Part 6 - Space-state representation (2)

In this part we’ll see how we can go from a given space-state representation to the systems transfer function and vice-versa.

Given this space-state representation:

\begin{cases} \dot{x} & = Ax + Bu & (1) \newline y & = Cx + Du & (2) \end{cases}

We want to find $G_{uy}(s) = \dfrac{Y(s)}{U(s)}$

Let’s Laplace our equation system:

\begin{cases} sX(s) & = AX(s) + BU(s) & (1) \newline Y(s) & = CX(s) + DU(S) & (2) \end{cases}

From $(1)$ we can find that:

sX(s) - AX(S) = BU(s) \ | \ \text{To factor out X(s) we need to do a little matrix trick} \newline X(s)(sI - A) = BU(S) \newline X(s) = (sI - A)^{-1})B\ U(s)

If we substitute this into $(2)$ :

Y(s) = C((sI - A)^{-1} B) U(s) + D U(s) \newline \boxed{G_{uy}(s) = C(sI - A)^{-1}B + D}

From our feather-mass example, we found that:

A = \begin{bmatrix} 0 & 1 \newline -\dfrac{k}{m} & -\dfrac{b}{m} \end{bmatrix} \quad B = \begin{bmatrix} 0 \newline \dfrac{1}{m} \end{bmatrix} \quad C = \begin{bmatrix} 1 & 0 \end{bmatrix} \quad D = \begin{bmatrix} 0 \end{bmatrix}

We therefore get:

\begin{bmatrix} 1 & 0 \end{bmatrix} \left( \begin{bmatrix} s & 0 \newline 0 & s \end{bmatrix} - \begin{bmatrix} 0 & 1 \newline -\dfrac{k}{m} & -\dfrac{b}{m} \end{bmatrix}\right)^{-1} \begin{bmatrix} 0 \newline \dfrac{1}{m} \end{bmatrix}

\begin{bmatrix} 1 & 0 \end{bmatrix} \left( \begin{bmatrix} s & -1 \newline \dfrac{k}{m} & s+\dfrac{b}{m} \end{bmatrix}\right)^{-1} \begin{bmatrix} 0 \newline \dfrac{1}{m} \end{bmatrix}

\begin{bmatrix} 1 & 0 \end{bmatrix} \left( \dfrac{1}{s^2 + \dfrac{bs}{m} + \dfrac{k}{m}} \begin{bmatrix} s + \dfrac{b}{m} & 1 \newline -\dfrac{k}{m} & s \end{bmatrix}\right) \begin{bmatrix} 0 \newline \dfrac{1}{m} \end{bmatrix}

\dfrac{1}{s^2 + \dfrac{bs}{m} + \dfrac{k}{m}} \begin{bmatrix} s + \dfrac{b}{m} & 1 \end{bmatrix} \begin{bmatrix} 0 \newline \dfrac{1}{m} \end{bmatrix}

\dfrac{1}{m(s^2 + \dfrac{bs}{m} + \dfrac{k}{m})}

\boxed{\dfrac{1}{ms^2 + bs + k}}

Given:

G(s) = \dfrac{a}{s^3 + bs^2 + cs + d} = \dfrac{Y(s)}{U(s)}

Which means:

s^3 Y(s) + bs^2 Y(s) + cs Y(s) + d Y(s) = a U(s)

Inverse-Laplace yields:

y^{(3)}(t) + b \ddot{y}(t) + c \dot{y}(t) + dy(t) = a u(t)

Introduce state variables, $x_1 = y, x_2 = \dot{y}, x_3 = \ddot{y}$ .

This means:

\begin{cases} \dot{x_1} = x_2 \newline \dot{x_2} = x_3 \newline \dot{x_3} = au(t) - bx_3 - cx_2 - dx_1 \newline \end{cases}

Which yields:

\boxed{ \begin{cases} \dot{x} & = \begin{bmatrix} 0 & 1 & 0 \newline 0 & 0 & 1 \newline -d & -c & -b \end{bmatrix} \begin{bmatrix} x_1 \newline x_2 \newline x_3 \end{bmatrix} + \begin{bmatrix} 0 \newline 0 \newline a \end{bmatrix} u \newline y = x_1 \end{cases}}