Transfer Functions of a Two-Mass System Obtain the transfer functions [latex]X_1(s)/F(s)[/latex] and [latex]X_2(s)/F(s)[/latex] of the state-variable model obtained in Example 5.1.3 . The matrices and state vector of the model are [latex]A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & -\frac{12}{5} & \frac{4}{5} & \frac{8}{5} \\ 0 & 0 & 0 & 0 \\ \frac{4}{3} & \frac{8}{3} & -\frac{4}{3} & - \frac{8}{3} \end{bmatrix} B = \begin{bmatrix} 0\\ 0 \\ 0\\ \frac{1}{3} \end{bmatrix} [/latex] and [latex]z = \begin{bmatrix} z_{1}\\ z_{2} \\ z_{3}\\ z_{4} \end{bmatrix} = \begin{bmatrix} x_{1}\\ \dot{x}_{1} \\ x_{2}\\ \dot{x}_{2} \end{bmatrix} [/latex]

Because we want the transfer functions for [latex]x_{1}[/latex] and [latex]x_{2}[/latex] (which are the same as [latex]z_{1}[/latex] and [latex]z_{3}[/latex]), we must define the C and D matrices to indicate that [latex]z_{1}[/latex] and [latex]z_{3}[/latex] are the output variables [latex]y_{1}[/latex] and [latex]y_{2}[/latex]. Thus, [latex]C = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} D = \begin{bmatrix} 0 \\ 0 \end{bmatrix} [/latex]

Question 5.2.1: Transfer Functions of a Two-Mass System Obtain the transfer ...

System Dynamics

Transfer Functions of a Two-Mass System

Obtain the transfer functions $X_1(s)/F(s)$ and $X_2(s)/F(s)$ of the state-variable model obtained in Example 5.1.3. The matrices and state vector of the model are

A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & -\frac{12}{5} & \frac{4}{5} & \frac{8}{5} \\ 0 & 0 & 0 & 0 \\ \frac{4}{3} & \frac{8}{3} & -\frac{4}{3} & – \frac{8}{3} \end{bmatrix}           B = \begin{bmatrix} 0\\ 0 \\ 0\\ \frac{1}{3} \end{bmatrix}

and

z = \begin{bmatrix} z_{1}\\ z_{2} \\ z_{3}\\ z_{4} \end{bmatrix} = \begin{bmatrix} x_{1}\\ \dot{x}_{1} \\ x_{2}\\ \dot{x}_{2} \end{bmatrix}

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The MATLAB program is as follows.

A = [0, 1, 0, 0; -1, -12/5, 4/5, 8/5;...
0, 0, 0, 1; 4/3, 8/3, -4/3, -8/3];
B = [0; 0; 0; 1/3];
C = [1, 0, 0, 0; 0, 0, 1, 0]; D = [0; 0]
sys4 = ss(A, B, C, D);
tfsys4 = tf(sys4)

The results displayed on the screen are labeled #1 and #2. These correspond to the first and second transfer functions in order. The answers are

$\frac{X_{1}(s)}{F(s)} = \frac{0.5333s + 0.2667}{s^{4} + 5.067s^{3} + 4.467s^{2} + 1.6s + 0.2667}$

$\frac{X_{2}(s)}{F(s)} = \frac{0.3333s^{2} + 0.8s + 0.3333}{s^{4} + 5.067s^{3} + 4.467s^{2} + 1.6s + 0.2667}$