GATE EC · EE · IN

State-space models

State-space models describe an LTI system by n first-order differential equations in a state vector, $\overset{x}{˙} = A x + B u$ and $y = C x + D u$ , instead of by a single input-output transfer function. Solving GATE questions on this topic comes down to a small set of mechanical skills done accurately: writing a state model from a differential equation or a transfer function, recovering $G (s) = C (s I - A)^{- 1} B + D$ , computing $e^{A t}$ and the time response, applying the Kalman rank tests for controllability and observability, and using controllability to place the closed-loop poles with state feedback.

16 previous-year questions · 16 solved · 6 practice questions

State, state variables and the state model

The state of a dynamic system at time $t_{0}$ is the smallest set of variables that, together with the input $u (t)$ for $t \geq t_{0}$ , completely determines how the system behaves for all $t \geq t_{0}$ . These are the state variables $x_{1}$ to $x_{n}$ . Stacked into a column they form the state vector $x (t)$ . The number $n$ is the order of the state model. It equals the degree of the characteristic polynomial $det (s I - A)$ and the number of eigenvalues of $A$ counted with multiplicity. For a model built from a single $n$ th-order differential equation, it also equals the order of that equation. The transfer function can have a lower order than $n$ if a pole-zero cancellation occurs (see the section on controllability and observability).

In a lumped electrical network, $n$ is the number of independent energy-storing elements, and the natural state variables are inductor currents and capacitor voltages. These carry the stored energy and cannot change instantaneously as long as the sources are bounded (no impulses). Each loop made only of capacitors (and possibly voltage sources) reduces the count by one, because KVL fixes one capacitor voltage in that loop in terms of the others. Likewise, each cut-set made only of inductors (and possibly current sources) reduces the count by one, because KCL fixes one inductor current in terms of the others.

State variables are not unique. Any invertible linear combination of a valid set, $x = P z$ with $P$ invertible, is another valid set, so one system has infinitely many state models. All state models related by such a change of variables have the same transfer function and the same eigenvalues; only $A$ , $B$ and $C$ change, and $D$ stays the same. (A non-minimal realisation of the same $G (s)$ , with more states, has extra eigenvalues, so this statement is about models of the same order related by $x = P z$ .)

STATE MODEL OF A CONTINUOUS-TIME LTI SYSTEM

\overset{x}{˙} (t) y (t) = A x (t) + B u (t) (state equation) = C x (t) + D u (t) (output equation)

$n$ states, $m$ inputs, $p$ outputs. $A, B, C, D$ are constant matrices because the system is time-invariant.

The four matrices

Matrix	Name	Size	What it governs
$A$	system (state) matrix	$n \times n$	internal dynamics; its eigenvalues are the characteristic roots
$B$	input (control) matrix	$n \times m$	how the inputs drive the states; enters the controllability test
$C$	output matrix	$p \times n$	which combination of states is measured; enters the observability test
$D$	direct transmission (feed-through) matrix	$p \times m$	the part of the input that reaches the output directly; entry $D_{ij}$ is nonzero exactly when $G_{ij} (s)$ is biproper (numerator degree = denominator degree)

EXAM TIP

$D = lim_{s \to \infty} G (s)$ , because $(s I - A)^{- 1} \to 0$ as $s \to \infty$ . A strictly proper $G (s)$ (numerator degree lower than denominator degree) always gives $D = 0$ .

Building a state model: differential equations and canonical forms

From a differential equation with no derivatives of the input. Take $\frac{d ^{3} y}{d t ^{3}} + a_{2} \overset{y}{¨} + a_{1} \overset{y}{˙} + a_{0} y = b_{0} u$ . Choose the output and its successive derivatives as the states. These are the phase variables $x_{1} = y$ , $x_{2} = \overset{y}{˙}$ , $x_{3} = \overset{y}{¨}$ . Each state is the derivative of the one before it, and the last state equation is the ODE solved for the highest derivative.

PHASE-VARIABLE MODEL OF AN ODE (NO INPUT DERIVATIVES)

x_{1} = y, x_{2} = \overset{y}{˙}, x_{3} = \overset{y}{¨} : \overset{x}{˙}_{1} \overset{x}{˙}_{2} \overset{x}{˙}_{3} = 00 - a_{0} 10 - a_{1} 01 - a_{2} x_{1} x_{2} x_{3} + 00 b_{0} u, y = [100] x

The coefficient of the highest derivative must be 1 (divide through first). The same pattern holds for any order $n$ .

When derivatives of the input appear, for example $\overset{y}{¨} + 3 \overset{y}{˙} + 2 y = \overset{u}{˙} + 4 u$ , choosing $x_{1} = y$ , $x_{2} = \overset{y}{˙}$ would put $\overset{u}{˙}$ into the state equation, and $\overset{x}{˙} = A x + B u$ does not allow that. Go through the transfer function instead and use a canonical form. The forms below are written for the third-order strictly proper $G (s) = \frac{b _{2} s ^{2} + b _{1} s + b _{0}}{s ^{3} + a _{2} s ^{2} + a _{1} s + a _{0}}$ , and the pattern extends to any order. For the example above, $G (s) = \frac{s + 4}{s ^{2} + 3 s + 2}$ , so the controllable canonical form is $A = [0 - 2 1 - 3]$ , $B = [01]$ , $C = [41]$ . In this model the states are not $y$ and $\overset{y}{˙}$ : the output is $y = 4 x_{1} + x_{2}$ .

CONTROLLABLE CANONICAL (PHASE-VARIABLE) FORM

A = 00 - a_{0} 10 - a_{1} 01 - a_{2}, B = 001, C = [b_{0} b_{1} b_{2}], D = 0

The denominator must be monic. The last row of $A$ holds the negated denominator coefficients in ascending powers of $s$ , and $C$ holds the numerator coefficients in ascending powers. This form is always controllable. Some texts number the states in reverse, which puts the coefficients in the first row; $G (s)$ is the same either way.

OBSERVABLE CANONICAL FORM (DUAL OF THE CONTROLLABLE FORM)

A = 010001 - a_{0} - a_{1} - a_{2}, B = b_{0} b_{1} b_{2}, C = [001], D = 0

Convention used here: $A_{o} = A_{c}^{T}$ , $B_{o} = C_{c}^{T}$ , $C_{o} = B_{c}^{T}$ . This form is always observable. Reversing the state order flips rows and columns without changing $G (s)$ .

DIAGONAL (PARALLEL) FORM: DISTINCT POLES

G (s) = d + i = 1 \sum 3 \frac{r _{i}}{s - λ _{i}} \Rightarrow A = λ_{1} 00 0 λ_{2} 0 00 λ_{3}, B = 111, C = [r_{1} r_{2} r_{3}], D = d

Only for distinct poles; the $r_{i}$ are the partial-fraction residues. The residues can sit in $B$ instead of $C$ , because only each product $b_{i} c_{i} = r_{i}$ is fixed. Complex poles give complex entries.

JORDAN FORM: REPEATED POLE

G (s) = \frac{r _{1}}{( s - λ ) ^{2}} + \frac{r _{2}}{s - λ} + \frac{r _{3}}{s - μ} \Rightarrow A = λ 00 1 λ 0 00 μ, B = 011, C = [r_{1} r_{2} r_{3}], D = 0

A pole of multiplicity $k$ gives a $k \times k$ Jordan block, with $λ$ on the diagonal and 1 on the superdiagonal. With this convention ( $B$ has a 1 only in the last row of each block), the coefficient of the highest power $1/ (s - λ)^{k}$ goes in the first column of that block in $C$ . If $G (s)$ is biproper, add its constant term as $D$ .

Procedure: transfer function to controllable canonical form

Make the denominator monic by dividing the numerator and denominator by its leading coefficient.
If the numerator degree equals the denominator degree, long-divide first: $G (s) = d + N_{r} (s) / Den (s)$ . Then $D = d$ , and you realise the strictly proper remainder.
Put 1s on the superdiagonal of $A$ and $- a_{0}, - a_{1}, \dots, - a_{n - 1}$ in its last row. Every other entry is zero.
Set $B = [00 \dots 1]^{T}$ .
Set $C = [b_{0} b_{1} \dots b_{n - 1}]$ from the remainder numerator, in ascending powers of $s$ .

EXAM TIP

Check the companion matrix by expanding $det (s I - A)$ . It must return the original denominator. Writing the last row as $- a_{2}, - a_{1}, - a_{0}$ (descending order) is an easy slip that this check catches at once.

EXAM TIP

Make the denominator monic before you read off coefficients. For $G (s) = \frac{4 s + 8}{2 s ^{2} + 6 s + 4}$ , first rewrite it as $\frac{2 s + 4}{s ^{2} + 3 s + 2}$ . That gives last row $[- 2 - 3]$ and $C = [42]$ , not $[84]$ .

From state model to transfer function: eigenvalues and similarity

Take the Laplace transform of the state model with zero initial state. This gives $s X (s) = A X (s) + B U (s)$ , so $X (s) = (s I - A)^{- 1} B U (s)$ and $Y (s) = [C (s I - A)^{- 1} B + D] U (s)$ . Before any cancellation, the denominator of every entry is $det (s I - A)$ , so every pole comes from an eigenvalue of $A$ .

TRANSFER FUNCTION FROM A STATE MODEL

G (s) = \frac{Y ( s )}{U ( s )} = C (s I - A)^{- 1} B + D = \frac{C adj ( s I - A ) B}{det ( s I - A )} + D

Assumes zero initial conditions. For MIMO systems $G (s)$ is a $p \times m$ matrix with $G_{ij} = Y_{i} / U_{j}$ . The order of multiplication is fixed: $C$ , then $(s I - A)^{- 1}$ , then $B$ .

CHARACTERISTIC EQUATION

det (s I - A) = 0

Its roots are the eigenvalues of $A$ . Every pole of $G (s)$ is an eigenvalue. If $C adj (s I - A) B$ shares a factor with $det (s I - A)$ , that eigenvalue cancels and does not appear as a pole. $det (A - λ I)$ differs from $det (λ I - A)$ only by $(- 1)^{n}$ , so the two have the same roots.

INVERSE OF A 2×2 MATRIX

[a c b d]^{- 1} = \frac{1}{a d - b c} [d - c - b a], a d - b c \neq = 0

Swap the diagonal entries, negate the off-diagonal ones and divide by the determinant. You use this in nearly every 2×2 $(s I - A)^{- 1}$ calculation.

EIGENVALUE SHORTCUTS

i \sum λ_{i} = tr (A), i \prod λ_{i} = det A, n = 2 : s^{2} - tr (A) s + det A = 0

Holds for any square $A$ , with eigenvalues counted by multiplicity. The eigenvalues of a triangular or diagonal matrix are its diagonal entries.

SIMILARITY TRANSFORMATION

x = P z \Rightarrow \overset{ˉ}{A} = P^{- 1} A P, \overset{ˉ}{B} = P^{- 1} B, \overset{ˉ}{C} = C P, \overset{ˉ}{D} = D

$P$ is any invertible $n \times n$ matrix. The transformation does not change the eigenvalues, the characteristic equation, $G (s)$ , controllability or observability. With $P$ = modal matrix (eigenvectors as columns), $\overset{ˉ}{A}$ is diagonal. This needs $n$ linearly independent eigenvectors, which is guaranteed when the eigenvalues are distinct.

MODAL MATRIX OF A COMPANION MATRIX

P = 1 λ_{1} λ_{1}^{2} 1 λ_{2} λ_{2}^{2} 1 λ_{3} λ_{3}^{2} \Rightarrow P^{- 1} A P = λ_{1} 00 0 λ_{2} 0 00 λ_{3}

For $A$ in controllable canonical (companion) form, with the coefficients in the last row, the eigenvector for $λ_{i}$ is $[1 λ_{i} λ_{i}^{2}]^{T}$ . This $P$ (a Vandermonde matrix) is invertible only when the eigenvalues are distinct.

EXAM TIP

A system has many state models but only one transfer function. If two options have different $A$ matrices with the same characteristic polynomial, and both give the same $G (s)$ , both are valid realisations. Compare $det (s I - A)$ and $G (s)$ , not the matrices entry by entry.

The state transition matrix

The unforced system $\overset{x}{˙} = A x$ has solution $x (t) = e^{A t} x (0)$ . The matrix $Φ (t) = e^{A t}$ is the state transition matrix (STM): it carries the state from time 0 to time $t$ . It is a matrix exponential defined by a power series. It is not the matrix of element-wise exponentials $e^{a_{ij} t}$ . A reliable way to compute it by hand is the inverse Laplace transform of the resolvent $(s I - A)^{- 1}$ .

STATE TRANSITION MATRIX

Φ (t) = e^{A t} = k = 0 \sum \infty \frac{A ^{k} t ^{k}}{k !} = I + A t + \frac{A ^{2} t ^{2}}{2 !} + \dots = L^{- 1} [(s I - A)^{- 1}]

Valid for any constant square $A$ . The series stops after finitely many terms if $A$ is nilpotent, e.g. $A = [0010]$ gives $e^{A t} = [10 t 1]$ .

Method: the STM by inverse Laplace transform

Form $s I - A$ .
Compute $det (s I - A)$ , which is the characteristic polynomial, and factor it.
Write $(s I - A)^{- 1} = adj (s I - A) / det (s I - A)$ .
Split every entry into partial fractions.
Take the inverse transform of each entry. Check that $Φ (0) = I$ before you use the result.

PROPERTIES OF THE STM

Φ (0) = I Φ^{- 1} (t) = Φ (- t) Φ (t_{1} + t_{2}) = Φ (t_{1}) Φ (t_{2}) Φ (t_{2} - t_{0}) = Φ (t_{2} - t_{1}) Φ (t_{1} - t_{0}) [Φ (t)]^{k} = Φ (k t) (k any integer) \dot{Φ} (t) = A Φ (t) = Φ (t) A, A = \dot{Φ} (0) det Φ (t) = e^{tr (A) t} \neq = 0

True for every constant $A$ , so $Φ (t)$ is always invertible. Use $Φ (0) = I$ and $\dot{Φ} (0) = A$ to test whether a given matrix can be an STM, and to recover $A$ from a given STM.

STMS TO RECOGNISE ON SIGHT

A = [λ_{1} 0 0 λ_{2}] A = [λ 0 1 λ] A = [σ - ω ω σ] A = P Λ P^{- 1} \Rightarrow e^{A t} = [e^{λ_{1} t} 0 0 e^{λ_{2} t}] \Rightarrow e^{A t} = e^{λ t} [10 t 1] \Rightarrow e^{A t} = e^{σ t} [cos ω t - sin ω t sin ω t cos ω t] \Rightarrow e^{A t} = P e^{Λ t} P^{- 1}

The third matrix has eigenvalues $σ \pm j ω$ . The last line is the diagonalisation method: $Λ$ is the diagonal matrix of eigenvalues and $P$ is the modal matrix, which exists when $A$ has $n$ linearly independent eigenvectors (always the case for distinct eigenvalues).

CAYLEY-HAMILTON METHOD

e^{A t} = k = 0 \sum n - 1 α_{k} (t) A^{k}, e^{λ_{i} t} = k = 0 \sum n - 1 α_{k} (t) λ_{i}^{k} for each distinct eigenvalue λ_{i}

Basis: the Cayley-Hamilton theorem says every square matrix satisfies its own characteristic equation, so $A^{n}$ and all higher powers reduce to combinations of $I, A, \dots, A^{n - 1}$ . The same reduction evaluates any polynomial in $A$ , such as $A^{10}$ . For $n = 2$ with distinct eigenvalues: $e^{A t} = α_{0} I + α_{1} A$ , with $e^{λ_{1} t} = α_{0} + α_{1} λ_{1}$ and $e^{λ_{2} t} = α_{0} + α_{1} λ_{2}$ . For a repeated eigenvalue, the distinct-eigenvalue equations are too few; add the equation you get by differentiating with respect to $λ$ : $t e^{λ t} = α_{1} + 2 α_{2} λ + \dots$ .

EXAM TIP

To test whether a matrix can be an STM, check $Φ (0) = I$ first; it is a quick filter. A matrix can pass that check and still fail. Then take $A = \dot{Φ} (0)$ and confirm $\dot{Φ} (t) = A Φ (t)$ , or check $Φ (t) Φ (- t) = I$ . Example: $[cos t sin t sin t cos t]$ equals $I$ at $t = 0$ but is not an STM.

EXAM TIP

If the unforced responses $x_{a} (t)$ and $x_{b} (t)$ of a second-order system are known for two linearly independent initial states $x_{a} (0)$ and $x_{b} (0)$ , linearity gives $Φ (t) [x_{a} (0) x_{b} (0)] = [x_{a} (t) x_{b} (t)]$ , so $Φ (t) = [x_{a} (t) x_{b} (t)] [x_{a} (0) x_{b} (0)]^{- 1}$ , and then $A = \dot{Φ} (0)$ . An $n$ th-order system needs $n$ independent initial states. The response to any other initial state follows without finding $A$ .

EXAM TIP

$e^{(A_{1} + A_{2}) t} = e^{A_{1} t} e^{A_{2} t}$ holds for all $t$ if and only if $A_{1} A_{2} = A_{2} A_{1}$ . It does work for $σ I + ω J$ , because $σ I$ commutes with every matrix. That is why the $e^{σ t}$ factor splits off in the rotation-type STM above.

Solving the state equation

The complete response is the sum of two parts. The zero-input response comes from the initial state with no input. The zero-state response comes from the input with zero initial state, and is the convolution of $e^{A t} B$ with the input. In the $s$ -domain both parts use the same resolvent $(s I - A)^{- 1}$ , so compute it once and reuse it.

COMPLETE SOLUTION OF THE STATE EQUATION

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

Initial time 0. For initial time $t_{0}$ : $x (t) = Φ (t - t_{0}) x (t_{0}) + \int_{t_{0}}^{t} Φ (t - τ) B u (τ) d τ$ .

OUTPUT RESPONSE

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

Substitute $x (t)$ into the output equation. The $D u (t)$ term appears instantly and has no dynamics.

SOLUTION IN THE S-DOMAIN

X (s) = (s I - A)^{- 1} x (0) + (s I - A)^{- 1} B U (s), Y (s) = C X (s) + D U (s)

Often the quickest route by hand. Take the inverse transform of each entry at the end.

IMPULSE RESPONSE

g (t) = C e^{A t} B + D δ (t) (t \geq 0), G (s) = L [g (t)]

Zero initial state and a unit impulse input; $g (t) = 0$ for $t < 0$ .

UNIT-STEP RESPONSE AND FINAL VALUES

x_{z s} (t) = A^{- 1} (e^{A t} - I) B, x (\infty) = - A^{- 1} B, y (\infty) = - C A^{- 1} B + D = G (0)

The first formula (unit-step input, zero initial state) needs $A$ invertible. The final values hold, for any initial state, when every eigenvalue of $A$ has a negative real part; $A$ is then automatically invertible, because 0 is not an eigenvalue. You can also get $x (\infty)$ by setting $\overset{x}{˙} = 0$ : $0 = A x + B$ .

WORKED EXAMPLE · WORKED EXAMPLE 1: STM AND COMPLETE RESPONSE

For $\overset{x}{˙} = [0 - 2 1 - 3] x + [01] u$ with $x (0) = [10]$ and $u (t)$ a unit step, find $e^{A t}$ and $x (t)$ for $t \geq 0$ .

Solution

Form $s I - A = [s 2 - 1 s + 3]$ , so $det (s I - A) = s (s + 3) + 2 = s^{2} + 3 s + 2 = (s + 1) (s + 2)$ . The eigenvalues are $- 1$ and $- 2$ .
By the 2×2 inverse rule, $(s I - A)^{- 1} = \frac{1}{( s + 1 ) ( s + 2 )} [s + 3 - 2 1 s]$ .
Partial fractions: $\frac{s + 3}{( s + 1 ) ( s + 2 )} = \frac{2}{s + 1} - \frac{1}{s + 2}$ , $\frac{1}{( s + 1 ) ( s + 2 )} = \frac{1}{s + 1} - \frac{1}{s + 2}$ , $\frac{- 2}{( s + 1 ) ( s + 2 )} = \frac{- 2}{s + 1} + \frac{2}{s + 2}$ , $\frac{s}{( s + 1 ) ( s + 2 )} = \frac{- 1}{s + 1} + \frac{2}{s + 2}$ .
Take the inverse transform of each entry: $e^{A t} = [2 e^{- t} - e^{- 2 t} - 2 e^{- t} + 2 e^{- 2 t} e^{- t} - e^{- 2 t} - e^{- t} + 2 e^{- 2 t}]$ . Check: at $t = 0$ this is $I$ .
Zero-input part: $e^{A t} x (0)$ is the first column, $[2 e^{- t} - e^{- 2 t} - 2 e^{- t} + 2 e^{- 2 t}]$ .
Zero-state part: $(s I - A)^{- 1} B U (s) = \frac{1}{( s + 1 ) ( s + 2 )} [1 s] \frac{1}{s} = [\frac{1}{s ( s + 1 ) ( s + 2 )} \frac{1}{( s + 1 ) ( s + 2 )}]$ . Its inverse transform is $[\frac{1}{2} - e^{- t} + \frac{1}{2} e^{- 2 t} e^{- t} - e^{- 2 t}]$ .
Add the two parts: $x_{1} (t) = \frac{1}{2} + e^{- t} - \frac{1}{2} e^{- 2 t}$ and $x_{2} (t) = - e^{- t} + e^{- 2 t}$ .
Checks: $x (0) = [10]^{T}$ . $x_{2} = \overset{x}{˙}_{1}$ , as the first row of $A$ requires. $x (\infty) = [\frac{1}{2} 0]^{T}$ , which matches $- A^{- 1} B$ with $A^{- 1} = \frac{1}{2} [- 3 2 - 1 0]$ .

Answer: $e^{A t}$ is the matrix in step 4. For $t \geq 0$ : $x_{1} (t) = \frac{1}{2} + e^{- t} - \frac{1}{2} e^{- 2 t}$ and $x_{2} (t) = - e^{- t} + e^{- 2 t}$ .

EXAM TIP

Internal (asymptotic) stability requires every eigenvalue of $A$ to have $Re λ < 0$ . BIBO stability depends only on the poles of $G (s)$ that remain after cancellation. An unstable mode that is uncontrollable or unobservable cancels out of $G (s)$ . The system can then be BIBO stable yet internally unstable. So read the question: for internal or asymptotic stability use the eigenvalues of $A$ ; for BIBO stability use the poles of $G (s)$ in lowest terms.

EXAM TIP

To recover an earlier state of an unforced system, run the STM backwards: $x (0) = Φ (- t) x (t) = Φ^{- 1} (t) x (t)$ . You do not need to invert a matrix numerically if you already have $Φ (t)$ in closed form. Replace $t$ by $- t$ .

DISCRETE-TIME STATE MODEL AND ITS SOLUTION

x [k + 1] x [k] A^{k} = A x [k] + B u [k], y [k] = C x [k] + D u [k] = A^{k} x [0] + i = 0 \sum k - 1 A^{k - 1 - i} B u [i] = Z^{- 1} [z (z I - A)^{- 1}], G (z) = C (z I - A)^{- 1} B + D

Here $A, B, C, D$ are the discrete-time matrices, and $A^{k}$ plays the role of the STM. The system is asymptotically stable exactly when every eigenvalue of $A$ has $∣ λ ∣ < 1$ . The Kalman matrices $[B A B \dots A^{n - 1} B]$ and $[C; C A; \dots; C A^{n - 1}]$ take the same form: full rank of the first means every state can be reached from the origin, and full rank of the second means $x [0]$ can be recovered from the output. Sampling $\overset{x}{˙} = A x + B u$ with a zero-order hold of period $T$ gives $A_{d} = e^{A T}$ and $B_{d} = (\int_{0}^{T} e^{A σ} d σ) B$ , which equals $A^{- 1} (e^{A T} - I) B$ when $A$ is invertible.

Controllability, observability and state feedback

A system is completely state controllable if, for any initial state and any target state, some input $u (t)$ moves the state to the target in finite time. It is completely observable if the initial state $x (t_{0})$ can be found uniquely from the output $y (t)$ (and the known input) over a finite interval. Controllability asks whether the input reaches every mode. Observability asks whether every mode appears in the output. Neither property depends on $D$ , and a similarity transformation changes neither.

KALMAN CONTROLLABILITY TEST

Q_{c} = [B A B A^{2} B \dots A^{n - 1} B], controllable ⟺ rank Q_{c} = n

Depends only on $(A, B)$ . For a single input, $Q_{c}$ is $n \times n$ and the system is controllable exactly when $det Q_{c} \neq = 0$ . For $m$ inputs, $Q_{c}$ is $n \times nm$ and you must check its rank. Compute $A^{2} B$ as $A (A B)$ .

KALMAN OBSERVABILITY TEST

Q_{o} = C C A C A^{2} ⋮ C A^{n - 1}, observable ⟺ rank Q_{o} = n

Depends only on $(A, C)$ . An equivalent transposed convention is $[C^{T} A^{T} C^{T} \dots (A^{T})^{n - 1} C^{T}]$ , which has the same rank. For a single output, the system is observable exactly when $det Q_{o} \neq = 0$ .

DIAGONAL-FORM TEST

\overset{ˉ}{A} = diag (λ_{1}, \dots, λ_{n}) distinct : controllable ⟺ no zero row in \overset{ˉ}{B}, observable ⟺ no zero column in \overset{ˉ}{C}

Use only after transforming to a diagonal $\overset{ˉ}{A}$ with distinct eigenvalues. A zero row $i$ of $\overset{ˉ}{B}$ means the mode $e^{λ_{i} t}$ cannot be excited, and a zero column $i$ of $\overset{ˉ}{C}$ means that mode never reaches the output. In Jordan form (one block per distinct eigenvalue), check the row of $\overset{ˉ}{B}$ at the last row of each block and the column of $\overset{ˉ}{C}$ at the first column of each block. The test does not apply to a diagonal $\overset{ˉ}{A}$ with a repeated eigenvalue; see the last tip of this section.

EIGENVALUE RANK TEST (MODE BY MODE)

rank [λ_{i} I - A B] = n (controllable), rank [λ_{i} I - A C] = n (observable), for every eigenvalue λ_{i}

Equivalent to the Kalman tests. An eigenvalue at which the rank drops is the uncontrollable or unobservable mode.

Controllability versus observability

	Controllability	Observability
Question asked	Can the input steer every state anywhere in finite time?	Can the initial state be found from the output over a finite time?
Depends on	$A$ and $B$ only	$A$ and $C$ only
Kalman matrix	$Q_{c} = [B A B \dots A^{n - 1} B]$	$Q_{o} = [C; C A; \dots; C A^{n - 1}]$ (stacked rows)
Condition	$rank Q_{c} = n$	$rank Q_{o} = n$
Diagonal form, distinct $λ$	no zero row in $\overset{ˉ}{B}$	no zero column in $\overset{ˉ}{C}$
Form that guarantees it	controllable canonical form	observable canonical form
Duality	$(A, B)$ controllable $⟺$ $(A^{T}, B^{T})$ observable	$(A, C)$ observable $⟺$ $(A^{T}, C^{T})$ controllable

Pole-zero cancellation and minimal realisation (SISO)

If $C adj (s I - A) B$ and $det (s I - A)$ share a factor $(s - λ)$ , the mode $e^{λ t}$ is uncontrollable, unobservable, or both.
A SISO state model is both controllable and observable (a minimal realisation) exactly when $G (s)$ in lowest terms has order $n$ , i.e. when there is no cancellation.
The controllable canonical form of $N (s) / Den (s)$ is always controllable. It is observable exactly when $N (s)$ and $Den (s)$ have no common factor.
The observable canonical form is always observable. It is controllable exactly when there is no common factor.
A cancellation does not tell you which property fails. Run the Kalman tests to find out.

WORKED EXAMPLE · WORKED EXAMPLE 2: A HIDDEN MODE

A system has $A = [0 - 6 1 - 5]$ , $B = [01]$ , $C = [21]$ , $D = 0$ . Find $G (s)$ , test controllability and observability, and identify any hidden mode.

Solution

$det (s I - A) = s^{2} + 5 s + 6 = (s + 2) (s + 3)$ , so the eigenvalues are $- 2$ and $- 3$ .
$adj (s I - A) = [s + 5 - 6 1 s]$ . Multiplying by $B$ picks out the second column: $(s I - A)^{- 1} B = \frac{1}{( s + 2 ) ( s + 3 )} [1 s]$ .
$G (s) = C (s I - A)^{- 1} B = \frac{2 + s}{( s + 2 ) ( s + 3 )} = \frac{1}{s + 3}$ . The factor $(s + 2)$ cancels, so a second-order model gives a first-order $G (s)$ .
Controllability: $A B = [1 - 5]$ , $Q_{c} = [01 1 - 5]$ , $det Q_{c} = - 1 \neq = 0$ . The system is controllable, as expected for a controllable canonical form.
Observability: $C A = [21] A = [- 6 - 3]$ , $Q_{o} = [2 - 6 1 - 3]$ , $det Q_{o} = - 6 + 6 = 0$ . The rank is $1 < 2$ , so the system is not observable.
Hidden mode: for $λ = - 2$ , $(A + 2 I) v = 0$ gives $v = [1 - 2]^{T}$ , and $C v = 2 - 2 = 0$ . The mode $e^{- 2 t}$ never reaches the output.

Answer: $G (s) = \frac{1}{s + 3}$ . The system is controllable but not observable, and the mode at $s = - 2$ ( $e^{- 2 t}$ ) is the unobservable one.

EXAM TIP

For $n = 3$ the controllability matrix is $[B A B A^{2} B]$ . Stopping at $[B A B]$ gives a $3 \times 2$ matrix (single input) whose rank can never reach 3. Build the matrix out to $A^{n - 1} B$ (or $C A^{n - 1}$ for observability).

EXAM TIP

For a single-input system ( $Q_{c}$ square), when $A$ or $B$ contains a parameter, the roots of $det Q_{c} = 0$ are exactly the parameter values that make the system uncontrollable; for a single output, the roots of $det Q_{o} = 0$ are the values that make it unobservable. For a single input with $n \geq 2$ , if $B$ is an eigenvector of $A$ then $A B = λ B$ , the columns of $Q_{c}$ are dependent, and the system is uncontrollable.

EXAM TIP

If one eigenvalue has two or more Jordan blocks, a single-input system cannot be controllable and a single-output system cannot be observable. The simplest case is $A = λ I$ with $n \geq 2$ .

STATE FEEDBACK AND POLE PLACEMENT

u = - K x + r \Rightarrow \overset{x}{˙} = (A - B K) x + B r, closed-loop characteristic equation: det (s I - A + B K) = 0 single input: K = [0 \dots 01] Q_{c}^{- 1} α_{d} (A)

$K$ is $m \times n$ ( $1 \times n$ for a single input) and $r$ is the new reference input. The eigenvalues of $A - B K$ can be placed at any chosen locations (complex ones in conjugate pairs) if and only if $(A, B)$ is controllable. In the pole-placement formula, $α_{d} (s) = s^{n} + α_{n - 1} s^{n - 1} + \dots + α_{0}$ is the desired closed-loop characteristic polynomial and $α_{d} (A) = A^{n} + α_{n - 1} A^{n - 1} + \dots + α_{0} I$ . State feedback does not change controllability and, for a SISO system, does not move the zeros of the transfer function; it can destroy observability if a closed-loop pole is placed on a zero.

EXAM TIP

Shortcut in controllable canonical form: if $A$ has last row $[- a_{0} - a_{1} \dots - a_{n - 1}]$ , $B = [0 \dots 01]^{T}$ and $K = [k_{1} k_{2} \dots k_{n}]$ , then $A - B K$ is still in that form with last row $[- (a_{0} + k_{1}) - (a_{1} + k_{2}) \dots - (a_{n - 1} + k_{n})]$ . Match coefficients with the desired polynomial: $k_{i + 1} = α_{i} - a_{i}$ . For any other form, expand $det (s I - A + B K)$ and match coefficients.

PREVIOUS-YEAR QUESTIONS

16 questions from the EC, EE and IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2026 · GATE ECE 2025 · GATE ECE 2024 · GATE ECE 2023 · GATE ECE 2020 · GATE ECE 2018 · GATE EE 2026 · GATE EE 2025 · GATE EE 2023 · GATE EE 2018 · GATE IN 2026 · GATE IN 2024 · GATE IN 2022 · GATE IN 2020 · GATE IN 2018

GATE 2026 EC Q.41

GATE 2026 · ECQ.41 · 2 marks · MCQ

The state and output equations for a control system are:

\overset{x}{˙} = [- 4 4 - 1.5 0] x + [20] u

y = [1.5 0.625] x

Which of the following expressions correctly represents the transfer function $\frac{Y ( s )}{U ( s )}$ of the system with zero initial conditions?

(A)
$\frac{3 s}{s ^{2} + 4 s - 6}$
(B)
$\frac{3 s + 5}{s ^{2} + 4 s - 6}$
(C)
$\frac{3 s + 5}{s ^{2} + 4 s + 6}$
(D)
$\frac{3 s}{s ^{2} + 4 s + 6}$

Answer and solution

GATE 2026 EC Q.41: answer and solution

See the question in the paper

State-space models

State, state variables and the state model

Building a state model: differential equations and canonical forms

From state model to transfer function: eigenvalues and similarity

The state transition matrix

Solving the state equation

Controllability, observability and state feedback

16 questions from the EC, EE and IN papers

6 questions to test yourself