Control system fundamentals — plant, actuator, sensor, controller: Exercises

We reuse three tools relentlessly, so let us name why each exists before we lean on it:

Definition The three tools of this page

Block multiplication. In the Laplace ( $s$ ) domain, cascaded blocks multiply. Why this and not addition? Because a signal passing through block $A$ then block $B$ is scaled by $A$ then by $B$ — two scalings compose by multiplying. See Transfer Functions and Laplace Domain.
The closed-loop formula $T (s) = \frac{G}{1 + G H}$ . Why? It collapses the whole feedback loop into ONE input→output map so we can read behaviour directly.
The Final Value Theorem (FVT): $t \to \infty lim f (t) = s \to 0 lim s F (s)$ . Why this tool? It answers "where does the signal settle?" without inverting the Laplace transform — but it is only valid when the loop is stable (all closed-loop poles in the left half-plane). Every time we use FVT below we first check stability — skip that check and the theorem can hand you a wrong (or infinite) answer.

The feedback loop we keep referring to is drawn once here — every problem is a special case of this picture (red is the forward path $G = C A P$ , blue is the sensor feedback $H$ , the yellow circle is where the error $e = r - y_{m}$ is formed):

Figure — Control system fundamentals — plant, actuator, sensor, controller

Level 1 — Recognition

L1.1 — Name the organ

For a self-driving car holding a lane:

(a) the camera reading lane position,
(b) the software computing "steer 3° right",
(c) the steering motor turning the wheels,
(d) the car's mass-and-tyre dynamics.

Match each to plant / actuator / sensor / controller.

Recall Solution

(a) camera → sensor ( $H$ ): it measures the output and turns it into a signal.
(b) software → controller ( $C$ ): the math/brain producing the command number $u$ (the "control signal" — a plain number the software outputs).
(c) steering motor → actuator ( $A$ ): converts $u$ into real physical force.
(d) car dynamics → plant ( $P$ ): the fixed physics you command but cannot redesign.

L1.2 — Which equation is the error?

Given $r$ (reference), $y$ (true output), $y_{m}$ (measured output), pick the correct error definition and say why: $(A) e = y - r (B) e = r - y_{m} (C) e = r + y_{m}$

Recall Solution

(B) $e = r - y_{m}$ . Error is "where I want to be minus where I appear to be." We subtract the measured output $y_{m}$ (not the true $y$ ), because the controller can only see what the sensor reports. (A) has the sign flipped (would drive us away from target); (C) adds, which never reaches zero.

Level 2 — Application

L2.1 — Build the forward path and the closed loop

A thermal system has controller $C (s) = 4$ , actuator $A (s) = 2$ , plant $P (s) = \frac{1}{s + 1}$ , sensor $H (s) = 1$ . Find (a) the forward path $G (s)$ , (b) the closed-loop transfer function $T (s)$ .

Recall Solution

(a) Multiply cascaded blocks (that is why we multiply — the signal is scaled by each block in turn): $G (s) = C A P = 4 \cdot 2 \cdot \frac{1}{s + 1} = \frac{8}{s + 1} .$ (b) Apply $T = \frac{G}{1 + G H}$ with $H = 1$ : $T (s) = \frac{\frac{8}{s + 1}}{1 + \frac{8}{s + 1}} .$ Now clear the small fractions — why? to turn a "fraction over a fraction" into one clean ratio we can read poles off. Multiply top and bottom by $(s + 1)$ : $T (s) = \frac{8}{( s + 1 ) + 8} = \frac{8}{s + 9} .$ The single pole moved from $s = - 1$ (open loop) to $s = - 9$ (closed loop): feedback made it 9× faster.

L2.2 — DC gain of the closed loop

For the $T (s) = \frac{8}{s + 9}$ above, what is the steady-state output for a unit-step reference?

Recall Solution

First check stability: the only pole is $s = - 9$ , in the left half-plane, so FVT is legal. Steady state of a step response = DC gain $T (0)$ (set $s = 0$ ; why? FVT: $lim_{s \to 0} s \cdot T (s) \cdot \frac{1}{s} = T (0)$ ): $T (0) = \frac{8}{0 + 9} = \frac{8}{9} \approx 0.889.$ So the output settles at $0.889$ , not $1$ — a steady-state error of $1 - \frac{8}{9} = \frac{1}{9}$ because there is no integrator.

Level 3 — Analysis

L3.1 — Position servo: read off $ω_{n}$ and $ζ$

A servo has $P (s) = \frac{1}{s ( s + 4 )}$ , $A = 1$ , $H = 1$ , proportional controller $C = K$ . (a) Find $T (s)$ . (b) Express natural frequency $ω_{n}$ and damping ratio $ζ$ in terms of $K$ . (c) What $K$ gives critical damping ( $ζ = 1$ )?

Recall Solution

(a) $G = \frac{K}{s ( s + 4 )}$ , so $T = \frac{G}{1 + G} = \frac{K}{s ( s + 4 ) + K} = \frac{K}{s ^{2} + 4 s + K} .$ (We multiplied top and bottom by $s (s + 4)$ to clear the inner fraction, then expanded $s (s + 4) = s^{2} + 4 s$ .) (b) Compare with the standard second-order denominator $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ (see Second-order System Response). Matching term by term — the constant term gives $ω_{n}^{2}$ , the $s$ -term gives $2 ζ ω_{n}$ : $ω_{n}^{2} = K \Rightarrow ω_{n} = K, 2 ζ ω_{n} = 4 \Rightarrow ζ = \frac{4}{2 K} = \frac{2}{K} .$ (c) $ζ = 1$ : $\frac{2}{K} = 1 \Rightarrow K = 2 \Rightarrow K = 4.$ Below $K = 4$ the system is overdamped (sluggish, no overshoot); above $K = 4$ it is underdamped (overshoots). $K = 4$ is the fast-but-clean boundary.

L3.2 — Where do the poles sit?

For the same servo with $K = 8$ , find the closed-loop poles and classify the response.

Recall Solution

Characteristic equation $s^{2} + 4 s + 8 = 0$ . Solve (quadratic formula, why? poles are the roots of the denominator): $s = \frac{- 4 \pm 16 - 32}{2} = \frac{- 4 \pm - 16}{2} = - 2 \pm 2 j .$ Complex conjugate pair with negative real part ( $- 2$ ) → stable, underdamped, oscillates at damped frequency $ω_{d} = 2$ rad/s while decaying. Check: $ζ = 2/ 8 = 0.707$ — the classic "nicely damped" value. See Poles Zeros and Stability.

The figure below (left) plots these two poles as red crosses in the complex plane — both sit in the shaded stable (left-half) region, with equal-and-opposite imaginary parts. The right panel shows what that pole pair does in time: the output rises, overshoots the reference once, and rings down — exactly the "underdamped" signature.

Level 4 — Synthesis

L4.1 — Kill the steady-state error

Unity-feedback plant $P (s) = \frac{2}{s + 2}$ , $A = 1$ , $H = 1$ , unit-step reference. (a) With proportional control $C = K$ , find the steady-state error $e_{ss}$ as a function of $K$ . (b) Now use integral control $C (s) = \frac{K _{i}}{s}$ . Find $e_{ss}$ . Explain the difference.

Recall Solution

(a) First, derive the error transfer function $\frac{E}{R}$ — why? the FVT needs $E (s)$ , so we must build the map from $R$ to $E$ from the loop equations. Start from the two loop facts (unity feedback, $H = 1$ ): $E = R - Y, Y = \frac{G}{1 + G} R .$ Substitute the second into the first (eliminate $Y$ ): $E = R - \frac{G}{1 + G} R = R (\frac{1 + G - G}{1 + G}) = \frac{1}{1 + G} R \Rightarrow \frac{E}{R} = \frac{1}{1 + G} .$ Check stability before FVT: $G = \frac{2 K}{s + 2}$ gives closed-loop pole at $s = - (2 + 2 K) < 0$ for $K > 0$ → stable, FVT legal. Then: $e_{ss} = lim_{s \to 0} s \cdot \frac{1}{1 + G ( s )} \cdot \frac{1}{s} = \frac{1}{1 + G ( 0 )} = \frac{1}{1 + \frac{2 K}{2}} = \frac{1}{1 + K} .$ Finite for any finite $K$ — you can shrink it but never zero it with proportional gain alone. (b) Now $G (s) = \frac{K _{i}}{s} \cdot \frac{2}{s + 2} = \frac{2 K _{i}}{s ( s + 2 )}$ . The same $\frac{E}{R} = \frac{1}{1 + G}$ holds. Closed-loop characteristic $s^{2} + 2 s + 2 K_{i} = 0$ has both roots in the left half-plane for $K_{i} > 0$ → stable, FVT legal. As $s \to 0$ , $G (0) \to \infty$ , so $e_{ss} = \frac{1}{1 + \infty} = 0.$ Why: the integrator adds a pole at $s = 0$ . Physically, integral action keeps accumulating the error and pushing until the error is exactly zero — it never stops adjusting while any error remains. This is the "I" in PID Control.

L4.2 — Sensor lag destabilizes an "accurate" loop

Forward path $G (s) = \frac{10}{s + 1}$ . Compare two sensors, both DC-accurate ( $H (0) = 1$ ):

Sensor A (fast): $H_{A} (s) = 1$ .
Sensor B (laggy): $H_{B} (s) = \frac{1}{0.1 s + 1}$ .

Find the closed-loop poles in each case and comment on stability.

Recall Solution

Sensor A: $1 + G H_{A} = 1 + \frac{10}{s + 1} = 0$ . Multiply through by $(s + 1)$ (why? to clear the fraction and get a polynomial we can solve): $s + 1 + 10 = 0 \Rightarrow s = - 11.$ One stable pole, very fast.

Sensor B: characteristic equation $1 + \frac{10}{s + 1} \cdot \frac{1}{0.1 s + 1} = 0$ . Multiply through by the common denominator $(s + 1) (0.1 s + 1)$ (why? same reason — clear both fractions): $(s + 1) (0.1 s + 1) + 10 = 0.$ Expand $(s + 1) (0.1 s + 1) = 0.1 s^{2} + s + 0.1 s + 1 = 0.1 s^{2} + 1.1 s + 1$ , then add $10$ : $0.1 s^{2} + 1.1 s + 11 = 0.$ Divide every term by $0.1$ (why? to make the leading coefficient $1$ — a monic polynomial — so we can compare against $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ and read roots cleanly). Dividing $0.1, 1.1, 11$ by $0.1$ gives $1, 11, 110$ : $s^{2} + 11 s + 110 = 0.$ Roots: $s = \frac{- 11 \pm 121 - 440}{2} = \frac{- 11 \pm - 319}{2} = - 5.5 \pm 8.93 j .$ Both poles have negative real part ( $- 5.5$ ) → still stable, but now a complex oscillatory pair where sensor A gave a calm single real pole. The lag injected ringing even though both sensors read perfectly accurate at DC.

Moral: DC accuracy ( $H (0) = 1$ ) does not guarantee good closed-loop behaviour — the sensor's dynamics live inside $G H$ . (With a larger gain or slower sensor, that real part can go positive → instability.) See the mistake callout below and Kalman Filter and Navigation for how estimators mitigate sensor imperfection.

Level 5 — Mastery

L5.1 — Design a controller to a spec (from scratch)

Plant $P (s) = \frac{1}{s ( s + 6 )}$ , $A = 1$ , $H = 1$ . Design a proportional gain $K$ so the closed loop has damping ratio $ζ = 0.5$ . Then report the resulting $ω_{n}$ , the closed-loop poles, and the approximate percent overshoot.

Recall Solution

Step 1 — closed loop. $G = \frac{K}{s ( s + 6 )}$ , so $T = \frac{K}{s ^{2} + 6 s + K}$ . Denominator $s^{2} + 6 s + K$ vs $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ . Step 2 — match damping. $2 ζ ω_{n} = 6$ and $ω_{n}^{2} = K$ , so $ω_{n} = \frac{6}{2 ζ} = \frac{6}{2 ( 0.5 )} = 6$ rad/s. Step 3 — solve for $K$ . $K = ω_{n}^{2} = 6^{2} = 36.$ Step 4 — poles. Solve $s^{2} + 6 s + 36 = 0$ with the quadratic formula (why? poles = roots of the denominator): $s = \frac{- 6 \pm 6 ^{2} - 4 \cdot 36}{2} = \frac{- 6 \pm 36 - 144}{2} = \frac{- 6 \pm - 108}{2} .$ Now simplify $- 108$ : since $108 = 36 \cdot 3$ , $- 108 = 363 j = 63 j$ . Divide numerator by $2$ : $s = \frac{- 6 \pm 6 3 j}{2} = - 3 \pm 33 j \approx - 3 \pm 5.196 j .$ (Sanity check: real part $- 3 = ζ ω_{n} = 0.5 \cdot 6$ . ✓) Step 5 — overshoot. $M_{p} = 100 e^{- π (0.5) / 1 - 0.25} = 100 e^{- π / (2 \cdot 0.866)} = 100 e^{- 1.814} \approx 16.3%.$ So $K = 36$ delivers a well-damped, ~16% overshoot response — a textbook GNC design point.

L5.2 — Integral windup and why anti-windup exists

An integral controller $C (s) = K_{i} / s$ drives an actuator that saturates at $\pm u_{m a x}$ . During a large step, the actuator hits $u_{m a x}$ and stays there while the integrator keeps accumulating error. Qualitatively: (a) what happens to the output, (b) why does overshoot get worse, (c) what fix removes it?

Recall Solution

(a) While saturated, the actuator delivers a constant $u_{m a x}$ regardless of the growing controller command — the loop is momentarily open (feedback can't push harder). The output slews toward the target at maximum effort. (b) Meanwhile the integrator keeps integrating the (still large) error, so its stored value overshoots the level it needs. When the output finally reaches target, the integrator is "over-charged" and keeps commanding a large push — the output sails past the reference → big overshoot and slow recovery. This is integrator windup. (c) Anti-windup: stop or back-off integration whenever the actuator is saturated (e.g. clamp the integrator, or back-calculate from the difference between commanded and actual actuator output). Then the integrator never over-charges. Full treatment: Actuator Saturation and Anti-Windup.

Active recall

Recall One-line self-checks

Closed-loop pole location for L2.1 (thermal loop)? ::: $s = - 9$ (was $- 1$ open loop). $ζ$ for the servo $s^{2} + 4 s + K$ in terms of $K$ ? ::: $ζ = 2/ K$ . Steady-state step error with $G (0) = K$ and unity feedback? ::: $e_{ss} = 1/ (1 + K)$ . Why does integral control zero the steady-state error? ::: It adds a pole at $s = 0$ , so $G (0) \to \infty$ and $1/ (1 + G (0)) \to 0$ . Does a DC-accurate sensor guarantee stability? ::: No — its full dynamics sit inside $G H$ and reshape every pole. What must you verify before applying the Final Value Theorem? ::: That the loop is stable (all closed-loop poles in the left half-plane).