Control system fundamentals — plant, actuator, sensor, controller

In GNC, Guidance decides where to go (reference), Navigation figures out where you are (state estimate), and Control makes the vehicle actually get there. This note is about the machinery of that last block: the four organs of any feedback loop.

The four building blocks

WHY these four? Because control is fundamentally information → action → measurement. You need something to act (actuator), something to be acted on (plant), something to see the result (sensor), and something to decide (controller). Remove any one and the loop breaks.

Open loop vs. closed loop — WHY feedback?

HOW the signal flows (closed loop):

$r(t) \xrightarrow{\ \text{compare}\ } e(t) = r(t) - y_m(t) \xrightarrow{\text{controller } C} u(t) \xrightarrow{\text{actuator } A} \text{force} \xrightarrow{\text{plant } P} y(t) \xrightarrow{\text{sensor } H} y_m(t)$

• $r$ = reference (desired output, from Guidance)
• $e$ = error
• $u$ = control signal (controller output)
• $y$ = true plant output
• $y_m$ = measured output (fed back)

Deriving the closed-loop transfer function from first principles

WHAT we want: a single relation from $R(s)$ (what we ask) to $Y(s)$ (what we get), in the Laplace domain where cascaded blocks multiply.

Let the transfer functions be $C(s)$ (controller), $A(s)$ (actuator), $P(s)$ (plant), $H(s)$ (sensor). Lump the forward path $G(s) = C(s)A(s)P(s)$.

Step 1 — write the error. Why? Error is what drives the controller. $E(s) = R(s) - H(s)\,Y(s)$

Step 2 — the forward path produces the output. Why? Signals through cascaded LTI blocks multiply in the $s$-domain. $Y(s) = G(s)\,E(s)$

Step 3 — substitute Step 1 into Step 2. Why? To eliminate the internal signal $E$ and keep only $R,Y$. $Y = G\big(R - H Y\big) = GR - GHY$

Step 4 — collect $Y$. Why? Solve algebraically for the input–output map. $Y + GHY = GR \;\Rightarrow\; Y(1 + GH) = GR$

Unity feedback special case ($H=1$, i.e. the sensor is perfect and scaled to 1): $T(s) = \frac{G(s)}{1+G(s)}$

Worked example 1 — DC motor angular position

Setup: Plant (motor to angle) $P(s)=\dfrac{1}{s(s+2)}$, actuator gain $A=1$, sensor $H=1$, proportional controller $C(s)=K$.

Step 1 — forward path. Why? Multiply cascaded blocks. $G(s)=K\cdot 1\cdot \frac{1}{s(s+2)} = \frac{K}{s(s+2)}$

Step 2 — closed loop. Why? Apply the boxed formula. $T(s)=\frac{G}{1+G}=\frac{K/[s(s+2)]}{1+K/[s(s+2)]}=\frac{K}{s(s+2)+K}=\frac{K}{s^2+2s+K}$

Step 3 — read stability & response. Why? Compare denominator to $s^2+2\zeta\omega_n s+\omega_n^2$. $\omega_n=\sqrt{K},\qquad 2\zeta\omega_n=2 \;\Rightarrow\; \zeta=\frac{1}{\sqrt{K}}$ Bigger $K$ → faster ($\omega_n\uparrow$) but less damped ($\zeta\downarrow$) → more overshoot. This is the eternal control trade-off.

Worked example 2 — steady-state error with a step

Setup: unity feedback, $G(s)=\dfrac{5}{s+3}$, input a unit step $R(s)=1/s$.

Step 1 — error transfer function. Why? We want $E$, not $Y$. $\frac{E}{R}=\frac{1}{1+G}\quad\text{(from }E=R-Y,\ Y=\tfrac{G}{1+G}R)$

Step 2 — Final Value Theorem. Why? Steady-state error = $\lim_{t\to\infty}e(t)=\lim_{s\to0}sE(s)$ (valid since the loop is stable). $e_{ss}=\lim_{s\to0} s\cdot \frac{1}{1+G(s)}\cdot\frac{1}{s}=\frac{1}{1+G(0)}=\frac{1}{1+5/3}=\frac{3}{8}=0.375$

Interpretation: finite error because the loop has no integrator (no pole at $s=0$). Add integral action ($C=K/s$) → $G(0)\to\infty$ → $e_{ss}\to0$. That's why the "I" in PID kills steady-state error.

Common mistakes (Steel-manned)

Active recall

Connections

• PID Control — how P, I, D shape $C(s)$
• Transfer Functions and Laplace Domain — why cascaded blocks multiply
• Poles Zeros and Stability — roots of $1+GH=0$
• Second-order System Response — $\zeta,\omega_n$ trade-offs
• State-Space Representation — alternative to transfer functions for MIMO GNC
• Kalman Filter and Navigation — the sensor/estimator side of GNC
• Actuator Saturation and Anti-Windup

Concept Map

Hinglish (regional understanding)