Lead Compensation for a DC Motor As an example of designing a lead compensator, let us repeat the design of compensation for the DC motor with the transfer function G(s) = 1 / s(s + 1) that was carried out in Section 5.4.1. This also represents the model of a satellite tracking antenna (see Fig

Question

Lead Compensation for a DC Motor
As an example of designing a lead compensator, let us repeat the design of compensation for the DC motor with the transfer function

[latex] G(s) = \frac{1}{s(s + 1)} [/latex]

that was carried out in Section 5.4.1. This also represents the model of a satellite tracking antenna (see Fig. 3.61). This time we wish to obtain a steady-state error of less than [latex] 0.1 [/latex] for a unit-ramp input. Furthermore, we desire an overshoot [latex] M_p < 25%. [/latex]
1. Determine the lead compensation satisfying the specifications.
2. Determine the digital version of the compensation with [latex] T_s = 0.05 ext{ sec }. [/latex]
3. Compare the step and ramp responses of both implementations.

Accepted Answer

1. The steady-state error is given by

[latex] e_{ss} = \underset{s ightarrow 0}{ \lim} s \left[\frac{1}{1 + KD(s)G(s)} ight] R(s), [/latex] (6.45)

where [latex] R(s) = 1/s^2 [/latex] for a unit ramp, so Eq. (6.45) reduces to

[latex] e_{ss} = \underset{s ightarrow 0}{ \lim} \left\{\frac{1}{s+ KD(s) [1 / (s+1)]} ight\} = \frac{1}{KD(0)} [/latex] .

Therefore, we find that [latex] KD(0), [/latex] the steady-state gain of the compensation, cannot be less than [latex] 10 (K_v ≥ 10) [/latex] if it is to meet the error criterion, so we pick [latex] K = 10.[/latex] To relate the overshoot requirement to [latex] PM, [/latex] Fig. 6.38 shows that a [latex] PM [/latex] of [latex] 45°[/latex] should suffice. The frequency response of [latex] KG(s) [/latex] in Fig. 6.55 shows that the [latex] PM = 20° [/latex] if no phase lead is added by compensation. If it were possible to simply add phase without affecting the magnitude, we would need an additional phase of only [latex] 25° [/latex] at the [latex] KG(s) [/latex] crossover frequency of [latex] ω = 3 ext{ rad/sec }. [/latex] However, maintaining the same low-frequency gain and adding a compensator zero would increase the crossover frequency; hence more than a [latex] 25° [/latex] phase contribution will be required from the lead compensation. To be safe, we will design the lead compensator so that it supplies a maximum phase lead of [latex] 40°. [/latex] Fig. 6.54 shows that [latex] 1/α = 5 [/latex] will accomplish that goal. We will derive the greatest benefit from the compensation if the maximum phase lead from the compensator occurs at the crossover frequency. With some trial and error, we determine that placing the zero at [latex] ω = 2 ext{ rad/sec } [/latex] and the pole at [latex] ω = 10 ext{ rad/sec } [/latex] causes the maximum phase lead to be at the crossover frequency. The compensation, therefore, is

[latex] KD(s) = 10 \frac{s/2 + 1}{s/10 + 1}. [/latex]

The frequency-response characteristics of [latex] L(s) = KD(s)G(s) [/latex] in Fig. 6.55 can be seen to yield a [latex] PM [/latex] of [latex] 53°, [/latex] which satisfies the design goals.

The root locus for this design, originally given as Fig. 5.24, is repeated here as Fig. 6.56, with the root locations marked for [latex] K = 10. [/latex] The locus is not needed for the frequency-response design procedure; it is presented here only for comparison with the root locus design method presented in Chapter 5. The entire process can be expedited by the use of MATLAB’s SISOTOOL routine, which simultaneously provides the root locus and the Bode plot through an interactive GUI interface. For this example, the MATLAB statements

Question 6.EX.15: Lead Compensation for a DC Motor As an example of designing ...

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