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

3.5.29 · D5 · Physics › Guidance, Navigation & Control (GNC) › State-space representation — x' = Ax + Bu, y = Cx + Du

Definition Scope aur symbol reminder (pehle yeh padho — yeh page self-contained hai)

Yahan sab kuch continuous-time, linear, time-invariant systems ke baare mein hai: time $t$ smoothly flow karta hai (discrete steps mein nahi), isliye matrix exponential $e^{A t}$ aur derivative-dot $\overset{x}{˙} = \frac{d x}{d t}$ aate hain. Do governing equations hain: $\overset{x}{˙} = A x + B u, y = C x + D u .$

$x \in R^{n}$ — state vector: numbers ki woh minimal list (system ki memory) jo future predict karne ke liye chahiye. $n$ = system order.
$u \in R^{m}$ — input (jo controls tum apply karte ho, jaise force, thrust).
$y \in R^{p}$ — output (jo tumhare sensors actually padhte hain).
$A$ ( $n \times n$ ) — internal dynamics (state apne aap par feed back karta hai).
$B$ ( $n \times m$ ) — inputs state ko kaise push karte hain.
$C$ ( $p \times n$ ) — kaun se state combinations tum measure karte ho.
$D$ ( $p \times m$ ) — direct input→output shortcut (aksar $0$ hota hai).

Definition Neeche baar baar use hone wale words

Memory / energy-storing element: ek physical quantity jise agli instant jaanne ke liye yaad rakhna padta hai — jaise ek mass kinetic energy (uski velocity) store karta hai aur ek spring potential energy (uski stretch) store karti hai. Har independent element ko apna state variable chahiye, isliye memory elements ki sankhya = system order $n$ (mass–spring example dekhein, neeche figure mein).
Free / zero-input response: $e^{A t} x (0)$ wali motion jab input band ho ( $u = 0$ ) — pure memory ka kholna.
Forced / zero-state response: $\int_{0}^{t} e^{A (t - τ)} B u (τ) d τ$ wali motion jo rest se shuru hoti hai ( $x (0) = 0$ ) — input par pure reaction.
Poora solution dono ka sum hai: $x (t) = e^{A t} x (0) + \int_{0}^{t} e^{A (t - τ)} B u (τ) d τ$ .
Laplace transform: ek tarika jo time-signal $f (t)$ ko ek frequency-jaisi variable $s$ ki function $F (s)$ mein badalta hai, likha jaata hai $F (s) = \int_{0}^{\infty} f (t) e^{- s t} d t$ . Yeh derivatives ko $s$ se multiply karne mein badal deta hai, isliye $\overset{x}{˙} = A x + B u$ algebraic $G (s) = C (s I - A)^{- 1} B + D$ ban jaata hai. Yahan $Y (s)$ aur $U (s)$ ka matlab sirf "output $y (t)$ aur input $u (t)$ ke Laplace transforms" hai.

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

ek projection hai (figure s01:

C

2-D state ko ek measured axis par flatten karta hai, velocity dimension drop kar ke);

C

velocity jaisi states chhupa sakta hai jo motion drive karti hain lekin kabhi sense nahi hoti. State = woh sab kuch jo future predict karne ke liye chahiye, measured ho ya na ho.

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

True. System order

n

independent memory (energy-storing) elements ki sankhya ke barabar hota hai; mass–spring mein do hain (mass mein velocity, spring mein stretch), isliye

n = 2

: function aur uska pehla derivative.

D = 0

, then the input can never affect the output.

False.

D = 0

ke saath bhi input state ke zariye output ko shape karta hai

\overset{x}{˙} = A x + B u

ke through;

D

sirf instantaneous shortcut hataata hai, delayed influence nahi.

Two state-space models with the same eigenvalues of

A

describe the same physical system.

False in general. Same eigenvalues ka matlab same natural modes/poles hai, lekin

B

aur

C

alag ho sakte hain, alag input coupling aur outputs dete hain — identical dynamics ke bawajood alag behaviour.

A change of coordinates

\tilde{x} = T x

changes the system's stability.

False.

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

A

ke similar hai, isliye exact same eigenvalues hain. Stability asli system ki property hai, chosen coordinates ki nahi.

The matrix exponential

e^{A t}

equals the element-wise exponential of

A t

False. Yeh series

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

se define hota hai; tum poori matrix ko exponentiate karte ho, entries ko couple karte hue (figure s03: yeh poore state ko saath mein carry karta hai ek swirling flow ki tarah). Element-wise exp almost kabhi sahi nahi hota jab tak

A

diagonal na ho.

Every stable linear system oscillates as it decays.

False. Oscillation ke liye nonzero imaginary part wale eigenvalues chahiye. Real negative eigenvalues pure exponential decay dete hain bina kisi ringing ke (figure s02 mein real axis par green ×).

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

False.

B

mein

m

columns hain aur

C

mein

p

rows hain, isliye MIMO native hai — yahi precise reason hai ki GNC (Guidance, Navigation and Control) scalar transfer functions ke upar state-space use karta hai.

The eigenvalues of

A

are exactly the poles of the transfer function

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

False in general (yahi trap hai). Woh tab coincide karte hain jab koi pole–zero cancellation na ho. Agar koi mode uncontrollable ya unobservable hai toh woh cancel ho jaata hai, isliye woh eigenvalue

G (s)

ka pole nahi hai — sahi statement hai "poles eigenvalues ka ek subset hain."

Spot the error

"The system

\overset{x}{˙} = A x + B u

is stable because all entries of

A

are negative."

Galat test hai. Stability

A

ke eigenvalues ke negative real part hone par depend karti hai (figure s02 mein plane ka left half), entries ke signs par nahi. Negative entries wali matrix ke phir bhi unstable eigenvalues ho sakte hain.

"To convert

m \overset{q}{¨} + k q = F

I set

x_{1} = q

and

x_{2} = q

, giving two states."

Error: dono states identical hain, isliye yeh valid state nahi hai. Tumhe

x_{2} = \overset{q}{˙}

(derivative) lena chahiye, taaki

\overset{x}{˙}_{1} = x_{2}

unhe link kare aur second-order memory capture ho.

"Since

u

appears in

\overset{x}{˙} = A x + B u

, the input matrix must be square."

Error.

B

n \times m

hai:

n

rows (ek per state),

m

columns (ek per input). Yeh square tabhi hota hai jab accidentally

n = m

ho.

"The free response is

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

Galat label hai. Yeh full response hai. Free (zero-input) part sirf

e^{A t} x (0)

hai; integral forced (zero-state) part hai jo

u

se driven hai.

"Because

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

captures everything, state-space adds nothing."

Error.

Y (s), U (s)

output aur input ke Laplace transforms hain (definition callout dekhein);

G (s)

zero initial conditions assume karta hai aur internal/unmeasured states chhupa leta hai. State-space nonzero

x (0)

, MIMO, aur internal variables handle karta hai — ek tumbling vehicle ke liye zaroori.

e^{A t}

satisfies

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

, so order never matters."

Identity sahi hai lekin lesson galat hai. Yeh isliye hold karta hai kyunki

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

$A$ ki khud ki power series hai, aur koi bhi matrix apni khud ki powers ke saath commute karta hai, isliye

A

kisi bhi side slide ho sakta hai. Yeh

A

aur

e^{A t}

ke liye special hai; general matrix products commute nahi karte, isliye iska assumption kabhi mat karo.

Why questions

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

Solution machinery (

e^{A t}

, eigenvalue analysis, controllers) first-order vector form

\overset{x}{˙} = A x + B u

ke liye built hai; derivatives ko nayi states ki tarah naam dena kisi bhi

n

th-order ODE ko us standard shape mein badal deta hai.

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

Kyunki dynamics continuous-time hain:

\overset{x}{˙} = A x

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

ka vector version hai. Exponential woh unique function hai jiska derivative khud ko times

A

reproduce karta hai (matrix powers discrete-time steps ke liye hote).

Why does stability live entirely in

A

and not in

B

C

, or

D

Unforced motion

e^{A t} x (0)

hai, sirf

A

ke eigenvalues se governed.

B, C, D

inputs aur outputs route karte hain lekin internal modes ke grow ya decay hone ko change nahi kar sakte.

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

Kyunki ek baar

x (t_{0})

aur saare future inputs jaano, future poori tarah determine ho jaata hai — pehle ki koi history nahi chahiye. Koi bhi chhota set future uniquely predict karne mein fail ho jaata.

Why can two engineers write different

A

matrices for the same rocket and both be correct?

Unhone alag state coordinates choose kiye. Models

\tilde{x} = T x

se related hain,

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

dete hue — same physics, alag bookkeeping, identical eigenvalues.

Why do we need

C

at all instead of just outputting the whole state?

Real sensors state ke kuch hi combinations padhte hain (jaise position, velocity nahi — figure s01).

C

exactly model karta hai ki hardware kya measure kar sakta hai, isliye Kalman filter jaisi estimators baaki recover karne ke liye exist karti hain.

Why does a repeated eigenvalue sometimes make

e^{A t}

contain a factor of

t

Jab

A

defective hota hai (uske repeated eigenvalue ki demand se kam independent eigenvectors),

e^{A t}

mein polynomial-in-

t

terms aate hain jaise

t e^{λ t}

Jordan block se — neeche edge-case item dekhein.

Edge cases

What happens to the eigenvalues when there is zero damping (

c = 0

) in the mass–spring system?

c = 0

ke saath,

A = [0 - k / m 10]

ke eigenvalues

λ = \pm i k / m

hain — purely imaginary, isliye koi real part nahi: undamped oscillation forever (figure s02 mein imaginary axis par).

What does an eigenvalue with positive real part mean?

Mode

e^{λ t}

bina bound ke grow karta hai, isliye system unstable hai — koi bhi choti disturbance blow up ho jaati hai. Figure s02 mein yeh poora right half-plane hai; ek bhi aisa eigenvalue system ko unstable banata hai.

What does an eigenvalue exactly at

λ = 0

mean physically?

Ek aisa mode jo na grow karta hai na decay, balki drift karta hai — jaise ek free integrator (bina restoring force ke position). Yeh stability boundary par baitha hai: marginally stable, asymptotically stable nahi.

What if

A

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

e^{A t}

Haan. Hum

A

ko Jordan form mein daalte hain: repeated eigenvalue

λ

ke liye jab eigenvectors kam hon,

e^{A t}

mein

t e^{λ t}

\frac{t ^{2}}{2} e^{λ t}

, … jaisi terms aati hain Jordan block se. Stability abhi bhi

Re (λ)

par depend karti hai, lekin imaginary axis par repeated eigenvalue ab unstable hai kyunki

t

factor unbounded grow karta hai chahe

∣ e^{λ t} ∣ = 1

ho.

A

is the zero matrix, what is the free response?

\overset{x}{˙} = 0

ka matlab state constant rehta hai, aur indeed

e^{0 \cdot t} = I

, isliye

x (t) = x (0)

— ek pure memory jo input ke bina kabhi nahi badlata. (Note karo

A = 0

ek repeated eigenvalue

0

hai jo defective nahi hai, isliye koi

t

terms nahi aati.)

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

Haan, woh

A

mein rehta hai aur evolve karta hai, lekin

u

ki koi bhi choice use steer nahi kar sakti kyunki woh direction

B

ki reach mein missing hai. Woh

G (s)

se cancellation ke zariye gayab ho sakta hai — Controllability and Observability dekhein.

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

Haan. Agar ek unstable eigenvalue zero se cancel ho jaaye (unobservable/uncontrollable mode),

G (s)

stable lagta hai jabki hidden internal state blow up ho jaata hai — ek classic aur dangerous trap.

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

y = D u

and no dynamics?

Zero. Koi memory elements nahi hone par state vector khaali hota hai (

n = 0

A

aur

B

gayab ho jaate hain,

C

absent hota hai, aur sirf algebraic map

y = D u

rehta hai — ek valid

0 \times 0

-state model, bas

y = D u

bina kisi dynamics ke.

Recall One-line self-test

Q: Woh ek invariant batao jo state-space model ke kisi bhi coordinate change mein bachta hai. A: $A$ ke eigenvalues (hence stability aur poles), kyunki $T A T^{- 1}$ $A$ ke similar hai.

Connections

Eigenvalues and stability — in adhe se zyada traps ka recurring theme.
Transfer functions and G(s) — pole–zero cancellation wahan hai jahan hidden-mode traps chhupte hain.
Matrix exponential — series kyun, element-wise exp kyun nahi; defective $A$ ke liye Jordan blocks.
Controllability and Observability — reachability/visibility edge cases.
Kalman filter — unmeasured states estimation ko force kyun karte hain.
Linearization of nonlinear systems — real GNC mein $A, B$ kahaan se aate hain.