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Before the invention of linear motors, the only way to produce linear motion was to use pneumatic or hydraulic cylinders, or to translate rotary motion to linear motion with ballscrews or belts and pulleys. A linear motor consists of two mechanical assemblies: coil and magnet, as shown in Figure 1-27. Current flowing in a winding in a magnetic flux field produces a force. The copper windings conduct current (I ), and the assembly generates magnetic flux density (B). When the current and flux density interact, a force (F ) is generated in the direction shown in Figure 1-27, where F = I B. Even a small motor will run efficiently, and large forces can be created if a large number of turns are wound in the coil and the magnets are powerful rare-earth magnets. The windings are phased 120 electrical degrees apart, and they must be continually switched or commutated to sustain motion. Only brushless linear motors for closed-loop servomotor applications are discussed here. Two types of these motors are available commercially steel-core (also called iron-core) and epoxy-core (also called ironless). Each of these linear servomotors has characteristics and features that are optimal in different applications The coils of steel-core motors are wound on silicon steel to maximize the generated force available with a single-sided magnet assembly or way. Figure 1-28 shows a steel-core brushless linear motor. The steel in these motors focuses the magnetic flux to produce very high force density. The magnet assembly consists of rare-earth bar magnets mounted on the upper surface of a steel base plate arranged to have alternating polarities (i.e., N, S, N, S)



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To nd the unit-step response, we rst solve the following equation: dv R2 C t>0 v2 R 1 C 2 0 t<0 dt The solution of the preceding equation is R2 C 1 e t= R1 C u t . derivative of the preceding solution. The unit-step response of the circuit is the time-

CHAP. 7]





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The unit step response may also be found by the Laplace transform method (see 16). EXAMPLE 7.17 Passive phase shifter. Find the relationship between v2 and v1 in the circuit of Fig. 7-45(a). Let node D be the reference node. Apply KCL at nodes A and B to nd KCL at node A: KCL at node B: C dvA vA v1 0 R dt d vB v1 vB C 0 dt R

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Subtracting the second equation from the rst and noting that v2 vA vB we get v2 RC dv2 dv v1 RC 1 dt dt

EXAMPLE 7.18 Active phase shifter. Show that the relationship between v2 and v1 in the circuit of Fig. 7-45(b) is the same as in Fig. 7-45(a). Apply KCL at the inverting (node A) and non-inverting (node B) inputs of the op amp. KCL at node A: KCL at node B: vA v1 vA v2 0 R1 R1 vB v1 dv C B 0 R dt Substituting

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From the op amp we have vA vB and from the KCL equation for node A, we have vA v1 v2 =2. the preceding values in the KCL at node B, we nd v2 RC dv2 dv v1 RC 1 dt dt

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Figure 1-28 A linear iron-core linear servomotor consists of a magnetic way and a mating coil assembly.

7.1 At t 0 , just before the switch is closed in Fig. 7-20, vC 100 V. charge transients. Obtain the current and

Fig. 7-20 With the polarities as indicated on the diagram, vR vC for t > 0, and 1=RC 62:5 s 1 . vC 0 vC 0 100 V. Thus, vR vC 100e 62:5t V i vR 0:25e 62:5t R A q CvC 4000e 62:5t mC Also,

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In Problem 7.1, obtain the power and energy in the resistor, and compare the latter with the initial energy stored in the capacitor.

In other words, all the stored energy in the capacitor is eventually delivered to the resistor, where it is converted into heat.

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