3.5.26 · D5Guidance, Navigation & Control (GNC)
Question bank — Control system fundamentals — plant, actuator, sensor, controller
The word loop gain () below means the total factor a signal picks up going once all the way around the loop; characteristic equation means , whose roots are the closed-loop poles. Both are built in the parent note — revisit it if either feels unfamiliar.
True or false — justify
A perfect (unity-gain, zero-lag) sensor removes from the stability analysis.
False. Only if exactly does the loop gain reduce to ; any real dynamics or scaling in stays inside and can move the poles — accuracy at DC does not equal at all frequencies. See Poles Zeros and Stability.
Adding an integrator always improves the system.
False. An integrator (pole at ) kills steady-state step error but adds of phase lag, eroding stability margin and inviting oscillation — it is a trade, not a free win.
In the Laplace domain, cascaded blocks add their transfer functions.
False. Cascaded LTI blocks multiply: . Addition happens at summing junctions (like ), not along a chain. See Transfer Functions and Laplace Domain.
A stable open-loop plant guarantees a stable closed loop.
False. Closing the loop moves the poles to the roots of ; high gain can push those roots into the right half-plane even when the bare plant was fine.
The reference is part of the plant's dynamics.
False. is an external command from Guidance; the plant is the fixed physics being commanded. They enter the loop at completely different points.
Raising proportional gain always makes the response faster and better.
False. It raises (faster) but lowers damping in the motor example, so you buy speed with overshoot and eventual instability. See PID Control.
Steady-state error to a step can be zero even with a purely proportional controller.
True — but only if the plant itself already has an integrator (a pole at ), making ; without one, .
The controller and actuator can be merged into one block with no loss of fidelity.
False. The controller outputs an abstract number ; the actuator has hardware limits (saturation, rate limit, lag) that the controller math never sees. Merging them hides windup. See Actuator Saturation and Anti-Windup.
Feedback makes a system robust to not knowing the plant exactly.
True. Because the loop measures the actual output and corrects the error, moderate modelling errors in get partly suppressed — that robustness is the whole point of closing the loop.
Spot the error
" because feedback subtracts."
The sign is wrong: subtraction at the summing junction becomes in the denominator after solving . Correct: .
"For unity feedback , so and ."
The unity refers to feedback, not to dropping the term. With , , not .
" for the forward path."
They added instead of multiplying: . Series blocks multiply in the -domain.
"The characteristic equation is the numerator of set to zero."
No — it is the denominator ; numerator roots are zeros (they shape response but don't decide stability).
"Steady-state error works for any input."
Only for a step. For a ramp or when the Final Value Theorem's stability condition fails, this formula is invalid.
"Increasing sensor gain is harmless because it just scales the reading."
A larger increases loop gain , shifting poles and possibly destabilizing — it is not a passive rescaling.
"An open-loop system corrects for wind disturbance if pre-planned well."
Open loop cannot react to an unmeasured disturbance; only feedback measures the resulting error and pushes back.
Why questions
Why does the loop-gain term appear as and not just ?
The "" is the direct path from error to output; the geometric series of the loop chasing itself sums to .
Why is the plant the one block you cannot redesign?
Its dynamics are set by physics (mass, inertia, aerodynamics); you can only choose what surrounds it — controller, actuator, sensor.
Why does integral action specifically kill steady-state step error?
Its pole at sends , so — the integrator keeps accumulating error until it is exactly zero.
Why must the loop stability be checked with , not just ?
The poles live at the roots of ; leaving out analyses a different loop than the one you actually built.
Why does higher gain reduce steady-state error but threaten stability at the same time?
Big shrinks (less error) but also drags the closed-loop poles toward and across the imaginary axis (less stability). See Second-order System Response.
Why do we work in the Laplace -domain at all?
Because convolution in time becomes multiplication in , turning a chain of dynamic blocks into simple algebra . See Transfer Functions and Laplace Domain.
Why is Navigation (state estimation) a separate concern from Control?
Control acts on the estimated state; if the estimate is noisy or biased the loop chases a wrong target — which is why filtering matters. See Kalman Filter and Navigation.
Edge cases
What happens to as loop gain ?
: the output tracks the reference divided by the sensor gain, essentially independent of the plant — the ideal high-gain limit.
What happens when (loop gain very small)?
: feedback does almost nothing and the system behaves open-loop, losing its disturbance rejection.
What does a pole exactly on the imaginary axis () mean physically?
Sustained, undamped oscillation — the boundary between stable and unstable; the marginal case you must never sit on in practice.
If the actuator saturates, is the linear closed-loop formula still valid?
No — saturation is nonlinear, so the transfer-function analysis breaks and integrator windup can occur. See Actuator Saturation and Anti-Windup.
What is the steady-state error if the plant already contains one pure integrator and the input is a step?
Zero, because makes — even without an integral controller term.
What happens to the error signal in a perfectly tracking loop at steady state?
It settles toward zero (for a step with sufficient system type); a non-zero residual signals a missing integrator or a finite loop gain.
Can a controller with zero gain () ever stabilize an unstable plant?
No — with the forward path is dead (), no corrective push reaches the plant, and its unstable dynamics run free.
What is the closed-loop behaviour if the sensor fails and reads a constant ?
The loop thinks the error is always , driving the actuator hard regardless of true output — a runaway, since feedback is effectively broken.
Recall One-line self-test before you leave
Denominator of the closed loop? ::: , whose roots are the poles that decide stability. The single trade behind "just crank the gain"? ::: Speed () versus damping ().