State Space Analysis MCQ Quiz - Objective Question with Answer for State Space Analysis - Download Free PDF

Concept:

To convert a second-order differential equation into state-space form, define:

Then: \( \dot{x}_1 = x_2,\quad \dot{x}_2 = \ddot{x} \)

Given differential equation:

\(\ddot{x} + 3\dot{x} - 9x = 3\sin(\omega t)\)

Rewriting:

\(\ddot{x} = -3\dot{x} + 9x + 3\sin(\omega t)\)

In terms of state variables:

• \(\dot{x}_1 = x_2\)
• \(\dot{x}_2 = 9x_1 - 3x_2 + 3\sin(\omega t)\)

So the state-space equation \(\dot{X} = AX + BU\) has:

A matrix:

\(\left(\begin{array}{cc} 0 & 1 \\ 9 & -3 \end{array}\right)\)

Controllability Matrix in Control Systems

Definition: Controllability is a fundamental concept in control system theory, which determines whether a system's state can be driven to a desired state using appropriate control inputs over a finite duration. The controllability matrix is a mathematical tool used to check the controllability of a linear time-invariant (LTI) system.

State-Space Representation:

A linear time-invariant (LTI) system can be represented in state-space form as:

dx/dt = Ax + Bu

y = Cx + Du

• Here, x is the state vector, u is the input vector, y is the output vector, A is the system matrix, B is the input matrix, C is the output matrix, and D is the feedthrough matrix.

Controllability Matrix:

To determine the controllability of a system, the controllability matrix is defined as:

[B AB A²B ... A^n-1B]

• Here, A is the system matrix and B is the input matrix.
• The columns of the controllability matrix represent the influence of the input u through the dynamics of the system.

Explanation of Correct Option:

The correct option is:

Option 1: [B AB A²B...]

This representation is the standard form of the controllability matrix used in control system theory. The matrix is constructed by appending B, AB, A²B, and so on, up to A^n-1B, where n is the order of the system.

Explanation:

System Controllability and Observability

In control system theory, the concepts of controllability and observability are fundamental for understanding the behavior and capabilities of a system. These properties determine whether a system can be controlled to achieve desired states and whether the internal states of the system can be inferred from external outputs, respectively.

Statement 1 Analysis:

Statement 1: "A system is said to be completely controllable if it is possible to transfer the system state from any initial state X(to) to any desired state X(t) in specified finite time by a control vector u(t)."

This statement defines the concept of controllability. A system is considered completely controllable if, given any initial state, it is possible to drive the system to any final state using an appropriate control input within a finite time. This property ensures that the system can be maneuvered to follow desired trajectories or reach specific states, which is crucial for the effective control and operation of the system.

The given statement accurately captures the essence of controllability. Therefore, Statement 1 is true.

Statement 2 Analysis:

Statement 2: "A system is said to be completely observable if every state X(to) can be completely identified by measurements of the outputs y(t) over the time interval (0 ≤ t ≤ ∞)."

This statement defines the concept of observability. A system is considered completely observable if the current state of the system can be determined from its outputs over a given time interval. This property is essential for system monitoring, state estimation, and feedback control, as it ensures that the internal states of the system can be inferred from external measurements.

The given statement accurately describes the concept of observability. Therefore, Statement 2 is true.

Correct Option Analysis:

The correct option is:

Option 2: Both Statement 1 and Statement 2 are true.

This option correctly identifies that both statements accurately describe the fundamental properties of controllability and observability in control system theory.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Statement 1 is true and Statement 2 is false.

This option is incorrect because both statements accurately describe the concepts of controllability and observability, so Statement 2 is not false.

Option 3: Both Statement 1 and Statement 2 are false.

This option is incorrect because both statements are true and provide accurate definitions of controllability and observability.

Option 4: Statement 2 is true and Statement 1 is false.

This option is incorrect because Statement 1 is true, as it correctly defines the concept of controllability.

Conclusion:

Understanding the concepts of controllability and observability is essential for the design and analysis of control systems. These properties ensure that the system can be effectively controlled and monitored. Both statements provided in the question accurately describe these concepts, making Option 2 the correct choice.

Concept

The zero-input response of a system is calculated as:

\(ZIR=ϕ(t)x(0)\)

where, ϕ(t) = State transition matrix

The state transition matrix is calculated as:

\(\phi(t)=L^{-1}[(sI-A)^{-1}]\)

Calculation

Given, \(A=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]\)

\(x(0)=\left[\begin{array}{c} 1 \\ 0 \end{array}\right]\)

\((sI-A)^{-1}=\left[\begin{array}{ll} s-1 & 0 \\ -1 & s-1 \end{array}\right]^{-1}\)

\((sI-A)^{-1}={1\over (s-1)^2}\left[\begin{array}{ll} s-1 & 0 \\ +1 & s-1 \end{array}\right]\)

\(\phi(t)x(0)={1\over (s-1)^2}\left[\begin{array}{c} s-1 \\ 1 \end{array}\right]\)

\(ZIR=\left[\begin{array}{c} {1\over s-1} \\ {1\over (s-1)^2} \end{array}\right]\)

Taking inverse Laplace, we get:

\(ZIR=\left[\begin{array}{c} \mathrm{e}^{\mathrm{t}} \\ \mathrm{te}^{\mathrm{t}} \end{array}\right]\)

Given:

\(A =\rm \begin{bmatrix}0&1\\\ -1&-2\end{bmatrix}, B=\rm \begin{bmatrix}0\\\ 1\end{bmatrix}, C=\rm \begin{bmatrix}3&-2\end{bmatrix}, D = 1\)

Transfer function is given by,

T.F. = C[sI - A]-1B + D

\( {[s I-A]=\left[\begin{array}{cc} s & -1 \\ 1 & s+2 \end{array}\right]} \)

\([s I-A]^{-1}=\left[\begin{array}{cc} s & -1 \\ 1 & s+2 \end{array}\right]^{-1} \)

\( {[s I-A]^{-1}=\frac{1}{s(s+2)+1}\left[\begin{array}{cc} s+2 & 1 \\ -1 & s \end{array}\right]}\)

\(T . F .=\left[\begin{array}{ll} 3 & -2 \end{array}\right] \frac{1}{s(s+2)+1}\left[\begin{array}{cc} s+2 & 1 \\ -1 & s \end{array}\right]\left[\begin{array}{l} 0 \\ 1 \end{array}\right]+1 \)

\(T . F .=\frac{1}{s^2+2 s+1}\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{ll} s+2 & 1 \\ -1 & s \end{array}\right]\left[\begin{array}{l} 0 \\ 1 \end{array}\right]+1 \)

\( T . F .=\frac{1}{s^2+2 s+1}\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{l} 1 \\ s \end{array}\right]+1 \)

\(T . F .=\frac{3-2 s}{s^2+2 s+1}+1 \)

\(T . F=\frac{s^2+4}{s^2+2 s+1} \)

\(H(s)=\frac{s^2+4}{s^2+2 s+1}\)

s = jω

\(H(\mathrm{j} ω)=\frac{4-ω^2}{1+2 j ω-ω^2}\)

Steady state output of system is zero

4 - ω² = 0

4 = ω²

ω = 2 rad/sec

Hence, the correct answer is 2.

State space representation:

ẋ(t) = A(t)x(t) + B(t)u(t)

y(t) = C(t)x(t) + D(t)u(t)

y(t) is output

u(t) is input

x(t) is a state vector

A is a system matrix

This representation is continuous time-variant.

Controllability:

A system is said to be controllable if it is possible to transfer the system state from any initial state x(t0) to any desired state x(t) in a specified finite time interval by a control vector u(t)

Kalman’s test for controllability:

ẋ = Ax + Bu

Qc = {B AB A2B … An-1 B]

Qc = controllability matrix

If |Qc| = 0, system is not controllable

If |Qc|≠ 0, the system is controllable

Observability:

A system is said to be observable if every state x(t0) can be completely identified by measurement of output y(t) over a finite time interval.

Kalman’s test for observability:

Q0 = [CT ATCT (AT)2CT …. (AT)n-1 CT]

Q0 = observability testing matrix

If |Q0| = 0, system is not observable

If |Q0| ≠ 0, system is observable.

Duality property of controllability and observability:

• If the pair (A, B) is controllable then the pair (AT, BT) is observable.
• If the pair of (A, C) is observable then the pair of (AT, CT) is controllable.

If there are no pole-zero cancellations in the transfer function then the system is completely controllable and observable.

Concept:

A transfer function (TF) is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output] / L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, transfer function is also known as impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Calculation:

Given differential equation is,

\(3\dfrac{dy}{dt}+2y=u\)

Laplace transform of the above equation is given by

\(\dfrac{dy}{dt} = sY(s)\), y(t) = Y(s), u(t) = U(s)

⇒ 3s Y(s) + 2 Y(s) = U(s)

⇒ Y(s) (3s + 2) = U(s)

∴ Transfer function is given by

\( \Rightarrow TF = \frac{{Y\left( s \right)}}{{U\left( s \right)}} = \frac{1}{{ 3s + 2}}\)

The given system is

Ẋ = AX + Bu

Where,

\(A = \left[ {\begin{array}{*{20}{c}} { - 1}&2\\ 0&2 \end{array}} \right]\:and\:B = \left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right]\)

We determine stability using characteristic equation (i.e. poles or eigenvalues of the system).

|sI - A| = 0

\(\left| {\left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} { - 1}&2\\ 0&2 \end{array}} \right]} \right| = 0\)

\(\left| {\begin{array}{*{20}{c}} {s + 1}&{ - 2}\\ 0&{s - 2} \end{array}} \right| = 0\)

(s + 1)(s - 2) = 0

s = -1 and s = 2

i.e. the system has one pole in right half of the s-plane.

Hence, the system is unstable.

Now, the controllability matrix is given by:

C_M = [B AB]

Where

\(\left[ {AB} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&2\\ 0&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2\\ 2 \end{array}} \right]\)

So,

\({C_M} = \left[ {\begin{array}{*{20}{c}} 0&2\\ 1&2 \end{array}} \right]\)

|C_M| = -2 ≠ 0

Hence, the system is Controllable.

• In the transfer function model, initial conditions are assumed to be zero. In the state-space model, transfer function is not the starting point for the system analysis, and it considers the initial conditions of the system.
• The transfer function for the system is unique.
• The transfer function is applicable only for linear time-invariant systems. The state-space model is applicable to linear time-variant systems also.
• State-space representation of a system is not unique, it may have more than one representation.
• State model can be derived from transfer function of the system.

Calculation:

Given  state model,

ẋ = - x + u

u = - Kx

ẋ = - x - Kx

ẋ = - (K + 1)x = A x

A = - (K + 1)

Charactestic equation of a state model is given by

|sI - A| = 0

⇒ sI = s

s - [- (K + 1)] = 0

s + K + 1 = 0

s = - 1 - K

Given that a pole is located at -2

⇒ s = -2

-2 = - 1 - K

K = 1

State transition matrix:

It is defined as inverse Laplace transform of |sI - A|-1

⇒ L-1 |sI - A|-1 = eAt = ϕ(t)

Properties:

State transition matrix, ϕ(t) = eAt

• ϕ(0) = e(A0) = I, Identity matrix
• \({\phi ^{ - 1}}\left( t \right) = \phi \left( { - t} \right)\)
• ϕ(t1 + t2) = ϕ(t1) ϕ(t2)
• [ϕ(t)]n = ϕ(nt)
• ϕ(t2 – t1) ϕ (t2 – t0) = ϕ (t2 – t0) = ϕ (t1 – t0) ϕ (t2 – t1)

Application:

Given as state model, ẋ = A x(t)

Comparing with standard equation ẋ = A x(t) + B u(t)

\(\;A = \left[ {\begin{array}{*{20}{c}} 0&{ - 2}\\ 1&{ - 3} \end{array}} \right]\), B = 0

\( \Rightarrow \left( {sI - A} \right) = s\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0&{ - 2}\\ 1&{ - 3} \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} s&{0 + 2}\\ {0 - 1}&{s + 3} \end{array}} \right]\)

\({\left( {sI - A} \right)^{ - 1}} = \frac{{adj\left( {sI - A} \right)}}{{\left| {sI - A} \right|}} = \frac{{\left[ {\begin{array}{*{20}{c}} {s + 3}&{ - 2}\\ 1&s \end{array}} \right]}}{{s\left( {s + 3} \right) + 2}}\)

\({\left[ {sI - A} \right]^{ - 1}} = \frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 3}&{ - 2}\\ 1&s \end{array}} \right]\)

\({L^{ - 1}}{\left[ {sI - A} \right]^{ - 1}} = {L^{ - 1}}\left[ {\begin{array}{*{20}{c}} {\frac{{s + 3}}{{\left( {s + 1} \right)\left( {s + 2} \right)}}}&{\frac{{ - 2}}{{\left( {s + 1} \right)\left( {s + 2} \right)}}}\\ {\frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}}&{\frac{s}{{\left( {s + 1} \right)\left( {s + 2} \right)}}} \end{array}} \right]\)

\(= {L^{ - 1}}\left[ {\begin{array}{*{20}{c}} {\frac{2}{{s + 1}} - \frac{1}{{s + 2}}}&{\frac{{ - 2}}{{\left( {s + 1} \right)}} + \frac{2}{{\left( {s + 2} \right)}}}\\ {\frac{1}{{s + 1}} - \frac{1}{{s + 2}}}&{\frac{{ - 1}}{{\left( {s + 1} \right)}} + \frac{2}{{\left( {s + 2} \right)}}} \end{array}} \right]\)

\(\phi \left( t \right) = \left[ {\begin{array}{*{20}{c}} {2{e^{ - t}} - {e^{ - 2t}}}&{ - 2{e^{ - t}} + 2{e^{ - 2t}}}\\ {{e^{ - t}} - {e^{ - 2t}}}&{ - {e^{ - t}} + 2{e^{ - 2t}}} \end{array}} \right]\)

Shortcut TrickShortcut Method to finding Adj (A)

Let A = \(\left[ {\begin{array}{*{20}{c}} {\rm{a}}&b\\ {\rm{c}}&{\rm{d}} \end{array}} \right]\)

Adj A = \(\left[ {\begin{array}{*{20}{c}} {\rm{d}}&{ - b}\\ { - {\rm{c}}}&{\rm{a}} \end{array}} \right]\)

Note: Interchange the diagonal elements and change the sign of remaining elements.

Concept:

Solution of state equation is given as:

x(t) = ϕ(t) x(0)

Where,

ϕ(t) = state transition matrix and

ϕ(t) = L^-1 {(sI-A)^-1}

Analysis:

\(A = \left[ {\begin{array}{*{20}{c}} { - 1}&0\\ 0&{ - 2} \end{array}} \right]\)

\(sI - A = \left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} { - 1}&0\\ 0&{ - 2} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} {s + 1}&0\\ 0&{s + 2} \end{array}} \right]\)

\({\left( {sI - A} \right)^{ - 1}} = \frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 2}&0\\ 0&{s + 1} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{s + 1}}}&0\\ 0&{\frac{1}{{s + 2}}} \end{array}} \right]\)

\(\phi \left( t \right) = {L^{ - 1}}\left\{ {{{\left( {sI - A} \right)}^{ - 1}}} \right\}\)

\( = \left[ {\begin{array}{*{20}{c}} {{e^{ - t}}}&0\\ 0&{{e^{ - 2t}}} \end{array}} \right]\)

x(t) = ϕ(t) x(0)

\(x\left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^{ - t}}}&0\\ 0&{{e^{2t}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ { - 1} \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} {{x_1}\left( t \right)}\\ {{x_2}\left( t \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{e^{ - t}}}\\ { - {e^{ - 2t}}} \end{array}} \right]\)

x₁(t) = e^-t and x₂(t) = -e^-2t

State transition matrix:

It is defined as inverse Laplace transform of |sI - A|-1

⇒ L-1 |sI - A|-1 = eAt = ϕ(t)

Properties:

State transition matrix, ϕ(t) = eAt

• ϕ(0) = e(A0) = I, Identity matrix
• \({\phi ^{ - 1}}\left( t \right) = \phi \left( { - t} \right)\)
• ϕ(t1 + t2) = ϕ(t1) ϕ(t2)
• [ϕ(t)]n = ϕ(nt)
• ϕ(t2 – t1) ϕ (t2 – t0) = ϕ (t2 – t0) = ϕ (t1 – t0) ϕ (t2 – t1)

Application:

Given as state model, ẋ = A x(t)

Comparing with standard equation ẋ = A x(t) + B u(t)

\(A = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 2}&{ - 3} \end{array}} \right],B = 0\)

\(\left[ {sI - A} \right] = \left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 2}&{ - 3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} s&{ - 1}\\ 2&{s + 3} \end{array}} \right]\)

\({\left[ {sI - A} \right]^{ - 1}} = \frac{1}{{s\left( {s + 3} \right) + 2}}\left[ {\begin{array}{*{20}{c}} {s + 3}&1\\ { - 2}&s \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} {\frac{{\left( {s + 3} \right)}}{{\left( {s + 1} \right)\left( {s + 2} \right)}}}&{\frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}}\\ { - \frac{2}{{\left( {s + 1} \right)\left( {s + 2} \right)}}}&{\frac{s}{{\left( {s + 1} \right)\left( {s + 2} \right)}}} \end{array}} \right]\)

Taking inverse Laplace transform,

\({e^{At}} = \left[ {\begin{array}{*{20}{c}} {2{e^{ - t}} - {e^{ - 2t}}}&{{e^{ - t}} - {e^{ - 2t}}}\\ { - 2{e^{ - t}} + 2{e^{ - 2t}}}&{ - {e^{ - t}} + 2{e^{ - 2t}}} \end{array}} \right]\)

\({e^{ - At}} = \left[ {\begin{array}{*{20}{c}} {2{e^t} - {e^{2t}}}&{{e^t} - {e^{2t}}}\\ { - 2{e^t} + 2{e^{2t}}}&{ - {e^t} + 2{e^{2t}}} \end{array}} \right]\)

The total response of the system is given by

X(t) = ZIR + ZSR

Where,

ZIR = Zero input response

ZSR = Zero state response

ZIR = e^t x(0)

ZSR = L^-1 [ϕ(s) B U(s)]

Where

Φ(s) = (SI - A)^-1

Calculations:

ZIR = e^at x(0)

X(0) = \(\left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\;\)

\(\Rightarrow ZIR = \left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\)

ZSR = L^-1 [ϕ (s) B U(s)]

\(SI\;A = \left[ {\begin{array}{*{20}{c}} s&0\\ 0&{s + 9} \end{array}} \right]\)

\(Adj\;\left( {sI - A} \right) = \left[ {\begin{array}{*{20}{c}} {s+9}&0\\ 0&{s} \end{array}} \right]\)

\(\phi \left( S \right) = {\left( {sI - A} \right)^{ - 1}} = \frac{{Adj\;\left[ {sI - A} \right]}}{{\left| {sI - a} \right|}}\)

\(= \left[ {\begin{array}{*{20}{c}} {\frac{1}{s}}&0\\ 0&{\frac{1}{{s + 9}}} \end{array}} \right]\)

\(ZSR = {L^{ - 1}}\left[ {\left( {\begin{array}{*{20}{c}} {\frac{1}{s}}&0\\ 0&{\frac{1}{{s + 9}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0\\ {45} \end{array}} \right)\left( {\frac{1}{s}} \right)} \right]\)

CI \(= {L^{ - 1}}\left[ {\begin{array}{*{20}{c}} 0\\ {\frac{{45}}{{s\left( {s + 9} \right)}}\;} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ {5\left( {1 - {e^{ - 9t}}} \right)} \end{array}} \right]\)

x1(t) = 0

x2 (t) = 5 (1 – e-9t)

\(\mathop {\lim }\limits_{t \to \infty } \left| {\sqrt {x_1^2\left( t \right) + x_2^2\left( t \right)} } \right|\)

\(= \sqrt {0 + {5^2}}\)

= 5

\(\frac{{d{x_1}\left( t \right)}}{{dt}} = \mathop {\mathop x\nolimits_1 }\limits^. \left( t \right) = {x_2}\left( t \right) - {x_1}\left( t \right)\)

\(\frac{{d{x_2}\left( t \right)}}{{dt}} = \mathop {\mathop x\nolimits_2 }\limits^. \left( t \right) = {x_1}\left( t \right) - {x_2}\left( t \right)\)

\(\left[ {\begin{array}{*{20}{c}} {\mathop {\mathop x\nolimits_1 }\limits^. \left( t \right)}\\ {\mathop {\mathop x\nolimits_2 }\limits^. \left( t \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} {{x_1}\left( t \right)}\\ {{x_2}\left( t \right)} \end{array}} \right]\)

\(A = \left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\)

\(\phi \left( t \right) = {e^{at}} = {L^{ - 1}}\left[ {{{\left( {sI - A} \right)}^{ - 1}}} \right]\)

\(sI - A = \left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {s + 1}&{ - 1}\\ { - 1}&{s + 1} \end{array}} \right]\)

\({\left[ {sI - A} \right]^{ - 1}} = \frac{1}{{{{\left( {s + 1} \right)}^2} - 1}}\left[ {\begin{array}{*{20}{c}} {s + 1}&1\\ 1&{s + 1} \end{array}} \right]\)

\(= \frac{1}{{s\left( {s + 2} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 1}&1\\ 1&{s + 1} \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {\frac{{s + 1}}{{s\left( {s + 2} \right)}}}&{\frac{1}{{s\left( {s + 2} \right)}}}\\ {\frac{1}{{s\left( {s + 2} \right)}}}&{\frac{{s + 1}}{{s\left( {s + 2} \right)}}} \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {\frac{1}{{2s}} + \frac{1}{{2\left( {s + 2} \right)}}}&{\frac{1}{{2s}} - \frac{1}{{2\left( {s + 2} \right)}}}\\ {\frac{1}{{2s}} - \frac{1}{{2\left( {s + 2} \right)}}}&{\frac{1}{{2s}} - \frac{1}{{2\left( {s + 2} \right)}}} \end{array}} \right]\)

\(\phi \left( t \right) = {L^{ - 1}}\left[ {{{\left( {sI - A} \right)}^{ - 1}}} \right]\)

\(= \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {1 + {e^{ - 2t}}}&{1 - {e^{ - 2t}}}\\ {1 - {e^{ - 2t}}}&{1 + {e^{ - 2t}}} \end{array}} \right]\)

\(x\left( t \right) = \phi \left( t \right)\;x\left( 0 \right) = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} {1 + {e^{ - 2t}}}&{1 - {e^{ - 2t}}}\\ {1 - {e^{ - 2t}}}&{1 + {e^{ - 2t}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}\left( 0 \right)}\\ {{x_2}\left( 0 \right)} \end{array}} \right]\)

\(x\left( t \right) = \frac{1}{2}\;\left[ {\begin{array}{*{20}{c}} {{x_1}\left( 0 \right) + {x_2}\left( 0 \right) + {e^{ - 2t}}\left( {{x_1}\left( 0 \right) - {x_2}\left( 0 \right)} \right)}\\ {{x_1}\left( 0 \right) + {x_2}\left( 0 \right) + {e^{ - 2t}}\left( {{x_2}\left( 0 \right) - {x_1}\left( 0 \right)} \right)} \end{array}} \right]\)

\({x_{1f}} = \mathop {{\rm{lt}}}\limits_{t \to \infty } {x_1}\left( t \right) = \frac{1}{2}\left( {{x_1}\left( 0 \right) + {x_2}\left( 0 \right)} \right)\)

\({x_{2f}} = \mathop {{\rm{lt}}}\limits_{t \to \infty } {x_2}\left( t \right) = \frac{1}{2}\left( {{x_1}\left( 0 \right) + {x_2}\left( 0 \right)} \right)\)

Given that, x₁(0) < x₂ (0) < ∞

⇒ x_1f = x_2f < ∞