Question 5.2.2: Free Response of the Two-Mass Model Compute the free respons...
Free Response of the Two-Mass Model
Compute the free response x_{1}(t) of the state model derived in Example 5.1.3, for x_{1}(0) = 5, \dot{x}_{1}(0) = −3, x_{2}(0) = 4, and \dot{x}_{2}(0) = 2. The model is
\dot{z}_{1} = z_{2}\dot{z}_{2} = \frac{1}{5} (−5z_{1} − 12z_{2} + 4z_{3} + 8z_{4})
\dot{z}_{3} = z_{4}
\dot{z}_{4} = \frac{1}{3} [4z_{1} + 8z_{2} − 4z_{3} − 8z_{4} + f (t)]
or
\dot{z} = Az + B f (t)where
A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & -\frac{12}{5} & \frac{4}{5} & \frac{8}{5} \\ 0 & 0 & 0 & 1 \\ \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}We must first relate the initial conditions given in terms of the original variables to the state variables. From the definition of the state vector z, we see that z_{1}(0) = x_{1}(0) = 5, z_{2}(0) = x_{1}(0)=−3, z_{3}(0) = x_{2}(0) = 4, and z_{4}(0) = \dot{x}_{2}(0) = 2. Next we must define the model in state-variable form.
The system sys4 created in Example 5.2.1 specified two outputs, x_{1} and x_{2}. Because we want to obtain only one output here (x_{1}), we must create a new state model using the same values for the A and B matrices, but now using
C = [1 0 0 0] D = [0]
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]; D = [0];
sys5 = ss(A, B, C, D);
initial(sys5, [5, -3, 4, 2])
The plot of x_{1}(t) will be displayed on the screen.