Question: A speed control system using an armature-controlled motor with proportional control action was discussed in Section 10.3. Its block diagram is shown in Figure 10.3.8

A speed control system using an armature-controlled motor with proportional control action was discussed in Section 10.3. Its block diagram is shown in Figure 10.3.8 with a simplified version given in Figure 10.3.9. The given parameter values for a certain motor, load, and tachometer are
KT = Kb = 0.04 N.m/A
cm = 0
cL = 10-3 N-m s/rad
Ra = 0.6Ω
La = 2 x 10-3 H
lm = 2 x 10-5
lt = 10-5
lL. = 4 x 10-3 kg-m2:
N = 1.5
Ka = 5 V/V
Ktach = 10 V/(rad/s)
Kpot = 5 V/rad
Kd = 2 rad/(rad/s)
Where the subscript m refers to the motor, L refers to the load, and t refers to the tachometer,
(a) Determine the value of the proportional gain KP required for the load speed to be within 10% of the desired speed of 1000 rpm at steady-state, and plot the resulting transient response,
(b) For this value of Kp plot the resulting deviation of the load speed caused by a load torque TL = 1N.m.

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a First we reflect the load and tachometer inertias back to the motor shaft to obtain the equivalent inertia I e I e I m 1N 2 I L I t 2 10 5 1152 4 10 ... View full answer

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