Do you need feedback if a voltage already gives the right speed?

A sensor alone does not close the loop

Open-loop control chooses its command without feeding the measured output back into that decision. Holding a voltage is one example; a preplanned time-varying command is another. A logger may measure speed and draw a beautiful trace while the actuator continues to follow exactly the same schedule.

Closed-loop control makes the measured output part of the command calculation. In a speed-feedback loop, the controller compares the sensor reading with the requested speed. A person watching a graph and manually correcting voltage also creates a feedback path, although that person's timing and actions differ from an automatic controller.

Calculate what 7 V actually buys

The lab motor obeys L di/dt=V−Ri−Keω and J dω/dt=Kt i−bω−τL. At equilibrium both derivatives are zero. Eliminating current gives V=(Ke+Rb/Kt)ω+(R/Kt)τL. With R=2 Ω, Ke=Kt=0.05 in the appropriate SI units and b=0.0005 N·m·s/rad, the result is V=0.07ω+40τL.

Without load, 7 V corresponds to 100 rad/s. Add 0.04 N·m and the same voltage corresponds to about 77.14 rad/s. Inertia determines how the transient evolves but disappears from this equilibrium equation. These are model-specific calculations, not specifications for a motor you might buy.

Compare one load event, two command rules

Open the linked experiment. Both runs start from rest, use the same nominal motor and 100 rad/s target, and receive a persistent 0.04 N·m load at 4 s. First pin the closed-loop result. Switch only the operating mode to Open loop at the preset 7 V, then pin that result. Inspect speed and Control effort at the same replay time.

In the held-voltage run, applied voltage stays at 7 V after the speed falls. In the feedback run, the error changes the command and integral action supplies the lasting correction. At equilibrium the loaded motor needs 8.6 V to maintain 100 rad/s. Startup behavior also differs, so use the post-4 s interval to answer the specific question about load recovery.

Now change the target in open loop. The line and displayed error move, but voltage stays fixed. That observation identifies which signals actually enter the command calculation. Next change manual voltage, then restore the original settings before comparing again.

Choose from repeatability, uncertainty and limits

Open-loop operation can meet a requirement when conditions are known and repeatable, or when no suitable output measurement exists. It avoids a feedback design and its sensor-noise path, but an unknown load or changed parameter is not automatically corrected. A command computed from a measured disturbance or known reference is feedforward; it can be combined with output feedback.

Feedback also has a cost. Noise can reach the actuator, delay can reduce stability margins, and aggressive gains can create oscillation or saturation. For this motor at 100 rad/s, a 0.15 N·m constant load requires 13 V. The ±12 V supply cannot meet that target even with integral action. Try that condition and compare requested with applied voltage rather than blaming every remaining error on poor tuning.

Why a closed-loop design uses an open-loop Bode plot

In linear analysis, the loop transfer L=CP describes what travels around the specified feedback path. With unity negative feedback and zero initial conditions, the reference-to-output transfer is T=L/(1+L). Bode and Nyquist examine L to reason about the denominator 1+L. They are analysis tools for a closed-loop design, not a switch that disconnects the motor's sensor.

The Lab therefore shows its loop-frequency workbench during feedback operation. In manual-voltage mode there is no active speed-feedback loop to assign those controller margins to. The motor still has a voltage-to-speed plant response; a future plant-analysis view can display that separately with an explicit input and output.