For a paper processing plant, it is important to maintain a constant tension on the continuous sheet of paper between the wind-off and wind-up rolls. The tension varies as the widths of the rolls change, and an adjustment in the take-up motor speed is necessary, as shown in Figure P4.10. If the wind-up motor speed is uncontrolled, as the paper transfers from the wind-off roll to the wind-up roll, the velocity $v_0$ decreases and the tension of the paper drops $[10,14]$. The threeroller and spring combination provides a measure of the tension of the paper. The spring force is equal to $k_1 y$, and the linear differential transformer, rectifier, and amplifier may be represented by $e_0=-k_2 y$. Therefore, the measure of the tension is described by the relation $2 T(s)=k_1 y$, where $y$ is the deviation from the equilibrium condition, and $T(s)$ is the vertical component of the deviation in tension from the equilibrium condition. The time constant of the motor is $\tau=L_a / R_a$, and the linear velocity of the wind-up roll is twice the angular velocity of the motor, that is, $v_0(t)=2 \omega_0(t)$. The equation of the motor is then
$$
E_0(s)=\frac{1}{K_m}\left[\tau s \omega_0(s)+\omega_0(s)\right]+k_3 \Delta T(s),
$$
where $\Delta T=$ a tension disturbance. (a) Draw the closed-loop block diagram for the system, including the disturbance $\Delta T(s)$. (b) Add the effect of a disturbance in the wind-off roll velocity $\Delta V_1(s)$ to the block diagram. (c) Determine the sensitivity of the system to the motor constant $K_m$. (d) Determine the steadystate error in the tension when a step disturbance in the input velocity, $\Delta V_1(s)=A / s$, occurs.