A Single-Link Robot Arm Driven by a DC Motor Consider the dynamic system shown in Figure 6.46, in which a single-link robot arm is driven by a DC motor. The differential equation of the robot arm in terms of the motor variable θm was determined in Example 5.17 to be

Question

A Single-Link Robot Arm Driven by a DC Motor

Consider the dynamic system shown in Figure 6.46, in which a single-link robot arm is driven by a DC motor. The differential equation of the robot arm in terms of the motor variable [latex] heta_{\mathrm{m}}[/latex] was determined in Example 5.17 to be

[latex]\left(I_{\mathrm{m}}+N^{2}I ight) \ddot{ heta}_{\mathrm{m}}+\left(B_{\mathrm{m}}+N^{2} B ight) \dot{ heta}_{\mathrm{m}}= au_{\mathrm{m}^{\prime}}[/latex]

where [latex]I_{\mathrm{m}}[/latex] and [latex]I[/latex] are the mass moments of inertia of the motor and the load, respectively, [latex]B_{\mathrm{m}}[/latex] and [latex]B[/latex] are the coefficients of the torsional viscous damping of the motor and the load, respectively, [latex] au_{\mathrm{m}}[/latex] is the torque generated by the motor, and [latex]N[/latex] is the gear ratio. Assume that the armature inductance is negligibly small, that is, [latex]L_{\mathrm{a}} \approx 0[/latex]. The torque and the back emf constants of the motor are [latex]K_{\mathrm{t}}[/latex] and [latex]K_{\mathrm{e}}[/latex], respectively.

a. Derive the differential equation relating the applied voltage [latex]v_{\mathrm{a}}[/latex] and the link angular displacement [latex] heta[/latex].

b. Determine the transfer function [latex]\Theta(s) / V_{\mathrm{a}}(s)[/latex] using the differential equation obtained in Part (a). Assume that all initial conditions are zero.

Accepted Answer

a. For the electrical circuit, applying Kirchhoff's voltage law gives

[latex]R_{\mathrm{a}} i_{\mathrm{a}}+\mathrm{e}_{\mathrm{b}}-v_{\mathrm{a}}=0,[/latex]

where [latex]e_{\mathrm{b}}=K_{\mathrm{e}} \dot{ heta}_{\mathrm{m}}[/latex]. With the given gear ratio,

[latex] heta=N heta_{m},[/latex]

we have

[latex]R_{\mathrm{a}} i_{\mathrm{a}}=v_{\mathrm{a}}-K_{\mathrm{e}} \frac{1}{N} \dot{ heta}.[/latex]

Thus, the current [latex]i_{\mathrm{a}}[/latex] can be expressed as

[latex]i_{\mathrm{a}}=\frac{1}{R_{\mathrm{a}}} v_{\mathrm{a}}-\frac{K_{\mathrm{e}}}{R_{\mathrm{a}}} \frac{1}{N} \dot{ heta}.[/latex]

The model of the mechanical part in terms of [latex] heta[/latex] is given by

[latex] \left(I_{\mathrm{m}}+N^{2} I ight) \frac{1}{N} \ddot{ heta}+\left(B_{\mathrm{m}}+N^{2} B ight) \frac{1}{N} \dot{ heta}= au_{\mathrm{m}^{\prime}} [/latex]

where [latex] au_{\mathrm{m}}=K_{\mathrm{t}} i_{\mathrm{a}}[/latex]. Substituting the expression of the current [latex]i_{\mathrm{a}^{\prime}}[/latex] we obtain

[latex] \left(I_{\mathrm{m}}+N^{2} I ight) \frac{1}{N} \ddot{ heta}+\left(B_{\mathrm{m}}+N^{2} B ight) \frac{1}{N} \dot{ heta}=K_{\mathrm{t}}\left(\frac{1}{R_{\mathrm{a}}} V_{\mathrm{a}}-\frac{K_{\mathrm{e}}}{R_{\mathrm{a}}} \frac{1}{N} \dot{ heta} ight) . [/latex]

Rearranging the equation yields

[latex]\left(I_{\mathrm{m}}+N^{2} I ight) \ddot{ heta}+\left(B_{\mathrm{m}}+N^{2} B+\frac{K_{\mathrm{t}} K_{\mathrm{e}}}{R_{\mathrm{a}}} ight) \dot{ heta}=\frac{N K_{\mathrm{t}}}{R_{\mathrm{a}}} V_{\mathrm{a}} .[/latex]

b. Taking the Laplace transform results in

[latex]\frac{\Theta(s)}{V_{\mathrm{a}}(s)}=\frac{N K_{\mathrm{t}} / R_{\mathrm{a}}}{\left(I_{\mathrm{m}}+N^{2} I ight) s^{2}+\left(B_{\mathrm{m}}+N^{2} B+K_{\mathrm{t}} K_{\mathrm{e}} / R_{\mathrm{a}} ight) s} .[/latex]

This electromechanical system can be modeled using Simulink and Simscape. The details will be discussed in Section 6.6.

Question 6.11: A Single-Link Robot Arm Driven by a DC Motor Consider the dy...

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