Question 10.CS.1: Antenna Control: Stability Design and Transient Performance ...
Antenna Control: Stability Design and Transient Performance
Our ongoing antenna position control system serves now as an example that summarizes the major objectives of the chapter. The case study demonstrates the use of frequency response methods to find the range of gain for stability and to design a value of gain to meet a percent overshoot requirement for the closed-loop step response.
Given the antenna azimuth position control system shown in Appendix A2, Configuration 1, use frequency response techniques to find the following:
a. The range of preamplifier gain, K, required for stability
b. Percent overshoot if the preamplifier gain is set to 30
c. The estimated settling time
d. The estimated peak time
e. The estimated rise time
Using the block diagram (Configuration 1) shown in Appendix A2 and performing block diagram reduction yields the loop gain, G(s)H(s), as
G (s) H (s) = \frac{6.63K}{s (s + 1.71) (s + 100) } = \frac{0.0388K}{s (\frac{s}{1.71} + 1 ) (\frac{s}{100} + 1 )} (10.91)
Letting K = 1, we have the magnitude and phase frequency response plots shown in Figure 10.61.
a. In order to find the range of K for stability, we notice from Figure 10.61 that the phase response is −180° at ω = 13.1 rad/s. At this frequency, the magnitude plot is −68.41 dB. The gain, K, can be raised by 68.41 dB. Thus, K = 2633 will cause the system to be marginally stable. Hence, the system is stable if 0 < K < 2633.
b. To find the percent overshoot if K = 30, we first make a second-order approximation and assume that the second-order transient response equations relating percent overshoot, damping ratio, and phase margin are true for this system. In other words, we assume that Eq. (10.73), which relates damping ratio to phase margin, is valid. If K = 30, the magnitude curve of Figure 10.61 is moved up by 20 log 30 = 29.54 dB. Therefore, the adjusted magnitude curve goes through zero dB at ω = 1. At this frequency, the phase angle is −120.9°, yielding a phase margin of 59.1°. Using Eq. (10.73) or Figure 10.48, ζ = 0.6, or 9.48% overshoot. A computer simulation shows 10%.
\Phi _{M} = 90 − tan^{-1} \frac{\sqrt{−2ζ² + \sqrt{1 + 4 ζ^{4} } } }{2ζ}
= tan^{-1} \frac{ 2ζ}{\sqrt{−2ζ² + \sqrt{1 + 4 ζ^{4} } }} (10.73)
c. To estimate the settling time, we make a second-order approximation and use Eq. (10.55). Since K = 30 (29.54 dB), the open-loop magnitude response is −7 dB when the normalized magnitude response of Figure 10.61 is −36.54 dB. Thus, the estimated bandwidth is 1.8 rad/s. Using Eq. (10.55), T_{s} = 4.25 seconds. A computer simulation shows a settling time of about 4.4 seconds.
ω_{BW} = \frac{4}{T_{s} ζ} \sqrt{(1 – 2 ζ²) + \sqrt{4 ζ^{4} – 4 ζ² + 2} } (10.55)
d. Using the estimated bandwidth found in Part c along with Eq. (10.56) and the damping ratio found in a, we estimate the peak time to be 2.5 seconds. A computer simulation shows a peak time of 2.8 seconds.
ω_{BW} = \frac{π}{T_{p} \sqrt{1 – ζ²}} \sqrt{(1 – 2 ζ²) + \sqrt{4 ζ^{4} – 4 ζ² + 2} } (10.56)
e. To estimate the rise time, we use Figure 4.16 and find that the normalized rise time for a damping ratio of 0.6 is 1.854. Using Eq. (10.54), the estimated bandwidth found in c, and ζ = 0.6, we find ω_{n} = 1.57. Using the normalized rise time and ω_{n} , we find T_{r} = 1.854/1.57 = 1.18 seconds. A simulation shows a rise time of 1.2 seconds.
ω_{BW} = ω_{n} \sqrt{(1 – 2 ζ²) + \sqrt{4 ζ^{4} – 4 ζ² + 2} } (10.54)
CHALLENGE:
You are now given a problem to test your knowledge of this chapter’s objectives. You are given the antenna azimuth position control system shown in Appendix A2, Configuration 3. Record the block diagram parameters in the table shown in Appendix A2 for Configuration 3 for use in subsequent case study challenge problems. Using frequency response methods, do the following:
a. Find the range of gain for stability.
b. Find the percent overshoot for a step input if the gain, K, equals 3.
c. Repeat Parts a and b using MATLAB.
