Question

Consider the system given below. The output is y(displacement from equilibrium position) and the input is V. (source voltage). The motor has an electrical constant Ke, a torque constant K, an armature inductance Lg and a resistance R. The rotor, shaft and disk together have inertia J and a viscous friction coefficient B. The disk has a radius ofr. (For the motor, assume that the torque is T = Ki,, and the back EMF is emf = KO). a. Derive the governing differential equations of motion. b. Represent the derived equations in state-space form. c. Represent the output equation as a function of the state variables W 1 Vin em) motor JB M

Answer 1

Given

$\\ emf=K_e \dot{\theta} \\ T=K_ti_a$

a)

The electrical circuit

R.NO -. Vio motor emf

Equation of the circuit

$\\ V_{in}=i_aR_a+L_a\frac{\mathrm{d} i_a}{\mathrm{d} t}+emf \\ V_{in}=i_aR_a+L_a \frac{\mathrm{d} i_a}{\mathrm{d} t}+K_e\dot{\theta_m} \\ \frac{\mathrm{d} i_a}{\mathrm{d} t}=- i_aR_a/L_a-K_e\dot{\theta_m}/L_a+V_{in}/L_a\ \ \ \ \ \ \ \ \ \ \ \ \ \ Eq1$

Torque on the motor

T a B om kr 6.7V)+ c(r Omn-)

Jom + Bom +r*k(rom - y) +r *crom - y) =T om Ktia/J - (B + crem/J – kr²0m/J + kyr/J + cry/J Eq2

Free body of the mass

k(y-r0.7 cly-rom м. у

$\\ M\ddot{y}=-k(y-r\theta_m)-c(\dot{y}-r\dot{\theta_m}) \\ \ddot{y}=-ky/M+kr\theta_m/M-c\dot{y}/M+cr\dot{\theta_m}/M\ \ \ Eq3$

The Eq1 Eq2 Eq3 gives the governing equation of the system

$\\ \frac{\mathrm{d} i_a}{\mathrm{d} t}=- i_aR_a/L_a-K_e\dot{\theta_m}/L_a+V_{in}/L_a \\\ddot{\theta_m}=K_ti_a/J-(B+cr^2)\dot{\theta_m}/J-kr^2\theta_m/J+kyr/J+cr\dot{y}/J \\ \ddot{y}=-ky/M+kr\theta_m/M-c\dot{y}/M+cr\dot{\theta_m}/M$

b)The equation of motion in the state-space model

$\\ \begin{bmatrix} \dot{y} \\ \ddot{y} \\ \dot{\theta_m} \\ \ddot{\theta_m} \\ \dot{i_A} \end{bmatrix} =\begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ -k/M &-c/M & kr/M & cr/M & 0 \\ 0 &0 &0 &1 &0 \\ kr/J & cr/J &-kr^2/J & -(B+cr^2)/J& K_t/J\\ 0&0 &0 & K_e/L_a & -R_a/La\end{bmatrix} \begin{bmatrix} y \\ \dot{y} \\ \theta_m \\ \dot{\theta_m} \\ i_A \end{bmatrix} +\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1/La \end{bmatrix}\begin{bmatrix} V_{in} \end{bmatrix} \\ X=\begin{bmatrix} y \\ \dot{y} \\ \theta_m \\ \dot{\theta_m} \\ i_A \end{bmatrix} \\ A=\begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ -k/M &-c/M & kr/M & cr/M & 0 \\ 0 &0 &0 &1 &0 \\ kr/J & cr/J &-kr^2/J & -(B+cr^2)/J& K_t/J\\ 0&0 &0 & K_e/L_a & -R_a/La\end{bmatrix} \\ B=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1/La \end{bmatrix} \\ U=\begin{bmatrix} V_{in} \end{bmatrix} \\ \dot{X}=AX+BU$

b)

Output equation can be written as

$\\ Y=\begin{bmatrix} y\\ \theta_m \\ i_a \end{bmatrix} \\ X=\begin{bmatrix} y \\ \dot{y} \\ \theta_m \\ \dot{\theta_m} \\ i_A \end{bmatrix} \\ \begin{bmatrix} y\\ \theta_m \\ i_a \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 0 &0 &1 & 0 & 0\\0 &0 &0 &0 & 1 \end{bmatrix}\begin{bmatrix} y \\ \dot{y} \\ \theta_m \\ \dot{\theta_m} \\ i_A \end{bmatrix} \\ C=\begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 0 &0 &1 & 0 & 0\\0 &0 &0 &0 & 1 \end{bmatrix} \\ Y=CX$