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

This page is a drill room. The parent note built the ideas; here we hit every kind of situation the state-space form can hand you, one worked example per cell.

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The scenario matrix

Every state-space problem lands in one of these cells. Each row is a case class; the last column says which worked example covers it.

#	Case class	The thing that varies	What you must not get wrong	Example
C1	Build $A,B,C,D$ from a physical ODE	order $n$, which state you pick	naming derivatives as states	Ex 1
C2	Eigenvalues: both real & 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	Eigenvalue with positive real part	$\lambda >0$	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	A 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	$D$ shows up in $G(s)$	Ex 9
C10	State-space $\to$ transfer function	invert $sI-A$	poles = eigenvalues	Ex 10
C11	Coordinate change (non-uniqueness)	$\tilde{x}=Tx$	all of $A,B,C,D$ transform	Ex 11
C12	Word problem / exam twist (MIMO)	$m>1$, $p>1$	$B$ columns, $C$ rows	Ex 12

Prerequisites we lean on: Eigenvalues and stability, Matrix exponential, Transfer functions and G(s).

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Ex 1 — Building the matrices (cell C1)

Step 1 — pick states. Why this step? The system's memory is where it is and how fast it moves, so $x_1=q,x_2=\dot{q}.$

Step 2 — normalize the ODE. Why? State-space needs $\ddot{q}$ alone on the left, so divide by the leading coefficient $2$: $\ddot{q}=-2q-3\dot{q}+4u.$

Step 3 — write each derivative. Why? We need ${\dot{x}}_1,{\dot{x}}_2$ purely in terms of $x$ and $u$: ${\dot{x}}_1=x_2,{\dot{x}}_2=-2x_1-3x_2+4u.$

Step 4 — stack. Why? One clean matrix object for the computer: $A=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right],B=\left[\begin{array}{c}0\\ 4\end{array}\right],C=\left[\begin{array}{cc}1& 0\end{array}\right],D=0.$

Here $n=2$ (two states), $m=1$ (one input $u$), $p=1$ (one measured output $q$) — a SISO system.

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Ex 2 — Two real negative eigenvalues (cell C2)

Step 1 — write the characteristic equation. Why? Eigenvalues $\lambda$ are the numbers where $\det (\lambda I-A)=0$, and they govern $e^{At}$: $\det \left[\begin{array}{cc}\lambda & -1\\ 2& \lambda +3\end{array}\right]=\lambda (\lambda +3)+2={\lambda }^2+3\lambda +2=0.$

Step 2 — factor. Why? We want the actual roots: $(\lambda +1)(\lambda +2)=0\Rightarrow {\lambda }_1=-1,{\lambda }_2=-2.$

Step 3 — read the physics. Why? Each eigenvalue $\lambda$ contributes $e^{\lambda t}$ to the motion. Both are real and negative → two decaying exponentials, no imaginary part means no oscillation. $x(t)\sim c_1{e}^{-t}+c_2{e}^{-2t}.$ In the figure below, the solid decay curve for $x_1(t)$ slides monotonically to zero and never crosses the dashed zero line — the legend labels it and both axes are marked in seconds and state units.

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

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

Step 2 — quadratic formula. Why? The discriminant tells us real vs complex: $\lambda =\frac{-1\pm \sqrt{1-8}}{2}=\frac{-1\pm \sqrt{-7}}{2}=-\frac{1}{2}\pm i\frac{\sqrt{7}}{2}.$ The discriminant is negative → complex roots.

Step 3 — decode $\sigma \pm i\omega$. Why? A complex pair $\sigma \pm i\omega$ means $e^{At}$ carries $e^{\sigma t}\cos (\omega t)$ and $e^{\sigma t}\sin (\omega t)$: an oscillation at frequency $\omega$ inside an envelope $e^{\sigma t}$. Here $\sigma =-\frac{1}{2}<0$ → envelope shrinks; $\omega =\frac{\sqrt{7}}{2}\approx 1.32$ rad/s.

In the figure, the solid oscillating curve for $x_1(t)$ wiggles but its peaks stay inside the two dashed envelope curves $\pm e^{-t/2}$ that squeeze toward zero; the legend names each and the axes are labelled.

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

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

Step 2 — read $\sigma$ and $\omega$. Why? Here $\sigma =0$ exactly. Envelope $e^{0\cdot t}=1$ never grows or shrinks. $\omega =2$ rad/s.

Step 3 — classify. Why? $\mathrm{Re}(\lambda )=0$ is neither $<0$ nor $>0$ → marginally stable (the razor's edge). This is the degenerate boundary between Ex 3 (stable) and Ex 5 (unstable).

In the figure the solid $x_1(t)$ curve keeps returning to the same peak height forever; the axes are labelled and an arrow marks that the amplitude never changes.

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

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

Step 2 — inspect signs. Why? Even one eigenvalue with $\mathrm{Re}(\lambda )>0$ makes the whole system unstable. Here ${\lambda }_1=+\sqrt{3}>0$.

Step 3 — physical meaning. Why? The $e^{+\sqrt{3}t}$ term grows without bound → the pendulum topples. No initial condition (except sitting exactly balanced) avoids it.

In the figure the solid $x_1(t)$ curve shoots upward off the top of the plot; the axes are labelled and an arrow marks "runs away."

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

Step 1 — characteristic equation. Why? 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. Why? Find the roots and notice they coincide: $(\lambda +1{)}^2=0\Rightarrow {\lambda }_1={\lambda }_2=-1.$ The discriminant is exactly $0$ — the razor's edge between real-distinct and complex.

Step 3 — read the physics. Why? When an eigenvalue repeats, $e^{At}$ carries not only $e^{\lambda t}$ but also $t{e}^{\lambda t}$ (one extra factor of $t$ for the repeated root). So $x_1(t)\sim (c_1+c_{2}t)e^{-t}.$ Both pieces still decay because $e^{-t}$ eventually beats the growing $t$. This is critical damping: the fastest return to rest with no oscillation.

In the figure, the solid critically-damped curve for $x_1(t)$ (with a starting velocity) rises to a single hump then decays to zero without ever crossing — the legend and axes are labelled, and it is compared against the pure $e^{-t}$ envelope.

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Ex 7 — A zero eigenvalue: a neutral (constant) mode (cell C7)

Step 1 — characteristic equation. Why? 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 — interpret $\lambda =0$. Why? The mode contributes $e^{0\cdot t}=1$: a constant that neither grows nor decays. The ${\lambda }_2=-2$ mode contributes $e^{-2t}$, which dies out. So velocity decays, but the final position is frozen at some nonzero value.

Step 3 — classify. Why? $\mathrm{Re}(\lambda )=0$ for one eigenvalue means the system is neutrally (marginally) stable: it does not blow up, but it also does not return to the origin. A constant offset lives forever.

In the figure, the solid position curve $x_1(t)$ climbs and then flattens at a nonzero constant (the frozen position), while the dashed velocity curve $x_2(t)$ decays to zero — axes and legend labelled.

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

Step 1 — recall the solution formula. Why? The parent derived $x(t)=e^{At}x(0)+{\int }_0^t{e}^{A(t-\tau )}Bud\tau$; for a scalar with $x(0)=0$ this is a plain integral: $x(t)={\int }_0^t{e}^{-2(t-\tau )}(4)(3)d\tau .$

Step 2 — do the integral. Why? We want an explicit $x(t)$. Pull $e^{-2t}$ out: $x(t)=12e^{-2t}{\int }_0^t{e}^{2\tau }d\tau =12e^{-2t}\cdot \frac{e^{2t}-1}{2}=6(1-e^{-2t}).$

Step 3 — take the limit. Why? Steady state is $t\to \infty$; since $e^{-2t}\to 0$: $x_{\infty }={\lim }_{t\to \infty }6(1-e^{-2t})=6.$ This matches the forecast. In the figure the solid $x(t)$ curve rises and flattens at the dashed line $x_{\infty }=6$; axes and legend labelled.

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

Step 1 — steady state of the state. Why? Set $\dot{x}=0$: $0=-5x+2u\Rightarrow x_{\infty }=\frac{2}{5}u=\frac{2}{5}$.

Step 2 — build the output. Why? $y=Cx+Du$ uses both the state and the direct shortcut: $y_{\infty }=C{x}_{\infty }+Du=1\cdot \frac{2}{5}+3\cdot 1=0.4+3=\mathrm{3.4.}$

Step 3 — the instant part. Why? At $t=0^{+}$ with $x(0)=0$, the state hasn't moved yet, but $y(0^{+})=C\cdot 0+D\cdot 1=3$ — the output jumps immediately because of $D$. Dropping $D$ would wrongly give $y(0^{+})=0$.

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

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

Step 2 — invert. Why? Need $(sI-A{)}^{-1}$; for a $2\times 2$ matrix, swap diagonal, negate off-diagonal, divide by determinant $\det =s^2+3s+2$: $(sI-A{)}^{-1}=\frac{1}{s^2+3s+2}\left[\begin{array}{cc}s+3& 1\\ -2& s\end{array}\right].$

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

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Ex 11 — Coordinate change: same system, new clothes (cell C11)

Step 1 — invert $T$. Why? Every transformed matrix needs $T^{-1}$: $T^{-1}=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right].$

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

Step 3 — transform $B$, $C$, $D$. Why? A coordinate change touches all four matrices, not just $A$:

\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.$$ **Step 4 — check eigenvalues.** *Why?* $\tilde A$ is upper-triangular, so eigenvalues are the diagonal: $-1$ and $-2$ — identical to $A$. > [!recall] Verify > $\mathrm{trace}(\tilde A)=-3=\mathrm{trace}(A)$ and $\det(\tilde A)=2=\det(A)$: same eigenvalues. And the transfer function is invariant — you can check $\tilde C(sI-\tilde A)^{-1}\tilde B+\tilde D$ gives the same $1/(s^2+3s+2)$. ✅ --- ## Ex 12 — MIMO word problem / exam twist (cell C12) > [!example] A tiny rocket: two inputs, two outputs > A rocket's lateral dynamics are modeled as > $$\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.$$ > Here $u_1$=main thrust deflection, $u_2$=reaction jet. (a) State the sizes $n,m,p$. (b) At steady state with $u_1=2,u_2=0$, find $x_\infty$. (c) Is the open-loop system stable? > > **Forecast:** $n=2,m=2,p=2$. With steady inputs, set $\dot x=0$ and solve a $2\times2$ system. Damping term $-2$ suggests stability. **Step 1 — sizes.** *Why?* Read them off the shapes using the definitions from the top of the page: state has 2 rows → $n=2$; $B$ has 2 columns → $m=2$ (two inputs); $C$ has 2 rows → $p=2$ (two outputs). MIMO is native — this is exactly the parent's point. **Step 2 — steady state (finish the algebra).** *Why?* Set $\dot x=0$ so $Ax_\infty = -Bu$. With $u=(2,0)^\top$, $$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}.$$ Top row: $x_2 = 0$. Substitute into the bottom row: $-4x_1 - 2x_2 = -2 \Rightarrow -4x_1 - 0 = -2 \Rightarrow x_1 = \tfrac{1}{2}=0.5$. $$\boxed{\,x_\infty = \begin{bmatrix}0.5\\0\end{bmatrix}\,}$$ **Step 3 — stability check.** *Why?* Stability lives in the eigenvalues of $A$: $$\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: $\lambda=\dfrac{-2\pm\sqrt{4-16}}{2}=-1\pm i\sqrt3$. Both have $\mathrm{Re}(\lambda)=-1<0$ → **stable** (a decaying oscillation, like Ex 3). > [!recall] Verify > Check $x_\infty=(0.5,0)$: $Ax_\infty=\begin{bmatrix}0\cdot0.5+1\cdot0\\-4\cdot0.5-2\cdot0\end{bmatrix}=\begin{bmatrix}0\\-2\end{bmatrix}=-Bu$ ✅. Eigenvalue real part $-1<0$ → stable ✅. Position output $y_1=0.5$ means the rocket settles offset by 0.5 under constant thrust. ✅ --- > [!mnemonic] The eigenvalue traffic light > **Real part negative → green** (decays, stable). **Real part zero → yellow** (rings or freezes, marginal). **Real part positive → red** (blows up, unstable). Imaginary part just adds the wiggle; a **repeated** root adds a $t\,e^{\lambda t}$ term (critical damping). ## Connections - [[Eigenvalues and stability]] — Ex 2–7 are the full eigenvalue traffic-light zoo. - [[Matrix exponential]] — powers the forced response in Ex 8 and the $t\,e^{\lambda t}$ term in Ex 6. - [[Transfer functions and G(s)]] — Ex 10 crosses the bridge; defines the Laplace variable $s$. - [[Controllability and Observability]] — whether $B,C$ in Ex 12 actually reach/see every state. - [[LQR optimal control]] — how you'd stabilize the unstable Ex 5. - [[Kalman filter]] — recovers unmeasured states behind $C$. - [[Linearization of nonlinear systems]] — where the rocket's $A,B$ in Ex 12 come from.