Control of a Single-Degree-of-Freedom Mass–Spring–Damper System Consider a single-degree-of-freedom mass–spring–damper system as shown in Figure 5.29, where m = 2 kg, b = 2 N·s/m, and k = 50 N/m. A PD controller, f = −kpx − kDx , is designed to adjust the input force f so that the mass block can be

Question

Control of a Single-Degree-of-Freedom Mass-Spring-Damper System

Consider a single-degree-of-freedom mass-spring-damper system as shown in Figure 5.29, where [latex]m=2 \mathrm{~kg}, b=2 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}[/latex], and [latex]k=50 \mathrm{~N} / \mathrm{m}[/latex]. A PD controller, [latex]f=-k_{\mathrm{p}} x-k_{\mathrm{D}} \dot{x}[/latex], is designed to adjust the input force [latex]f[/latex] so that the mass block can be maintained in the equilibrium position regardless of disturbance forces applied to the block. The performance requirements of the closed-loop system are overshoot [latex]M_{\mathrm{p}}<5 \%[/latex] and rise time [latex]t_{\mathrm{r}}<0.25 \mathrm{~s}[/latex].

a. Design a PD controller to meet the performance requirements.

b. A Build a block diagram of the feedback control system, in which the plant is constructed using Simscape blocks and the controller is constructed using Simulink blocks. Find the closed-loop response if the mass block is initially [latex]0.1 \mathrm{~m}[/latex] away from the equilibrium position.

Accepted Answer

a. The dynamics of the plant is described by

[latex]m \ddot{x}+b \dot{x}+k x=f[/latex]

where the control force is

[latex]f=-k_{\mathrm{p}} x-k_{\mathrm{D}} \dot{x}.[/latex]

Combining the two equations gives the dynamics of the closed-loop system,

[latex]m \ddot{x}+\left(b+k_{\mathrm{D}} ight) \dot{x}+\left(k+k_{\mathrm{p}} ight) x=0,[/latex]

which is a second-order system. Thus, the coefficients in the above differential equation can be related to the natural frequency and damping ratio of the closed-loop system via

[latex] \begin{aligned} \frac{b+k_{\mathrm{D}}}{m} & =2 \zeta \omega_{\mathrm{n}}, \ \frac{k+k_{\mathrm{p}}}{m} & =\omega_{\mathrm{n}}^{2} . \end{aligned} [/latex]

The requirement for overshoot indicates

[latex]\zeta>0.69.[/latex]

Pick [latex]\zeta=0.75[/latex] and substitute into the requirement for rise time to obtain

[latex]\omega_{\mathrm{n}}>9.26 \mathrm{rad} / \mathrm{s}.[/latex]

Pick [latex]\omega_{\mathrm{n}}=10 \mathrm{rad} / \mathrm{s}[/latex]. Simultaneous solution of the two relations given above yields [latex]k_{\mathrm{p}}=150[/latex] and [latex]k_{\mathrm{D}}=28[/latex].

b. Figure 10.85 is the block diagram of the resulting feedback control system built using Simulink and Simscape. The plant is constructed based on the physical mass-springdamper system and the details on Simscape modeling can be found in Example 5.4. The controller is constructed using Simulink blocks and its structure is similar to the PD control discussed in Section 10.4 with the reference signal [latex]r[/latex] set as 0. To specify a nonzero initial position, double-click on the Translational Spring block, type 0.1 for the Initial deformation, and choose the unit as [latex]\mathrm{m}[/latex]. This implies that the spring is initially elongated by [latex]0.1 \mathrm{~m}[/latex]. Also, double-click on the Ideal Translational Motion Sensor block, type 0.1 for the Initial position, and choose the unit as [latex]\mathrm{m}[/latex]. The corresponding displacement response of the system is shown in Figure 10.86.

Question 10.24: Control of a Single-Degree-of-Freedom Mass–Spring–Damper Sys...

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