Question 13.CS.1: Antenna Control: Transient Design via Gain We now demonstrat...
Antenna Control: Transient Design via Gain
We now demonstrate the objectives of this chapter by turning to our ongoing antenna azimuth position control system. We will show where the computer is inserted in the loop, model the system, and design the gain to meet a transient response requirement. Later, we will design a digital cascade compensator.
The computer will perform two functions in the loop. First, the computer will be used as the input device. It will receive digital signals from the keyboard in the form of commands and digital signals from the output for closed-loop control. The keyboard will replace the input potentiometer, and an analog-to-digital (A/D) converter along with a unity gain feedback transducer will replace the output potentiometer.
Figure 13.30(a) shows the original analog system, and Figure 13.30(b) shows the system with the computer in the loop. Here the computer is receiving digital signals from two sources: (1) the input via the keyboard or other tracking commands and (2) the output via an A/D converter. The plant is receiving signals from the digital computer via a digital-to-analog (D/A) converter and the sample-and-hold.
Figure 13.30(b) shows some simplifying assumptions we have made. The power amplifier’s pole is assumed to be far enough away from the motor’s pole that we can represent the power amplifier as a pure gain equal to its dc gain of unity. Also, we have absorbed any preamplifier and potentiometer gain in the computer and its associated D/A converter.
Design the gain for the antenna azimuth position control system shown in Figure 13.30(b) to yield a closed-loop damping ratio of 0.5. Assume a sampling interval of T = 0.1 second.
Modeling the System: Our first objective is to model the system in the z-domain. The forward transfer function, G(s), which includes the sample-and-hold, power amplifier, motor and load, and the gears, is
G (s) = \frac{1 − e^{−Ts}}{s} \frac{0.2083}{s (s + a)} = \frac{0.2083}{a} (1 − e^{Ts}) \frac{a}{s² (s + a)} (13.104)
where a = 1.71, and T = 0.1.
Since the z-transform of (1 − e^{−Ts}) is (1 – z^{−1}) and, from Example 13.6, the z-transform of a/[s² (s + a)] is
z \left\{\frac{a}{s² (s + a)} \right\} = \left[\frac{Tz}{(z − 1)²} – \frac{(1 − e^{−aT} ) z}{a (z − 1) (z − e^{−aT} )} \right] (13.105)
the z-transform of the plant, G(z), is
G (z) =\frac{0.2083}{a} (1 – z^{-1}) z \left\{\frac{a}{s² (s + a)} \right\} (13.106)
=\frac{0.2083}{a²} \left[\frac{[aT – (1 – e^{−aT})] z + [(1 – e^{−aT}) – aT e^{−aT}]}{(z − 1) (z − e^{−aT})} \right]
Substituting the values for a and T, we obtain
G (z) =\frac{9.846 × 10^{-4} (z + 0.945)}{(z − 1) (z − 0.843)} (13.107)
Figure 13.31 shows the computer and plant as part of the digital feedback control system.
Designing for Transient Response: Now that the modeling in the z-domain is complete, we can begin to design the system for the required transient response. We superimpose the root locus over the constant damping ratio curves in the z-plane, as shown in Figure 13.32. A line drawn from the origin to the intersection forms an 8.58° angle. Searching along this line for 180°, we find the intersection to be (0.915 + j 0.138), with a loop gain, 9.846 × 10^{-4}K, of 0.0135. Hence, K = 13.71.
Checking the design by finding the unit sampled step response of the closed-loop system yields the plot of Figure 13.33, which exhibits 20% overshoot (ζ = 0.456).
CHALLENGE:
We now give you a case study to test your knowledge of this chapter’s objectives: You are given the antenna azimuth position control system shown in Appendix A2, Configuration 2. Do the following:
a. Convert the system into a digital system with T = 0.1 second. For the purposes of the conversion, assume that the potentiometers are replaced with unity gain transducers. Neglect power amplifier dynamics.
b. Design the gain, K, for 16.3% overshoot.
c. For your designed value of gain, find the steady-state error for a unit ramp input.
d.Repeat Part b using MATLAB.
