3.5.38 · D5Guidance, Navigation & Control (GNC)
Question bank — PID control — proportional, integral, derivative terms
True or false — justify
A larger always makes the closed loop respond faster and is therefore always better.
False. It does speed up the initial response, but it shrinks — never kills — steady-state error, and beyond a point it drives the loop into oscillation and instability (see Stability & the Routh–Hurwitz criterion).
A pure proportional controller can achieve exactly zero steady-state error if you make large enough.
False. At rest must equal the disturbance , so ; this only goes to zero as , which is physically impossible and unstable long before then.
The integral term is what allows the loop to settle with zero steady-state error against a constant disturbance.
True. In steady state must stop changing, and , so the only resting state is — the integrator keeps pushing until the error vanishes.
The derivative term reacts to how big the error is right now.
False. It reacts to how fast the error is changing (), not its magnitude — a huge but steady error produces zero derivative action.
Adding a derivative term always improves stability.
Mostly true in ideal maths, but false in practice. It adds damping, yet it amplifies high-frequency sensor noise, so an unfiltered/large can make the actuator chatter and effectively destabilise the real system (see Sensor noise & filtering).
If the error signal is identically zero for all time, the integral term is also zero.
False in general. The integral holds the accumulated history; if past error was nonzero, can be large and constant even while the present error is zero — that stored value is exactly what cancels a disturbance.
Doubling doubles how fast the accumulated offset is erased and carries no downside.
False. More integral action erases offset faster but increases overshoot and worsens windup when the actuator saturates; there is a stability/overshoot cost.
The discrete PID running on the flight computer computes the same control as the continuous law.
False, it approximates it. The integral becomes a rectangle sum and the derivative a backward finite difference; these converge to the calculus versions only as .
Spot the error
"At steady state with pure P, , so a bigger gain gives a bigger leftover error."
Wrong direction. The correct relation is : bigger shrinks the error. The student inverted the fraction.
"The integral term removes offset because integrating a constant error gives zero."
Wrong mechanism. Integrating a constant error gives a growing ramp, not zero. That ramp keeps raising until the error itself is driven to zero — the offset dies because the output can only stop changing when .
"Derivative action helps because tells us the current position error."
Confuses value with rate. is the slope of the error, i.e. its trend. It says "the error is falling fast, brake now," which is prediction — it says nothing about the present magnitude.
"To fix integrator windup, just increase so the actuator saturates less."
Treats a symptom, not the cause. Windup is the integrator accumulating while the actuator is clamped at its limit; the cure is anti-windup (stop/clamp integration during saturation), not a different .
"A noisy sensor is a problem for the integral term because summing noise blows up."
Wrong term. Zero-mean noise tends to average out in a sum, so the integral is relatively robust. It's the derivative that amplifies noise, because differencing multiplies fast wiggles (see Sensor noise & filtering).
"Since D 'sees the future,' a system with only a D term can hold a setpoint."
Impossible. Pure D produces zero output whenever the error is constant (including a constant nonzero error), so it cannot supply the steady push needed to hold a target — it only shapes transients.
Why questions
Why must a pure-P controller output drop to zero exactly when the error is zero?
Because is directly tied to the present error; if then , leaving nothing to oppose a disturbance — which is precisely why an offset must persist.
Why does the integrator "wind up" specifically during actuator saturation and not during normal operation?
When saturated, the actuator can't deliver the commanded output, so the error stays large and the integral keeps accumulating a value the plant never actually acts on — that stored surplus then overshoots badly once the loop catches up.
Why is differentiation, rather than integration, the operation that amplifies noise?
Differentiation responds to slope; a small fast wiggle has a tiny amplitude but a huge slope, so its derivative is large — high frequencies get multiplied up, whereas integration smooths them.
Why do we describe P as "present," I as "past," and D as "future"?
P uses the error's value now, I accumulates its whole history, and D uses its rate of change to extrapolate where it's heading — value, integral, derivative are exactly present/past/future information.
Why can raising turn a stable loop unstable?
A larger gain feeds a bigger correction back through the plant's own lags and delays; past a threshold the correction arrives out of phase and reinforces the error instead of cancelling it — the boundary is what Stability & the Routh–Hurwitz criterion pins down.
Why does the derivative term reduce overshoot on the approach to target?
As rushes toward , the error falls so , giving ; this subtracts from the command early, braking the approach before the system barrels past the setpoint.
Edge cases
What does a PID controller command at the instant the system first reaches the setpoint after approaching fast?
so P contributes nothing, but (still moving) and the integral may hold a nonzero value — so is generally not zero; D and I still act to prevent overshoot and cancel disturbance.
If the disturbance is zero, does the integral term ever do anything at all?
Only during transients. Once the error settles to zero and stays there, stops changing; with no disturbance the loop can rest with , so integral action was only needed to clean up the transient offset.
What happens to a pure-integral controller () responding to a step?
It responds sluggishly and tends to oscillate — with no present-error or damping term, the accumulator overshoots and unwinds slowly, giving poor, ringing performance despite eventual zero offset.
What is the effect of the derivative term when the error is constant but nonzero (a stuck offset)?
Zero — means , so D is blind to standing offsets; only I can eliminate them. This is why a constant leftover error is a symptom pointing to the integral, not the derivative.
At with a sudden step in setpoint, what does the ideal derivative term momentarily command?
An impulse — a step in has infinite slope for an instant, so an unfiltered derivative spikes enormously. This "derivative kick" is why real controllers differentiate the measurement or filter the derivative (see Sensor noise & filtering).
If is set to exactly zero, what class of disturbance can the loop still reject with zero steady-state error?
None that are constant — without integral action a constant disturbance always leaves ; only a disturbance that itself decays to zero can be fully rejected (see Steady-state error and system type).
Recall One-line summary of every trap
Each PID term has exactly one job and exactly one failure mode: P is fast but leaves offset, I kills offset but winds up, D damps but amplifies noise. Almost every trap above is a case of blaming the wrong term for a symptom, or forgetting that D reads rate while P reads value and I reads history.