Control system fundamentals — plant, actuator, sensor, controller: Visual walkthrough

We assume you know only this: a signal is a number that changes with time (how far off you are, how hard you push). Everything else we build.

Step 0 — What is a "block", and why do things multiply?

WHAT. Draw a box. A wire carries a signal in; the box does something to it; a wire carries the signal out. The box is labelled with a letter like $G$ . The rule of the box is: output = (the box's factor) $\times$ (input).

WHY this matters. In the real world signals are wiggly functions of time $t$ , and combining them is messy (it needs convolution — a smearing operation). Engineers use a trick called the Laplace transform: it re-labels every signal by a new variable $s$ , and in that new language cascaded boxes just multiply. So a chain of three boxes becomes one multiplication. That is the only reason we work in $s$ — it turns "smear" into "times".

PICTURE. Two boxes in a row: the signal is multiplied once, then again.

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

Step 1 — Lay out the four blocks as one wire-loop

WHAT. We place the four organs from the parent note in the order a signal actually travels: controller $C$ → actuator $A$ → plant $P$ (these three push the vehicle), and sensor $H$ (this one watches and reports back).

WHY. Control is decide → push → watch → report → decide again. Drawing it as a loop (not a line) is the whole point: the output curls back to meet the input. Everything downstream lives in this ring.

PICTURE. Follow the arrow all the way around once. The four named signals ride on the wires:

$r$ — the reference: where Guidance wants us.
$e$ — the error: reference minus what the sensor reported.
$u$ — the control signal: the controller's decision.
$y$ — the true output: where the vehicle really is.
$y_{m}$ — the measured output: what the sensor says $y$ is.

Step 2 — Write the error (the subtraction that starts everything)

WHAT. At the very front sits a comparator: a circle that subtracts the fed-back measurement from the reference.

E (s) = what we ask for R (s) - what the sensor reports back H (s) Y (s)

WHY. A controller should only work when we are off target. If measurement equals reference, $E = 0$ and the controller relaxes. So the error is literally "gap between wish and reality." Note $Y$ passes through the sensor box $H$ before it is subtracted — the loop compares against the reported value, not the raw truth.

Term by term: $R (s)$ is the target signal; $Y (s)$ is the true output; $H (s)$ reshapes $Y$ into the report $y_{m} = H Y$ ; the minus sign is what makes it negative feedback — the correction always opposes the gap.

PICTURE. The comparator circle with a $+$ on the reference wire and a $-$ on the feedback wire.

Step 3 — Push the error through the forward path

WHAT. The error is the input to the big forward box $G$ . Its output is the true position:

Y (s) = controller \cdot actuator \cdot plant G (s) \cdot how far off we are E (s)

WHY. By Step 0, sending a signal through cascaded blocks is one multiplication. The error decides how hard to correct; $G$ turns that decision into an actual movement $y$ .

PICTURE. The straight run: $e$ enters the $G$ box, $y$ leaves it — the "action" half of the loop, drawn apart from the feedback for clarity.

Step 4 — Substitute to kill the internal signal $E$

WHAT. We have two equations sharing $E$ . Put Step 2 inside Step 3:

Y = G E = G (R - H Y) = GR - G H Y

WHY. $E$ is an internal wire — we can't measure it from outside and we don't care about it. We want a clean statement linking only what we ask ( $R$ ) to what we get ( $Y$ ). Substitution erases the middle-man.

Term by term: $GR$ = the "if there were no feedback" response (pure forward push on the reference). $- G H Y$ = the correction the loop feeds back — the output travels once around the ring (through $H$ , through $G$ ) and comes back to fight itself. That extra $G H Y$ term is feedback made visible.

PICTURE. The output $y$ traced backward through $H$ and forward through $G$ again — the same signal appearing on both sides of the equation.

Step 5 — Collect $Y$ and solve

WHAT. Gather every $Y$ on one side:

Y + G H Y = GR ⟹ Y (1 + G H) = GR

T (s) = \frac{Y ( s )}{R ( s )} = \frac{G ( s )}{1 + G ( s ) H ( s )}

WHY. This is just algebra now — factor out $Y$ , divide across. The reward is a single box $T (s)$ that swallows the whole loop: ask it $R$ , it hands you $Y$ .

Term by term:

Numerator $G$ = the forward push (controller·actuator·plant). Big $G$ → strong action.
Denominator $1 + G H$ = one, plus the loop gain. The " $1$ " is you asking; the " $G H$ " is the loop pushing back.

PICTURE. The whole ring collapses into one equivalent box labelled $T (s)$ .

Step 6 — WHY exactly " $1 + G H$ "? (the chasing picture)

WHAT. Follow a signal around the ring step by step instead of solving at once. Send in $R$ . The loop subtracts $G H$ of the last output, then that gets subtracted again, and again:

\frac{1}{1 + G H} = 1 - G H + (G H)^{2} - (G H)^{3} + \dots

WHY. This is a geometric series — each trip around the loop multiplies the leftover by $- G H$ (the minus from negative feedback, the $G H$ from one full lap). Each correction overshoots a little less than the last, and the infinite sum settles at $\frac{1}{1 + G H}$ . The formula is the loop converging on balance.

PICTURE. A staircase of shrinking corrections, each $- G H$ times the one before, piling up to the final value.

Step 7 — The edge cases (never leave a scenario unshown)

Every combination of "how big is the loop gain" tells a different story. Read all four.

Case A — $H = 1$ (perfect unity sensor). The report is the truth, so $T (s) = \frac{G}{1 + G} .$ This is the parent note's "unity feedback." Nothing special — just $H$ set to $1$ .

Case B — huge loop gain, $∣ G H ∣ ≫ 1$ . The " $1$ " is dwarfed: $T \approx \frac{G}{G H} = \frac{1}{H} .$ The output no longer depends on the messy plant $G$ at all — only on the sensor $H$ ! WHY it matters: strong feedback makes the loop immune to plant errors. This is the deep reason feedback is used.

Case C — tiny loop gain, $∣ G H ∣ ≪ 1$ . Now $1 + G H \approx 1$ , so $T \approx G$ — the loop behaves as if open, no self-correction. Feedback that is too weak does nothing.

Case D — degenerate/dangerous: $1 + G H = 0$ . The denominator hits zero and $T$ blows up. WHY this is the whole ballgame: the $s$ -values solving $1 + G (s) H (s) = 0$ are the closed-loop poles. If any sits in the right half of the $s$ -plane, the loop is unstable — the staircase in Step 6 grows instead of shrinking (because now $∣ G H ∣ > 1$ , the geometric series diverges). This single equation is why we can't just "crank up the gain": push $G$ too far and a root crosses over.

PICTURE. The four regimes side by side on a single map of loop gain, from "does nothing" to "blows up."

The one-picture summary

Everything above, on one canvas: the ring with its four blocks and five signals, the subtraction that makes the error, the forward push $G$ , the feedback path $H$ , and the collapse into $T = \frac{G}{1 + G H}$ with the shrinking-staircase reason for the denominator.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a wire bent into a ring. At the top sits a subtracting circle: it takes where you want to be ( $r$ ) and subtracts what your eyes reported ( $y_{m}$ ), giving the gap ( $e$ ). That gap flows into one big pusher-box $G$ — that's your brain, your muscles, and the vehicle all rolled into one — and out comes where you really are ( $y$ ). But $y$ doesn't just leave; it loops back around through your eyes (the sensor box $H$ ) and returns to the subtracting circle, so the loop is always comparing against itself. When we do the algebra to link only "asked" and "got," the loop's own pushback shows up as $G H$ sitting next to a $1$ on the bottom of the fraction. The $1$ is you asking once; the $G H$ is the loop answering, over and over, each answer a little smaller — a staircase of shrinking corrections that settles down to $\frac{1}{1 + G H}$ . And the one place it doesn't settle — where $1 + G H$ hits zero — is exactly where the loop goes unstable. That is the entire story of feedback in one bent wire.

Recall Quick self-test

Why do cascaded blocks multiply instead of add? ::: In the Laplace ( $s$ ) domain, sending a signal through blocks in series is one multiplication — that is the whole reason we switch to $s$ . What does the minus sign in $E = R - H Y$ create? ::: Negative feedback — the correction always opposes the error, so the staircase of corrections shrinks. In one line, why is the denominator $1 + G H$ ? ::: It is the geometric series of the loop feeding back on itself: $1 - G H + (G H)^{2} - \dots = \frac{1}{1 + G H}$ . With very large loop gain, $T \approx$ what, and why is that good? ::: $T \approx 1/ H$ — the output stops depending on the plant, so the loop is immune to plant errors. What equation gives the closed-loop poles? ::: $1 + G (s) H (s) = 0$ — its roots decide stability.

See also: State-Space Representation and Kalman Filter and Navigation for how "where am I" ( $y$ , $y_{m}$ ) is estimated when sensors are noisy.