Question 5.7.1: Set up the equations of motion for the actuator circuit of F...
Set up the equations of motion for the actuator circuit of Fig. 5.11 [1].
This is a magnetic circuit actuator. Choose a convention in which f_{fld} (force imparted to the moving member by the field) is positive in the direction of increasing displacement x as shown. For the equations of motion use the first-order system (5.86), which is written in terms of the independent variables (i, x),
\frac{di(t)}{dt} = \frac{1}{L(x)} [V_{s}(t) – i(t) (R+R_{coil}) – \frac{∂Λ(i,x)}{∂x} v(t)]
\frac{dv(t)}{dt} =\frac{1}{m} F(i,x)
\frac{dx(t)}{dt} = v(t). (5.140)
To solve Eq. (5.140) we need expressions for the terms L(x), ∂Λ(i, x)/∂x, and F(i, x). As the independent variables are i and x, we work with the coenergy W_{fld}^{c}(i, x). From Eq. (5.139) we have
Eq. (5.139): W_{fld}^{c} = \left\{\begin{matrix} \frac{Λ(i,x)i}{2}=\frac{L(x)i^{2}}{2} \text{(linear motion)} \\ \frac{Λ(i,θ)i}{2}=\frac{L(θ)i^{2}}{2} \text{(rotational motion)}\end{matrix} \right.
W_{fld}^{c} ( i,x) = \frac{L(x)i^{2}}{2}. (5.141)
The magnetic circuit of Fig. 5.11 was analyzed in Example 3.5.2 from Section 3.5.1. The inductance L(x) was found to be
L(x) = \frac{μ_{0}N^{2}A_{g}}{(A_{g}/A_{c})(μ_{0}/μ)l_{c}+ 2x} (5.142)
where l_{c} is the path length in the core, and A_{c} and A_{g} are the cross-sectional areas of the core and gap, respectively. Substitute Eq. (5.142) into Eq. (5.141) and obtain
W_{fld}^{c} ( i,x) = \frac{μ_{0}N^{2}A_{g} i^{2}}{2((A_{g}/A_{c})(μ_{0}/μ)l_{c}+ 2x)} (5.143)
Next, use Eqs. (5.137) and (5.138) to obtain ∧ and f_{fld}. We find that
∧( i,x) = \frac{∂W_{fld}^{c} ( i,x)}{∂i}
= \frac{μ_{0}N^{2}A_{g} i}{((A_{g}/A_{c})(μ_{0}/μ)l_{c}+ 2x)}, (5.144)
and that
f_{fld}( i,x) = \frac{∂W_{fld}^{c} ( i,x)}{∂x}
= – \frac{μ_{0}N^{2}A_{g} i^{2}}{((A_{g}/A_{c})(μ_{0}/μ)l_{c}+ 2x)^{2}}, (5.145)
The minus sign in Eq. (5.145) implies that the direction of f_{fld} is opposite to the direction of increasing x (i.e., toward the actuator and opposite to the direction of increasing air gap). The total force F(i, x) on the moving member is a sum of the forces due to the field and spring
F(i, x) = f_{fld}(i, x)+f_{s}(x), (5.146)
where f_{s}(x) is the force due to the spring. In its initial state, the actuator is unenergized with i(0) = 0, and the mass is at rest in an equilibrium position x_{0}. In this position the spring force is zero, f_{s}(x_{0}) = 0. Thus, when the mass is at a position x the spring force is
f_{s}(x)= -k(x- x_{0}) (5.147)
Notice that when x > x_{0} (x < x_{0} ) this force tends to move the mass towards (away from) the magnetic circuit. Substitute Eqs. (5.144), (5.145), (5.146), and (5.147) into Eq. (5.140) and obtain
\frac{di(t)}{dt} = \frac{((A_{g}/A_{c})(μ_{0}/μ)l_{c}+ 2x)}{μ_{0}N^{2}A_{g}}
× [V_{s}(t) – i(t) (R+R_{coil}) + \frac{2μ_{0}N^{2}A_{g} i}{((A_{g}/A_{c})(μ_{0}/μ)l_{c}+ 2x)^{2}} v(t) ]
\frac{dv(t)}{dt} = – \frac{1}{m} [\frac{μ_{0}N^{2}A_{g} i^{2}}{((A_{g}/A_{c})(μ_{0}/μ)l_{c}+ 2x)^{2}} + k(x- x_{0})]
\frac{dx(t)}{dt} = v(t). (5.148)
This nonlinear first-order system has to be solved subject to the initial conditions x(0) = x_{0}, v(0) = v_{0} and i(0) = i_{0}.