Feedback, stability and transient response

Follow a complete signal around the loop

Let r be the reference, y the measured output and e=r−y the indicated error after converting both signals to compatible units. A controller turns e into an actuator request. The actuator influences the plant, whose response reaches the sensor. Sensor scaling, filtering and delay are therefore part of the loop. A reference is a target, not a measurement proving that the physical output attained it.

For a plant whose output increases with positive input, a positive proportional gain applied to r−y produces a correcting action near the operating point. Reverse the sensor polarity and the action can reinforce the deviation. Even with correct static polarity, delayed corrections can arrive after the process has moved on. Stability must be assessed with the complete dynamic model.

Ask what happens to a small deviation

For an equilibrium, stability means sufficiently small initial deviations remain small; asymptotic stability additionally means they decay toward the equilibrium. A sustained bounded oscillation is not the same as decay. Stability alone places no useful bound on how long a response takes, how far it overshoots or how much actuator effort it consumes.

Transient measures need definitions. Rise time describes a chosen fraction of a transition; overshoot compares a peak with a declared final or target value; settling time specifies a band and remaining within it. A stable response may settle with a nonzero tracking error. Report that offset separately, especially when a faster response merely reaches the wrong final value sooner.

Check disturbances and measurement noise separately

Increasing loop gain can reduce certain low-frequency errors in a suitable model. It also changes actuator demand and the response to sensor noise. Unmodelled delays or resonances can limit the useful gain. A step response at one condition is evidence for that condition, not a universal robustness result. Track reference, measured output and applied actuator signal together to interpret what happened.

Two proportional loops that remain below their target

For 2dv/dt+v=u, use u=K(r−v), r=3 m/s and v(0)=0. Rearranging gives 2dv/dt+(1+K)v=Kr. K=1 gives equilibrium 1.5 m/s and time constant 1 s. K=3 gives 2.25 m/s and time constant 0.5 s. Both are asymptotically stable within this ideal continuous model.

At steady state their errors are 1.5 and 0.75 m/s. Higher gain improves this example’s speed and offset but does not eliminate the offset. No actuator limit, noise or delay has been included; the gain values are teaching inputs rather than tuning instructions.

Try the reasoning

1. Can a stable loop have a large steady error?

   Show the answer and reasoning

   Yes. The K=1 case decays toward 1.5 m/s although the target is 3 m/s. Decay and correct final value are distinct properties.

2. What happens to deviations if the K=3 control law is accidentally u=3(v−r)?

   Show the answer and reasoning

   The deviation equation is 2dδv/dt=2δv, so deviations grow as exp(t). The reversed static action is unstable in this model.

3. A plot stops shortly after crossing the target. Can it establish settling time?

   Show the answer and reasoning

   No. Settling requires subsequent behaviour to remain inside the stated band; a brief crossing may precede overshoot or oscillation.

What to carry forward

Assess decay, speed, offset and effort as separate requirements of the complete loop.