Question 6.5.2: Response of an Armature-Controlled dc Motor The parameter va...
Response of an Armature-Controlled dc Motor
The parameter values for a certain motor are
K_{T} = K_{b} = 0.05 N · m/Ac = 10^{−4} N · m ·s/rad R_{a} = 0.5 Ω
L_{a} = 2 × 10^{−3} H I = 9 × 10^{−5} kg · m²
where I includes the inertia of the armature and that of the load. The load torque T_{L} is zero.
Obtain the step response of i_{a}(t) and ω(t) if the applied voltage is v_{a} = 10 V
Substituting the given parameter values into (6.5.1) and (6.5.3), gives
\frac{I_{a}(s)}{V_{a}(s)} = \frac{I s + c}{L_{a} I s^{2} + (R_{a} I + c L_{a})s + cR_{a} + K_{b} K_{T}} (6.5.1)
\frac{\Omega(s)}{V_{a}(s)} = \frac{K_{T}}{L_{a} I s^{2} + (R_{a} I + cL_{a})s + cR_{a} + K_{b} K_{T}} (6.5.3)
\frac{I_{a}(s)}{V_{a}(s)} = \frac{9 × 10^{−5} s + 10^{−4}}{18 × 10^{−8} s^{2} + 4.52 × 10^{−5} s + 2.55 × 10^{−3}}\frac{\Omega(s)}{V_{a}(s)} = \frac{ 0.05}{18 × 10^{−8}s^{2} + 4.52 × 10^{−5} s + 2.55 × 10^{−3}}
If v_{a} is a step function of magnitude 10 V,
I_{a}(s) = \frac{5 × 10^{3} s + 5.555 × 10^{4}}{s(s + 165.52)(s + 85.59)} = \frac{C_{1}}{s} + \frac{C_{2}}{s + 165.52} + \frac{C_{3}}{s + 85.59}
Evaluating the partial-fraction coefficients by hand or with MATLAB, as described in Chapter 3, we obtain
i_{a}(t) = 0.39 − 61 e^{−165.52t} + 61.74 e^{−85.59t}
ω(t) = 196.1 + 210^{−165.52t} − 406 e^{−85.59t}
The plots are shown in Figure 6.5.1. Note the large overshoot in i_{a}, which is caused by the numerator dynamics. The plot shows that the steady-state calculation of i_{a} = 0.39 A greatly underestimates the maximum required current, which is approximately 15 A.
In practice, of course, a pure step input is impossible, and thus the required current will not be as high as 15 A. The real input would take some time to reach 10 V. The response to such an input is more easily investigated by computer simulation, so we will return to this topic in Section 6.7.
