Optimal guidance — ZEM - ZEV formulation

3.5.52 · Physics › Guidance, Navigation & Control (GNC)

HUM KYA solve kar rahe hain?

Hum ek thrust/acceleration command $\mathbf{a}(t)$ chahte hain jo vehicle ko is tarah fly kare ki final time $t_f$ par:

• position target hit kare: $\mathbf{r}(t_f)={\mathbf{r}}_f$
• (optionally) velocity match kare: $\mathbf{v}(t_f)={\mathbf{v}}_f$ (soft landing / rendezvous ke liye zaroori)

aur minimum control effort kharch ho: $J=\frac{1}{2}{\int }_t^{t_f}\Vert \mathbf{a}(\tau ){\Vert }^{2}d\tau .$

Dynamics (double integrator + gravity) coast ke dauran: ${\mathbf{r}}_{coast}(t_f)=\mathbf{r}(t)+\mathbf{v}(t)t_{go}+\frac{1}{2}\mathbf{g}{t}_{go}^2,$ ${\mathbf{v}}_{coast}(t_f)=\mathbf{v}(t)+\mathbf{g}{t}_{go}.$

Toh concretely: $\boxed{\mathbf{ZEM}={\mathbf{r}}_f-\mathbf{r}-\mathbf{v}{t}_{go}-\frac{1}{2}\mathbf{g}{t}_{go}^2}$ $\boxed{\mathbf{ZEV}={\mathbf{v}}_f-\mathbf{v}-\mathbf{g}{t}_{go}}$

KAISE: First principles se Derivation

Hum double-integrator model use karte hain (control acceleration hai): $\dot{\mathbf{r}}=\mathbf{v},\dot{\mathbf{v}}=\mathbf{a}+\mathbf{g}.$

Step 1 — Control included ke saath true final states likho

True dynamics ko abhi ($t$) se $t_f$ tak integrate karo. Kyun? Hume actual endpoint chahiye unknown $\mathbf{a}(\tau )$ ke function ke roop mein.

$\mathbf{v}(t_f)=\underset{\text{coast part}}{\underbrace{\mathbf{v}+\mathbf{g}{t}_{go}}}+{\int }_t^{t_f}\mathbf{a}(\tau )d\tau .$

Position ke liye double integral chahiye; integrating by parts karke, $\mathbf{r}(t_f)=\underset{\text{coast part}}{\underbrace{\mathbf{r}+\mathbf{v}{t}_{go}+\frac{1}{2}\mathbf{g}{t}_{go}^2}}+{\int }_t^{t_f}(t_f-\tau )\mathbf{a}(\tau )d\tau .$

$(t_f-\tau )$ weight kyun? Kyunki acceleration jo pehle apply hoti hai uske paas position move karne ke liye zyada time hota hai — woh do integrations se accumulate hoti hai, lever-arm $(t_f-\tau )$ chhodke.

Step 2 — ZEM/ZEV use karke boundary conditions impose karo

$\mathbf{r}(t_f)={\mathbf{r}}_f$, $\mathbf{v}(t_f)={\mathbf{v}}_f$ set karo. Coast parts subtract karo (= ZEM/ZEV definitions): ${\int }_t^{t_f}(t_f-\tau )\mathbf{a}(\tau )d\tau =\mathbf{ZEM},{\int }_t^{t_f}\mathbf{a}(\tau )d\tau =\mathbf{ZEV}.$

Yeh beautiful kyun hai: do constraints ab purely coast errors ke terms mein express ho gayi hain. Control ko bas ZEM aur ZEV "bharna" hai.

Step 3 — Un do integral constraints ke subject mein effort minimize karo

$J=\frac{1}{2}\int \Vert \mathbf{a}{\Vert }^{2}d\tau$ minimize karo do vector constraints ke saath. Constant Lagrange multipliers ${\mathbf{\lambda }}_1,{\mathbf{\lambda }}_2$ introduce karo: $\mathcal{L}={\int }_t^{t_f}[\frac{1}{2}\Vert \mathbf{a}{\Vert }^2-{\mathbf{\lambda }}_1\cdot (t_f-\tau )\mathbf{a}-{\mathbf{\lambda }}_2\cdot \mathbf{a}]d\tau .$ Pointwise stationarity $\partial \mathcal{L}/\partial \mathbf{a}=0$ optimal linear-in-time profile deta hai: ${\mathbf{a}}^{\ast }(\tau )={\mathbf{\lambda }}_1(t_f-\tau )+{\mathbf{\lambda }}_2.$ Linear kyun? Quadratic cost ke liye calculus of variations in constraints ke saath control ko remaining time ka affine function banne par majboor karta hai — optimal profile kabhi "wild" nahi hota.

Step 4 — Current instant par do moment equations solve karo

${\mathbf{a}}^{\ast }$ ko do constraints mein plug karo. Maano $s=t_f-\tau$, toh $ds=-d\tau$, $s:t_{go}\to 0$: ${\int }_0^{t_{go}}s({\mathbf{\lambda }}_{1}s+{\mathbf{\lambda }}_2)ds={\mathbf{\lambda }}_1\frac{t_{go}^3}{3}+{\mathbf{\lambda }}_2\frac{t_{go}^2}{2}=\mathbf{ZEM},$ ${\int }_0^{t_{go}}({\mathbf{\lambda }}_{1}s+{\mathbf{\lambda }}_2)ds={\mathbf{\lambda }}_1\frac{t_{go}^2}{2}+{\mathbf{\lambda }}_2{t}_{go}=\mathbf{ZEV}.$

Is $2\times 2$ system ko ${\mathbf{\lambda }}_1,{\mathbf{\lambda }}_2$ ke liye solve karo, phir command jo ABHI apply hoti hai woh hai ${\mathbf{a}}^{\ast }(t)={\mathbf{\lambda }}_1{t}_{go}+{\mathbf{\lambda }}_2$. Algebra karne par:

$6$ aur $2$ kyun? $2\times 2$ solve karne par: ${\mathbf{\lambda }}_2=(6\mathbf{ZEM}-4t_{go}\mathbf{ZEV})/t_{go}^2$ etc.; combine karne par coefficients $6/t_{go}^2$ aur $2/t_{go}$ milte hain. Signs alag hain kyunki ZEM (position error) aur ZEV (velocity error) ko opposite-tendency accelerations se correct karna padta hai end ke paas.

YEH coefficients blow up kyun karte hain ($t_{go}\to 0$ ki kahani)

Jab $t_{go}\to 0$, tab $1/t_{go}^2\to \infty$. Physically iska matlab: agar almost koi time nahi bacha aur abhi bhi miss hai, toh usse fix karne ke liye enormous acceleration chahiye — real thrusters saturate ho jaate hain, isliye ek achha guidance system ZEM→0 pehle drive karta hai taaki command bounded rahe. Yahi wajah hai ki ZEM/ZEV ko continuous feedback ke roop mein chalaya jaata hai: har cycle mein ZEM/ZEV recompute hota hai, toh tiny residuals tab khatam hote hain jab $t_{go}$ abhi bhi bada hai.

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Connections

• Proportional Navigation — $3/t_{go}^2\mathbf{ZEM}$ special case.
• Calculus of Variations & Pontryagin's Minimum Principle — linear costate profile ka source.
• Double Integrator Dynamics — plant model jo use hua.
• Time-to-go estimation — $t_{go}$ compute karna practically mushkil part hai.
• Powered Descent Guidance (Apollo E-Guidance) — historical ZEM/ZEV lander.
• Lambert's Problem — orbital mechanics mein related boundary-value targeting.
• Optimal Control — LQR — same quadratic-effort minimization philosophy.

Concept Map

Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions
• D5 · Question bank — concept traps