Response of an Armature-Controlled dc Motor

The parameter values for a certain motor are
$K_{T} = K_{b} = 0.05 N · m/A$

$c = 10^{−4} N · m ·s/rad R_{a} = 0.5 Ω$

$L_{a} = 2 × 10^{−3} H I = 9 × 10^{−5} kg · m²$
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 $v_{a} = 10 V$

Question

Response of an Armature-Controlled dc Motor

The parameter values for a certain motor are

[latex]K_{T} = K_{b} = 0.05 N · m/A [/latex]

[latex]c = 10^{−4} N · m ·s/rad R_{a} = 0.5 Ω[/latex]

[latex]L_{a} = 2 × 10^{−3} H I = 9 × 10^{−5} kg · m² [/latex]

where I includes the inertia of the armature and that of the load. The load torque [latex]T_{L}[/latex] is zero.

Obtain the step response of [latex]i_{a}(t)[/latex] and ω(t) if the applied voltage is [latex]v_{a} = 10 V[/latex]

Accepted Answer

Substituting the given parameter values into (6.5.1) and (6.5.3), gives

[latex]\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}} [/latex] (6.5.1)

[latex]\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}} [/latex] (6.5.3)

[latex]\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}} [/latex]

[latex]\frac{\Omega(s)}{V_{a}(s)} = \frac{ 0.05}{18 × 10^{−8}s^{2} + 4.52 × 10^{−5} s + 2.55 × 10^{−3}} [/latex]
If [latex]v_{a}[/latex] is a step function of magnitude 10 V,
[latex]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} [/latex]

[latex]\Omega(s) = \frac{2.777 × 10^{6}}{s(s + 165.52)(s + 85.59)} = \frac{D_{1}}{s} + \frac{D_{2}}{s + 165.52} + \frac{D_{3}}{s + 85.59} [/latex]

Evaluating the partial-fraction coefficients by hand or with MATLAB, as described in Chapter 3, we obtain
[latex]i_{a}(t) = 0.39 − 61 e^{−165.52t} + 61.74 e^{−85.59t} [/latex]
[latex]ω(t) = 196.1 + 210^{−165.52t} − 406 e^{−85.59t} [/latex]
The plots are shown in Figure 6.5.1. Note the large overshoot in [latex]i_{a}[/latex], which is caused by the numerator dynamics. The plot shows that the steady-state calculation of [latex]i_{a} = 0.39 A[/latex] 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.

Question 6.5.2: Response of an Armature-Controlled dc Motor The parameter va...

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