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

Definition Scope and symbol reminder (read first — this page is self-contained)

Everything here is about continuous-time, linear, time-invariant systems: time $t$ flows smoothly (not in discrete steps), which is exactly why the matrix exponential $e^{A t}$ and the derivative-dot $\overset{x}{˙} = \frac{d x}{d t}$ appear. The two governing equations are $\overset{x}{˙} = A x + B u, y = C x + D u .$

$x \in R^{n}$ — the state vector: the minimal list of numbers (the system's memory) needed to predict the future. $n$ = system order.
$u \in R^{m}$ — the input (controls you apply, e.g. force, thrust).
$y \in R^{p}$ — the output (what your sensors actually read).
$A$ ( $n \times n$ ) — internal dynamics (state feeds back on itself).
$B$ ( $n \times m$ ) — how inputs push the state.
$C$ ( $p \times n$ ) — which state combinations you measure.
$D$ ( $p \times m$ ) — direct input→output shortcut (often $0$ ).

Definition Words used repeatedly below

Memory / energy-storing element: a physical quantity that must be remembered to know the next instant — e.g. a mass stores kinetic energy (its velocity) and a spring stores potential energy (its stretch). Each independent one needs its own state variable, so number of memory elements = system order $n$ (see the mass–spring example, figure below).
Free / zero-input response: the motion $e^{A t} x (0)$ with the input switched off ( $u = 0$ ) — pure memory unwinding.
Forced / zero-state response: the motion $\int_{0}^{t} e^{A (t - τ)} B u (τ) d τ$ starting from rest ( $x (0) = 0$ ) — pure reaction to the input.
The full solution is their sum: $x (t) = e^{A t} x (0) + \int_{0}^{t} e^{A (t - τ)} B u (τ) d τ$ .
Laplace transform: a way to turn a time-signal $f (t)$ into a function $F (s)$ of a frequency-like variable $s$ , written $F (s) = \int_{0}^{\infty} f (t) e^{- s t} d t$ . It turns derivatives into multiplication by $s$ , which is why $\overset{x}{˙} = A x + B u$ becomes the algebraic $G (s) = C (s I - A)^{- 1} B + D$ . Here $Y (s)$ and $U (s)$ just mean "the Laplace transforms of the output $y (t)$ and input $u (t)$ ."

Read the pictures first (visual-first)

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

True or false — justify

The state vector is always equal to the measured output.

False.

y = C x

is a projection (figure s01:

C

flattens the 2-D state onto the one measured axis, dropping the velocity dimension);

C

can hide states like velocity that drive the motion but are never sensed. State = everything needed to predict the future, measured or not.

A 2nd-order ODE always needs a 2-dimensional state.

True. The system order

n

equals the number of independent memory (energy-storing) elements; the mass–spring has two (velocity in the mass, stretch in the spring), so

n = 2

: the function and its first derivative.

D = 0

, then the input can never affect the output.

False. With

D = 0

the input still shapes the output through the state via

\overset{x}{˙} = A x + B u

;

D

only removes the instantaneous shortcut, not the delayed influence.

Two state-space models with the same eigenvalues of

A

describe the same physical system.

False in general. Same eigenvalues means same natural modes/poles, but

B

and

C

can differ, giving different input coupling and outputs — different behaviour despite identical dynamics.

A change of coordinates

\tilde{x} = T x

changes the system's stability.

False.

\tilde{A} = T A T^{- 1}

is similar to

A

, so it has the exact same eigenvalues. Stability is a property of the real system, not the chosen coordinates.

The matrix exponential

e^{A t}

equals the element-wise exponential of

A t

False. It is defined by the series

\sum (A t)^{k} / k!

; you exponentiate the whole matrix, coupling entries (figure s03: it acts as a swirling flow carrying the whole state together). Element-wise exp is almost never right unless

A

is diagonal.

Every stable linear system oscillates as it decays.

False. Oscillation needs eigenvalues with nonzero imaginary part. Real negative eigenvalues give pure exponential decay with no ringing (the green × on the real axis in figure s02).

State-space is only for single-input single-output systems.

False.

B

has

m

columns and

C

has

p

rows, so MIMO is native — that is precisely why GNC (Guidance, Navigation and Control) uses state-space over scalar transfer functions.

The eigenvalues of

A

are exactly the poles of the transfer function

G (s) = C (s I - A)^{- 1} B + D

False in general (this is the trap). They coincide only when no pole–zero cancellation occurs. If a mode is uncontrollable or unobservable it cancels out, so that eigenvalue is not a pole of

G (s)

— the correct statement is "poles are a subset of the eigenvalues."

Spot the error

"The system

\overset{x}{˙} = A x + B u

is stable because all entries of

A

are negative."

Wrong test. Stability depends on the eigenvalues of

A

having negative real part (left half of the plane in figure s02), not on the signs of the entries. A matrix with negative entries can still have unstable eigenvalues.

"To convert

m \overset{q}{¨} + k q = F

I set

x_{1} = q

and

x_{2} = q

, giving two states."

Error: both states are identical, so it isn't a valid state. You must pick

x_{2} = \overset{q}{˙}

(the derivative), so that

\overset{x}{˙}_{1} = x_{2}

links them and captures the second-order memory.

"Since

u

appears in

\overset{x}{˙} = A x + B u

, the input matrix must be square."

Error.

B

n \times m

n

rows (one per state),

m

columns (one per input). It is square only in the accidental case

n = m

"The free response is

x (t) = e^{A t} x (0) + \int_{0}^{t} e^{A (t - τ)} B u d τ

Mislabelled. That is the full response. The free (zero-input) part is only

e^{A t} x (0)

; the integral is the forced (zero-state) part driven by

u

"Because

G (s) = Y (s) / U (s)

captures everything, state-space adds nothing."

Error.

Y (s), U (s)

are the Laplace transforms of output and input (see the definition callout);

G (s)

assumes zero initial conditions and hides internal/unmeasured states. State-space handles nonzero

x (0)

, MIMO, and internal variables — essential for a tumbling vehicle.

e^{A t}

satisfies

\frac{d}{d t} e^{A t} = e^{A t} A

, so order never matters."

The identity is true but the lesson is wrong. It holds because

e^{A t} = \sum (A t)^{k} / k!

is a power series in $A$ itself, and any matrix commutes with its own powers, so

A

can slide to either side. This is special to

A

and

e^{A t}

; general matrix products do not commute, so never assume it elsewhere.

Why questions

Why must we reduce a high-order ODE to first-order equations rather than keep it as one big equation?

The solution machinery (

e^{A t}

, eigenvalue analysis, controllers) is built for the first-order vector form

\overset{x}{˙} = A x + B u

; naming derivatives as new states turns any

n

th-order ODE into that standard shape.

Why does the matrix exponential appear in the solution and not, say, a matrix power?

Because the dynamics are continuous-time:

\overset{x}{˙} = A x

is the vector version of

\overset{x}{˙} = a x \Rightarrow x = e^{a t} x_{0}

. The exponential is the unique function whose derivative reproduces itself times

A

(matrix powers would be for discrete-time steps instead).

Why does stability live entirely in

A

and not in

B

C

, or

D

The unforced motion is

e^{A t} x (0)

, governed only by

A

's eigenvalues.

B, C, D

route inputs and outputs but cannot change whether the internal modes grow or decay.

Why is the state called the "minimal memory" of the system?

Because once you know

x (t_{0})

and all future inputs, the future is fully determined — no earlier history is needed. Any smaller set would fail to predict the future uniquely.

Why can two engineers write different

A

matrices for the same rocket and both be correct?

They chose different state coordinates. The models are related by

\tilde{x} = T x

, giving

(T A T^{- 1}, T B, C T^{- 1}, D)

— same physics, different bookkeeping, identical eigenvalues.

Why do we need

C

at all instead of just outputting the whole state?

Real sensors read only certain combinations of state (e.g. position, not velocity — figure s01).

C

models exactly what the hardware can measure, which is why estimators like the Kalman filter exist to recover the rest.

Why does a repeated eigenvalue sometimes make

e^{A t}

contain a factor of

t

When

A

is defective (fewer independent eigenvectors than its repeated eigenvalue demands),

e^{A t}

picks up polynomial-in-

t

terms like

t e^{λ t}

from the Jordan block — see the edge-case item below.

Edge cases

What happens to the eigenvalues when there is zero damping (

c = 0

) in the mass–spring system?

With

c = 0

A = [0 - k / m 10]

has eigenvalues

λ = \pm i k / m

— purely imaginary, so no real part: undamped oscillation forever (on the imaginary axis in figure s02).

What does an eigenvalue with positive real part mean?

The mode

e^{λ t}

grows without bound, so the system is unstable — any tiny disturbance blows up. In figure s02 this is the whole right half-plane; even one such eigenvalue makes the system unstable.

What does an eigenvalue exactly at

λ = 0

mean physically?

A mode that neither grows nor decays but drifts — e.g. a free integrator (position with no restoring force). It sits on the stability boundary: marginally stable, not asymptotically stable.

What if

A

is defective (non-diagonalizable) — can we still compute

e^{A t}

Yes. We put

A

in Jordan form: for a repeated eigenvalue

λ

with too few eigenvectors,

e^{A t}

gains terms like

t e^{λ t}

\frac{t ^{2}}{2} e^{λ t}

, … from the Jordan block. Stability still hinges on

Re (λ)

, but a repeated eigenvalue on the imaginary axis is now unstable because the

t

factor grows unbounded even though

∣ e^{λ t} ∣ = 1

A

is the zero matrix, what is the free response?

\overset{x}{˙} = 0

means the state holds constant, and indeed

e^{0 \cdot t} = I

, so

x (t) = x (0)

— a pure memory that never changes without input. (Note

A = 0

is a repeated eigenvalue

0

that is not defective, so no

t

terms appear.)

What if a state is uncontrollable — does it still appear in the model?

Yes, it stays in

A

and evolves, but no choice of

u

can steer it because that direction is missing from the reach of

B

. It may vanish from

G (s)

via cancellation — see Controllability and Observability.

Can a system be stable in the transfer-function sense yet unstable in state-space?

Yes. If an unstable eigenvalue is cancelled by a zero (unobservable/uncontrollable mode),

G (s)

looks stable while the hidden internal state blows up — a classic and dangerous trap.

What is the smallest possible state dimension for a system with a direct feedthrough

y = D u

and no dynamics?

Zero. With no memory elements the state vector is empty (

n = 0

A

and

B

vanish,

C

is absent, and only the algebraic map

y = D u

remains — a valid

0 \times 0

-state model, just

y = D u

with no dynamics to store.

Recall One-line self-test

Q: Give the single invariant that survives any coordinate change of a state-space model. A: The eigenvalues of $A$ (hence stability and the poles), because $T A T^{- 1}$ is similar to $A$ .

Connections

Eigenvalues and stability — the recurring theme behind half these traps.
Transfer functions and G(s) — pole–zero cancellation is where hidden-mode traps hide.
Matrix exponential — why the series, not element-wise exp; Jordan blocks for defective $A$ .
Controllability and Observability — the reachability/visibility edge cases.
Kalman filter — why unmeasured states force estimation.
Linearization of nonlinear systems — where $A, B$ come from in real GNC.