State Variables MCQ Quiz in मराठी - Objective Question with Answer for State Variables - मोफत PDF डाउनलोड करा

Given that $\dot{x}\left(t\right)=Ax\left(t\right)$ ………………(1)

Now let $A=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]$

putting in equation 1 for initial vector $\left[\begin{array}{c}1\\ -2\end{array}\right]$

$\left[\begin{array}{c}\frac{d}{dt}{e}^{-2t}\\ \frac{d}{dt}\left(-2e^{-2t}\right)\end{array}\right]=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]\left[\begin{array}{c}1\\ -2\end{array}\right]$

now at t=0, 

$\left[\begin{array}{c}-2\\ 4\end{array}\right]=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]\left[\begin{array}{c}1\\ -2\end{array}\right]$

Comparing, $p-2q=-2andr-2s=4$  …..(2)

putting in equation 1 for initial vector $\left[\begin{array}{c}1\\ -1\end{array}\right]$

$\left[\begin{array}{c}\frac{d}{dt}{e}^{-t}\\ \frac{d}{dt}\left(-e^{-t}\right)\end{array}\right]=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]\left[\begin{array}{c}1\\ -1\end{array}\right]$

now at t=0,

$\left[\begin{array}{c}-1\\ 1\end{array}\right]=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]\left[\begin{array}{c}1\\ -1\end{array}\right]$

Comparing, $p-q=-1andr-s=1$    ….(3)

Solving equation 2 and 3

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

Now characteristic equation is

$\left|\alpha I-A\right|=0$

$\left|\begin{array}{cc}\alpha & -1\\ 2& \alpha +3\end{array}\right|=0$

Solving,    $\alpha =-1,-2$

-1 and -2 are eigen values of the system. The corresponding eigen vectors are

$\left[\alpha I-A\right]\left[X\right]=0$

$\left[\begin{array}{cc}{\alpha }_1& -1\\ 2& {\alpha }_1+3\end{array}\right]\left[\begin{array}{c}{x}_{11}\\ x_{21}\end{array}\right]=0$

Now putting $\alpha =-1$ we get only one independent equation $x_{11}+x_{21}=0$

if $x_{11}=M$then $x_{21}=-M$ then eigen vector is

$\left[\begin{array}{c}{x}_{11}\\ x_{21}\end{array}\right]=\left[\begin{array}{c}M\\ -M\end{array}\right]=M\left[\begin{array}{c}1\\ -1\end{array}\right]$

again, putting $\alpha =-2$ we get only one independent equation $2x_{12}+x_{22}=0$ 

if $x_{12}=M$then $x_{22}=-2M$ then eigen vector is

$\left[\begin{array}{c}{x}_{12}\\ x_{22}\end{array}\right]=\left[\begin{array}{c}M\\ -2M\end{array}\right]=M\left[\begin{array}{c}1\\ -2\end{array}\right]$

From the above block diagram, we get:

ẋ₁ = x₂

ẋ₂ = x₃

ẋ₃ = -7x₃ - 19x₂ - 13x₁ + u(t)

y(t) = 26x₁ + 13x₂

∴ The matrix form representation is as follows:

$\left[\begin{array}{c}{\dot{x}}_1\\ {\dot{x}}_2\\ {\dot{x}}_3\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -13& -19& -7\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u\left(t\right)$

$y\left(t\right)=\left[26130\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]+\left[0\right]u\left(t\right)$

$A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -13& -19& -7\end{array}\right],B^T=\left[001\right],C=\left[26130\right]$

x = y1 & $\dot{x}=\frac{d{y}_1}{dx}$

$\underset{―}{y}=\left[y_2\right]=\left[\begin{array}{c}2\\ 2x\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right]x$

$y_1=\frac{1}{s+2}u$

${\dot{y}}_1+2y_1=u$

$\dot{x}+2x=u$

$\dot{x}+2x=u$

$\underset{―}{\dot{x}}=\left[-2\right]\underset{―}{x}+\left[1\right]u$

Alternate solution 

Assume x be a state variable

$[\mathrm{X}\left(\mathrm{s}\right)=\frac{1}{s+2}u\left(s\right)]$

sX(s)+2X(s) = U(S)

taking inverse laplace transform

x(t)’ + 2 x(t) = u(t)

x’(t)= u(t) – 2x(t)

output equation is given by

y1(t) = x

y2(t) = 2 x

$\begin{array}{l}{\dot{x}}_2\left(t\right)=ar\left(t\right)-p_2{x}_2\left(t\right)\\ {\dot{x}}_1\left(t\right)={\dot{x}}_2\left(t\right)-p_1{x}_1\left(t\right)+z_2{x}_2\left(t\right)\\ =ar\left(t\right)-p_2{x}_2\left(t\right)-p_1{x}_1\left(t\right)-p_1{x}_1\left(t\right)+z_2{x}_2\left(t\right)\\ \dot{x}\left(t\right)=-p_1{x}_1\left(t\right)+\left(z_2-p_2\right)x_2\left(t\right)+ar\left(t\right)\end{array}$

If we write the above equations in matrix form,

$\left[\begin{array}{c}{\dot{x}}_1\left(t\right)\\ {\dot{x}}_2\left(t\right)\end{array}\right]=\left[\begin{array}{cc}-p_1& z_1-p_2\\ 0& -p_2\end{array}\right]\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]+\left[\begin{array}{c}a\\ a\end{array}\right]r\left(t\right)$

Explanation:

Transfer function which is the ratio of Laplace output to the Laplace input when the initial conditions are zero in discrete is the same as continuous but in the z-domain.

Transfer function from state model is given as:

$\dot{X}=AX+BU$       ----(1)

Y = CX + DU       ----(2)

Where,

Y is single output 

X is single input

X(0) = 0, as initial condition is zero.

Taking Z transform of equation (1):

zX(z) = AX(z) + BU(z)

[zI - A] X(z) = BU(z)

Taking Z transform of equation (2):

Y(z) = CX(z) + DU(z)

putting value of X(z) from equation (1) in equation (2) we get:

$T(z)=\frac{Y(z)}{U(z)}$

T(z) = c[zI - A]^-1b + d

So, option (1) is correct answer.

Concept:

A standard form of State-space matrix:

$\dot{X}=AX+BU$

It is called a State equation or Dynamic equation

Y = CX + DU 

It is known as the output equation

$\dot{X}$: Differential State Vector

Y: Output vector

U: Input vector

A: State matrix

B: Input matrix

C: Output matrix

D: Transition matrix

Transfer Function:

For the state model defined by we have to write the Transfer function for the analysis.

It is given by

$TF=C{\left[sI-A\right]}^{-1}B+D$

The characteristic equation is defined by

|sI – A| = 0

Application:

The state equations in the question can be easily solved as follows:

ẋ(t) = - 2x(t) + 2u(t)

Convert the above equation into Laplace domain, we get

sX(s) = - 2X(s) + U(s)

(s+2) X(s) = U(s)

$X(s)=\frac{U(s)}{(s+2)}$        -----(1)

y(t) = 0.5x(t)

Y(s) = 0.5 X(s)        ----(2)

From (1)  and (2), we get

$Y(s)=0.5\frac{U(s)}{(s+2)}$

$\frac{Y(s)}{U(s)}=\frac{0.5}{(s+2)}$

Explanation:

Transfer function which is the ratio of Laplace output to the Laplace input when the initial conditions are zero in discrete is the same as continuous but in the z-domain.

Transfer function from state model is given as:

$\dot{X}=AX+BU$ -----(1)

Y = CX + DU -----(2)

Where,

Y is single output 

X is single input

X(0) = 0, as initial condition is zero.

Taking Z transform of equation (1):

zX(z) = AX(z) + BU(z)

[zI - A] X(z) = BU(z)

Taking Z transform of equation (2):

Y(z) = CX(z) + DU(z)

putting value of X(z) from equation (1) in equation (2) we get:

$T(z)=\frac{Y(z)}{U(z)}$

T(z) = c[zI - A]^-1b + d

So, option (1) is correct answer.

From the Diagram

$X_3={\dot{X}}_2$ 

$X_2={\dot{X}}_1$ 

X₄ = U(S) – 8X₃ – 16 X₂ – 6 X₁

$\dot{x}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -6& -16& -8\end{array}\right]x+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]U\left(t\right)$ 

y(s) = 6x₁ + 8x₂ + 2x₃

$y\left(t\right)=\left[\begin{array}{ccc}6& 8& 2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$ 

Hence option 3 is correct

$\begin{array}{l}{\dot{x}}_2\left(t\right)=ar\left(t\right)-p_2{x}_2\left(t\right)\\ {\dot{x}}_1\left(t\right)={\dot{x}}_2\left(t\right)-p_1{x}_1\left(t\right)+z_2{x}_2\left(t\right)\\ =ar\left(t\right)-p_2{x}_2\left(t\right)-p_1{x}_1\left(t\right)-p_1{x}_1\left(t\right)+z_2{x}_2\left(t\right)\\ \dot{x}\left(t\right)=-p_1{x}_1\left(t\right)+\left(z_2-p_2\right)x_2\left(t\right)+ar\left(t\right)\end{array}$

If we write the above equations in matrix form,

$\left[\begin{array}{c}{\dot{x}}_1\left(t\right)\\ {\dot{x}}_2\left(t\right)\end{array}\right]=\left[\begin{array}{cc}-p_1& z_1-p_2\\ 0& -p_2\end{array}\right]\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]+\left[\begin{array}{c}a\\ a\end{array}\right]r\left(t\right)$

${\dot{x}}_1=-4x_1+x_2$

${\dot{x}}_2=x_3+2u$

${\dot{x}}_3=-3x_3+u$

$\Rightarrow A=\left[\begin{array}{ccc}-4& 1& 0\\ 0& 0& 1\\ 0& 0& -3\end{array}\right]$