$\frac{Y(s)}{R(s)} = \frac{3s+4}{s^2 +4s +3}$

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

Topic : State Variable Models

Difficulty : High

Marks : 10M

The given transfer function can be written in the form of: $T(s)=\cfrac{X_{1}(s)}{R(s)}\cdot \cfrac{Y(s)}{X_{1}(s)}$

Let, $\cfrac{X_{1}(s)}{R(s)}=\cfrac{1}{s^{2}+4s+3}$ --------(1)

$\cfrac{Y(s)}{X_{1}(s)}=3s+4$ --------(2)

From equation (1), we get,

$X_{1}(s)[s^{2}+4s+3]=R(s)$

$s^{2}X_{1}+4s X_{1}+3X_{1}=R(s)$

Taking inverse laplace transform,

$\cfrac{d^{2}}{dt^{2}} x_{1}(t)+4 \cfrac{d}{dt} x_{1}(t)+3x_{1}(t)=r(t)$

$\ddot { { x }_{ 1 } } (t) +4 \dot { { x }_{ 1 } } (t)+x_{1}(t)=r(t)$ -------(3)

Let $\dot { { x }_{ 1 } } (t)=x_{2}(t)$

$\dot { { x }_{ 2 } } (t)=x_{3}(t)=\ddot { { x }_{ 1 } } $

Equation (3) becomes, $x_{3}(t)+4x_{2}(t)+3x_{1}(t)=r(t)$

This can be also written as

$o \dot{x_{3}} + x_{3}(t)+4x_{2}(t)+3x_{1}(t)=r(t)$

$o \dot{x_{3}} = -x_{3}(t)-4x_{2}(t)-3x_{1}(t)+r(t)$

In matrix form, $\begin{bmatrix} \dot { { x }_{ 1 } } (t) \\ \dot { { x }_{ 2 } } (t) \\ \dot { { x }_{ 3 } } (t) \end{bmatrix}=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -3 & -4 & -1 \end{bmatrix}\begin{bmatrix} { x }_{ 1 }(t) \\ { x }_{ 2 }(t) \\ { x }_{ 3 }(t) \end{bmatrix}+\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}r(t)-----(4)$

From equation (2), we get

$\cfrac{Y(s)}{X_{1}(s)}=3s+4$

$(3s+4)X_{1}(s)=Y(s)$

$3s X_{1}(s)+4X_{1}(s)=Y(s)$

Taking inverse laplace,

$3 \cfrac{d}{dt} x_{1}(t)+4x_{1}(t)=y(t)$

$3\dot{x_{1}}(t) +4 x_{1}(t)=y(t)$ ------(5)

Let, $\dot{x_{1}}(t)=x_{2}(t)$

Equation (5) becomes

$3x_{2}(t)+4x_{1}(t)=y(t)$

In matrix form, $y(t)=\begin{bmatrix} 4 & 3 & 0 \end{bmatrix}\begin{bmatrix} { x }_{ 1 }(t) \\ { x }_{ 2 }(t) \\ { x }_{ 3 }(t) \end{bmatrix}----(6)$

Equation (4) and (6) represents the state variable model.