Worked examples — Control system fundamentals — plant, actuator, sensor, controller

Prerequisites we lean on (all built in the parent topic): Transfer Functions and Laplace Domain, Poles Zeros and Stability, Second-order System Response, PID Control, and Actuator Saturation and Anti-Windup.

The scenario matrix

Before working anything, let us list every distinct case class a closed-loop problem can hand you. Each row is a "shape" of problem; each worked example below is tagged with the cell it fills.

Every numeric answer below is machine-checked in the verify block.

Forecast: Guess first — with one plant pole and positive gain, do you expect this to blow up or settle? Will the error to a step be zero or some leftover?

1. Forward path $G=CAP$. Why this step? Cascaded Laplace blocks multiply, so the whole forward chain is one product. $G(s)=2\cdot 1\cdot \frac{4}{s+1}=\frac{8}{s+1}$

2. Apply the boxed formula. Why? $H=1$, so $T=\frac{G}{1+G}$. $T(s)=\frac{8/(s+1)}{1+8/(s+1)}=\frac{8}{(s+1)+8}=\frac{8}{s+9}$

3. Read the pole. Why? The denominator root is the closed-loop pole; its sign decides stability. $s+9=0\Rightarrow s=-9(<0)\Rightarrow \text{stable}$

4. Steady-state error to a step. Why? $e_{ss}=\frac{1}{1+G(0)}$, valid because the loop is stable (final value theorem applies). $G(0)=\frac{8}{0+1}=8,e_{ss}=\frac{1}{1+8}=\frac{1}{9}\approx 0.111$

Verify: Pole moved from $-1$ (open loop) to $-9$ (closed loop) — feedback sped it up, which matches "more gain, faster." Error is small but nonzero because there is no integrator. ✓

Forecast: One value gives lazy no-overshoot motion, the other snappy-but-ringing. Guess which $K$ rings more before reading on.

1. Closed loop. Why? Same procedure as the parent's Ex 1. $T(s)=\frac{K}{s(s+2)+K}=\frac{K}{s^2+2s+K}$

2. Match to the standard second-order form. Why? Comparing to $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$ (see Second-order System Response) lets us read damping straight off. ${\omega }_n=\sqrt{K},2\zeta {\omega }_n=2\Rightarrow \zeta =\frac{1}{\sqrt{K}}$

3. Plug $K=1$. Why? Test the low-gain corner. ${\omega }_n=1,\zeta =1\Rightarrow \text{critically damped, no overshoot.}$

4. Plug $K=16$. Why? Test the high-gain corner. ${\omega }_n=4,\zeta =\frac{1}{4}=0.25\Rightarrow \text{underdamped, big overshoot.}$

Look at the figure: as $K$ grows the two poles slide off the real axis and up into the complex plane — that vertical climb is the ringing.

5. Overshoot for $K=16$. Why? The percent overshoot depends only on $\zeta$: $M_p=e^{-\pi \zeta /\sqrt{1-{\zeta }^2}}$. $M_p=e^{-\pi (0.25)/\sqrt{1-0.0625}}=e^{-0.811}\approx 0.444=44.4\%$

Verify: $\zeta =1$ sits exactly on the boundary between overshoot and none — consistent with $K=1$ being critically damped. Raising $K$ from 1 to 16 quadrupled ${\omega }_n$ and quartered $\zeta$. ✓

Forecast: A minus sign in feedback usually means "you push away from the target instead of toward it." Guess whether the pole ends up left or right of zero.

1. Forward path. Why? Same product, now with negative $K$. $G(s)=\frac{-8}{s+1}$

2. Closed loop. Why? Apply the formula and watch the sign land in the denominator. $T(s)=\frac{-8/(s+1)}{1-8/(s+1)}=\frac{-8}{(s+1)-8}=\frac{-8}{s-7}$

3. Pole. Why? Root of $s-7=0$. $s=+7(>0)\Rightarrow \text{UNSTABLE — right-half-plane pole.}$

Verify: The open-loop pole was at $-1$; positive feedback pushed it right across the imaginary axis. Physically: an error makes the controller push in the wrong direction, growing the error — exponential blow-up $e^{7t}$. This is why sign convention on the sensor/actuator is life-or-death. ✓ (Contrast Ex 1, same plant, correct sign → pole at $-9$.)

Forecast: There is a lonely $s$ in the denominator (a pole at $s=0$). Guess whether the step error is finite or exactly zero.

1. Forward path. Why? Multiply. $G(s)=\frac{18}{s(s+4)}$

2. DC gain. Why? Step error needs $G(0)$. $G(0)={\lim }_{s\to 0}\frac{18}{s(s+4)}=+\infty (\text{because of the }1/s)$

3. Steady-state error. Why? $e_{ss}=\frac{1}{1+G(0)}$. $e_{ss}=\frac{1}{1+\infty }=0$

Verify: A pole at $s=0$ (a "free integrator") means the loop keeps accumulating error until it is exactly wiped out. This is exactly why the "I" in PID Control kills steady-state error — it inserts that $1/s$ deliberately. Contrast Ex 1 (no integrator → $e_{ss}=1/9\ne 0$). ✓

Forecast: If the sensor under-reports by half, will the controller stop too early or overshoot the true target? Guess the direction of the mistake.

1. Closed loop with $H$. Why? Now use the full $T=\frac{G}{1+GH}$. $T(s)=\frac{10/(s+2)}{1+0.5\cdot 10/(s+2)}=\frac{10}{(s+2)+5}=\frac{10}{s+7}$

2. Final value of $y$. Why? $y_{ss}={\lim }_{s\to 0}sT(s)R(s)=T(0)$ for a unit step. $y_{ss}=T(0)=\frac{10}{7}\approx 1.429$

3. Interpret. Why? The loop drives the measured signal $y_m=Hy$ toward the reference, not $y$ itself. $y_m=0.5\times 1.429=0.714\text{(closer to 1, but the true }y\text{ overshot to }1.43)$

Verify: Because the sensor halves the reading, the controller must double the true output to make the measurement match — so the true $y$ ends above the target. A "1.0" command produced a physical 1.43. This is the mistake the parent warns about: an "accurate-looking" but mis-scaled $H$ silently biases the loop. ✓

Forecast: One limit turns the loop "off," the other cranks it as hard as physics allows. Guess what the poles do at each extreme.

1. $K\to 0^{+}$. Why? Turn the controller down to nothing. $s^2+2s+0=s(s+2)=0\Rightarrow s=0,-2$ The poles return to the open-loop locations — with a pole at $0$, the output no longer tracks (loop effectively opened).

2. $K\to \infty$. Why? Push gain to the ceiling. ${\omega }_n=\sqrt{K}\to \infty ,\zeta =\frac{1}{\sqrt{K}}\to 0$ Undamped oscillation: poles rocket up the imaginary direction, $M_p\to 100\%$. Response rings forever (marginally toward instability).

3. Middle ground. Why? Best trade at $\zeta =1/\sqrt{2}\approx 0.707$. $\frac{1}{\sqrt{K}}=\frac{1}{\sqrt{2}}\Rightarrow K=2$

Verify: As $K$ runs from $0\to \infty$, $\zeta$ runs $\infty \to 0$ monotonically — every intermediate damping is hit exactly once, and $K=2$ gives the textbook "nicest" damping. Consistent with Ex 2 ($K=1$: $\zeta =1$; $K=16$: $\zeta =0.25$). ✓

Forecast: There's an integrator in the plant again. Guess whether hover height locks exactly onto 10 m or drifts.

1. Translate words → blocks. Why? Reference $r=10$ (a step of height 10), $R(s)=10/s$.

2. DC gain & step error. Why? Integrator ⇒ $G(0)=\infty$ ⇒ zero error (Case D logic). $G(s)=\frac{2K}{s(s+3)},G(0)=\infty \Rightarrow e_{ss}=0\Rightarrow \text{holds exactly 10 m.}$

3. Damping for $K=4.5$. Why? Close the loop and match second-order form. $T(s)=\frac{2K}{s^2+3s+2K}\Rightarrow {\omega }_n=\sqrt{2K},2\zeta {\omega }_n=3\Rightarrow \zeta =\frac{3}{2\sqrt{2K}}$ $K=4.5:{\omega }_n=\sqrt{9}=3,\zeta =\frac{3}{2\cdot 3}=0.5$

Verify: $\zeta =0.5<1$ → underdamped: the drone rises past 10 m, dips, then settles at exactly 10 m (zero steady error thanks to the integrator). Units: ${\omega }_n$ in rad/s, $\zeta$ dimensionless. ✓

Forecast: Adding the integrator gives zero steady error — but integrators are destabilizing. Guess whether there's an upper limit on $K$.

1. Characteristic equation. Why? Stability lives in the roots of $1+GH=0$. $G=\frac{K}{s(s+1)(s+2)},H=1\Rightarrow 1+\frac{K}{s(s+1)(s+2)}=0$ $\Rightarrow s(s+1)(s+2)+K=0\Rightarrow s^3+3s^2+2s+K=0$

2. Routh array for $s^3+3s^2+2s+K$. Why? Routh–Hurwitz tells us sign changes = right-half poles without solving the cubic.

3. All first-column entries must be $>0$. Why? Any sign change ⇒ a right-half-plane pole ⇒ instability. $K>0\text{and}\frac{6-K}{3}>0\Rightarrow K<6$ $\boxed{0<K<6}$

4. Boundary check $K=6$. Why? At the edge the $s^1$ row vanishes → poles sit on the imaginary axis (sustained oscillation). $s^3+3s^2+2s+6=(s+3)(s^2+2)\Rightarrow s=-3,\pm j\sqrt{2}$ Purely imaginary pair ⇒ marginal, oscillating at $\omega =\sqrt{2}$ rad/s.

Verify: At $K=6$ we get exactly $\pm j\sqrt{2}$ (marginal) and one stable real pole at $-3$; for $K$ just below 6 those complex poles move left (stable), just above 6 they cross right (unstable). So $0<K<6$ is the full stable window. ✓

Active recall