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State and prove properties of state transition matrix. Obtain the state model for the system with Transfer function.

$\frac{Y(s)}{V(s)} = \frac{3s+4}{s^2 +5s +6}$

Mumbai University > Electronics Engineering > Sem 4 > Linear Control Systems

Topic : State Variable Models

Difficulty : High

Marks : 10M

1 Answer
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Properties of state Transition matrix =>

1) $\phi(o)$ = $e^{Ao}$ = I

2) $\phi(t) = e^{At} = [\phi(-t)]^{-1}$

i.e. $\phi^{-1}(t) = \phi(-t)$

3) $\phi(t_1 + t_2) = e^{A(t_1+t_2)}$

=$e^{At_1}.e^{At_2} = \phi(t_1).\phi(t_2)$

4) $[\phi(t)]^n = [e^{At}]^n$

= $\phi(nt)$

5) $\phi(t_2-t_1).\phi(t_1-t_o) = \phi(t_2-t_o)$

Given =>

$\frac{Y(s)}{V(s)} = \frac{3s+4}{s^2 + 5s + 6}$

The given transfer function can be written in the form,

T(s) = $\frac{Y(s)}{V(s)}.\frac{X_1(s)}{X_1(s)}$ = $\frac{X_1(s)}{V(s)}.\frac{Y(s)}{X_1(s)}$

let, $\frac{X_1(s)}{V(s)} = \frac{1}{s^2 + 5s + 6}$ -----(1)

and $\frac{Y(s)}{X_1(s)} = 3s+4$ ---------(2)

From equation (1),

$X_1(s) = [s^2 + 5s + 6] = V(s)$

$s^2 X_1(s) + 5sX_1(s) + 6X_1(s) = V(s)$

(Taking inverse laplace)

$\frac{d^2}{dt^2} x_1(t) + 5\frac{d}{dt}x_1(t) + 6x_1(t) = u(t)$

${\ddot{x}}_1(t) + 5{\dot{x}}_1(t) + 6x(t) = u(t)$ ----------(3)

let,

${\dot{x}}_1(t) = x_2(t)$

${\dot{x}}_2(t) = x_3(t) = x_1(t)$

equation (3) becomes,

$x_3(t) + 5x_2(t) + 6x_1(t) = u(t)$ --------(4)

equation (4) can be written as,

$0{\dot{x}}_3(t) + x_3(t) + 5 x_2(t) + 6x(t) = u(t)$

$0{\dot{x}}_3(t) = -x_3(t) - 5 x_2(t) - 6x(t) + u(t)$

writting in matrix form,

$\begin{bmatrix} {\dot{x}}_1(t) \\\ {\dot{x}}_2(t) \\\ {\dot{x}}_3(t) \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ -6 & -5 & -1 \end{bmatrix} . \begin{bmatrix} x_1(t) \\\ x_2(t) \\\ {x}_3(t) \end{bmatrix} + \begin{bmatrix} 0 \\\ 0 \\\ 1 \end{bmatrix} u(t)$ -----( 5 )

using equation (2)

$\frac{Y(s)}{X_1(s)} = 3s + 4$

$Y(s) = (3s+4)X_1(s)$

$Y(s) = 3sX_1(s) + 4X_1(s)$

Taking inverse laplace

$y(t) = 3\frac{d}{dt}x_1(t) + 4x_1(t)$

$\therefore 3\dot{x}_1(t) + 4x_1(t)$ = y(t)

let $x_1(t) = x_2(t)$

= $3x_2(t) + 4x_1(t) = y(t)$ ---------( 6 )

output can be written as

y(t) = $\begin{bmatrix} 4 & 3 & 0 \end{bmatrix}$.$\begin{bmatrix} x_1(t) \\\ x_2(t) \\\ x_3(t) \end{bmatrix} $ ----------( 7 )

Here, (5) and (7) represent the state variable model.

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