Exercises — PID control — proportional, integral, derivative terms

Har term ki ek signature disease-and-cure hoti hai:

• (a) Bahut slow → ==$K_p$== (Proportional) badhao. P present-error ka push hai; bada push matlab faster response.
• (b) Steady offset → ==$K_i$== (Integral) badhao. Sirf integral ek persistent error ko yaad rakhta hai aur tab tak ramping karta rehta hai jab tak offset khatam na ho jaaye.
• (c) Overshoot / ringing → ==$K_d$== (Derivative) badhao. D rate $\dot{e}$ par react karta hai, target se aage nikalne se pehle brake lagata hai.

Jawab: (a) P, (b) I, (c) D.

$e(t)=r(t)-y(t)$. Daalo: $e=30^{\circ }-34^{\circ }=-4^{\circ }.$ Sign negative hai, matlab hum target se overshoot kar gaye hain (hum us se upar hain). Controller $u$ ko us direction mein command karega jo $y$ ko wापas neeche kheenche.

KYA chahiye: woh resting error jab sab kuch change karna band ho jaaye. KYUN yeh formula: rest par control output ko disturbance exactly cancel karna chahiye, warna kuch na kuch abhi bhi move kar raha hoga. Pure P ke saath sirf output $u=K_{p}e$ hai, to equilibrium par $K_p{e}_{ss}=d\Rightarrow e_{ss}=\frac{d}{K_p}=\frac{3}{6}=0.5^{\circ }.$ Ek pure-P loop isse shrink kar sakta hai lekin kabhi zero nahi kar sakta — zero error zero output dega, aur phir kuch bhi torque se nahi ladega. Yahi woh steady-state error hai jisko khatam karne ke liye integral banaya gaya hai.

KYA: resting error aur resting integral value nikalo. KYUN: integrator apna output ${\dot{u}}_I=K_{i}e$ rate se change karta hai. Poore system ke liye rukne ke liye, yeh rate zero honi chahiye: ${\dot{u}}_I=K_i{e}_{ss}=0\Rightarrow e_{ss}=0.$ To sirf resting state mein zero error hota hai — integral quietly tab tak chadhta rehta hai jab tak ho na jaaye. Uski final value akele poori cancellation supply karni chahiye: $u_I=d=3.$ P rest par kuch contribute nahi karta (kyunki $e_{ss}=0$), to integrator akela disturbance ko rokta hai.

KYUN sum: computer sirf har $\Delta t$ par snapshots dekhta hai. Har snapshot ek patli rectangle of area $e_j\Delta t$ contribute karta hai; integral total area hai, yani rectangles ka sum. $I_2={\sum }_{j=0}^2{e}_j\Delta t=(1.0+0.6+0.4)(0.2)=(2.0)(0.2)=\mathrm{0.4.}$ Yeh $I_2$ wahi hai jo $K_i$ se multiply hoke $u_I$ banta hai. (Continuous view ke liye Laplace transforms & transfer functions dekho.)

Teen pieces banao, har ek apna sawaal answer karte hue. Integral (past): nayi rectangle add karo, phir scale karo. $I_k=I_{\text{prev}}+e_k\Delta t=0.3+(0.8)(0.1)=0.38,u_I=K_i{I}_k=0.5(0.38)=\mathrm{0.19.}$ Proportional (present): abhi ke error par react karo. $u_P=K_p{e}_k=2(0.8)=\mathrm{1.6.}$ Derivative (future): backward difference $\dot{e}$ estimate karta hai; error gir raha hai, to yeh negative hai — ek brake. $u_D=K_d\frac{e_k-e_{k-1}}{\Delta t}=1\cdot \frac{0.8-1.0}{0.1}=-\mathrm{2.0.}$ Total: $u_k=1.6+0.19-2.0=-\mathrm{0.21.}$ Interpretation: chahe error positive hai (abhi bhi target se neeche), error itni tezi se gir raha hai ki D dominate karta hai aur slight reversal command karta hai — yeh action mein anti-overshoot instinct hai.

D error ki rate, $\dot{e}$ dekhta hai. Yaad raho $e=r-y$, to agar $y$ badhta hai, $e$ girta hai.

• (a) $y$ target ki taraf upar ja raha hai ⇒ $e$ gir raha hai ⇒ $\dot{e}<0$ ⇒ $u_D<0$: brakes lagata hai approach ko overshoot se bachane ke liye.
• (b) $y$ overshoot ho gaya, ab neeche wapas aa raha hai ⇒ $e$ wapas 0 ki taraf badh raha hai ⇒ $\dot{e}>0$ ⇒ $u_D>0$: recovery mein help karta hai.
• (c) $y$ steady hai ⇒ $\dot{e}=0$ ⇒ $u_D=0$: D rest par kuch nahi kehta — isliye D akela kabhi steady offset fix nahi kar sakta.
• (d) $y$ door drift kar raha hai ⇒ $|e|$ badh raha hai ⇒ $\dot{e}$ ka same sign hai $e$ se ⇒ $u_D$ correction ko reinforce karta hai: runaway se pehle hi ladta hai. Zero case (c) key insight hai: D steady state mein exactly kuch contribute nahi karta, isliye yeh offset na cause kar sakta hai na cure.

• Flight A — persistent offset. ==$K_i$== badhao. Ek constant residual error matlab loop wahan rest karta hai jahan $u=K_p{e}_{ss}$, jo nonzero hai sirf tab jab $e_{ss}\ne 0$ ho. Integral action add karne/badhane se resting condition $K_i{e}_{ss}=0$ force hoti hai, to accumulator offset ko zero tak ramp karta hai. (Yeh type elevation hai — integrator system type ko ek se badha deta hai.)
• Flight B — overshoot & ringing. ==$K_d$== badhao. Ringing matlab approach under-damped hai. D $\dot{e}$ par react karta hai, exactly tab command subtract karta hai jab error sabse tezi se collapse ho raha hota hai, initial response slow kiye bina damping add karta hai.
• Flight C — motor chatter. ==$K_d$== kam karo aur/ya filter karo. Differentiation high-frequency sensor noise ko multiply karta hai: tiny fast wiggles ka slope bahut bada hota hai, to $\dot{e}$ explode karta hai aur D violent, useless twitches command karta hai. Fix: chota $K_d$ plus low-pass filtered derivative.

Saturation ke dauran, integrator error khata rehta hai chahe actuator kuch aur na kar sake. Constant $e=2$ par $3$s mein: $\Delta u_I=K_{i}eT=(0.8)(2)(3)=\mathrm{4.8.}$ Yeh $4.8$ jo already tha uske upar store hota hai — ek phantom command jo actuator ne actually deliver nahi ki. Jab $y$ finally target tak pahunchta hai, integrator ko pehle saara $4.8$ unwind karna padta hai uske output ke girne se pehle, to $y$ target se kaafi door nikal jaata hai. Yahi integrator windup hai. Fix: anti-windup — integrator ko freeze (ya clamp) karo jab bhi actuator saturated ho, taaki koi phantom charge na bane. (Related tuning wisdom: Ziegler–Nichols tuning $K_i$ ko modest isliye rakhta hai.)

Spec ke against term-by-term reason karo:

1. Requirement (i) zero steady-state error ⇒ hume I zaroor add karna hai. Sirf integral resting condition $K_i{e}_{ss}=0\Rightarrow e_{ss}=0$ deta hai. Is slow plant par windup se bachne ke liye $K_i$ modest rakhho.
2. Requirement (ii) mild overshoot ⇒ damping ke liye ek chhota D add karo — lekin requirement (iii) noisy sensor matlab heavy D noise amplify karega, to chhota, filtered derivative use karo. Agar overshoot P+I ke saath already acceptable hai, to D skip bhi kar sakte ho (PI controller slow, noisy thermal systems par common hai).
3. Baseline speed ke liye $K_p=4$ raho. To design hai PI (optionally ek filtered D), isliye choose kiya kyunki har requirement exactly ek term ke kaam se map karti hai.

Steady-state integral output: rest par $e_{ss}=0$, to P $K_p\cdot 0=0$ contribute karta hai aur D $K_d\cdot 0=0$ contribute karta hai. Integral ko isliye poori disturbance cancellation supply karni padti hai: $u_I=d=10.$ Integrator ki stored value wahan settle hoti hai jo $K_{i}I=10$ banaye; exact $I$ $K_i$ par depend karta hai, lekin output $u_I=10$ physics se fixed hai.

$u_P=K_p{e}_k=3(0.4)=1.2,$ $u_I=K_i{I}_k=0.5(1.2)=0.6,$ $u_D=K_d\frac{e_k-e_{k-1}}{\Delta t}=2\cdot \frac{0.4-0.7}{0.05}=2\cdot (-6)=-12.$ $u_k=1.2+0.6-12=-\mathrm{10.2.}$ Dominant term: D, aur strongly negative. Chahe abhi bhi positive error hai aur positive stored history hai, error bahut tezi se collapse ho raha hai ($-6$ per second), to derivative brakes jam deta hai — loop aggressively ek overshoot ko prevent kar raha hai. Yeh parent note se "pehle se un-turning shuru karo" wali instinct hai, quantified.

Kya tum, bina dekhe, ye kar sakte ho:

• $e=r-y$ state karo aur kyun sign matter karta hai (P1.2)?
• Pure P ke liye $e_{ss}=d/K_p$ compute karo aur explain karo kyun I use zero karta hai (P2.1, 2.2)?
• Haath se ek discrete PID update run karo (P3.1, 5.2)?
• Char motion cases mein $u_D$ ka sign do frozen case including (P3.2)?
• Windup ko ek number se explain karo aur uska fix batao (P4.2)?
• Ek spec se PI/PID design karo aur resting integral output nikalo (P5.1)? Agar koi shaky lage, aage badhne se pehle woh problem dobara kholo.