State Transition Matrix MCQ Quiz - Objective Question with Answer for State Transition Matrix - Download Free PDF

Concept

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

$ZIR=\varphi (t)x(0)$

where, ϕ(t) = State transition matrix

The state transition matrix is calculated as:

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

Calculation

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

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

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

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

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

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

Taking inverse Laplace, we get:

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

Concept:

The state transition matrix ϕ(t):

The state-transition matrix is defined as a matrix that satisfies the linear homogeneous state equation.

It represents the free response of the system.

The state-transition matrix ϕ(t) completely defines the transition of the states from the initial time t = 0 to any time t when the inputs are zero.

The state transition matrix is given by 

ϕ(t) = L-1 [sI - A]-1 = eAt

Where A = state matrix

The state-transition matrix is dependent only upon the matrix A and, therefore, is sometimes referred to as the state transition matrix of A.

Properties of ϕ(t):

1. ​ϕ(0) = I
2. ϕ-1(t) = ϕ(-t)
3. ϕ(t2 - t1) = ϕ(t2 - t0). ϕ(t0 - t1)
4. ϕ(t2 + t1) = ϕ(t2) . ϕ(t1)
5. [ϕ(t)]k = ϕ(kt)
6. At t =0, dϕ /dt = A

Calculation:

Given that

State transition matrix ϕ(t) = $\left[\begin{array}{cc}{e}^{-4t}& e^{-6t}-e^{-2t}\\ e^{-8t}-e^{-5t}& e^{-0.5t}\end{array}\right]$

Consider the sixth property of the state transition matrix

⇒ At t = 0, dϕ /dt = A

Where A is system or state matrix,

⇒ A = $\left[\begin{array}{cc}-4e^{-4t}& -6e^{-6t}+2e^{-2t}\\ -8e^{-8t}+5e^{-5t}& -0.5e^{-0.5t}\end{array}\right]$at t = 0

⇒ A = $\left[\begin{array}{cc}-4& -6+2\\ -8+5& -0.5\end{array}\right]$

⇒ A = $\left[\begin{array}{cc}-4& -4\\ -3& -0.5\end{array}\right]$

Concept:

The state transition matrix ϕ(t):

The state-transition matrix is defined as a matrix that satisfies the linear homogeneous state equation.

It represents the free response of the system.

The state-transition matrix ϕ(t) completely defines the transition of the states from the initial time t = 0 to any time t when the inputs are zero.

The state transition matrix is given by 

ϕ(t) = L^-1 [sI - A]^-1 = e^At

Where A = state matrix

The state-transition matrix is dependent only upon the matrix A and, therefore, is sometimes referred to as the state transition matrix of A.

Properties of ϕ(t):

1. ​ϕ(0) = I
2. ϕ^-1(t) = ϕ(-t)
3. ϕ(t2 - t1) = ϕ(t2 - t0). ϕ(t0 - t1)
4. ϕ(t2 + t1) = ϕ(t2) . ϕ(t1)
5. [ϕ(t)]^k = ϕ(kt)
6. At t =0, dϕ / dt = A

Application:

For statement 2,

ϕ(t2 - t1) = ϕ(t2) . ϕ-1(t1)

consider

ϕ(t₂ - t₁) = ϕ (t₂ + (-t₁)) 

By property 4

⇒ ϕ(t2 - t1) = ϕ (t2 + (-t1)) = ϕ(t2) . ϕ(-t1)

By property 2

⇒  ϕ(t2 - t1) = ϕ(t2) . ϕ(-t1) =  ϕ(t2) . ϕ-1(t1)

Concept:

The state transition matrix is given by:

ϕ(t) = e^-At = L^-1 [{sI - A}^-1]

Calculation:

[sI - A] = $\left[\begin{array}{cc}s+2& 0\\ 0& s+2\end{array}\right]$

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

[sI - A]-1 = $\left[\begin{array}{cc}\frac{1}{s+2}& 0\\ 0& \frac{1}{s+2}\end{array}\right]$

L-1 [{sI - A}-1] = $\left[\begin{array}{cc}{e}^{-2t}& 0\\ 0& e^{-2t}\end{array}\right]$

Additional Information Properties of state transition matrix:

1.) ϕ(0) = e-0t = 1

2.) $\frac{d\varphi (0)}{dt}=A$, where A is the system matrix

Concept:

The state transition matrix [ϕ(t)] is given by:

ϕ(t) = e^At = L-1 [(SI-A)]-1

where, A = System matrix

I = Identity matrix

Properties of state transition matrix:

1.) State transition matrix at t = 0 is always equal to the identity matrix. 

ϕ(0) = eA0 = I

2.) The differentiation of the state transition matrix at t = 0 is always equal to its system matrix.

$\frac{d}{dt}\varphi (0)=$ AeA0 = A

Concept:

State transition matrix:

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

⇒ L-1 |sI - A|-1 = eAt = ϕ(t)

General state equation: 

$\dot{x}\left(t\right)=Ax\left(t\right)+BU\left(t\right)$

Also, $\dot{x}\left(t\right)=\frac{dx}{dt}$

$\left[\begin{array}{c}{\dot{x}}_1\left(t\right)\\ {\dot{x}}_2\left(t\right)\\ ⋮\\ ⋮\\ {\dot{x}}_n\left(t\right)\end{array}\right]=\left[A\right]\left[\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ ⋮\\ x_n\end{array}\right]+\left[B\right]\left[\begin{array}{c}{U}_1\\ U_2\\ ⋮\\ U_n\end{array}\right]$

Where, x1, x2, x3 …. xn are state variables

A is state matrix

B is the input matrix

Calculation:

Given state equation

$\left[\begin{array}{c}{\dot{x}}_1\\ {\dot{x}}_2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]+\left[\begin{array}{c}1\\ 1\end{array}\right]u$

$A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],B=\left[\begin{array}{c}1\\ 1\end{array}\right]$

State transition matrix.

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

$\left[sI-A\right]=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$

$=\left[\begin{array}{cc}s-1& 0\\ -1& s-1\end{array}\right]$

${\left[sI-A\right]}^{-1}=\frac{1}{{\left(s-1\right)}^2}\left[\begin{array}{cc}s-1& 0\\ 1& s-1\end{array}\right]$

$=\left[\begin{array}{cc}\frac{1}{s-1}& 0\\ \frac{1}{{\left(s-1\right)}^2}& \frac{1}{s-1}\end{array}\right]$

$L^{-1}{\left[sI-A\right]}^{-1}=\left[\begin{array}{cc}{e}^t& 0\\ t{e}^t& e^t\end{array}\right]$

Concept:

State transition matrix:

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

⇒ L-1 |sI - A|-1 = eAt = ϕ(t)

Given as state model, ẋ = A x(t)

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

Calculation:

$A=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]$

$\left|sI-A\right|=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]=\left[\begin{array}{cc}s& -1\\ 2& s+3\end{array}\right]$

${\left|sI-A\right|}^{-1}=\frac{1}{s^2+3s+2}\left[\begin{array}{cc}s+3& 1\\ -2& s\end{array}\right]$

Taking inverse Laplace transform,

$\mathrm{\Phi }(t)=\left[\begin{array}{cc}2e^{-t}-e^{-2t}& e^{-t}-e^{-2t}\\ -2e^{-t}+2e^{-2t}& -e^{-t}+2e^{-2t}\end{array}\right]$

Additional Information

Properties:

State transition matrix, ϕ(t) = eAt

• ϕ(0) = eA0) = I, Identity matrix
• ${\varphi }^{-1}\left(t\right)=\varphi \left(-t\right)$
• 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)      

Trick: The above question can also be solved with the help of trick.

$\varphi \left(0\right)=e^{A^{\left(0\right)}}=I$, the Identity matrix.

We can put t = 0, in all the options and can be found the answer.

After putting t = 0, if the matrix becomes an identity matrix, that will be the answer.

Concept:

The state transition matrix [ϕ(t)] is given by:

ϕ(t) = e^At = L-1 [(SI-A)]-1

where, A = System matrix

I = Identity matrix

Properties of state transition matrix:

1.) State transition matrix at t = 0 is always equal to the identity matrix. 

ϕ(0) = eA0 = I

2.) The differentiation of the state transition matrix at t = 0 is always equal to its system matrix.

$\frac{d}{dt}\varphi (0)=$ AeA0 = A

Concept:

The state transition matrix is given by:

ϕ(t) = e^-At = L^-1 [{sI - A}^-1]

Calculation:

[sI - A] = $\left[\begin{array}{cc}s+2& 0\\ 0& s+2\end{array}\right]$

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

[sI - A]-1 = $\left[\begin{array}{cc}\frac{1}{s+2}& 0\\ 0& \frac{1}{s+2}\end{array}\right]$

L-1 [{sI - A}-1] = $\left[\begin{array}{cc}{e}^{-2t}& 0\\ 0& e^{-2t}\end{array}\right]$

Additional Information Properties of state transition matrix:

1.) ϕ(0) = e-0t = 1

2.) $\frac{d\varphi (0)}{dt}=A$, where A is the system matrix

State Transition Matrix ϕ(t):

The state-transition matrix is defined as a matrix that satisfies the linear homogeneous state equation.

It represents the free response of the system.

The state-transition matrix ϕ(t) completely defines the transition of the states from the initial time t = 0 to any time t when the inputs are zero.

The state transition matrix is given by 

ϕ(t) = L-1 [sI - A]-1 = eAt

Where A = state matrix, I = identity matrix, L-1 = Laplace inverse 

The state-transition matrix is dependent only upon the matrix A and, therefore, is sometimes referred to as the state transition matrix of A.

Properties of ϕ(t):

• ​ϕ(0) = I
• ϕ-1(t) = ϕ(-t)
• ϕ(t2 - t1) = ϕ(t2 - t0). ϕ(t0 - t1)
• ϕ(t2 + t1) = ϕ(t2) . ϕ(t1)
• [ϕ(t)]k = ϕ(kt)
• at t =0, dϕ / dt = A

Concept:

The inverse of 2 × 2 matrix

Let us consider a 2 × 2 matrix as:

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$A^{-1}=\frac{1}{\mathrm{d}\mathrm{e}\mathrm{t}\left[A\right]}adj\left[A\right]$

$A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

State transition matrix of for a given system is defined as:

ϕ(t) = e^At = L^-1[(sI – A)^-1]

Calculation:

Given matrix is:

 $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ 

$sI=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]$

$sI-A=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$

$sI-A=\left[\begin{array}{cc}s& -1\\ 0& s\end{array}\right]$

${\left(sI-A\right)}^{-1}=\frac{1}{s^2-\left(0\times -1\right)}\left[\begin{array}{cc}s& 1\\ 0& s\end{array}\right]$

${\left(sI-A\right)}^{-1}=\frac{1}{s^2}\left[\begin{array}{cc}s& 1\\ 0& s\end{array}\right]$

${\left(sI-A\right)}^{-1}=\left[\begin{array}{cc}\frac{1}{s}& \frac{1}{s^2}\\ 0& \frac{1}{s}\end{array}\right]$

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

$\varphi (t)=L^{-1}\left[\begin{array}{cc}\frac{1}{s}& \frac{1}{s^2}\\ 0& \frac{1}{s}\end{array}\right]$

$\varphi \left(t\right)=\left[\begin{array}{cc}u(t)& tu(t)\\ 0& u(t)\end{array}\right]$

Considering unit step signal

$\varphi \left(t\right)=\left[\begin{array}{cc}1& t\\ 0& 1\end{array}\right]$

Hence, option 4 is the correct answer. 

Extra concept:

$\dot{X}=AX+BU$

Two different methods are there to solve the above state equation.

Method 1: Laplace Transform method

sX(s) = AX(s) + BU(s)

sX(s) – AX(s) =  X(0) + BU(s)

X(s)[sI - A] = X(0) + BU(s)

X(s) = (sI - A)^-1 X(0) + (sI – A)^-1 BU(s)

x(t) = L^-1{(sI – A)^-1X(0)} + L^-1{(sI - A)^-1BU(s)}

Zero input response is defined as:

L^-1{(sI – A)^-1X(0)}

Zero state response is defined as:

 L^-1{(sI - A)^-1BU(s)}

Method 2: Classical method

$x\left(t\right)=e^{At}X\left(0\right)=\underset{0}{\overset{t}{\int }}{e}^{A\left(t-\tau \right)}BU\left(\tau \right)d\tau$

Zero input response is defined as:

$e^{At}X\left(0\right)$

Zero state response is defined as:

$\underset{0}{\overset{t}{\int }}{e}^{A\left(t-\tau \right)}BU\left(\tau \right)d\tau$

Compare ZIR

$\varphi \left(t\right)=e^{At}=L^{-1}\left[{\left(sI-A\right)}^{-1}\right]$

This is the state transition matrix

Compare ZSR

$\underset{0}{\overset{t}{\int }}{e}^{A\left(t-\tau \right)}BU\left(\tau \right)d\tau =L^{-1}\left[\varphi \left(s\right)BU\left(s\right)\right]$

$x\left(t\right)=e^{At}X\left(0\right)=L^{-1}\left[\varphi \left(s\right)BU\left(s\right)\right]$

For a Homogeneous state equation ( U = 0 )

$\dot{X}=AX$

$x\left(t\right)=ZIR=e^{At}X\left(0\right)$

$Y=x_1-x_2+U$

${\dot{x}}_1=-x_1-U$

${\dot{x}}_2=x_1-x_2+U$

$A=\left[\begin{array}{cc}-1& 0\\ 1& -1\end{array}\right];B=\left[\begin{array}{c}-1\\ 1\end{array}\right];C=\left[\begin{array}{cc}1& -1\end{array}\right];D=\left[1\right]$

$e^{At}={\mathcal{L}}^{-1}\left\{{\left(SI-A\right)}^{-1}\right\}$

$\left(SI-A\right)=\left[\begin{array}{cc}s+1& 0\\ -1& s+1\end{array}\right]$

${\left(SI-A\right)}^{-1}=\frac{1}{{\left(s+1\right)}^2}\left[\begin{array}{cc}s+1& 0\\ 1& s+1\end{array}\right]$

$=\left[\begin{array}{cc}\frac{1}{s+1}& 0\\ \frac{1}{{\left(s+1\right)}^2}& \frac{1}{s+1}\end{array}\right]$

$e^{At}=\left[\begin{array}{cc}{e}^{-t}& 0\\ t{e}^{-t}& e^{-t}\end{array}\right]$

Concept:

The state transition matrix ϕ(t):

The state-transition matrix is defined as a matrix that satisfies the linear homogeneous state equation.

It represents the free response of the system.

The state-transition matrix ϕ(t) completely defines the transition of the states from the initial time t = 0 to any time t when the inputs are zero.

The state transition matrix is given by 

ϕ(t) = L-1 [sI - A]-1 = eAt

Where A = state matrix

The state-transition matrix is dependent only upon the matrix A and, therefore, is sometimes referred to as the state transition matrix of A.

Properties of ϕ(t):

1. ​ϕ(0) = I
2. ϕ-1(t) = ϕ(-t)
3. ϕ(t2 - t1) = ϕ(t2 - t0). ϕ(t0 - t1)
4. ϕ(t2 + t1) = ϕ(t2) . ϕ(t1)
5. [ϕ(t)]k = ϕ(kt)
6. At t =0, dϕ /dt = A

Calculation:

Given that

State transition matrix ϕ(t) = $\left[\begin{array}{cc}{e}^{-4t}& e^{-6t}-e^{-2t}\\ e^{-8t}-e^{-5t}& e^{-0.5t}\end{array}\right]$

Consider the sixth property of the state transition matrix

⇒ At t = 0, dϕ /dt = A

Where A is system or state matrix,

⇒ A = $\left[\begin{array}{cc}-4e^{-4t}& -6e^{-6t}+2e^{-2t}\\ -8e^{-8t}+5e^{-5t}& -0.5e^{-0.5t}\end{array}\right]$at t = 0

⇒ A = $\left[\begin{array}{cc}-4& -6+2\\ -8+5& -0.5\end{array}\right]$

⇒ A = $\left[\begin{array}{cc}-4& -4\\ -3& -0.5\end{array}\right]$

From the properties of state transition matrix,

e^A0 = I

$\Rightarrow \left[\begin{array}{cc}1& 2\left(a-b\right)\\ 0& a^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

⇒ a = b and a = ± 1

For a > 0, a = b = 1

$e^{At}=\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right]$

X(t) = e^At x[0]

$=\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}{e}^{-t}\\ e^{-2t}\end{array}\right]$

Concept:

The state transition matrix ϕ(t):

The state-transition matrix is defined as a matrix that satisfies the linear homogeneous state equation.

It represents the free response of the system.

The state-transition matrix ϕ(t) completely defines the transition of the states from the initial time t = 0 to any time t when the inputs are zero.

The state transition matrix is given by 

ϕ(t) = L^-1 [sI - A]^-1 = e^At

Where A = state matrix

The state-transition matrix is dependent only upon the matrix A and, therefore, is sometimes referred to as the state transition matrix of A.