Design of a PD Control System For the system shown in Figure 10.7.1, we are given that I = 10 and c = 2. The dominant time constant τ is specified to be 2 s, and the damping ratio is specified to be ζ = 1. a. Compute the required values for KP and K D. Evaluate the steady-state command error and

Question

Design of a PD Control System

For the system shown in Figure 10.7.1, we are given that I = 10 and c = 2. The dominant time constant [latex] au[/latex] is specified to be 2 s, and the damping ratio is specified to be ζ = 1.

a. Compute the required values for [latex]K_{P}[/latex] and [latex]K_{D}[/latex]. Evaluate the steady-state command error and the steady-state disturbance error given that both the command input [latex]\Theta_{r}(t) [/latex] and the disturbance [latex]T_{d} (t)[/latex] are unit-step functions.
b. Evaluate the steady-state command error for each case given that the command input [latex]\Theta_{r}(t)[/latex] is a unit-ramp function and the disturbance [latex]T_{d} (t)[/latex] is zero.
c. Evaluate the frequency response characteristics of the disturbance transfer function.
d. Discuss the actuator output as a function of time when the command input [latex]\Theta_{r}(t)[/latex] is a unit-step function and the disturbance [latex]T_{d} (t)[/latex] is zero.

Accepted Answer

a. From the figure we obtain the following output, error, and actuator equations.
[latex]\Theta(s) = \frac{K_{P} + K_{D} s}{10 s^{2} + (2 + K_{D})s + K_{P}} \Theta_{r}(s) − \frac{1}{10s^{2} + (2 + K_{D})s + K_{P}} T_{d} (s) [/latex] (1)
[latex]E(s) = \frac{10s^{2} + 2s}{10 s^{2} + (2 + K_{D})s + K_{P}} \Theta_{r}(s) + \frac{1}{10 s^{2} + (2 + K_{D})s + K_{P}} T_{d} (s) [/latex] (2)
[latex]T (s) = \frac{(10s^{2} + 2s)(K_{P} + K_{D}s)}{10 s^{2} + (2 + K_{D})s + K_{P}} \Theta_{r}(s) + \frac{K_{D} s + K_{P}}{10 s^{2} + (2 + K_{D})s + K_{P}} T_{d} (s) [/latex] (3)
The characteristic equation is
[latex]10 s^{2} + (2 + K_{D})s + K_{P} = 0 [/latex]
and the damping ratio is
[latex]ζ = \frac{2 + K_{D}}{2 \sqrt{10K_{P}}} = 1 [/latex] (4)

Since ζ = 1, the expression for the time constant is
[latex] au = \frac{20}{2 + K_{D}} = 2 sec [/latex]
which gives [latex]K_{D} = 8[/latex]. Using this value in equation (4) gives
[latex]\frac{2 + 8}{2 \sqrt{10K_{P}}} = 1 [/latex]
which gives [latex]K_{P} = 2.5 [/latex].
Applying the final value theorem to the error equation (2) with [latex]\Theta_{r}(s) = 1/s [/latex] and [latex]T_{d} (s) = 0[/latex] gives
[latex]e_{ss} = \lim_{s→0} s E(s) = \lim_{s→0} s \frac{10s^{2} + 2s}{10s^{2} + (2 + K_{D})s + K_{P}} \frac{1}{s} = 0 [/latex]
Thus the system has zero command error for a step command.
Applying the final value theorem to the error equation (2) with [latex]\Theta_{r}(s) = 0[/latex] and [latex]T_{d} (s) = 1/s [/latex] gives
[latex]e_{ss} = \lim_{s→0} s E(s) = \lim_{s→0} s \frac{1}{10s^{2} + (2 + K_{D})s + K_{P}} \frac{1}{s} = \frac{1}{K_{P}} = \frac{1}{2.5} = 0.4 rad [/latex]
Thus the system has a nonzero disturbance error for a step disturbance.
b. Applying the final value theorem to the error equation (2) with [latex]\Theta_{r}(s) = 1/s^{2}[/latex] and [latex] T_{d} (s) = 0[/latex] gives
[latex]e_{ss} = \lim_{s→0} s E(s) = \lim_{s→0} s \frac{s 10s^{2} + 2s}{10s^{2} + (2 + K_{D})s + K_{P}} \frac{1}{s^{2}} = \frac{2}{K_{P}} = 0.8 rad [/latex]
Thus the ramp command error is nonzero.
c. The disturbance transfer function is
[latex]\frac{\Theta_{r}(s)}{T_{d} (s)} = − \frac{1}{10s^{2} + (2 + K_{D})s + K_{P}} = − \frac{1}{10s^{2} + 10s + 2.5} [/latex]
Its bandwidth is the frequency range 0 ≤ ω ≤ 0.32 rad/s. Its low-frequency gain is 1/2.5 = 0.4 or 20 log 0.4 = −7.86 dB. Thus the system will not respond very much to disturbances that have a frequency content higher than 0.32 rad/s.
d. Substituting the values of [latex]K_{P}[/latex] and [latex]K_{D} [/latex] into the actuator equation (3) and using [latex]\Theta_{r}(s) = 1/s[/latex] and [latex]T_{d} (s) = 0[/latex], we obtain
[latex]T (s) = \frac{(10s^{2} + 2s)(8s + 2.5)}{10s^{2} + 10s + 2.5} \frac{1}{s} = \frac{80s^{2} + 41s + 5}{10s^{2} + 10s + 2.5} [/latex] (5)
Because the orders of the numerator and denominator are equal, this can be expressed as
[latex]T (s) = C_{1} + \frac{C_{2}s + C_{3}}{10s^{2} + 10s + 2.5} [/latex]
which may be arranged as a single fraction as follows.
[latex]T (s) = \frac{10C_{1}s^{2} + (10C_{1} + C_{2})s + 2.5C_{1} + C_{3}}{10s^{2} + 10s + 2.5} [/latex] (6)
Comparing the numerators of equations (5) and (6), we find that [latex]C_{1} = 8, C_{2} = −39[/latex], and [latex]C_{3} = −15[/latex], so that
[latex]T (s) = 8 − \frac{39s + 15}{10s^{2} + 10s + 2.5} = 8 − \frac{3.9s + 1.5}{(s + 0.5)^{2}} [/latex]

This could also have been obtained with synthetic division. Using the inverse transform we obtain
[latex]T (t) = 8δ(t) − 3.9e^{−0.5t} + 0.45te^{−0.5t }[/latex]
where δ(t) is the unit impulse function. Therefore, the actuator output predicted by the model will contain an impulse of strength 8 at t = 0. Of course, this is impossible physically, and so we must view these results with skepticism. The impulse is caused by the derivative term [latex]K_{D^{s}}[/latex] in the control algorithm.

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