is in contact with a surface and its controller stiffness normal to the surface is K = 5000 N/m. To press with 20 N, command a position Δx = F/K = 20/5000 = 4 mm below the surface.

Common mistakes

• Reflecting inertia with N instead of N² (or reflecting it the wrong way).
• Expecting PD alone to hold position exactly under gravity — it leaves e_ss = τ_L/K_p.
• Trying to control force and position in the same direction — hybrid control splits directions.
• Using τ = J⁻¹·F instead of τ = Jᵀ·F.
• Tuning for fast response with overshoot — an underdamped joint overshoots into the workpiece.
• Ignoring that high gains on a stiff contact cause chatter and instability.

For GATE ME

Questions draw on control-systems basics applied to a joint: second-order response (ω_n, ζ, critical damping gain), steady-state error under load, reflected inertia through a gearbox, and conceptual distinctions between position, force, impedance and hybrid control. Practise the J_eff = J_m + J_L/N² relation and ζ from PD gains.

Quick check

1. With PD control, what is the steady-state error for K_p = 500 N·m/rad and a 5 N·m load?
2. A link inertia of 1 kg·m² through a 100:1 gearbox appears as what inertia at the motor?
3. In hybrid control, which variable is controlled normal to a rigid surface?
4. Write the critical-damping condition for a PD joint.
5. What does impedance control regulate?

Answers: 1. 0.01 rad; 2. 1 × 10⁻⁴ kg·m²; 3. Force; 4. K_d = 2√(K_p·J); 5. The dynamic relation between tool force and motion (desired mass, damping, stiffness).