Optimal guidance — ZEM - ZEV formulation: Exercises

3.5.52 · D4 · Physics › Guidance, Navigation & Control (GNC) › Optimal guidance — ZEM - ZEV formulation

Shuru karne se pehle, ek shared cheat-sheet taaki koi symbol bina samjhe na aaye.

Definition Domain rule:

$t_{g o} > 0$ , aur thrust saturate hoti hai Har command mein $t_{g o}$ denominator mein hai, isliye ye law sirf $t_{g o} > 0$ (strictly positive time left) ke liye defined hai. Divide karne se pehle do edge cases handle karne zaroori hain:

$t_{g o} = 0$ — koi time nahi bacha; fraction undefined hai. Practice mein guidance $t_{g o} < 0.5$ s jaise ek chhote cutoff par last command freeze kar deta hai (ya terminal/touchdown controller ko hand off kar deta hai).
$t_{g o} < 0$ — aap intended final time se aage nikal gaye; coast prediction meaningless hai. Guidance ko re-plan karna hoga ya engagement over declare karna hoga.
Saturation: ek real thruster ka ek maximum $a_{ma x}$ hota hai. Commanded $a_{c}$ hamesha clip hota hai: agar $∥ a_{c} ∥ > a_{ma x}$ toh aap $a_{ma x} a_{c} /∥ a_{c} ∥$ apply karte ho. Isliye (L3.2) $t_{g o} \to 0$ par theoretical blow-up dangerous hai: $a_{ma x}$ ke baad law jo maangta hai woh deliver nahi kar sakta.

Yahan "coast" ka matlab hai: engine band kar do, sirf jaani-pehchani gravity $g$ act karne do. Neeche sab kuch un chaar boxes aur double-integrator model $\dot{r} = v, \dot{v} = a + g$ par arithmetic hai.

Level 1 — Recognition

L1.1 — Time-to-go aur coast position padhna

$t = 4$ s par ek vehicle ko $t_{f} = 19$ s par apne target par pahunchna hai. Abhi $r = 500$ m, $v = - 30$ m/s, aur $g = - 9.81$ m/s² (Earth, seedha neeche). $t_{g o}$ aur coast position $r_{co a s t} (t_{f}) = r + v t_{g o} + \frac{1}{2} g t_{g o}^{2}$ find karo (jahan aap thrust off hone par land karte).

Recall Solution

KYA: $t_{g o}$ bas remaining time hai. KYU: har formula isko use karta hai, isliye pehle ishe compute karo. $t_{g o} = t_{f} - t = 19 - 4 = 15 s .$ Domain rule check karo: $t_{g o} = 15 > 0$ , toh law well-defined hai. Coast position — $r = 500$ , $v = - 30$ , $g = - 9.81$ , $t_{g o} = 15$ daalo: $r_{co a s t} = 500 + (- 30) (15) + \frac{1}{2} (- 9.81) (1 5^{2}) = 500 - 450 - 1103.6 = - 1053.6 m .$ Negative number ka matlab hai: unpowered coast karo aur aap target level se 1053.6 m neeche crash karo — gravity ne aapko bahut aage kheench liya. Yahi woh error hai jo ZEM measure karega.

L1.2 — Kaun sa law?

Har mission ke liye batao ki $6/ t_{g o}^{2}, 2/ t_{g o}$ soft law use hoga ya $3/ t_{g o}^{2}$ intercept law: (a) ek missile aircraft ko hit karta hai, (b) ek Moon lander zero speed par touch down karta hai, (c) ek station ke saath docking, (d) ek shell jo sirf ek map coordinate tak pahunchne ki koshish kar rahi hai.

Recall Solution

Rule: kya aapko final velocity ki parwah hai? Agar haan → soft law (dono terms). Agar sirf wahan pahunchne ki parwah hai → intercept law. (a) intercept ( $3/ t_{g o}^{2}$ ) — bas hit karo. (b) soft — zero speed par pahunchna zaroori hai. (c) soft — station ki velocity match karni hai, isi idea ke liye landing mein Powered Descent Guidance (Apollo E-Guidance) dekho. (d) intercept.

Level 2 — Application

L2.1 — Full 1-D soft-landing command

Ek Mars lander: altitude $r = 200$ m, velocity $v = - 40$ m/s (neeche), target $r_{f} = 0$ , $v_{f} = 0$ , $g = - 3.71$ m/s², $t_{g o} = 8$ s. ZEM, ZEV, aur commanded acceleration $a_{c}$ compute karo.

Recall Solution

Step 1 — ZEM (coast karte toh kahan land karte?). Kyun pehle: command ko iske zaroorat hai. $ZEM = 0 - 200 - (- 40) (8) - \frac{1}{2} (- 3.71) (8^{2}) = - 200 + 320 + 118.72 = 238.72 m .$ Step 2 — ZEV (hamari speed kitni galat hoti?). $ZEV = 0 - (- 40) - (- 3.71) (8) = 40 + 29.68 = 69.68 m/s .$ Step 3 — Command (soft law apply karo). $a_{c} = \frac{6}{8 ^{2}} (238.72) - \frac{2}{8} (69.68) = \frac{6}{64} (238.72) - 0.25 (69.68) .$ $a_{c} = 22.38 - 17.42 = 4.96 m/s^{2} .$ Yeh acceleration gravity ke upar hai (gravity ZEM/ZEV ke andar pehle se hai). Upward braking thrust ko $a_{c}$ supply karna hai aur gravity se bhi ladna hai.

L2.2 — 2-D intercept command

Ek interceptor compute karta hai $ZEM = (600, - 800)$ m $t_{g o} = 5$ s ke saath. $a_{c}$ aur uska magnitude find karo.

Recall Solution

Position-only (intercept) law — kyun: hum sirf hit karna chahte hain. $a_{c} = \frac{3}{5 ^{2}} (600, - 800) = \frac{3}{25} (600, - 800) = (72, - 96) m/s^{2} .$ Magnitude: $7 2^{2} + 9 6^{2} = 5184 + 9216 = 14400 = 120 m/s^{2} .$ Yeh law navigation ratio $N = 3$ ke saath Proportional Navigation hai — same physics, ZEM-flavoured.

Level 3 — Analysis

L3.1 — ZEV term ka sign

L2.1 mein ZEM term ne $+ 22.38$ aur ZEV term ne $- 17.42$ diya. Explain karo kyun ZEV term subtract karta hai, aur agar aap (galti se) ise add karte toh physically kya hota. $a_{c}$ ko $+$ sign ke saath recompute karo aur interpret karo.

Recall Solution

Kyun subtract karta hai: ZEM keh raha hai "coast error ki wajah se aap target se 238.7 m upar overshoot karoge, toh use cut karne ke liye push karo," jabki ZEV keh raha hai "aap 69.7 m/s bahut fast pahunchoge, toh aapko brake karna hai — yaani position push ke opposite direction mein accelerate karo end ke paas." $2 \times 2$ Lagrange solution (parent derivation ka Step 4) ise opposite signs ke roop mein encode karta hai. Galat ( $+$ ) version: $a_{c}^{wrong} = 22.38 + 17.42 = 39.80 m/s^{2} .$ Yeh correct command $4.96$ ka 8× ke aas paas hai. Physically aap braking-hunger ko position-hunger mein add kar rahe ho, over-decelerate karte, short aur high ruk jaate, phir gir jaate — ek failed landing. Minus sign hi "gently pahuncho" ko "hard pahuncho" se alag banata hai.

L3.2 — $t_{g o} \to 0$ blow-up

Soft law ke liye, $ZEM$ aur $ZEV$ ko fixed small residuals $ZEM = 1$ m, $ZEV = 0.5$ m/s maano. $t_{g o} = 5, 1, 0.5, 0.1$ s par $a_{c}$ tabulate karo. Yeh kya dikhata hai kab aapko errors khatam karni chahiye?

Recall Solution

$a_{c} (t_{g o}) = \frac{6}{t _{g o}^{2}} (1) - \frac{2}{t _{g o}} (0.5) = \frac{6}{t _{g o}^{2}} - \frac{1}{t _{g o}} .$

$t_{g o}$ (s)	$6/ t_{g o}^{2}$	$1/ t_{g o}$	$a_{c}$ (m/s²)
5	0.24	0.20	0.04
1	6	1	5
0.5	24	2	22
0.1	600	10	590

Jaise $t_{g o} \to 0$ hota hai $1/ t_{g o}^{2}$ term dominate karta hai aur $a_{c}$ explode ho jaata hai. Sabak: $t_{g o} = 5$ par jo residual harmless tha woh $t_{g o} = 0.1$ par monstrous, thruster-saturating command maangta hai. Isliye aapko ZEM/ZEV zero karna tab hai jab $t_{g o}$ abhi bhi bada ho — yahi wajah hai ki ZEM/ZEV continuous feedback ke roop mein run hota hai.

Neeche ki figure ko aise padhein: magenta curve $a_{c} (t_{g o})$ hai; chaar violet dots table ke rows hain. Apni aankhein right se left sweep karo (yahi direction hai jisme time actually chalta hai). Notice karo curve $t_{g o} > 2$ ke liye nearly flat aur tiny hai, phir left axis ke paas aate hi sharply upar hook karti hai — orange arrow us blow-up ko mark karta hai. Message visual hai: saari danger last fraction of a second mein rehti hai, jahan saturation ceiling $a_{ma x}$ (ek horizontal cap jo aap upar draw kar sakte ho) smash ho jaati.

Figure — Optimal guidance — ZEM - ZEV formulation

Level 4 — Synthesis

L4.1 — Forecast karo phir verify karo (intercept self-shrinking)

Ek interceptor perfect collision course par hai. $t_{g o} = 6$ s par uska $ZEM = (300, - 450)$ m hai. Forecast: agar woh sach mein on track hai, toh $t_{g o} = 3$ s par $ZEM$ bada hoga ya chota? Verify karo ideal on-track relation $ZEM \propto t_{g o}^{2}$ assume karke (jo tab hold karta hai jab applied command exactly $3 ZEM / t_{g o}^{2}$ ho aur koi naya disturbance na aaye), aur $t_{g o} = 3$ par $a_{c}$ recompute karo.

Recall Solution

Forecast: on track, gap close ho raha hai, ZEM chota hona chahiye. Guess: chota. Verify: $ZEM \propto t_{g o}^{2}$ ke saath, $t_{g o}$ ko 6 se 3 tak halve karna ZEM ko quarter kar deta hai: $ZEM (3) = (\frac{3}{6})^{2} (300, - 450) = \frac{1}{4} (300, - 450) = (75, - 112.5) m .$ Naya command: $a_{c} = \frac{3}{3 ^{2}} (75, - 112.5) = \frac{1}{3} (75, - 112.5) = (25, - 37.5) m/s^{2} .$ Interpretation: command magnitude $t_{g o} = 6$ jaise hi same hai (wahan bhi $\frac{3}{36} (300, - 450) = (25, - 37.5)$ tha). Ek ideal course par ZEM/ZEV self-consistently command ko constant aur bounded rakhta hai — koi blow-up nahi. ZEM mein $t_{g o}^{2}$ shrink coefficient mein $1/ t_{g o}^{2}$ growth ko exactly cancel karta hai. Yeh L3.2 ka healthy counterpart hai.

L4.2 — Gravity + full vector soft law combine karo

Ek lunar lander, sab kuch ek horizontal-vertical frame mein: $r = (0, 150)$ m, $v = (10, - 25)$ m/s, target $r_{f} = (0, 0)$ , $v_{f} = (0, 0)$ , $g = (0, - 1.62)$ m/s², $t_{g o} = 10$ s. Full 2-D command $a_{c}$ compute karo.

Recall Solution

Har axis par kaam karo; gravity sirf $y$ par act karti hai. ZEM:

$x$ : $0 - 0 - (10) (10) - \frac{1}{2} (0) (100) = - 100$ m.
$y$ : $0 - 150 - (- 25) (10) - \frac{1}{2} (- 1.62) (100) = - 150 + 250 + 81 = 181$ m.
$ZEM = (- 100, 181)$ m.

ZEV:

$x$ : $0 - 10 - 0 = - 10$ m/s.
$y$ : $0 - (- 25) - (- 1.62) (10) = 25 + 16.2 = 41.2$ m/s.
$ZEV = (- 10, 41.2)$ m/s.

Command $\frac{6}{100} = 0.06$ , $\frac{2}{10} = 0.2$ ke saath:

$x$ : $0.06 (- 100) - 0.2 (- 10) = - 6 + 2 = - 4$ m/s².
$y$ : $0.06 (181) - 0.2 (41.2) = 10.86 - 8.24 = 2.62$ m/s².
$a_{c} = (- 4, 2.62) m/s^{2} .$ $x$ -command negative hai — yeh sideways drift brake karta hai taaki hum pad se bahar na utarein; $y$ -command ek mild upward push hai, (hamesha ki tarah) subtracted ZEV term se soften kiya hua.

Neeche ki figure ko aise padhein: navy dot woh hai jahan hum abhi hain, $(0, 150)$ ; orange star origin par target hai. Magenta arrow ZEM vector hai — yeh coasting-landing spot (magenta dot, down-and-left) se wapas target tak point karta hai, woh miss dikhata hai jo hume mitani hai. Violet arrow hamare current position par command $a_{c} = (- 4, 2.62)$ hai (scaled-up draw kiya taaki visible ho). Uski direction dekho: yeh up-aur-left ki taraf tilt karta hai — leftward tilt $x$ -brake hai jo $+ 10$ m/s sideways drift ko maar raha hai, exactly wahi sign jo humne compute kiya.

Level 5 — Mastery

Definition Do helper objects jo aap L5 mein use karoge

$s = t_{f} - τ$ — ek backward clock: $τ$ calendar time hai jo forward chalta hai, toh $s$ instant $τ$ par remaining seconds count karta hai. "Now" ( $τ = t$ ) par hamein $s = t_{g o}$ milta hai; final time ( $τ = t_{f}$ ) par $s = 0$ milta hai. $s$ use karna messy integrals ko $t_{g o}$ ki clean powers mein badal deta hai.
$λ_{1}, λ_{2}$ — Lagrange multipliers (jo Calculus of Variations & Pontryagin's Minimum Principle mein costate variables bhi kehlaate hain). Yeh constant vectors hain jinka kaam sirf do constraints enforce karna hai — "ZEM hit karo" aur "ZEV fill karo." Inhe optimizer ka woh price samjho jo ek constraint satisfy karne ke liye pay karta hai. Kyunki exactly do constraints hain, exactly do multipliers hain, aur woh time ke saath change nahi karte.

L5.1 — Interception coefficient re-derive karo

Do moment equations se shuru karo ( $λ_{1}, λ_{2}$ constant hain aur $s = t_{f} - τ$ ), $λ_{1} \frac{t _{g o}^{3}}{3} + λ_{2} \frac{t _{g o}^{2}}{2} = ZEM, λ_{1} \frac{t _{g o}^{2}}{2} + λ_{2} t_{g o} = ZEV,$ derive karo position-only command $a_{c} = 3 ZEM / t_{g o}^{2}$ . (Intercept ke liye koi velocity constraint nahi hai, toh ZEV equation side mein rakh do aur require karo ki effort-minimizing profile mein $λ_{2} = 0$ ho, yaani affine control purely $λ_{1} s$ hai.) Abhi apply command hai $a_{c} = λ_{1} t_{g o}$ .

Recall Solution

KYU $λ_{2} = 0$ : sirf ek constraint (ZEM) ke saath, aapko sirf ek price chahiye — ek multiplier. Velocity constraint jo $λ_{2}$ produce karta tha woh drop ho gaya, toh Calculus of Variations & Pontryagin's Minimum Principle se optimal control $a^{*} (τ) = λ_{1} (t_{f} - τ) = λ_{1} s$ hai. ZEM moment equation use karo $λ_{2} = 0$ ke saath: $λ_{1} \frac{t _{g o}^{3}}{3} = ZEM \Rightarrow λ_{1} = \frac{3 ZEM}{t _{g o}^{3}} .$ Command now ( $s = t_{f} - t = t_{g o}$ ): $a_{c} = λ_{1} t_{g o} = \frac{3 ZEM}{t _{g o}^{3}} \cdot t_{g o} = \frac{3 ZEM}{t _{g o}^{2}} . ■$ Woh $3$ PN ka navigation ratio $N = 3$ hai — yaad nahi, kamaya hua.

L5.2 — Full $2 \times 2$ soft coefficients ke liye solve karo

Usi do moment equations ko dono $λ_{1}, λ_{2}$ rakh ke solve karo, phir $a_{c} = λ_{1} t_{g o} + λ_{2}$ banao aur confirm karo ki $\frac{6}{t _{g o}^{2}} ZEM - \frac{2}{t _{g o}} ZEV$ milta hai.

Recall Solution

Har axis identical hai, toh bold drop karo aur $Z_{M}, Z_{V}$ ko scalars maano. $T = t_{g o}$ brevity ke liye. Do moment equations hain: $\frac{T ^{3}}{3} λ_{1} + \frac{T ^{2}}{2} λ_{2} = Z_{M}, \frac{T ^{2}}{2} λ_{1} + T λ_{2} = Z_{V} .$

Step A — simpler (doosri) equation se $λ_{2}$ isolate karo. Kyun yahi: yeh $λ_{2}$ mein linear hai sabse chhote powers ke saath, toh iske liye solve karna sabse kam algebra leta hai. $λ_{2} = \frac{Z _{V}}{T} - \frac{T}{2} λ_{1} .$

Step B — pehli equation mein substitute karo. Kyun: yeh $λ_{2}$ eliminate karta hai aur ek equation ek unknown $λ_{1}$ mein chhodata hai. $\frac{T ^{3}}{3} λ_{1} + \frac{T ^{2}}{2} (\frac{Z _{V}}{T} - \frac{T}{2} λ_{1}) = Z_{M} .$ Bracket term by term expand karo: $\frac{T ^{2}}{2} \cdot \frac{Z _{V}}{T} = \frac{T Z _{V}}{2}$ aur $\frac{T ^{2}}{2} \cdot (- \frac{T}{2} λ_{1}) = - \frac{T ^{3}}{4} λ_{1} .$ Toh $\frac{T ^{3}}{3} λ_{1} - \frac{T ^{3}}{4} λ_{1} + \frac{T Z _{V}}{2} = Z_{M} .$

Step C — $λ_{1}$ terms group karo. Kyun: like terms collect karna $λ_{1}$ padhne deta hai. Common denominator $12$ ke saath fractions combine karo: $\frac{T ^{3}}{3} - \frac{T ^{3}}{4} = \frac{4 T ^{3} - 3 T ^{3}}{12} = \frac{T ^{3}}{12} .$ $\frac{T ^{3}}{12} λ_{1} = Z_{M} - \frac{T Z _{V}}{2} .$

Step D — $λ_{1}$ get karne ke liye denominator clear karo. Kyun: dono sides ko $12/ T^{3}$ se multiply karo taaki $λ_{1}$ akela reh jaaye. $λ_{1} = \frac{12}{T ^{3}} (Z_{M} - \frac{T Z _{V}}{2}) = \frac{12 Z _{M}}{T ^{3}} - \frac{6 Z _{V}}{T ^{2}} .$

Step E — $λ_{2}$ get karne ke liye back-substitute karo. Kyun: Step A ne pehle se $λ_{2}$ ko $λ_{1}$ ke terms mein diya hai; result plug karo. $λ_{2} = \frac{Z _{V}}{T} - \frac{T}{2} (\frac{12 Z _{M}}{T ^{3}} - \frac{6 Z _{V}}{T ^{2}}) = \frac{Z _{V}}{T} - \frac{6 Z _{M}}{T ^{2}} + \frac{3 Z _{V}}{T} = - \frac{6 Z _{M}}{T ^{2}} + \frac{4 Z _{V}}{T} .$

Step F — abhi apply command $a_{c} = λ_{1} T + λ_{2}$ banao. Kyun: control profile $a^{*} (τ) = λ_{1} s + λ_{2}$ hai, aur "now" matlab $s = t_{g o} = T$ hai. $a_{c} = (\frac{12 Z _{M}}{T ^{2}} - \frac{6 Z _{V}}{T}) + (- \frac{6 Z _{M}}{T ^{2}} + \frac{4 Z _{V}}{T}) .$ $Z_{M}$ terms group karo: $\frac{12 Z _{M}}{T ^{2}} - \frac{6 Z _{M}}{T ^{2}} = \frac{6 Z _{M}}{T ^{2}} .$ $Z_{V}$ terms group karo: $- \frac{6 Z _{V}}{T} + \frac{4 Z _{V}}{T} = - \frac{2 Z _{V}}{T} .$ Isliye $a_{c} = \frac{6 Z _{M}}{T ^{2}} - \frac{2 Z _{V}}{T} = \frac{6}{t _{g o}^{2}} ZEM - \frac{2}{t _{g o}} ZEV ■$ Yeh raha: $6$ , $2$ , aur crucial minus sign — sab do-price Lagrange solution se seedha nikal rahe hain.

Recall Self-test cloze

Intercept law coefficient hai ::: $3/ t_{g o}^{2}$ (navigation ratio $N = 3$ ) Soft law ZEV term ko coefficient ke saath subtract karta hai ::: $2/ t_{g o}$ Fixed residuals ke saath $t_{g o} \to 0$ par, command ::: $1/ t_{g o}^{2}$ ki tarah blow up hota hai, isliye errors jaldi khatam karo Ideal intercept course par ZEM scale karta hai ::: $t_{g o}^{2}$ ki tarah, command bounded rakhta hai Gravity ZEM mein enter karti hai ::: $\frac{1}{2} g t_{g o}^{2}$ ke roop mein aur ZEV mein $g t_{g o}$ ke roop mein Law sirf valid hai ::: $t_{g o} > 0$ ke liye; $t_{g o} \leq 0$ par command freeze karo ya re-plan karo, aur hamesha $a_{ma x}$ tak clip karo