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

3.5.29 · D3 · Physics › Guidance, Navigation & Control (GNC) › State-space representation — x' = Ax + Bu, y = Cx + Du

Yeh page ek drill room hai. Parent note mein ideas banaye gaye the; yahan hum har tarah ki situation ko tackle karte hain jo state-space form de sakta hai, ek worked example per cell.

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The scenario matrix

Har state-space problem in cells mein se kisi ek mein aata hai. Har row ek case class hai; last column batata hai kaunsa worked example use cover karta hai.

#	Case class	Jo cheez vary karti hai	Jo galat nahi hona chahiye	Example
C1	Ek physical ODE se $A,B,C,D$ banana	order $n$, kaun sa state choose kiya	derivatives ko states name karna	Ex 1
C2	Eigenvalues: dono real aur negative	$A$ real, distinct $\lambda <0$	stable, non-oscillating decay	Ex 2
C3	Eigenvalues: complex, negative real part	$\lambda =\sigma \pm i\omega ,\sigma <0$	decaying oscillation	Ex 3
C4	Eigenvalues: pure imaginary ($\sigma =0$)	degenerate boundary	undamped ring, marginal	Ex 4
C5	Positive real part wala eigenvalue	$\lambda >0$	unstable — blows up	Ex 5
C6	Repeated real negative eigenvalue	${\lambda }_1={\lambda }_2<0$	critically damped, $t{e}^{\lambda t}$ term	Ex 6
C7	Zero eigenvalue ($\lambda =0$)	neutral mode	constant, non-decaying direction	Ex 7
C8	Forced response, constant input $u$	$u\ne 0$, steady state	zero-state term, $x_{\infty }$	Ex 8
C9	Nonzero feedthrough $D\ne 0$	direct shortcut	$D$ dikhta hai $G(s)$ mein	Ex 9
C10	State-space $\to$ transfer function	$sI-A$ ko invert karo	poles = eigenvalues	Ex 10
C11	Coordinate change (non-uniqueness)	$\tilde{x}=Tx$	$A,B,C,D$ sab transform hote hain	Ex 11
C12	Word problem / exam twist (MIMO)	$m>1$, $p>1$	$B$ columns, $C$ rows	Ex 12

Jo prerequisites hum use karte hain: Eigenvalues and stability, Matrix exponential, Transfer functions and G(s).

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Ex 1 — Matrices banana (cell C1)

Step 1 — states choose karo. Kyun? System ki memory yahi hai ki woh kahan hai aur kitni tez chal raha hai, isliye $x_1=q,x_2=\dot{q}.$

Step 2 — ODE normalize karo. Kyun? State-space ko $\ddot{q}$ akele left side pe chahiye, toh leading coefficient $2$ se divide karo: $\ddot{q}=-2q-3\dot{q}+4u.$

Step 3 — har derivative likho. Kyun? Humein ${\dot{x}}_1,{\dot{x}}_2$ purely $x$ aur $u$ ke terms mein chahiye: ${\dot{x}}_1=x_2,{\dot{x}}_2=-2x_1-3x_2+4u.$

Step 4 — stack karo. Kyun? Computer ke liye ek clean matrix object: $A=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right],B=\left[\begin{array}{c}0\\ 4\end{array}\right],C=\left[\begin{array}{cc}1& 0\end{array}\right],D=0.$

Yahan $n=2$ (do states), $m=1$ (ek input $u$), $p=1$ (ek measured output $q$) — ek SISO system.

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Ex 2 — Do real negative eigenvalues (cell C2)

Step 1 — characteristic equation likho. Kyun? Eigenvalues $\lambda$ woh numbers hain jahan $\det (\lambda I-A)=0$, aur yahi $e^{At}$ govern karte hain: $\det \left[\begin{array}{cc}\lambda & -1\\ 2& \lambda +3\end{array}\right]=\lambda (\lambda +3)+2={\lambda }^2+3\lambda +2=0.$

Step 2 — factor karo. Kyun? Hum actual roots chahte hain: $(\lambda +1)(\lambda +2)=0\Rightarrow {\lambda }_1=-1,{\lambda }_2=-2.$

Step 3 — physics padho. Kyun? Har eigenvalue $\lambda$ motion mein $e^{\lambda t}$ contribute karta hai. Dono real aur negative hain → do decaying exponentials, imaginary part nahi matlab koi oscillation nahi. $x(t)\sim c_1{e}^{-t}+c_2{e}^{-2t}.$ Neeche figure mein, $x_1(t)$ ke liye solid decay curve monotonically zero ki taraf jaati hai aur dashed zero line ko kabhi cross nahi karti — legend mein label hai aur dono axes seconds aur state units mein marked hain.

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Ex 3 — Complex eigenvalues, negative real part (cell C3)

Step 1 — characteristic equation. Kyun? Same reason: $\det (\lambda I-A)$ ke roots dynamics set karte hain: ${\lambda }^2+\lambda +2=0.$

Step 2 — quadratic formula. Kyun? Discriminant humein real vs complex batata hai: $\lambda =\frac{-1\pm \sqrt{1-8}}{2}=\frac{-1\pm \sqrt{-7}}{2}=-\frac{1}{2}\pm i\frac{\sqrt{7}}{2}.$ Discriminant negative hai → complex roots.

Step 3 — $\sigma \pm i\omega$ decode karo. Kyun? Ek complex pair $\sigma \pm i\omega$ ka matlab hai ki $e^{At}$ mein $e^{\sigma t}\cos (\omega t)$ aur $e^{\sigma t}\sin (\omega t)$ hota hai: frequency $\omega$ pe ek oscillation envelope $e^{\sigma t}$ ke andar. Yahan $\sigma =-\frac{1}{2}<0$ → envelope shrink hoti hai; $\omega =\frac{\sqrt{7}}{2}\approx 1.32$ rad/s.

Figure mein, $x_1(t)$ ke liye solid oscillating curve wiggle karti hai lekin uske peaks do dashed envelope curves $\pm e^{-t/2}$ ke andar rehte hain jo zero ki taraf squeeze hoti hain; legend mein har cheez name hai aur axes labelled hain.

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Ex 4 — Pure imaginary eigenvalues, boundary case (cell C4)

Step 1 — characteristic equation. Kyun? Same tool: ${\lambda }^2+4=0\Rightarrow \lambda =\pm 2i.$

Step 2 — $\sigma$ aur $\omega$ padho. Kyun? Yahan exactly $\sigma =0$ hai. Envelope $e^{0\cdot t}=1$ kabhi badhti ya ghatti nahi. $\omega =2$ rad/s.

Step 3 — classify karo. Kyun? $\mathrm{Re}(\lambda )=0$ na $<0$ hai na $>0$ → marginally stable (razor's edge). Yeh Ex 3 (stable) aur Ex 5 (unstable) ke beech ka degenerate boundary hai.

Figure mein solid $x_1(t)$ curve hamesha same peak height pe return karti hai; axes labelled hain aur ek arrow mark karta hai ki amplitude kabhi nahi badlti.

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Ex 5 — Positive-real-part eigenvalue: instability (cell C5)

Step 1 — characteristic equation. Kyun? Roots fate decide karte hain: ${\lambda }^2-3=0\Rightarrow \lambda =\pm \sqrt{3}\approx \pm \mathrm{1.732.}$

Step 2 — signs inspect karo. Kyun? Ek bhi eigenvalue $\mathrm{Re}(\lambda )>0$ ke saath poore system ko unstable bana deta hai. Yahan ${\lambda }_1=+\sqrt{3}>0$.

Step 3 — physical meaning. Kyun? $e^{+\sqrt{3}t}$ term bina bound ke badhta hai → pendulum gir jaata hai. Koi bhi initial condition (siwaaye exactly balance pe baithne ke) isse avoid nahi kar sakti.

Figure mein solid $x_1(t)$ curve plot ke upar se shoot karti hai; axes labelled hain aur ek arrow "runs away" mark karta hai.

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Ex 6 — Repeated real negative eigenvalue: critical damping (cell C6)

Step 1 — characteristic equation. Kyun? Same eigenvalue tool: $\det \left[\begin{array}{cc}\lambda & -1\\ 1& \lambda +2\end{array}\right]=\lambda (\lambda +2)+1={\lambda }^2+2\lambda +1=0.$

Step 2 — factor karo. Kyun? Roots dhundho aur notice karo ki yeh coincide karte hain: $(\lambda +1{)}^2=0\Rightarrow {\lambda }_1={\lambda }_2=-1.$ Discriminant exactly $0$ hai — real-distinct aur complex ke beech razor's edge.

Step 3 — physics padho. Kyun? Jab koi eigenvalue repeat hota hai, $e^{At}$ mein sirf $e^{\lambda t}$ nahi balki $t{e}^{\lambda t}$ bhi hota hai (repeated root ke liye $t$ ka ek extra factor). Toh $x_1(t)\sim (c_1+c_{2}t)e^{-t}.$ Dono pieces phir bhi decay karte hain kyunki $e^{-t}$ eventually badhte $t$ ko beat kar leta hai. Yeh hai critical damping: koi oscillation ke bina rest pe sabse tezi se return.

Figure mein, $x_1(t)$ ke liye solid critically-damped curve (starting velocity ke saath) ek single hump tak uthti hai phir zero tak decay hoti hai bina kabhi cross kiye — legend aur axes labelled hain, aur isse pure $e^{-t}$ envelope se compare kiya gaya hai.

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Ex 7 — Zero eigenvalue: ek neutral (constant) mode (cell C7)

Step 1 — characteristic equation. Kyun? Same tool: $\det \left[\begin{array}{cc}\lambda & -1\\ 0& \lambda +2\end{array}\right]=\lambda (\lambda +2)=0\Rightarrow {\lambda }_1=0,{\lambda }_2=-2.$

Step 2 — $\lambda =0$ interpret karo. Kyun? Mode contribute karta hai $e^{0\cdot t}=1$: ek constant jo na badhta hai na ghatta hai. ${\lambda }_2=-2$ mode contribute karta hai $e^{-2t}$, jo khatam ho jaata hai. Toh velocity decay karti hai, lekin final position kisi nonzero value pe frozen hai.

Step 3 — classify karo. Kyun? Ek eigenvalue ke liye $\mathrm{Re}(\lambda )=0$ matlab system neutrally (marginally) stable hai: yeh blow up nahi karta, lekin origin pe bhi return nahi karta. Ek constant offset hamesha ke liye rehta hai.

Figure mein, solid position curve $x_1(t)$ uthti hai phir ek nonzero constant pe flatten ho jaati hai (frozen position), jabki dashed velocity curve $x_2(t)$ zero tak decay karti hai — axes aur legend labelled hain.

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Ex 8 — Forced response with constant input (cell C8)

Step 1 — solution formula yaad karo. Kyun? Parent ne derive kiya tha $x(t)=e^{At}x(0)+{\int }_0^t{e}^{A(t-\tau )}Bud\tau$; $x(0)=0$ ke saath scalar ke liye yeh ek plain integral hai: $x(t)={\int }_0^t{e}^{-2(t-\tau )}(4)(3)d\tau .$

Step 2 — integral karo. Kyun? Hum explicit $x(t)$ chahte hain. $e^{-2t}$ bahar nikalo: $x(t)=12e^{-2t}{\int }_0^t{e}^{2\tau }d\tau =12e^{-2t}\cdot \frac{e^{2t}-1}{2}=6(1-e^{-2t}).$

Step 3 — limit lo. Kyun? Steady state hai $t\to \infty$; kyunki $e^{-2t}\to 0$: $x_{\infty }={\lim }_{t\to \infty }6(1-e^{-2t})=6.$ Yeh forecast se match karta hai. Figure mein solid $x(t)$ curve uthti hai aur dashed line $x_{\infty }=6$ pe flatten ho jaati hai; axes aur legend labelled hain.

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Ex 9 — Nonzero feedthrough $D\ne 0$ (cell C9)

Step 1 — state ka steady state. Kyun? $\dot{x}=0$ set karo: $0=-5x+2u\Rightarrow x_{\infty }=\frac{2}{5}u=\frac{2}{5}$.

Step 2 — output banao. Kyun? $y=Cx+Du$ dono state aur direct shortcut use karta hai: $y_{\infty }=C{x}_{\infty }+Du=1\cdot \frac{2}{5}+3\cdot 1=0.4+3=\mathrm{3.4.}$

Step 3 — instant part. Kyun? $t=0^{+}$ pe $x(0)=0$ ke saath, state abhi nahi hila, lekin $y(0^{+})=C\cdot 0+D\cdot 1=3$ — output turant jump karta hai $D$ ki wajah se. $D$ ko drop karne par galti se $y(0^{+})=0$ milega.

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Ex 10 — State-space se transfer function (cell C10)

Step 1 — $sI-A$ form karo. Kyun? Parent formula hai $G(s)=C(sI-A{)}^{-1}B+D$: $sI-A=\left[\begin{array}{cc}s& -1\\ 2& s+3\end{array}\right].$

Step 2 — invert karo. Kyun? $(sI-A{)}^{-1}$ chahiye; ek $2\times 2$ matrix ke liye, diagonal swap karo, off-diagonal negate karo, determinant $\det =s^2+3s+2$ se divide karo: $(sI-A{)}^{-1}=\frac{1}{s^2+3s+2}\left[\begin{array}{cc}s+3& 1\\ -2& s\end{array}\right].$

Step 3 — assemble karo. Kyun? $C(\cdots )B$ multiply karo (aur $D=0$): $G(s)=\left[\begin{array}{cc}1& 0\end{array}\right]\frac{1}{s^2+3s+2}\left[\begin{array}{cc}s+3& 1\\ -2& s\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\frac{1}{s^2+3s+2}.$

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Ex 11 — Coordinate change: same system, naye kapde (cell C11)

Step 1 — $T$ invert karo. Kyun? Har transformed matrix ko $T^{-1}$ chahiye: $T^{-1}=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right].$

Step 2 — $A$ transform karo. Kyun? $\tilde{A}=TA{T}^{-1}$ new dynamics hai: $TA=\left[\begin{array}{cc}0& 1\\ -2& -2\end{array}\right],\tilde{A}=(TA)T^{-1}=\left[\begin{array}{cc}-1& 1\\ 0& -2\end{array}\right].$

Step 3 — $B$, $C$, $D$ transform karo. Kyun? Ek coordinate change charon matrices ko touch karta hai, sirf $A$ ko nahi:

\tilde C = CT^{-1} = \begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1&0\\-1&1\end{bmatrix}=\begin{bmatrix}1&0\end{bmatrix},\quad \tilde D = D = 0.$$ **Step 4 — eigenvalues check karo.** *Kyun?* $\tilde A$ upper-triangular hai, toh eigenvalues diagonal par hain: $-1$ aur $-2$ — $A$ ke identical. > [!recall] Verify karo > $\mathrm{trace}(\tilde A)=-3=\mathrm{trace}(A)$ aur $\det(\tilde A)=2=\det(A)$: same eigenvalues. Aur transfer function invariant hai — aap check kar sakte ho $\tilde C(sI-\tilde A)^{-1}\tilde B+\tilde D$ same $1/(s^2+3s+2)$ deta hai. ✅ --- ## Ex 12 — MIMO word problem / exam twist (cell C12) > [!example] Ek chhota rocket: do inputs, do outputs > Ek rocket ki lateral dynamics ko model kiya gaya hai: > $$\dot x = \begin{bmatrix}0&1\\-4&-2\end{bmatrix}x + \begin{bmatrix}0&0\\1&0.5\end{bmatrix}\begin{bmatrix}u_1\\u_2\end{bmatrix},\qquad y=\begin{bmatrix}1&0\\0&1\end{bmatrix}x.$$ > Yahan $u_1$=main thrust deflection, $u_2$=reaction jet. (a) Sizes $n,m,p$ batao. (b) Steady state pe $u_1=2,u_2=0$ ke saath, $x_\infty$ dhundho. (c) Kya open-loop system stable hai? > > **Forecast:** $n=2,m=2,p=2$. Steady inputs ke saath, $\dot x=0$ set karo aur ek $2\times2$ system solve karo. Damping term $-2$ stability suggest karta hai. **Step 1 — sizes.** *Kyun?* Page ke upar definitions use karke shapes se padho: state ki 2 rows hain → $n=2$; $B$ ki 2 columns hain → $m=2$ (do inputs); $C$ ki 2 rows hain → $p=2$ (do outputs). MIMO native hai — yahi parent ka point tha. **Step 2 — steady state (algebra khatam karo).** *Kyun?* $\dot x=0$ set karo toh $Ax_\infty = -Bu$. $u=(2,0)^\top$ ke saath, $$Bu=\begin{bmatrix}0&0\\1&0.5\end{bmatrix}\begin{bmatrix}2\\0\end{bmatrix}=\begin{bmatrix}0\\2\end{bmatrix},\qquad \begin{bmatrix}0&1\\-4&-2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=-\begin{bmatrix}0\\2\end{bmatrix}.$$ Top row: $x_2 = 0$. Bottom row mein substitute karo: $-4x_1 - 2x_2 = -2 \Rightarrow -4x_1 - 0 = -2 \Rightarrow x_1 = \tfrac{1}{2}=0.5$. $$\boxed{\,x_\infty = \begin{bmatrix}0.5\\0\end{bmatrix}\,}$$ **Step 3 — stability check.** *Kyun?* Stability $A$ ke eigenvalues mein rehti hai: $$\det(\lambda I - A)=\det\!\begin{bmatrix}\lambda&-1\\4&\lambda+2\end{bmatrix}=\lambda(\lambda+2)+4=\lambda^2+2\lambda+4=0.$$ Quadratic formula: $\lambda=\dfrac{-2\pm\sqrt{4-16}}{2}=-1\pm i\sqrt3$. Dono ka $\mathrm{Re}(\lambda)=-1<0$ hai → **stable** (ek decaying oscillation, Ex 3 jaise). > [!recall] Verify karo > $x_\infty=(0.5,0)$ check karo: $Ax_\infty=\begin{bmatrix}0\cdot0.5+1\cdot0\\-4\cdot0.5-2\cdot0\end{bmatrix}=\begin{bmatrix}0\\-2\end{bmatrix}=-Bu$ ✅. Eigenvalue real part $-1<0$ → stable ✅. Position output $y_1=0.5$ matlab rocket constant thrust ke under 0.5 offset pe settle hota hai. ✅ --- > [!mnemonic] Eigenvalue traffic light > **Real part negative → green** (decay karta hai, stable). **Real part zero → yellow** (ring karta hai ya freeze karta hai, marginal). **Real part positive → red** (blow up, unstable). Imaginary part sirf wiggle add karta hai; ek **repeated** root ek $t\,e^{\lambda t}$ term add karta hai (critical damping). ## Connections - [[Eigenvalues and stability]] — Ex 2–7 poora eigenvalue traffic-light zoo hai. - [[Matrix exponential]] — Ex 8 mein forced response aur Ex 6 mein $t\,e^{\lambda t}$ term ko power deta hai. - [[Transfer functions and G(s)]] — Ex 10 bridge cross karta hai; Laplace variable $s$ define karta hai. - [[Controllability and Observability]] — kya Ex 12 mein $B,C$ actually har state tak reach/see kar sakte hain. - [[LQR optimal control]] — unstable Ex 5 ko kaise stabilize karoge. - [[Kalman filter]] — $C$ ke peeche unmeasured states recover karta hai. - [[Linearization of nonlinear systems]] — jahan se Ex 12 mein rocket ka $A,B$ aata hai.