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

3.5.26 · D4 · Physics › Guidance, Navigation & Control (GNC) › Control system fundamentals — plant, actuator, sensor, contr

Hum teen tools baar baar use karte hain, toh chaliye pehle name karte hain kyun har ek exist karta hai, phir use karte hain:

Definition Is page ke teen tools

Block multiplication. Laplace ( $s$ ) domain mein, cascaded blocks multiply karte hain. Kyun addition nahi? Kyunki ek signal jo block $A$ se phir block $B$ se guzarta hai, woh $A$ se scale hota hai phir $B$ se — do scalings compose karte hain multiply karke. Dekho Transfer Functions and Laplace Domain.
Closed-loop formula $T (s) = \frac{G}{1 + G H}$ . Kyun? Yeh poore feedback loop ko EK input→output map mein compress kar deta hai taaki hum behaviour seedha padh sakein.
Final Value Theorem (FVT): $t \to \infty lim f (t) = s \to 0 lim s F (s)$ . Yeh tool kyun? Yeh jawaab deta hai "signal kahan settle karega?" Laplace transform invert kiye bina — lekin yeh sirf tab valid hai jab loop stable ho (saare closed-loop poles left half-plane mein). Har baar jab hum FVT use karte hain neeche, pehle stability check karte hain — woh check skip karo toh theorem galat (ya infinite) jawaab de sakta hai.

Jo feedback loop hum baar baar refer karte hain woh ek baar yahan draw kiya gaya hai — har problem is picture ka ek special case hai (red hai forward path $G = C A P$ , blue hai sensor feedback $H$ , yellow circle hai jahan error $e = r - y_{m}$ banta hai):

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

Level 1 — Recognition

L1.1 — Organ pehchano

Ek self-driving car lane hold kar rahi hai, uske liye:

(a) camera jo lane position padh raha hai,
(b) software jo compute kar raha hai "3° right steer karo",
(c) steering motor jo wheels ghuma raha hai,
(d) car ki mass-aur-tyre dynamics.

Har ek ko plant / actuator / sensor / controller se match karo.

Recall Solution

(a) camera → sensor ( $H$ ): yeh output measure karta hai aur use signal mein convert karta hai.
(b) software → controller ( $C$ ): math/brain jo command number $u$ produce karta hai (yeh "control signal" hai — ek plain number jo software output karta hai).
(c) steering motor → actuator ( $A$ ): $u$ ko real physical force mein convert karta hai.
(d) car dynamics → plant ( $P$ ): fixed physics jo tum command karte ho lekin redesign nahi kar sakte.

L1.2 — Error equation kaun sa hai?

$r$ (reference), $y$ (true output), $y_{m}$ (measured output) diye hain, sahi error definition choose karo aur kyun batao: $(A) e = y - r (B) e = r - y_{m} (C) e = r + y_{m}$

Recall Solution

(B) $e = r - y_{m}$ . Error matlab hai "main jahan jaana chahta hun minus jahan main appear kar raha hun." Hum measured output $y_{m}$ subtract karte hain (true $y$ nahi), kyunki controller sirf wahi dekh sakta hai jo sensor report karta hai. (A) mein sign ulta hai (hume target se door drive karega); (C) add karta hai, jo kabhi zero nahi hoga.

Level 2 — Application

L2.1 — Forward path aur closed loop banao

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

Recall Solution

(a) Cascaded blocks multiply karo (isliye hum multiply karte hain — signal har block se scale hota hai baari baari): $G (s) = C A P = 4 \cdot 2 \cdot \frac{1}{s + 1} = \frac{8}{s + 1} .$ (b) $T = \frac{G}{1 + G H}$ apply karo, $H = 1$ ke saath: $T (s) = \frac{\frac{8}{s + 1}}{1 + \frac{8}{s + 1}} .$ Ab chhoti fractions clear karo — kyun? "fraction ke upar fraction" ko ek clean ratio mein convert karne ke liye taaki hum poles padh sakein. Top aur bottom dono ko $(s + 1)$ se multiply karo: $T (s) = \frac{8}{( s + 1 ) + 8} = \frac{8}{s + 9} .$ Single pole $s = - 1$ (open loop) se $s = - 9$ (closed loop) move ho gaya: feedback ne isse 9 guna faster bana diya.

L2.2 — Closed loop ka DC gain

Upar wale $T (s) = \frac{8}{s + 9}$ ke liye, unit-step reference ke liye steady-state output kya hai?

Recall Solution

Pehle stability check karo: single pole $s = - 9$ hai, left half-plane mein, toh FVT legal hai. Step response ka steady state = DC gain $T (0)$ ( $s = 0$ set karo; kyun? FVT: $lim_{s \to 0} s \cdot T (s) \cdot \frac{1}{s} = T (0)$ ): $T (0) = \frac{8}{0 + 9} = \frac{8}{9} \approx 0.889.$ Toh output $0.889$ par settle karta hai, $1$ nahi — steady-state error $1 - \frac{8}{9} = \frac{1}{9}$ hai kyunki koi integrator nahi hai.

Level 3 — Analysis

L3.1 — Position servo: $ω_{n}$ aur $ζ$ padho

Ek servo mein $P (s) = \frac{1}{s ( s + 4 )}$ , $A = 1$ , $H = 1$ , proportional controller $C = K$ hai. (a) $T (s)$ nikalo. (b) Natural frequency $ω_{n}$ aur damping ratio $ζ$ ko $K$ ke terms mein express karo. (c) Kaunsa $K$ critical damping deta hai ( $ζ = 1$ )?

Recall Solution

(a) $G = \frac{K}{s ( s + 4 )}$ , toh $T = \frac{G}{1 + G} = \frac{K}{s ( s + 4 ) + K} = \frac{K}{s ^{2} + 4 s + K} .$ (Humne top aur bottom dono ko $s (s + 4)$ se multiply kiya inner fraction clear karne ke liye, phir $s (s + 4) = s^{2} + 4 s$ expand kiya.) (b) Standard second-order denominator $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ se compare karo (dekho Second-order System Response). Term by term match karo — constant term $ω_{n}^{2}$ deta hai, $s$ -term $2 ζ ω_{n}$ deta hai: $ω_{n}^{2} = K \Rightarrow ω_{n} = K, 2 ζ ω_{n} = 4 \Rightarrow ζ = \frac{4}{2 K} = \frac{2}{K} .$ (c) $ζ = 1$ : $\frac{2}{K} = 1 \Rightarrow K = 2 \Rightarrow K = 4.$ $K = 4$ se neeche system overdamped hai (sluggish, no overshoot); $K = 4$ se upar underdamped hai (overshoots). $K = 4$ fast-but-clean boundary hai.

L3.2 — Poles kahan baithe hain?

Same servo mein $K = 8$ ke saath, closed-loop poles nikalo aur response classify karo.

Recall Solution

Characteristic equation $s^{2} + 4 s + 8 = 0$ . Solve karo (quadratic formula, kyun? poles denominator ke roots hote hain): $s = \frac{- 4 \pm 16 - 32}{2} = \frac{- 4 \pm - 16}{2} = - 2 \pm 2 j .$ Complex conjugate pair with negative real part ( $- 2$ ) → stable, underdamped, damped frequency $ω_{d} = 2$ rad/s par oscillate karta hai decay hote hue. Check: $ζ = 2/ 8 = 0.707$ — classic "nicely damped" value. Dekho Poles Zeros and Stability.

Neeche wali figure (left) in do poles ko complex plane mein red crosses ke roop mein plot karta hai — dono shaded stable (left-half) region mein baithe hain, equal-and-opposite imaginary parts ke saath. Right panel dikhata hai woh pole pair time mein kya karta hai: output rise karta hai, reference ko ek baar overshoot karta hai, aur ring down hota hai — exactly "underdamped" signature.

Level 4 — Synthesis

L4.1 — Steady-state error khatam karo

Unity-feedback plant $P (s) = \frac{2}{s + 2}$ , $A = 1$ , $H = 1$ , unit-step reference. (a) Proportional control $C = K$ ke saath, $K$ ke function ke roop mein steady-state error $e_{ss}$ nikalo. (b) Ab integral control $C (s) = \frac{K _{i}}{s}$ use karo. $e_{ss}$ nikalo. Difference explain karo.

Recall Solution

(a) Pehle error transfer function $\frac{E}{R}$ derive karo — kyun? FVT ko $E (s)$ chahiye, toh hume $R$ se $E$ tak ka map loop equations se banana hoga. Do loop facts se shuru karo (unity feedback, $H = 1$ ): $E = R - Y, Y = \frac{G}{1 + G} R .$ Doosre ko pehle mein substitute karo ( $Y$ eliminate karo): $E = R - \frac{G}{1 + G} R = R (\frac{1 + G - G}{1 + G}) = \frac{1}{1 + G} R \Rightarrow \frac{E}{R} = \frac{1}{1 + G} .$ FVT se pehle stability check karo: $G = \frac{2 K}{s + 2}$ closed-loop pole $s = - (2 + 2 K) < 0$ deta hai $K > 0$ ke liye → stable, FVT legal. Phir: $e_{ss} = lim_{s \to 0} s \cdot \frac{1}{1 + G ( s )} \cdot \frac{1}{s} = \frac{1}{1 + G ( 0 )} = \frac{1}{1 + \frac{2 K}{2}} = \frac{1}{1 + K} .$ Kisi bhi finite $K$ ke liye finite — tum ise shrink kar sakte ho lekin proportional gain akele se kabhi zero nahi kar sakte. (b) Ab $G (s) = \frac{K _{i}}{s} \cdot \frac{2}{s + 2} = \frac{2 K _{i}}{s ( s + 2 )}$ . Wahi $\frac{E}{R} = \frac{1}{1 + G}$ hold karta hai. Closed-loop characteristic $s^{2} + 2 s + 2 K_{i} = 0$ ke dono roots $K_{i} > 0$ ke liye left half-plane mein hain → stable, FVT legal. Jaise $s \to 0$ , $G (0) \to \infty$ , toh $e_{ss} = \frac{1}{1 + \infty} = 0.$ Kyun: integrator $s = 0$ par ek pole add karta hai. Physically, integral action error ko accumulate karta rehta hai aur push karta rehta hai jab tak error exactly zero na ho jaye — jab tak koi bhi error baaki ho, adjust karna band nahi karta. Yahi hai PID Control mein "I".

L4.2 — Sensor lag ek "accurate" loop ko destabilize karta hai

Forward path $G (s) = \frac{10}{s + 1}$ . Do sensors compare karo, dono DC-accurate ( $H (0) = 1$ ):

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

Dono cases mein closed-loop poles nikalo aur stability par comment karo.

Recall Solution

Sensor A: $1 + G H_{A} = 1 + \frac{10}{s + 1} = 0$ . $(s + 1)$ se multiply karo (kyun? fraction clear karne ke liye aur ek polynomial lene ke liye jo hum solve kar sakein): $s + 1 + 10 = 0 \Rightarrow s = - 11.$ Ek stable pole, bahut fast.

Sensor B: characteristic equation $1 + \frac{10}{s + 1} \cdot \frac{1}{0.1 s + 1} = 0$ . Common denominator $(s + 1) (0.1 s + 1)$ se multiply karo (kyun? same reason — dono fractions clear karo): $(s + 1) (0.1 s + 1) + 10 = 0.$ Expand karo $(s + 1) (0.1 s + 1) = 0.1 s^{2} + s + 0.1 s + 1 = 0.1 s^{2} + 1.1 s + 1$ , phir $10$ add karo: $0.1 s^{2} + 1.1 s + 11 = 0.$ Har term ko $0.1$ se divide karo (kyun? leading coefficient $1$ banane ke liye — ek monic polynomial — taaki hum $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ se compare karein aur roots cleanly padh sakein). $0.1, 1.1, 11$ ko $0.1$ se divide karo toh $1, 11, 110$ milta hai: $s^{2} + 11 s + 110 = 0.$ Roots: $s = \frac{- 11 \pm 121 - 440}{2} = \frac{- 11 \pm - 319}{2} = - 5.5 \pm 8.93 j .$ Dono poles ka negative real part hai ( $- 5.5$ ) → phir bhi stable, lekin ab ek complex oscillatory pair hai jahan sensor A ne ek calm single real pole diya tha. Lag ne ringing inject kar di jabki dono sensors DC par perfectly accurate padh rahe the.

Moral: DC accuracy ( $H (0) = 1$ ) closed-loop acche behaviour ki guarantee nahi deta — sensor ki dynamics $G H$ ke andar rehti hain. (Zyada gain ya slower sensor ke saath, woh real part positive ho sakta hai → instability.) Neeche ka mistake callout dekho aur Kalman Filter and Navigation dekho jahan estimators sensor imperfection ko mitigate karte hain.

Level 5 — Mastery

L5.1 — Ek spec ke liye controller design karo (scratch se)

Plant $P (s) = \frac{1}{s ( s + 6 )}$ , $A = 1$ , $H = 1$ . Ek proportional gain $K$ design karo taaki closed loop ka damping ratio $ζ = 0.5$ ho. Phir resulting $ω_{n}$ , closed-loop poles, aur approximate percent overshoot report karo.

Recall Solution

Step 1 — closed loop. $G = \frac{K}{s ( s + 6 )}$ , toh $T = \frac{K}{s ^{2} + 6 s + K}$ . Denominator $s^{2} + 6 s + K$ vs $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ . Step 2 — damping match karo. $2 ζ ω_{n} = 6$ aur $ω_{n}^{2} = K$ , toh $ω_{n} = \frac{6}{2 ζ} = \frac{6}{2 ( 0.5 )} = 6$ rad/s. Step 3 — $K$ solve karo. $K = ω_{n}^{2} = 6^{2} = 36.$ Step 4 — poles. $s^{2} + 6 s + 36 = 0$ quadratic formula se solve karo (kyun? poles = denominator ke roots): $s = \frac{- 6 \pm 6 ^{2} - 4 \cdot 36}{2} = \frac{- 6 \pm 36 - 144}{2} = \frac{- 6 \pm - 108}{2} .$ Ab $- 108$ simplify karo: kyunki $108 = 36 \cdot 3$ , $- 108 = 363 j = 63 j$ . Numerator ko $2$ se divide karo: $s = \frac{- 6 \pm 6 3 j}{2} = - 3 \pm 33 j \approx - 3 \pm 5.196 j .$ (Sanity check: real part $- 3 = ζ ω_{n} = 0.5 \cdot 6$ . ✓) Step 5 — overshoot. $M_{p} = 100 e^{- π (0.5) / 1 - 0.25} = 100 e^{- π / (2 \cdot 0.866)} = 100 e^{- 1.814} \approx 16.3%.$ Toh $K = 36$ ek well-damped, ~16% overshoot response deliver karta hai — ek textbook GNC design point.

L5.2 — Integral windup aur anti-windup kyun exist karta hai

Ek integral controller $C (s) = K_{i} / s$ ek actuator drive karta hai jo $\pm u_{m a x}$ par saturate karta hai. Ek bade step ke dauran, actuator $u_{m a x}$ hit karta hai aur wahan rehta hai jabki integrator error accumulate karta rehta hai. Qualitatively: (a) output ka kya hota hai, (b) overshoot worse kyun hota hai, (c) ise kaunsa fix remove karta hai?

Recall Solution

(a) Saturation ke dauran, actuator growing controller command ki parwah kiye bina ek constant $u_{m a x}$ deliver karta hai — loop momentarily open ho jaata hai (feedback zyada push nahi kar sakta). Output maximum effort se target ki taraf slew karta hai. (b) Iss dauran integrator (abhi bhi bade) error ko integrate karta rehta hai, toh uski stored value overshoot kar jaati hai us level se jo use chahiye. Jab output finally target reach karta hai, integrator "over-charged" hai aur continue karta hai ek bada push command karne ke liye — output reference se bahut aage sail kar jaata hai → bada overshoot aur slow recovery. Yahi integrator windup hai. (c) Anti-windup: jab bhi actuator saturated ho tab integration stop karo ya back-off karo (jaise integrator clamp karo, ya commanded aur actual actuator output ke difference se back-calculate karo). Phir integrator kabhi over-charge nahi hoga. Full treatment: Actuator Saturation and Anti-Windup.

Active recall

Recall One-line self-checks

L2.1 (thermal loop) ke liye closed-loop pole location? ::: $s = - 9$ (open loop mein $- 1$ tha). Servo $s^{2} + 4 s + K$ ke liye $K$ ke terms mein $ζ$ ? ::: $ζ = 2/ K$ . $G (0) = K$ aur unity feedback ke saath step ka steady-state error? ::: $e_{ss} = 1/ (1 + K)$ . Integral control steady-state error zero kyun karta hai? ::: Yeh $s = 0$ par ek pole add karta hai, toh $G (0) \to \infty$ aur $1/ (1 + G (0)) \to 0$ . Kya ek DC-accurate sensor stability guarantee karta hai? ::: Nahi — uski full dynamics $G H$ ke andar baithe hain aur har pole ko reshape karte hain. Final Value Theorem apply karne se pehle tumhe kya verify karna chahiye? ::: Ki loop stable hai (saare closed-loop poles left half-plane mein hain).