State-space representation — x' = Ax + Bu, y = Cx + Du: Exercises

3.5.29 · D4 · Physics › Guidance, Navigation & Control (GNC) › State-space representation — x' = Ax + Bu, y = Cx + Du

Parent note se kuch objects ki reminder jo aage bhi kaam aayenge:

Level 1 — Recognition

Goal: matrices padho aur jo dikhe use naam do. Counting se zyada koi calculation nahi.

Ek system is tarah likha gaya hai: $\overset{x}{˙} = - 2 01 0 - 3 0 10 - 1 x + 102 u, y = [100001] x .$ Batao (a) order $n$ , (b) inputs ki sankhya $m$ , (c) outputs ki sankhya $p$ , aur (d) kya direct feedthrough hai.

Recall Solution 1.1

Hum kya karte hain: bas har box ki shapes padho.

$A$ hai $3 \times 3$ , toh $x$ mein 3 entries hain $\Rightarrow$ ==order $n = 3$ ==.
$B$ mein 1 column hai $\Rightarrow$ $m = 1$ (ek input).
$C$ mein 2 rows hain $\Rightarrow$ $p = 2$ (do outputs).
Koi $D$ matrix nahi likha, toh $D = 0$ : koi direct feedthrough nahi.

Kyun: $B$ ke columns inputs count karte hain (har column = "ek input kahan push karta hai"); $C$ ki rows outputs count karti hain (har row = "ek sensor ki recipe").

Exercise 1.2

Inn mein se kaun sa matrix legally ek state-space system ka $A$ matrix ho sakta hai, aur kaun sa $B$ matrix, agar $n = 2$ aur $m = 1$ ho? $P = [0 - 4 1 - 5], Q = [03], R = [102101] .$

Recall Solution 1.2

$A$ ko $n \times n = 2 \times 2$ hona chahiye. Sirf $P$ hi $2 \times 2$ hai $\Rightarrow$ $P$ hai == $A$ matrix==.
$B$ ko $n \times m = 2 \times 1$ hona chahiye. Sirf $Q$ hi $2 \times 1$ hai $\Rightarrow$ $Q$ hai == $B$ matrix==.
$R$ hai $2 \times 3$ — dono ke liye galat shape; yeh $A$ nahi ho sakta (square nahi) aur na hi $B$ (1 column chahiye). Yeh $C$ matrix fit kar sakta tha sirf tab agar $n = 3$ hota, jo $n = 2$ se contradict karta hai. Toh $R$ yahan invalid hai.

Level 2 — Application

Goal: derivation recipe aur standard formulas ko end-to-end chalao.

Exercise 2.1

Third-order ODE ko state-space form $\overset{x}{˙} = A x + B u$ mein convert karo jahan output $y = q$ ho: $q ... + 6 \overset{q}{¨} + 11 \overset{q}{˙} + 6 q = 4 u$

Recall Solution 2.1

Step 1 — state chunna (KYU: memory wo function hai aur uske derivatives jo highest se ek neeche tak jaate hain). 3rd-order ODE ke liye 3 states chahiye: $x_{1} = q, x_{2} = \overset{q}{˙}, x_{3} = \overset{q}{¨} .$ Step 2 — har derivative ko states ke terms mein likho (KYU: $\overset{x}{˙}$ sirf $x$ aur $u$ ke terms mein chahiye). $\overset{x}{˙}_{1} = \overset{q}{˙} = x_{2}, \overset{x}{˙}_{2} = \overset{q}{¨} = x_{3},$ $\overset{x}{˙}_{3} = q ... = - 6 \overset{q}{¨} - 11 \overset{q}{˙} - 6 q + 4 u = - 6 x_{3} - 11 x_{2} - 6 x_{1} + 4 u .$ Step 3 — stack karo (KYU: ek clean matrix object). $A = 00 - 6 10 - 11 01 - 6, B = 004,$ $C = [100], D = 0.$ Yeh layout (super-diagonal par identity ones, bottom row par coefficients) controllable canonical form hai — jab bhi koi single high-order ODE mile, ise pehchano.

Exercise 2.2

$A = [0 - 2 1 - 3]$ , $B = [01]$ , $C = [10]$ , $D = 0$ ke liye transfer function $G (s) = C (s I - A)^{- 1} B + D$ compute karo.

Recall Solution 2.2

Step 1 — $s I - A$ banao (KYU: yahi wo matrix hai jiska inverse formula ko chahiye). $s I - A = [s 2 - 1 s + 3] .$ Step 2 — $2 \times 2$ ko invert karo (KYU: $(s I - A)^{- 1}$ dikhata hai ki state $s$ par kaise respond karta hai). $[a c b d]$ ka inverse hai $\frac{1}{a d - b c} [d - c - b a]$ . Yahan $det = s (s + 3) - (- 1) (2) = s^{2} + 3 s + 2$ . $(s I - A)^{- 1} = \frac{1}{s ^{2} + 3 s + 2} [s + 3 - 2 1 s] .$ Step 3 — $C$ aur $B$ ke saath sandwich karo. $(s I - A)^{- 1} B = \frac{1}{s ^{2} + 3 s + 2} [s + 3 - 2 1 s] [01] = \frac{1}{s ^{2} + 3 s + 2} [1 s] .$ $G (s) = C (s I - A)^{- 1} B = [10] \frac{1}{s ^{2} + 3 s + 2} [1 s] = \frac{1}{s ^{2} + 3 s + 2} .$ Denominator $s^{2} + 3 s + 2 = (s + 1) (s + 2)$ — iske roots $- 1, - 2$ bilkul $A$ ke eigenvalues hain. Dekho Transfer functions and G(s).

Level 3 — Analysis

Goal: matrices se behaviour extract karo — stability, oscillation, response.

Exercise 3.1

$A = [0 - 2 1 - 3]$ ke eigenvalues nikalo aur system ki stability aur motion classify karo.

Recall Solution 3.1

Step 1 — characteristic polynomial (KYU: $e^{A t}$ $e^{λ t}$ ki tarah grow karta hai, toh eigenvalues $λ$ sab kuch dictate karte hain). $det (λ I - A) = det [λ 2 - 1 λ + 3] = λ (λ + 3) + 2 = λ^{2} + 3 λ + 2.$ Step 2 — $λ^{2} + 3 λ + 2 = 0$ solve karo. Factor karo $(λ + 1) (λ + 2) = 0 \Rightarrow λ = - 1, - 2$ . Step 3 — classify karo (KYU: stability real parts mein rehti hai). Dono eigenvalues real aur negative hain $\Rightarrow$ $Re (λ) < 0$ $\Rightarrow$ system asymptotically stable hai. Real (complex nahi) roots $\Rightarrow$ koi oscillation nahi: response ek decaying sum hai $c_{1} e^{- t} + c_{2} e^{- 2 t}$ (ek overdamped settle). Dekho Eigenvalues and stability.

Exercise 3.2

$A = [0 - 4 10]$ (mass–spring jahan $k / m = 4$ , no damping) ke eigenvalues nikalo aur motion describe karo. Yeh kis angular frequency par ring karta hai?

Recall Solution 3.2

$det (λ I - A) = λ^{2} + 4 = 0 \Rightarrow λ = \pm 2 i .$ Interpretation: purely imaginary eigenvalues $\Rightarrow$ marginally stable — response hai $e^{\pm 2 i t}$ , jo Euler's identity se $cos (2 t)$ aur $sin (2 t)$ hai: yeh bina grow ya decay kiye hamesha oscillate karta rehta hai. Imaginary part $2$ angular frequency hai $ω = 2$ rad/s. Physically: koi damping nahi ( $c = 0$ ) toh koi energy lost nahi hoti — yeh hamesha ring karta rehta hai, jo parent note ki undamped forecast se match karta hai.

Figure — State-space representation — x' = Ax + Bu, y = Cx + Du

Level 4 — Synthesis

Goal: state-space models banao aur charon boxes ko ek designed purpose ke liye combine karo.

Exercise 4.1

Ek cart jiska mass $m$ hai, force $u$ se drive hoti hai aur viscous friction coefficient $b$ hai: $m \overset{p}{¨} + b \overset{p}{˙} = u$ , jahan $p$ position hai. Tumhare paas do sensors hain: ek position $p$ read karta hai, ek velocity $\overset{p}{˙}$ read karta hai. $m = 2, b = 1$ ke saath full $(A, B, C, D)$ banao.

Recall Solution 4.1

Step 1 — state: $x_{1} = p$ , $x_{2} = \overset{p}{˙}$ (memory = kahan hai + kitni tezi se). Step 2 — derivatives: $\overset{x}{˙}_{1} = x_{2}$ ; aur $\overset{p}{¨} = \frac{1}{m} (u - b \overset{p}{˙}) = \frac{1}{2} u - \frac{1}{2} x_{2}$ , toh $\overset{x}{˙}_{2} = - \frac{1}{2} x_{2} + \frac{1}{2} u$ . Step 3 — matrices: $A = [00 1 - \frac{1}{2}], B = [0 \frac{1}{2}] .$ Step 4 — output (KYU: do sensors $\Rightarrow$ $C$ ki do rows, toh $p = 2$ ). Hum dono states directly read karte hain: $C = [1001] = I, D = [00] .$ Stability check: eigenvalues solve karte hain $λ (λ + \frac{1}{2}) = 0 \Rightarrow λ = 0, - \frac{1}{2}$ . $λ = 0$ free-drift mode hai (bina spring ke cart coasting karta rehta hai) — marginally stable, bilkul physics ke saath match.

Exercise 4.2

Parent note ka mass–spring–damper lo, $m = 1, k = 1, c = 0$ , jisse $A = [0 - 1 10]$ milta hai. Coordinate change $\tilde{x} = T x$ apply karo jahan $T = [2001]$ . Naya $\tilde{A} = T A T^{- 1}$ compute karo aur confirm karo ki uske eigenvalues unchanged hain.

Recall Solution 4.2

Step 1 — $T^{- 1}$ : kyunki $T$ diagonal hai, $T^{- 1} = [\frac{1}{2} 0 01]$ . Step 2 — $\tilde{A} = T A T^{- 1}$ (KYU: coordinate changes boxes ko reshape karte hain par physics ko nahi). $A T^{- 1} = [0 - 1 10] [\frac{1}{2} 0 01] = [0 - \frac{1}{2} 10],$ $\tilde{A} = T (A T^{- 1}) = [2001] [0 - \frac{1}{2} 10] = [0 - \frac{1}{2} 20] .$ Step 3 — $\tilde{A}$ ke eigenvalues: $det (λ I - \tilde{A}) = λ^{2} - (2) (- \frac{1}{2}) = λ^{2} + 1 = 0 \Rightarrow λ = \pm i$ . Original ke jaisa hi ( $\pm i$ )! Matrices alag dikhte hain par identical ringing system describe karte hain — eigenvalues wo invariant hain jo coordinate changes ke baad bhi survive karte hain.

Level 5 — Mastery

Goal: poore framework ko chain karo — solution, invariance, feedback, feedthrough — pressure mein.

Exercise 5.1

$A = [- 1 0 0 - 2]$ (already diagonal) aur $x (0) = [35]$ ke liye $u = 0$ ke saath, exact free-response $x (t) = e^{A t} x (0)$ likho aur $t = 1$ par $x (t)$ evaluate karo (4 decimals tak).

Recall Solution 5.1

Kyun matrix exponential (parent se): $\overset{x}{˙} = A x$ ka free response hai $x (t) = e^{A t} x (0)$ , jo scalar $e^{a t} x_{0}$ ko generalize karta hai. Diagonal $A$ ke liye exponential bas har diagonal entry ka exponential hota hai (modes decoupled hain): $e^{A t} = [e^{- t} 0 0 e^{- 2 t}] .$ Solution: $x (t) = [e^{- t} 0 0 e^{- 2 t}] [35] = [3 e^{- t} 5 e^{- 2 t}] .$ $t = 1$ par: $x_{1} = 3 e^{- 1} \approx 1.1036$ , $x_{2} = 5 e^{- 2} \approx 0.6767$ . Dono decay karte hain (eigenvalues $- 1, - 2 < 0$ ): stable. Dekho Matrix exponential.

Exercise 5.2

$A = [0010]$ , $B = [01]$ (double integrator) ke liye state-feedback $u = - K x$ design karo taaki closed-loop matrix $A - B K$ ke eigenvalues $- 2$ aur $- 2$ (dono) par hon. $K = [k_{1} k_{2}]$ nikalo.

Recall Solution 5.2

Step 1 — closed loop (KYU: feedback $u = - K x$ dynamics ko $A$ se $A - B K$ mein reshape karta hai). $B K = [01] [k_{1} k_{2}] = [0 k_{1} 0 k_{2}], A - B K = [0 - k_{1} 1 - k_{2}] .$ Step 2 — characteristic polynomial: $det (λ I - (A - B K)) = λ^{2} + k_{2} λ + k_{1}$ . Step 3 — desired polynomial se match karo. Hum roots $- 2, - 2$ chahte hain: $(λ + 2)^{2} = λ^{2} + 4 λ + 4$ . Toh $k_{2} = 4$ , $k_{1} = 4$ . $K = [44] .$ Yahi exactly hai jo pole placement / LQR optimal control automate karta hai: $K$ choose karo taaki $A - B K$ ke eigenvalues stable left half-plane mein move ho jayein.

Exercise 5.3

Ex 4.1 ki cart mein ek sensor commanded force se directly acceleration read karta hai (toh $y = \overset{p}{¨}$ ). $m = 2, b = 1$ aur $x_{2} = \overset{p}{˙}$ ke saath, $y$ ko $C x + D u$ ke roop mein express karo aur dikhaao ki $D \neq = 0$ hai.

Recall Solution 5.3

Kyun $D$ appear hota hai (parent se): agar ek input instantly ek output ko affect kare, toh algebraic shortcut $D$ nonzero hota hai. Yahan $\overset{p}{¨} = \frac{1}{m} (u - b \overset{p}{˙}) = \frac{1}{2} u - \frac{1}{2} x_{2}$ . Row padho (KYU: $y = C x + D u$ , state-part ko input-part se alag karo): $y = \overset{p}{¨} = C [0 - \frac{1}{2}] [x_{1} x_{2}] + D \frac{1}{2} u .$ Toh $C = [0 - \frac{1}{2}]$ aur == $D = \frac{1}{2} \neq = 0$ ==. Yahan $D$ drop karna transfer function ke high-frequency behaviour ko corrupt kar deta — input immediately feed through ho jaata hai.

Wrap-up recall

Recall Ek-line self-test

Kisi bhi $(A, B, C, D)$ ko diya jaaye, ab tum kya teen cheezein compute kar sakte ho? ::: (1) $A$ ke eigenvalues se stability; (2) transfer function $G (s) = C (s I - A)^{- 1} B + D$ ; (3) time response $x (t) = e^{A t} x (0) + \int_{0}^{t} e^{A (t - τ)} B u d τ$ .

Connections

Transfer functions and G(s) — Ex 2.2 ne $(A, B, C)$ se ek banaya.
Eigenvalues and stability — Ex 3.1–3.2 ne $λ$ se stability padhi.
Matrix exponential — Ex 5.1 ne free response ke liye $e^{A t}$ use kiya.
Controllability and Observability — Ex 5.2 mein $K$ koi bhi pole place kar sakta hai ya nahi, yeh $(A, B)$ par depend karta hai.
LQR optimal control — Ex 5.2 ka feedback automate karta hai.
Kalman filter — tab chahiye jab $C$ (Ex 4.1) har state nahi read karta.
Linearization of nonlinear systems — inhi problems ka $A, B$ Jacobians ke roop mein kahan se aata hai.

Exercises — State-space representation — x' = Ax + Bu, y = Cx + Du