PID tuning — Ziegler-Nichols, loop shaping: Exercises

3.5.39 · D4 · Physics › Guidance, Navigation & Control (GNC) › PID tuning — Ziegler-Nichols, loop shaping

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Ek first-order factor ka phase. Factor $(1 + j ω τ)$ ka phase hai $arctan (ω τ)$ ; $\frac{1}{1 + j ω τ}$ ka phase hai $- arctan (ω τ)$ . Arctan kyun? Kyunki $1 + j ω τ$ complex plane mein ek aisa point hai jiska horizontal part $1$ hai aur vertical part $ω τ$ hai; yeh horizontal axis ke saath jo angle banata hai, usका tangent (opposite $ω τ$ over adjacent $1$ ) barabar hai $ω τ$ ke — aur "kis angle ka yeh tangent hai?" exactly yahi $arctan$ answer karta hai. Branch caution: $arctan$ hamesha sirf $- 90°$ aur $+ 90°$ ke beech ka angle return karta hai, jo yahan theek hai kyunki real part $+ 1$ hai (point right half-plane mein baitha hai). Agar kabhi aisa factor mile jiska real part negative ho jaye (jaise $(1 + j ω τ_{1}) (1 + j ω τ_{2})$ multiply out karne par, ya $(- 1 + j ω τ)$ jaisa factor), to asli angle $\pm 90°$ se bahar chali jaati hai aur tumhe raw $arctan$ mein $180°$ add ya subtract karna hoga — calculator ki principal value use khamoshi se wapas fold kar deti hai. Is page ke har factor ka real part positive rehta hai, isliye yahan seedha $arctan$ bilkul safe hai.
Ek first-order factor ki magnitude. $∣1 + j ω τ ∣ = 1 + (ω τ)^{2}$ — usi chhote triangle ka hypotenuse.
Ek pure integrator $\frac{1}{j ω}$ . Magnitude $1/ ω$ , phase exactly $- 90°$ . Exactly $- 90°$ kyun? $j$ se divide karna kisi bhi complex number ko ek quarter turn clockwise rotate karta hai; arctan ki zaroorat nahi kyunki point seedha negative imaginary axis par neeche baitha hai.

Closed-loop Ziegler–Nichols standard form mein full PID ke liye rules deta hai: $K_{p} = 0.6 K_{u}$ , $T_{i} = T_{u} /2$ , $T_{d} = T_{u} /8$ . Ek test mein mila $K_{u} = 10$ aur $T_{u} = 4$ s. $K_{p}$ , $T_{i}$ , $T_{d}$ likhो aur $K_{i}$ aur $K_{d}$ mein convert karo.

Recall Solution 1.1

KYA karna hai: table padho, phir $K_{i} = K_{p} / T_{i}$ aur $K_{d} = K_{p} T_{d}$ se convert karo (parent note se).

$K_{p} = 0.6 \times 10 = 6$ .
$T_{i} = 4/2 = 2$ s $\Rightarrow K_{i} = K_{p} / T_{i} = 6/2 = 3$ .
$T_{d} = 4/8 = 0.5$ s $\Rightarrow K_{d} = K_{p} T_{d} = 6 \times 0.5 = 3$ .

Conversion kyun: table times mein bolta hai ( $T_{i}$ , $T_{d}$ ); tumhara code gains chahta hai ( $K_{i}$ , $K_{d}$ ). $K_{p}$ se multiply/divide karna dono forms ke beech ka dictionary hai.

Answer: $K_{p} = 6, K_{i} = 3, K_{d} = 3$ .

Problem 1.2

Ek Bode plot par open loop $L (j ω)$ $0$ dB (magnitude $= 1$ ) cross karta hai $ω_{c} = 5$ rad/s par, aur wahan $∠ L = - 140°$ hai. Phase margin batao. Kya yeh healthy $30°$ – $60°$ band ke andar hai?

Recall Solution 1.2

Definition (parent): $PM = 180° + ∠ L (j ω_{c})$ . $PM = 180° + (- 140°) = 40°.$ $40°$ $30°$ – $60°$ ke andar hai, isliye haan, healthy hai.

Yeh formula kyun: instability tab hoti hai jab phase $- 180°$ tak pahunch jaaye jabki gain abhi bhi $1$ ho. Phase margin woh bacha hua angle measure karta hai jo tumhare paas hai us cliff se pehle. Yahan tum crossover se pehle aur $40°$ phase kho sakte ho.

Level 2 — Application

Problem 2.1

Plant $G (s) = \frac{1}{( s + 1 ) ^{3}}$ (parent ka test plant). Ultimate frequency, gain aur period confirm karo, phir ZN PID gains $K_{p}, K_{i}, K_{d}$ do.

Recall Solution 2.1

Step A — $ω_{u}$ dhundo (phase $- 180°$ hit kare). Pehle phase kyun? Stability edge phase $= - 180°$ se define hoti hai; us equation mein sirf $ω$ hai, isliye use akele solve karo. $∠ G (j ω) = - 3 arctan ω = - 180° \Rightarrow arctan ω_{u} = 60° \Rightarrow ω_{u} = tan 60° = 3 .$ Step B — wahan magnitude se $K_{u}$ milta hai. $∣ G (j ω_{u}) ∣ = (ω_{u}^{2} + 1)^{- 3/2} = (3 + 1)^{- 3/2} = 4^{- 3/2} = \frac{1}{8}, K_{u} = \frac{1}{∣ G ∣} = 8.$ $K_{u} = 1/∣ G ∣$ kyun? Stability edge par hum pure P gain itna badhate hain jab tak open loop $L = K_{u} G$ ki magnitude exactly $1$ na ho us frequency $ω_{u}$ par jahan phase $- 180°$ hai — yeh woh "signal wapas inverted aur same size mein reinforced hamesha ke liye" condition hai parent se. $∣ L (j ω_{u}) ∣ = K_{u} ∣ G (j ω_{u}) ∣ = 1$ set karke $K_{u}$ ke liye solve karne par milta hai $K_{u} = 1/∣ G (j ω_{u}) ∣$ : ultimate gain exactly woh number hai jo plant ki simi hui magnitude ko wapas $1$ tak uthata hai. Step C — period. $T_{u} = \frac{2 π}{ω _{u}} = \frac{2 π}{3} \approx 3.628$ s. Step D — ZN table.

$K_{p} = 0.6 \times 8 = 4.8$ .
$T_{i} = T_{u} /2 = 1.814 \Rightarrow K_{i} = 4.8/1.814 = 2.646$ .
$T_{d} = T_{u} /8 = 0.4535 \Rightarrow K_{d} = 4.8 \times 0.4535 = 2.177$ .

Answer: $K_{p} = 4.8, K_{i} \approx 2.65, K_{d} \approx 2.18$ — parent ke example se match karta hai.

Problem 2.2 (open-loop / reaction-curve method)

Ek heater par open-loop step test ek FOPDT model fit karta hai jisme static gain $K = 2$ , time constant $T = 10$ s, aur dead time $L = 3$ s hai. Yaad karo $R = K / T$ . ZN open-loop PID parameters $K_{p}, T_{i}, T_{d}$ do aur phir $K_{i}, K_{d}$ .

Recall Solution 2.2

Step A — slope parameter. $R = K / T = 2/10 = 0.2$ (per second). Step B — ZN open-loop formulas (parent). $K_{p} = \frac{1.2}{R L} = \frac{1.2}{0.2 \times 3} = \frac{1.2}{0.6} = 2.$ $T_{i} = 2 L = 6 s, T_{d} = 0.5 L = 1.5 s .$ Step C — gains. $K_{i} = K_{p} / T_{i} = 2/6 = 0.333$ , $K_{d} = K_{p} T_{d} = 2 \times 1.5 = 3$ .

Yeh ratios kyun: dead time $L$ villain hai — yeh pure phase lag hai bina kisi gain drop ke, isliye zyada $L$ se chhota $K_{p}$ force hota hai (note karo $K_{p} \propto 1/ L$ ). Integral aur derivative times $L$ se pin hain kyunki aise plants ke liye natural oscillation period roughly $4 L$ hoti hai.

Answer: $K_{p} = 2, T_{i} = 6, T_{d} = 1.5, K_{i} \approx 0.333, K_{d} = 3$ .

Level 3 — Analysis

Problem 3.1

Usi plant $G (s) = \frac{1}{( s + 1 ) ^{3}}$ ke liye, maano hum pure P control use karte hain $K_{p} = 4$ ke saath (jo $K_{u} = 8$ se neeche hai). Gain margin dB mein dhundho. Interpret karo.

Recall Solution 3.1

Gain margin kya hai: us frequency $ω_{180}$ par jahan phase $= - 180°$ hai, aur kitna gain badhe tab $∣ L ∣ = 1$ ho jaaye. Problem 2.1 se, woh frequency hai $ω_{u} = 3$ aur wahan $∣ G ∣ = 1/8$ hai. $K_{p} = 4$ ke saath: $∣ L (j ω_{180}) ∣ = 4 \times \frac{1}{8} = \frac{1}{2}$ . $GM = \frac{1}{∣ L ( j ω _{180} ) ∣} = \frac{1}{0.5} = 2 ⟹ 20 lo g_{10} 2 = 6.02 dB .$ Interpretation: tum gain double kar sakte ho (4 se 8= $K_{u}$ tak) sustained oscillation se pehle — exactly consistent hai ultimate gain ke $8$ hone se. " $6$ dB minimum" kahan se aata hai: yeh ek design criterion hai, nature ka koi law nahi. Standard robustness rule of thumb (dekho Stability margins (gain & phase margin) aur Nyquist stability criterion) maangta hai ki gain margin kam se kam factor of $2$ ho — yaani $\geq 6.02$ dB — taaki ordinary modelling errors aur gain drift (component tolerances, temperature) poora margin nahi kha sake aur Nyquist curve ko $- 1$ point ke paas na dhakele. Isliye $K_{p} = 4$ exactly us recommended lower bound par baitha hai: acceptable hai, lekin koi extra room nahi — yahi wajah hai ki ek designer aur peeche hat jaayega.

Problem 3.2

Phase use karke explain karo kyun integral action $K_{i} / s$ add karna ek loop mein jo pehle se $PM = 45°$ tha, phase margin reduce karta hai — aur estimate karo loss kitna hoga agar integral corner $ω_{i} = K_{i} / K_{p}$ ko $ω_{c} /10$ par rakha jaaye.

Recall Solution 3.2

Integral phase ko kya karta hai: factor $\frac{1}{j ω}$ exactly $- 90°$ contribute karta hai; lekin ek real PI controller $K_{p} (1 + \frac{1}{T _{i} s})$ frequency $ω$ par sirf $- arctan (ω_{i} / ω)$ contribute karta hai, kyunki apne corner $ω_{i}$ se kaafi upar integral term negligible ho jaati hai aur phase $0°$ ki taraf wapas aati hai. $ω = ω_{c}$ par estimate karo jab $ω_{i} = ω_{c} /10$ ho: $Δ ϕ = - arctan (\frac{ω _{i}}{ω _{c}}) = - arctan (0.1) = - 5.71°.$ Naya phase margin $\approx 45° - 5.7° = 39.3°$ . Yeh loss kyun hona hi hai: integral sirf lag add karta hai (lead kabhi nahi), isliye yeh phase ko crossover par sirf neeche kheench sakta hai; design trick yeh hai ki iska corner $ω_{c}$ se kaafi neeche rakho taaki $ω_{c}$ tak pahunchte tak lag almost decay ho jaaye — isliye " $ω_{i} \approx ω_{c} /10$ " rule loss ko tolerable few degrees mein rakhta hai.

Level 4 — Synthesis

Problem 4.1 (lead design by phase margin)

Plant $G (s) = \frac{1}{s ( s + 1 )}$ . Ek PD controller $C (s) = K_{p} (1 + T_{d} s)$ design karo taaki gain crossover $ω_{c} = 2$ rad/s par ho aur phase margin $PM = 45°$ ho. $T_{d}$ aur $K_{p}$ dhundho.

Figure s01 (neeche) Step C ka geometric core hai. Yeh single complex number $1 + j T_{d} ω_{c}$ ko ek right triangle ke roop mein draw karta hai: adjacent side $= 1$ (kaali), opposite side $= T_{d} ω_{c}$ (kaali), hypotenuse laal arrow. Laal arrow dekho — woh horizontal axis ke upar jo angle banata hai, WOH phase lead hai jo PD zero donate karta hai, aur is triangle se "opposite over adjacent" padhna exactly yahi wajah hai ki $tan (lead) = T_{d} ω_{c}$ .

Figure — PID tuning — Ziegler-Nichols, loop shaping

Recall Solution 4.1

Step A — $ω_{c} = 2$ par plant ka phase. $G$ mein ek integrator hai ( $- 90°$ flat) aur ek lag $(s + 1)$ ( $- arctan ω$ ). $∠ G (j 2) = - 90° - arctan 2 = - 90° - 63.43° = - 153.43°.$ Step B — controller ko kitna phase supply karna chahiye? Hum chahte hain ki $ω_{c}$ par total open-loop phase $- 135°$ ho, kyunki $PM = 180° + ∠ L = 45°$ ka matlab hai $∠ L = - 135°$ . $ϕ_{lead} = ∠ L - ∠ G = - 135° - (- 153.43°) = + 18.43°.$ Step C — ek PD zero lead supply karta hai $arctan (T_{d} ω_{c})$ (figure s01 mein yeh triangle hai). Arctan phir se kyun? Factor $1 + j T_{d} ω$ woh chhota triangle hai vertical side $T_{d} ω$ ke saath; axis ke upar uska angle $arctan (T_{d} ω)$ hai — positive, yaani phase lead. $arctan (T_{d} \cdot 2) = 18.43° \Rightarrow T_{d} \cdot 2 = tan 18.43° = 0.3333 \Rightarrow T_{d} = 0.1667.$ Step D — $ω_{c}$ par magnitude $1$ set karo (iska matlab hi "crossover" hai). $∣ G (j 2) ∣ = \frac{1}{2 2 ^{2} + 1} = \frac{1}{2 5} = 0.2236, ∣1 + j T_{d} \cdot 2∣ = 1 + 0.333 3^{2} = 1.0541.$ $∣ L (j 2) ∣ = K_{p} \cdot 0.2236 \cdot 1.0541 = 1 \Rightarrow K_{p} = \frac{1}{0.2236 \times 1.0541} = 4.243.$

Answer: $T_{d} \approx 0.167$ s, $K_{p} \approx 4.24$ . Parent example se match karta hai.

Figure s02 (neeche) Step B ko phase budget ke roop mein visible banata hai. Har kaali bar $ω_{c} = 2$ par ek factor ka phase contribution hai; woh stack hote hain. Laal "total" bar dekho — woh exactly laal dotted $- 135°$ target line par land karta hai — yahi phase-margin condition meet ho rahi hai, aur dashed $- 180°$ line jiske upar laal bar kabhi nahi pahunchna chahiye woh instability cliff hai.

Problem 4.2 (us lead design se full PID banao)

Problem 4.1 se $K_{p} = 4.24$ , $T_{d} = 0.167$ lo aur integral action add karo corner $ω_{i} = ω_{c} /10 = 0.2$ rad/s ke saath (rule of thumb). $K_{i}$ do, aur batao ki isse phase margin kitna trim hota hai.

Recall Solution 4.2

Step A — corner se $K_{i}$ . $ω_{i} = K_{i} / K_{p} \Rightarrow K_{i} = ω_{i} K_{p} = 0.2 \times 4.243 = 0.8485$ . Step B — $ω_{c}$ par phase penalty (jaise 3.2 mein): $Δ ϕ = - arctan (ω_{i} / ω_{c}) = - arctan (0.2/2) = - arctan (0.1) = - 5.71°.$ Isliye phase margin $45°$ se girhkar lagbhag $39.3°$ ho jaata hai — abhi bhi healthy hai. Answer: $K_{i} \approx 0.849$ , PM $\approx 39.3°$ . (Practice mein tum $T_{d}$ thoda upar nudge karoge taaki khoye hue $5.7°$ recover ho sakein.)

Level 5 — Mastery

Problem 5.1 (do designs judge karo)

$G (s) = \frac{1}{( s + 1 ) ^{3}}$ ke liye tumhare paas do candidate PID controllers hain:

A (raw ZN): $K_{p} = 4.8, K_{i} = 2.65, K_{d} = 2.18$ (2.1 se).
B (detuned): wahi $K_{p}$ $30%$ reduce kiya, integral aur derivative times unchanged.

B ke gains compute karo, aur ek line mein argue karo ki position servo ke liye kaunsa ship karoge jisme zyada overshoot nahi hona chahiye.

Recall Solution 5.1

Step A — B ka $K_{p}$ . $K_{p}^{B} = 0.7 \times 4.8 = 3.36$ . Step B — times fixed rakho ( $T_{i} = 1.814$ , $T_{d} = 0.4535$ ), isliye: $K_{i}^{B} = K_{p}^{B} / T_{i} = 3.36/1.814 = 1.852, K_{d}^{B} = K_{p}^{B} T_{d} = 3.36 \times 0.4535 = 1.524.$ Step C — judgement. Raw ZN quarter-amplitude decay aim karta hai $\approx 25%$ overshoot — position servo ke liye bahut bouncy hai. $K_{p}$ ko $30%$ detune karna crossover neeche karta hai, phase margin khareedhta hai, aur overshoot thodi speed ki keemat par kam karta hai. B ship karo.

Answer: $K_{p}^{B} = 3.36, K_{i}^{B} \approx 1.85, K_{d}^{B} \approx 1.52$ .

Problem 5.2 (mastery synthesis: full loop-shaping check)

Sab combine karo. Problem 4.2 ke lead-plus-integral controller ke liye ( $K_{p} = 4.243$ , $T_{d} = 0.1667$ , $ω_{i} = 0.2$ ) on $G (s) = \frac{1}{s ( s + 1 )}$ , $ω_{c} = 2$ par final open-loop phase verify karo aur resulting phase margin numerically confirm karo.

Recall Solution 5.2

$ω_{c} = 2$ par phases assemble karo (sab superpose hoti hain):

Plant integrator: $- 90°$ .
Plant lag $(s + 1)$ : $- arctan 2 = - 63.43°$ .
Controller lead zero $(1 + T_{d} s)$ : $+ arctan (0.1667 \times 2) = + arctan (0.3333) = + 18.43°$ .
Controller integral lag: $- arctan (ω_{i} / ω_{c}) = - arctan (0.1) = - 5.71°$ . $∠ L (j 2) = - 90 - 63.43 + 18.43 - 5.71 = - 140.71°.$ $PM = 180° + (- 140.71°) = 39.29°.$ Interpretation: exactly wahi $\approx 39.3°$ jo 4.2 mein predict kiya tha. Design self-consistent hai — integral ki loss exactly wahan dikhi jahan humne expect ki thi, aur loop safely stable hai fast $ω_{c} = 2$ rad/s ke saath. Poore $45°$ recover karne ke liye, $T_{d}$ badhao taaki lead $18.43° + 5.71° \approx 24°$ ho jaaye.

Answer: $∠ L (j 2) \approx - 140.7°$ , $PM \approx 39.3°$ .

Recall Final self-check (chhupao aur jawab do)

ZN closed loop ne $(s + 1)^{- 3}$ par measure kiya $K_{u} = 8$ , $T_{u} = 3.63$ — $3$ kahan se aaya? ::: $arctan ω_{u} = 60°$ se, isliye $ω_{u} = tan 60° = 3$ , woh frequency jahan phase $- 180°$ hit karti hai.
Ultimate gain $∣ G (j ω_{u}) ∣$ ka reciprocal kyun hai? ::: Kyunki edge par $∣ L ∣ = K_{u} ∣ G ∣ = 1$ hai, isliye $K_{u} = 1/∣ G (j ω_{u}) ∣$ plant ki magnitude ko wapas $1$ tak lift karta hai.
Ek PD zero phase margin kyun raise karta hai? ::: Yeh crossover ke paas $+ arctan (T_{d} ω)$ lead contribute karta hai, $∠ L$ ko $- 180°$ se door karta hai.
$6.02$ dB gain margin literally kya allow karta hai, aur yeh minimum kyun hai? ::: Oscillation se pehle loop gain double karna; $6$ dB (factor $2$ ) standard robustness criterion hai taaki gain drift margin erase na kar sake.
$ω_{i} = ω_{c} /10$ kyun rakhte hain? ::: Taaki integral ki lag $- arctan (ω_{i} / ω_{c}) \approx - 5.7°$ almost koi phase margin cost na kare jabki low-frequency gain boost karta rahe.

Related vault notes: PID controller basics · Bode plot & frequency response · Stability margins (gain & phase margin) · Nyquist stability criterion · Integral windup and anti-windup · Root locus method · State-space control & LQR.

Exercises — PID tuning — Ziegler-Nichols, loop shaping

Level 1 — Recognition

Problem 1.1