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

Reminder of the objects you will keep using (all from the parent):

Level 1 — Recognition

Goal: read the matrices and name what you see. No calculation heavier than counting.

A system is written as $\overset{x}{˙} = - 2 01 0 - 3 0 10 - 1 x + 102 u, y = [100001] x .$ State (a) the order $n$ , (b) the number of inputs $m$ , (c) the number of outputs $p$ , and (d) whether there is direct feedthrough.

Recall Solution 1.1

What we do: just read the shapes of each box.

$A$ is $3 \times 3$ , so $x$ has 3 entries $\Rightarrow$ ==order $n = 3$ ==.
$B$ has 1 column $\Rightarrow$ $m = 1$ (one input).
$C$ has 2 rows $\Rightarrow$ $p = 2$ (two outputs).
No $D$ matrix is written, so $D = 0$ : no direct feedthrough.

Why: columns of $B$ count inputs (each column is "where one input pushes"); rows of $C$ count outputs (each row is "one sensor's recipe").

Exercise 1.2

Which of these matrices could legally be an $A$ matrix for a state-space system, and which could be a $B$ matrix, if $n = 2$ and $m = 1$ ? $P = [0 - 4 1 - 5], Q = [03], R = [102101] .$

Recall Solution 1.2

$A$ must be $n \times n = 2 \times 2$ . Only $P$ is $2 \times 2$ $\Rightarrow$ $P$ is the == $A$ matrix==.
$B$ must be $n \times m = 2 \times 1$ . Only $Q$ is $2 \times 1$ $\Rightarrow$ $Q$ is the == $B$ matrix==.
$R$ is $2 \times 3$ — wrong shape for both; it could not be $A$ (not square) nor $B$ (needs 1 column). It would fit a $C$ matrix only if $n = 3$ , which contradicts $n = 2$ . So $R$ is invalid here.

Level 2 — Application

Goal: run the derivation recipe and the standard formulas end-to-end.

Exercise 2.1

Convert the third-order ODE $q ... + 6 \overset{q}{¨} + 11 \overset{q}{˙} + 6 q = 4 u$ into state-space form $\overset{x}{˙} = A x + B u$ with output $y = q$ .

Recall Solution 2.1

Step 1 — pick the state (WHY: the memory is the function and its derivatives up to one below the highest). For a 3rd-order ODE we need 3 states: $x_{1} = q, x_{2} = \overset{q}{˙}, x_{3} = \overset{q}{¨} .$ Step 2 — write each derivative in terms of the states (WHY: we need $\overset{x}{˙}$ only in terms of $x$ and $u$ ). $\overset{x}{˙}_{1} = \overset{q}{˙} = x_{2}, \overset{x}{˙}_{2} = \overset{q}{¨} = x_{3},$ $\overset{x}{˙}_{3} = q ... = - 6 \overset{q}{¨} - 11 \overset{q}{˙} - 6 q + 4 u = - 6 x_{3} - 11 x_{2} - 6 x_{1} + 4 u .$ Step 3 — stack (WHY: one clean matrix object). $A = 00 - 6 10 - 11 01 - 6, B = 004,$ $C = [100], D = 0.$ This layout (identity ones on the super-diagonal, coefficients along the bottom row) is the controllable canonical form — recognise it whenever you meet a single high-order ODE.

Exercise 2.2

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

Recall Solution 2.2

Step 1 — form $s I - A$ (WHY: this is the matrix whose inverse the formula needs). $s I - A = [s 2 - 1 s + 3] .$ Step 2 — invert a $2 \times 2$ (WHY: $(s I - A)^{- 1}$ is how the state responds to $s$ ). For $[a c b d]$ the inverse is $\frac{1}{a d - b c} [d - c - b a]$ . Here $det = s (s + 3) - (- 1) (2) = s^{2} + 3 s + 2$ . $(s I - A)^{- 1} = \frac{1}{s ^{2} + 3 s + 2} [s + 3 - 2 1 s] .$ Step 3 — sandwich with $C$ and $B$ . $(s I - A)^{- 1} B = \frac{1}{s ^{2} + 3 s + 2} [s + 3 - 2 1 s] [01] = \frac{1}{s ^{2} + 3 s + 2} [1 s] .$ $G (s) = C (s I - A)^{- 1} B = [10] \frac{1}{s ^{2} + 3 s + 2} [1 s] = \frac{1}{s ^{2} + 3 s + 2} .$ The denominator $s^{2} + 3 s + 2 = (s + 1) (s + 2)$ — its roots $- 1, - 2$ are exactly the eigenvalues of $A$ . See Transfer functions and G(s).

Level 3 — Analysis

Goal: extract behaviour — stability, oscillation, response — from the matrices.

Exercise 3.1

Find the eigenvalues of $A = [0 - 2 1 - 3]$ and classify the system's stability and motion.

Recall Solution 3.1

Step 1 — characteristic polynomial (WHY: $e^{A t}$ grows like $e^{λ t}$ , so eigenvalues $λ$ dictate everything). $det (λ I - A) = det [λ 2 - 1 λ + 3] = λ (λ + 3) + 2 = λ^{2} + 3 λ + 2.$ Step 2 — solve $λ^{2} + 3 λ + 2 = 0$ . Factor $(λ + 1) (λ + 2) = 0 \Rightarrow λ = - 1, - 2$ . Step 3 — classify (WHY: stability lives in the real parts). Both eigenvalues are real and negative $\Rightarrow$ $Re (λ) < 0$ $\Rightarrow$ the system is asymptotically stable. Real (not complex) roots $\Rightarrow$ no oscillation: the response is a decaying sum $c_{1} e^{- t} + c_{2} e^{- 2 t}$ (an overdamped settle). See Eigenvalues and stability.

Exercise 3.2

For $A = [0 - 4 10]$ (a mass–spring with $k / m = 4$ , no damping), find the eigenvalues and describe the motion. What angular frequency does it ring at?

Recall Solution 3.2

$det (λ I - A) = λ^{2} + 4 = 0 \Rightarrow λ = \pm 2 i .$ Interpretation: purely imaginary eigenvalues $\Rightarrow$ marginally stable — the response is $e^{\pm 2 i t}$ , which by Euler's identity is $cos (2 t)$ and $sin (2 t)$ : it oscillates forever without growing or decaying. The imaginary part $2$ is the angular frequency $ω = 2$ rad/s. Physically: no damping ( $c = 0$ ) so no energy is lost — it rings forever, matching the parent note's undamped forecast.

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

Level 4 — Synthesis

Goal: build state-space models and combine the four boxes for a designed purpose.

Exercise 4.1

A cart of mass $m$ is driven by force $u$ with viscous friction coefficient $b$ : $m \overset{p}{¨} + b \overset{p}{˙} = u$ , where $p$ is position. You have two sensors: one reads position $p$ , one reads velocity $\overset{p}{˙}$ . Build the full $(A, B, C, D)$ with $m = 2, b = 1$ .

Recall Solution 4.1

Step 1 — state: $x_{1} = p$ , $x_{2} = \overset{p}{˙}$ (memory = where it is + how fast). Step 2 — derivatives: $\overset{x}{˙}_{1} = x_{2}$ ; and $\overset{p}{¨} = \frac{1}{m} (u - b \overset{p}{˙}) = \frac{1}{2} u - \frac{1}{2} x_{2}$ , so $\overset{x}{˙}_{2} = - \frac{1}{2} x_{2} + \frac{1}{2} u$ . Step 3 — matrices: $A = [00 1 - \frac{1}{2}], B = [0 \frac{1}{2}] .$ Step 4 — output (WHY: two sensors $\Rightarrow$ two rows of $C$ , so $p = 2$ ). We read both states directly: $C = [1001] = I, D = [00] .$ Check stability: eigenvalues solve $λ (λ + \frac{1}{2}) = 0 \Rightarrow λ = 0, - \frac{1}{2}$ . The $λ = 0$ is the free-drift mode (a cart with no spring keeps coasting) — marginally stable, exactly the physics.

Exercise 4.2

Take the mass–spring–damper of the parent note, $m = 1, k = 1, c = 0$ , giving $A = [0 - 1 10]$ . Apply the coordinate change $\tilde{x} = T x$ with $T = [2001]$ . Compute the new $\tilde{A} = T A T^{- 1}$ and confirm its eigenvalues are unchanged.

Recall Solution 4.2

Step 1 — $T^{- 1}$ : since $T$ is diagonal, $T^{- 1} = [\frac{1}{2} 0 01]$ . Step 2 — $\tilde{A} = T A T^{- 1}$ (WHY: coordinate changes reshape the boxes but not the physics). $A T^{- 1} = [0 - 1 10] [\frac{1}{2} 0 01] = [0 - \frac{1}{2} 10],$ $\tilde{A} = T (A T^{- 1}) = [2001] [0 - \frac{1}{2} 10] = [0 - \frac{1}{2} 20] .$ Step 3 — eigenvalues of $\tilde{A}$ : $det (λ I - \tilde{A}) = λ^{2} - (2) (- \frac{1}{2}) = λ^{2} + 1 = 0 \Rightarrow λ = \pm i$ . Same as the original ( $\pm i$ )! The matrices look different but describe the identical ringing system — eigenvalues are the invariant that survives coordinate changes.

Level 5 — Mastery

Goal: chain the whole framework — solution, invariance, feedback, feedthrough — under pressure.

Exercise 5.1

For $A = [- 1 0 0 - 2]$ (already diagonal) and $x (0) = [35]$ with $u = 0$ , write the exact free-response $x (t) = e^{A t} x (0)$ and evaluate $x (t)$ at $t = 1$ (to 4 decimals).

Recall Solution 5.1

Why the matrix exponential (from parent): the free response of $\overset{x}{˙} = A x$ is $x (t) = e^{A t} x (0)$ , generalising the scalar $e^{a t} x_{0}$ . For a diagonal $A$ the exponential is just the exponential of each diagonal entry (the modes are decoupled): $e^{A t} = [e^{- t} 0 0 e^{- 2 t}] .$ Solution: $x (t) = [e^{- t} 0 0 e^{- 2 t}] [35] = [3 e^{- t} 5 e^{- 2 t}] .$ At $t = 1$ : $x_{1} = 3 e^{- 1} \approx 1.1036$ , $x_{2} = 5 e^{- 2} \approx 0.6767$ . Both decay (eigenvalues $- 1, - 2 < 0$ ): stable. See Matrix exponential.

Exercise 5.2

Design a state-feedback $u = - K x$ for $A = [0010]$ , $B = [01]$ (a double integrator) so that the closed-loop matrix $A - B K$ has eigenvalues at $- 2$ and $- 2$ (both). Find $K = [k_{1} k_{2}]$ .

Recall Solution 5.2

Step 1 — closed loop (WHY: feedback $u = - K x$ reshapes the dynamics from $A$ to $A - B K$ ). $B K = [01] [k_{1} k_{2}] = [0 k_{1} 0 k_{2}], A - B K = [0 - k_{1} 1 - k_{2}] .$ Step 2 — characteristic polynomial: $det (λ I - (A - B K)) = λ^{2} + k_{2} λ + k_{1}$ . Step 3 — match the desired polynomial. We want roots at $- 2, - 2$ : $(λ + 2)^{2} = λ^{2} + 4 λ + 4$ . So $k_{2} = 4$ , $k_{1} = 4$ . $K = [44] .$ This is exactly what pole placement / LQR optimal control automates: choose $K$ to move the eigenvalues of $A - B K$ into the stable left half-plane.

Exercise 5.3

A sensor reads acceleration directly from a commanded force in the cart of Ex 4.1 (so $y = \overset{p}{¨}$ ). With $m = 2, b = 1$ , and $x_{2} = \overset{p}{˙}$ , express $y$ as $C x + D u$ and show $D \neq = 0$ .

Recall Solution 5.3

Why $D$ appears (from parent): if an input instantly affects an output, the algebraic shortcut $D$ is nonzero. Here $\overset{p}{¨} = \frac{1}{m} (u - b \overset{p}{˙}) = \frac{1}{2} u - \frac{1}{2} x_{2}$ . Read off the row (WHY: $y = C x + D u$ , split state-part from input-part): $y = \overset{p}{¨} = C [0 - \frac{1}{2}] [x_{1} x_{2}] + D \frac{1}{2} u .$ So $C = [0 - \frac{1}{2}]$ and == $D = \frac{1}{2} \neq = 0$ ==. Dropping $D$ here would corrupt the transfer function's high-frequency behaviour — the input feeds through immediately.

Wrap-up recall

Recall One-line self-test

Given any $(A, B, C, D)$ , what three things can you now compute? ::: (1) stability via eigenvalues of $A$ ; (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 built one from $(A, B, C)$ .
Eigenvalues and stability — Ex 3.1–3.2 read stability off $λ$ .
Matrix exponential — Ex 5.1 used $e^{A t}$ for the free response.
Controllability and Observability — whether $K$ in Ex 5.2 can place any pole depends on $(A, B)$ .
LQR optimal control — automates the feedback of Ex 5.2.
Kalman filter — needed when $C$ (Ex 4.1) does not read every state.
Linearization of nonlinear systems — where the $A, B$ of these problems come from as Jacobians.

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