Introduction to Controllability and Observability The concepts of controllability and observability are fundamental to modern control system theory, having been introduced by R.E. Kalman in 1960.
While originally theoretical, these concepts are particularly important for practical implementations in engineering.
Controllability: This property determines if it is possible to manipulate the dynamic system to achieve a desired behavior under a specific control input.
Observability: This property determines if the internal state of a system can be identified by observing its external outputs.
Definitions and Core Principles Controllability Definition: A system is said to be controllable if an input function u ( t ) u(t) u ( t ) can transform the initial state x ( t 0 ) x(t_0) x ( t 0 ) of the system to any required final state x ( t f ) x(t_f) x ( t f ) within a finite time interval (t_f - t_0), t > 0.
Observability Definition: A system is said to be observable if, from a finite measurement of output carried over an interval of time ( 0 ≤ t ≤ t f ) (0 \le t \le t_f) ( 0 ≤ t ≤ t f ) , the initial state of the system can be determined.
Uncontrollability: If there is no connection between a specific state and the input, the system is called uncontrollable.
Unobservability: If there is no contact/connection between a certain state and the output, the system is deemed unobservable.
Observability of Discrete-Time Systems Consider a linear, time-invariant, discrete-time system in state-space form:
x ( k + 1 ) = A d x ( k ) x(k+1) = A_d x(k) x ( k + 1 ) = A d x ( k ) x ( 0 ) = x 0 x(0) = x_0 x ( 0 ) = x 0 (unknown)
With output measurement:
y ( k ) = C d x ( k ) y(k) = C_d x(k) y ( k ) = C d x ( k )
Where:
x ( k ) ∈ R n x(k) \in \mathbb{R}^n x ( k ) ∈ R n
y ( k ) ∈ R p y(k) \in \mathbb{R}^p y ( k ) ∈ R p
A d A_d A d and C d C_d C d are constant matrices of appropriate dimensions.
Since the n n n -dimensional vector x ( 0 ) x(0) x ( 0 ) has n n n unknown components, it is expected that n n n measurements are sufficient to determine it. By taking k = 0 , 1 , … , n − 1 k = 0, 1, \dots, n-1 k = 0 , 1 , … , n − 1 , a sequence of equations is generated:
y ( 0 ) = C d x ( 0 ) y(0) = C_d x(0) y ( 0 ) = C d x ( 0 ) y ( 1 ) = C d x ( 1 ) = C d A d x ( 0 ) y(1) = C_d x(1) = C_d A_d x(0) y ( 1 ) = C d x ( 1 ) = C d A d x ( 0 ) y ( 2 ) = C d x ( 2 ) = C d A d 2 x ( 0 ) y(2) = C_d x(2) = C_d A_d^2 x(0) y ( 2 ) = C d x ( 2 ) = C d A d 2 x ( 0 ) … \dots … y ( n − 1 ) = C d x ( n − 1 ) = C d A d n − 1 x ( 0 ) y(n-1) = C_d x(n-1) = C_d A_d^{n-1} x(0) y ( n − 1 ) = C d x ( n − 1 ) = C d A d n − 1 x ( 0 )
In matrix form, this is represented as:
( y ( 0 ) y ( 1 ) y ( 2 ) … y ( n − 1 ) ) = ( C d C d A d C d A d 2 … C d A d n − 1 ) x ( 0 ) \begin{pmatrix} y(0) \\ y(1) \\ y(2) \\ \dots \\ y(n-1) \end{pmatrix} = \begin{pmatrix} C_d \\ C_d A_d \\ C_d A_d^2 \\ \dots \\ C_d A_d^{n-1} \end{pmatrix} x(0) y ( 0 ) y ( 1 ) y ( 2 ) … y ( n − 1 ) = C d C d A d C d A d 2 … C d A d n − 1 x ( 0 )
From linear algebra, a unique solution for the unknown x ( 0 ) x(0) x ( 0 ) exists if and only if the system matrix has rank n n n . This leads to the definition of the observability matrix Q o Q_o Q o :
Q o = [ C d T , ( C d A d ) T , ( C d A d 2 ) T , … , ( C d A d n − 1 ) T ] T Q_o = [C_d^T, (C_d A_d)^T, (C_d A_d^2)^T, \dots, (C_d A_d^{n-1})^T]^T Q o = [ C d T , ( C d A d ) T , ( C d A d 2 ) T , … , ( C d A d n − 1 ) T ] T
Theorem: The linear discrete-time system is observable if and only if the observability matrix Q o Q_o Q o has rank equal to n n n .
Controllability of Discrete-Time Systems Consider a linear discrete-time invariant control system:
x ( k + 1 ) = A d x ( k ) + B d u ( k ) x(k+1) = A_d x(k) + B_d u(k) x ( k + 1 ) = A d x ( k ) + B d u ( k ) x ( 0 ) = x 0 x(0) = x_0 x ( 0 ) = x 0
Assuming the input u ( k ) u(k) u ( k ) is a scalar, the input matrix B d B_d B d is a vector denoted by b d b_d b d . Taking k = 0 , 1 , 2 , … k = 0, 1, 2, \dots k = 0 , 1 , 2 , … :
x ( 1 ) = A d x ( 0 ) + b d u ( 0 ) x(1) = A_d x(0) + b_d u(0) x ( 1 ) = A d x ( 0 ) + b d u ( 0 ) x ( 2 ) = A d x ( 1 ) + b d u ( 1 ) = A d 2 x ( 0 ) + A d b d u ( 0 ) + b d u ( 1 ) x(2) = A_d x(1) + b_d u(1) = A_d^2 x(0) + A_d b_d u(0) + b_d u(1) x ( 2 ) = A d x ( 1 ) + b d u ( 1 ) = A d 2 x ( 0 ) + A d b d u ( 0 ) + b d u ( 1 ) … \dots … x ( n ) = A d n x ( 0 ) + A d n − 1 b d u ( 0 ) + ⋯ + b d u ( n − 1 ) x(n) = A_d^n x(0) + A_d^{n-1} b_d u(0) + \dots + b_d u(n-1) x ( n ) = A d n x ( 0 ) + A d n − 1 b d u ( 0 ) + ⋯ + b d u ( n − 1 )
Reorganized into matrix form:
x ( n ) − A d n x ( 0 ) = [ b d , A d b d , … , A d n − 1 b d ] ( u ( n − 1 ) u ( n − 2 ) … u ( 0 ) ) x(n) - A_d^n x(0) = [b_d, A_d b_d, \dots, A_d^{n-1} b_d] \begin{pmatrix} u(n-1) \\ u(n-2) \\ \dots \\ u(0) \end{pmatrix} x ( n ) − A d n x ( 0 ) = [ b d , A d b d , … , A d n − 1 b d ] u ( n − 1 ) u ( n − 2 ) … u ( 0 )
The term [ b d , A d b d , … , A d n − 1 b d ] [b_d, A_d b_d, \dots, A_d^{n-1} b_d] [ b d , A d b d , … , A d n − 1 b d ] is the controllability matrix, denoted as Q c Q_c Q c . For the general vector input case, the matrix is defined as:
Q c = [ B d , A d B d , … , A d n − 1 B d ] Q_c = [B_d, A_d B_d, \dots, A_d^{n-1} B_d] Q c = [ B d , A d B d , … , A d n − 1 B d ]
Theorem: The linear discrete-time system is controllable if and only if the rank of Q c = n Q_c = n Q c = n .
Controllability and Observability of Continuous Systems For continuous systems defined by the state-space model:
x ˙ = A x + B u \dot{x} = Ax + Bu x ˙ = A x + B u y = C x + D u y = Cx + Du y = C x + D u
Where:
x x x = State vector
u u u = Input vector
y y y = Output vector
A , B , C , D A, B, C, D A , B , C , D = System matrices
Controllability Matrix (Q c Q_c Q c ): Q c = [ B , A B , A 2 B , … , A n − 1 B ] Q_c = [B, AB, A^2 B, \dots, A^{n-1} B] Q c = [ B , A B , A 2 B , … , A n − 1 B ]
Observability Matrix (Q o Q_o Q o ): Q o = [ C T , A T C T , ( A T ) 2 C T , … , ( A T ) n − 1 C T ] T Q_o = [C^T, A^T C^T, (A^T)^2 C^T, \dots, (A^T)^{n-1} C^T]^T Q o = [ C T , A T C T , ( A T ) 2 C T , … , ( A T ) n − 1 C T ] T
Theorems:
The linear continuous-time system is controllable if and only if r a n k ( Q c ) = n rank(Q_c) = n r ank ( Q c ) = n .
The linear continuous-time system is observable if and only if r a n k ( Q o ) = n rank(Q_o) = n r ank ( Q o ) = n .
Kalman's Tests Controllability Test Step 1: Construct the matrix Q c = [ B , A B , A 2 B , … , A n − 1 B ] Q_c = [B, AB, A^2 B, \dots, A^{n-1} B] Q c = [ B , A B , A 2 B , … , A n − 1 B ] .
Step 2: Find the determinant of Q c Q_c Q c . If det ( Q c ) ≠ 0 \det(Q_c) \neq 0 det ( Q c ) = 0 , the control system is controllable (system is non-singular).
Observability Test Step 1: Construct the matrix Q o = [ C T , A T C T , … , ( A T ) n − 1 C T ] T Q_o = [C^T, A^T C^T, \dots, (A^T)^{n-1} C^T]^T Q o = [ C T , A T C T , … , ( A T ) n − 1 C T ] T .
Step 2: Find the determinant of Q o Q_o Q o . If det ( Q o ) ≠ 0 \det(Q_o) \neq 0 det ( Q o ) = 0 , the system is observable.
Frequency Domain Perspective In the frequency domain, controllability and observability are related to the transfer function H ( s ) H(s) H ( s ) for a single-input single-output (SISO) system:
H ( s ) = C ( s I − A ) − 1 B H(s) = C(sI - A)^{-1} B H ( s ) = C ( s I − A ) − 1 B
Theorem:
If there are no pole-zero cancellations in the transfer function H ( s ) H(s) H ( s ) , then the system is both controllable and observable.
If a pole-zero cancellation occurs in H ( s ) H(s) H ( s ) , then the system is either uncontrollable, unobservable, or both.
Worked Examples Example 1: Discrete-Time System Given: x ( k + 1 ) = ( 1 a m p ; 2 3 a m p ; 4 ) x ( k ) x(k+1) = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} x(k) x ( k + 1 ) = ( 1 3 am p ; 2 am p ; 4 ) x ( k ) y ( k ) = [ 1 , 2 ] x ( k ) y(k) = [1, 2] x(k) y ( k ) = [ 1 , 2 ] x ( k ) B d = ( 1 0 ) B_d = \begin{pmatrix} 1 \\ 0 \end{pmatrix} B d = ( 1 0 )
Observability Check:
n = 2 n = 2 n = 2 .
C d = [ 1 , 2 ] C_d = [1, 2] C d = [ 1 , 2 ] .
C d A d = [ 1 , 2 ] ( 1 a m p ; 2 3 a m p ; 4 ) = [ 7 , 10 ] C_d A_d = [1, 2] \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = [7, 10] C d A d = [ 1 , 2 ] ( 1 3 am p ; 2 am p ; 4 ) = [ 7 , 10 ] .
Q o = ( 1 a m p ; 2 7 a m p ; 10 ) Q_o = \begin{pmatrix} 1 & 2 \\ 7 & 10 \end{pmatrix} Q o = ( 1 7 am p ; 2 am p ; 10 ) .
det ( Q o ) = ( 1 × 10 ) − ( 2 × 7 ) = 10 − 14 = − 4 \det(Q_o) = (1 \times 10) - (2 \times 7) = 10 - 14 = -4 det ( Q o ) = ( 1 × 10 ) − ( 2 × 7 ) = 10 − 14 = − 4 .
Since det ( Q o ) ≠ 0 \det(Q_o) \neq 0 det ( Q o ) = 0 , the system is observable.
Controllability Check:
B d = ( 1 0 ) B_d = \begin{pmatrix} 1 \\ 0 \end{pmatrix} B d = ( 1 0 ) .
A d B d = ( 1 a m p ; 2 3 a m p ; 4 ) ( 1 0 ) = ( 1 3 ) A_d B_d = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} A d B d = ( 1 3 am p ; 2 am p ; 4 ) ( 1 0 ) = ( 1 3 ) .
Q c = ( 1 a m p ; 1 0 a m p ; 3 ) Q_c = \begin{pmatrix} 1 & 1 \\ 0 & 3 \end{pmatrix} Q c = ( 1 0 am p ; 1 am p ; 3 ) .
det ( Q c ) = 3 − 0 = 3 \det(Q_c) = 3 - 0 = 3 det ( Q c ) = 3 − 0 = 3 . (Note: A separate calculation in notes suggested B d = ( 0 0 ) B_d = \begin{pmatrix} 0 \\ 0 \end{pmatrix} B d = ( 0 0 ) leading to uncontrollability; however, using the provided B d = ( 1 0 ) B_d = \begin{pmatrix} 1 \\ 0 \end{pmatrix} B d = ( 1 0 ) yields rank 2).
Example 2: Continuous-Time System Given: x ˙ = ( 3 a m p ; 2 − 1 a m p ; 1 ) x + ( 1 2 ) u \dot{x} = \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix} x + \begin{pmatrix} 1 \\ 2 \end{pmatrix} u x ˙ = ( 3 − 1 am p ; 2 am p ; 1 ) x + ( 1 2 ) u y = [ 1 , 1 ] x y = [1, 1] x y = [ 1 , 1 ] x
Controllability:
B = ( 1 2 ) B = \begin{pmatrix} 1 \\ 2 \end{pmatrix} B = ( 1 2 ) .
A B = ( 3 a m p ; 2 − 1 a m p ; 1 ) ( 1 2 ) = ( 7 1 ) AB = \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \end{pmatrix} A B = ( 3 − 1 am p ; 2 am p ; 1 ) ( 1 2 ) = ( 7 1 ) .
Q c = ( 1 a m p ; 7 2 a m p ; 1 ) Q_c = \begin{pmatrix} 1 & 7 \\ 2 & 1 \end{pmatrix} Q c = ( 1 2 am p ; 7 am p ; 1 ) .
det ( Q c ) = 1 − 14 = − 13 ≠ 0 \det(Q_c) = 1 - 14 = -13 \neq 0 det ( Q c ) = 1 − 14 = − 13 = 0 . System is controllable.
Observability:
C = [ 1 , 1 ] C = [1, 1] C = [ 1 , 1 ] implies C T = ( 1 1 ) C^T = \begin{pmatrix} 1 \\ 1 \end{pmatrix} C T = ( 1 1 ) .
A T C T = ( 3 a m p ; − 1 2 a m p ; 1 ) ( 1 1 ) = ( 2 3 ) A^T C^T = \begin{pmatrix} 3 & -1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} A T C T = ( 3 2 am p ; − 1 am p ; 1 ) ( 1 1 ) = ( 2 3 ) .
Q o = ( 1 a m p ; 1 2 a m p ; 3 ) Q_o = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} Q o = ( 1 2 am p ; 1 am p ; 3 ) .
det ( Q o ) = 3 − 2 = 1 ≠ 0 \det(Q_o) = 3 - 2 = 1 \neq 0 det ( Q o ) = 3 − 2 = 1 = 0 . System is observable.
Example 3: 3rd Order System Given: A = ( − 3 a m p ; 1 a m p ; 0 2 a m p ; − 3 a m p ; 2 0 a m p ; 1 a m p ; − 3 ) , B = ( 0 1 0 ) , C = [ 0 , 1 , 0 ] A = \begin{pmatrix} -3 & 1 & 0 \\ 2 & -3 & 2 \\ 0 & 1 & -3 \end{pmatrix}, B = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, C = [0, 1, 0] A = − 3 2 0 am p ; 1 am p ; − 3 am p ; 1 am p ; 0 am p ; 2 am p ; − 3 , B = 0 1 0 , C = [ 0 , 1 , 0 ]
Controllability:
B = ( 0 a m p ; 1 a m p ; 0 ) T B = \begin{pmatrix} 0 & 1 & 0 \end{pmatrix}^T B = ( 0 am p ; 1 am p ; 0 ) T .
A B = ( 1 − 3 1 ) AB = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} A B = 1 − 3 1 .
A 2 B = A ( A B ) = ( − 3 a m p ; 1 a m p ; 0 2 a m p ; − 3 a m p ; 2 0 a m p ; 1 a m p ; − 3 ) ( 1 − 3 1 ) = ( − 6 13 − 6 ) A^2 B = A(AB) = \begin{pmatrix} -3 & 1 & 0 \\ 2 & -3 & 2 \\ 0 & 1 & -3 \end{pmatrix} \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} -6 \\ 13 \\ -6 \end{pmatrix} A 2 B = A ( A B ) = − 3 2 0 am p ; 1 am p ; − 3 am p ; 1 am p ; 0 am p ; 2 am p ; − 3 1 − 3 1 = − 6 13 − 6 .
Q c = ( 0 a m p ; 1 a m p ; − 6 1 a m p ; − 3 a m p ; 13 0 a m p ; 1 a m p ; − 6 ) Q_c = \begin{pmatrix} 0 & 1 & -6 \\ 1 & -3 & 13 \\ 0 & 1 & -6 \end{pmatrix} Q c = 0 1 0 am p ; 1 am p ; − 3 am p ; 1 am p ; − 6 am p ; 13 am p ; − 6 .
det ( Q c ) = 0 \det(Q_c) = 0 det ( Q c ) = 0 . (Two columns/rows are identical). The system is uncontrollable.
Observability:
C T = ( 0 1 0 ) C^T = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} C T = 0 1 0 .
A T C T = ( 2 − 3 2 ) A^T C^T = \begin{pmatrix} 2 \\ -3 \\ 2 \end{pmatrix} A T C T = 2 − 3 2 .
( A T ) 2 C T = ( − 12 13 − 12 ) (A^T)^2 C^T = \begin{pmatrix} -12 \\ 13 \\ -12 \end{pmatrix} ( A T ) 2 C T = − 12 13 − 12 .
Q o = ( 0 a m p ; 1 a m p ; 0 2 a m p ; − 3 a m p ; 2 − 12 a m p ; 13 a m p ; − 12 ) Q_o = \begin{pmatrix} 0 & 1 & 0 \\ 2 & -3 & 2 \\ -12 & 13 & -12 \end{pmatrix} Q o = 0 2 − 12 am p ; 1 am p ; − 3 am p ; 13 am p ; 0 am p ; 2 am p ; − 12 .
det ( Q o ) = − 1 ( − 24 − ( − 24 ) ) = 0 \det(Q_o) = -1( -24 - (-24)) = 0 det ( Q o ) = − 1 ( − 24 − ( − 24 )) = 0 . The system is unobservable.
Example 4: Transfer Function to State-Space Given: H ( s ) = 2 s 3 + 6 s 2 + 11 s + 6 H(s) = \frac{2}{s^3 + 6s^2 + 11s + 6} H ( s ) = s 3 + 6 s 2 + 11 s + 6 2
Equation derivation:
( s 3 + 6 s 2 + 11 s + 6 ) y ( s ) = 2 u ( s ) (s^3 + 6s^2 + 11s + 6)y(s) = 2u(s) ( s 3 + 6 s 2 + 11 s + 6 ) y ( s ) = 2 u ( s ) .
Inverse Laplace: y ... ( t ) + 6 y ¨ ( t ) + 11 y ˙ ( t ) + 6 y ( t ) = 2 u ( t ) \dddot{y}(t) + 6\ddot{y}(t) + 11\dot{y}(t) + 6y(t) = 2u(t) y ... ( t ) + 6 y ¨ ( t ) + 11 y ˙ ( t ) + 6 y ( t ) = 2 u ( t ) .
Let x 1 = y , x 2 = y ˙ , x 3 = y ¨ x_1 = y, x_2 = \dot{y}, x_3 = \ddot{y} x 1 = y , x 2 = y ˙ , x 3 = y ¨ .
x ˙ 1 = x 2 \dot{x}_1 = x_2 x ˙ 1 = x 2
x ˙ 2 = x 3 \dot{x}_2 = x_3 x ˙ 2 = x 3
x ˙ 3 = − 6 x 1 − 11 x 2 − 6 x 3 + 2 u \dot{x}_3 = -6x_1 - 11x_2 - 6x_3 + 2u x ˙ 3 = − 6 x 1 − 11 x 2 − 6 x 3 + 2 u
State Matrices:
A = ( 0 a m p ; 1 a m p ; 0 0 a m p ; 0 a m p ; 1 − 6 a m p ; − 11 a m p ; − 6 ) , B = ( 0 0 2 ) , C = [ 1 , 0 , 0 ] A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -11 & -6 \end{pmatrix}, B = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix}, C = [1, 0, 0] A = 0 0 − 6 am p ; 1 am p ; 0 am p ; − 11 am p ; 0 am p ; 1 am p ; − 6 , B = 0 0 2 , C = [ 1 , 0 , 0 ] .
Verification:
Q c = ( 0 a m p ; 0 a m p ; 2 0 a m p ; 2 a m p ; − 12 2 a m p ; − 12 a m p ; 50 ) Q_c = \begin{pmatrix} 0 & 0 & 2 \\ 0 & 2 & -12 \\ 2 & -12 & 50 \end{pmatrix} Q c = 0 0 2 am p ; 0 am p ; 2 am p ; − 12 am p ; 2 am p ; − 12 am p ; 50 .
det ( Q c ) = 8 ≠ 0 \det(Q_c) = 8 \neq 0 det ( Q c ) = 8 = 0 . System is fully controllable.
Q o = ( 1 a m p ; 0 a m p ; 0 0 a m p ; 1 a m p ; 0 0 a m p ; 0 a m p ; 1 ) Q_o = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} Q o = 1 0 0 am p ; 0 am p ; 1 am p ; 0 am p ; 0 am p ; 0 am p ; 1 .
det ( Q o ) = 1 ≠ 0 \det(Q_o) = 1 \neq 0 det ( Q o ) = 1 = 0 . System is fully observable.
Verification Problems Verify the controllability of a system represented by: A = ( 2 a m p ; − 4 1 a m p ; − 1 ) , B = ( 3 1 ) , C = [ 1 , − 1 ] A = \begin{pmatrix} 2 & -4 \\ 1 & -1 \end{pmatrix}, B = \begin{pmatrix} 3 \\ 1 \end{pmatrix}, C = [1, -1] A = ( 2 1 am p ; − 4 am p ; − 1 ) , B = ( 3 1 ) , C = [ 1 , − 1 ] .
Determine if the system is completely controllable and observable given: A = ( 0 a m p ; 1 − 2 a m p ; − 3 ) , B = ( 0 1 ) , C = [ 2 , 1 ] A = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}, B = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, C = [2, 1] A = ( 0 − 2 am p ; 1 am p ; − 3 ) , B = ( 0 1 ) , C = [ 2 , 1 ] .
Find controllability for: A = ( 1 a m p ; 4 − 5 a m p ; 10 ) , B = ( 1 2 ) , C = [ 1 , 2 ] A = \begin{pmatrix} 1 & 4 \\ -5 & 10 \end{pmatrix}, B = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, C = [1, 2] A = ( 1 − 5 am p ; 4 am p ; 10 ) , B = ( 1 2 ) , C = [ 1 , 2 ] .
Test controllability and observability for the system: H ( s ) = 1 s 2 + 3 s + 2 H(s) = \frac{1}{s^2 + 3s + 2} H ( s ) = s 2 + 3 s + 2 1 .