Our ongoing antenna azimuth position control system serves now as an example to summarize the major objectives of the chapter. The following cases demonstrate the use of frequency response methods to (1) design a value of gain to meet a percent overshoot requirement for the closed-loop step response and (2) design cascade compensation to
meet both transient and steady-state error requirements.

Question

Antenna Control: Gain Design
PROBLEM: Given the antenna azimuth position control system shown on the front endpapers, Configuration 1, use frequency response techniques to do the following:
a. Find the preamplifier gain required for a closed-loop response of 20% overshoot for a step input.
b. Estimate the settling time.

Accepted Answer

The block diagram for the control system is shown on the inside front cover (Configuration 1). The loop gain, after block diagram reduction, is

[latex] G(s)=\frac{6.63K}{s(s+1.71)(s+100)}=\frac{0.0388K}{s\left(\frac{s}{1.71}+1 ight) \left(\frac{s}{100} +1 ight)} [/latex]

Letting K = 1, the magnitude and phase frequency response plots are shown in Figure 10.61.

a. To find K to yield a 20% overshoot, 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. Thus, a 20% overshoot implies a damping ratio of 0.456. Using Eq. (10.73),

[latex] \Phi _{M}=90- an^{-1}\frac{\sqrt{-2\zeta ^{2}+\sqrt{1+4\zeta ^{4}}} }{2\zeta } [/latex]

[latex]= an^{-1} \frac{2\zeta }{\sqrt{-2\zeta ^{2}+\sqrt{1+4\zeta ^{4}}}}[/latex]

this damping ratio implies a phase margin of 48.1°. The phase angle should therefore be (-180° + 48.1°) = -131.9°. The phase angle is -131.9° at ω = 1.49 rad/s, where the gain is -34.1 dB. Thus K = 34.1 dB = 50.7
for a 20% overshoot. Since the system is third-order, the second-order approximation should be checked.A computer simulation shows a 20% overshoot for the step response.

b. Adjusting the magnitude plot of Figure 10.61 for K = 50.7, we find -7 dB at ω = 2.5 rad/s, which yields a closed-loop bandwidth of 2.5 rad/s. Using Eq. (10.55)

[latex]\omega _{BW}=\frac{4}{T_{s}\zeta } \sqrt{(1-2\zeta ^{2} )+\sqrt{4\zeta ^{4}-4\zeta ^{2}+2 } }[/latex]

with ζ = 0.456 and [latex]\omega _{BW}[/latex] = 2.5, we find [latex] T_{s}[/latex] = 4.63 seconds. A computer simulation shows a settling time of approximately 5 seconds.

CHALLENGE: We now give you a problem to test your knowledge of this chapter’s objectives. You are given the antenna azimuth position control system shown on the inside front cover (Configuration 3). Using frequency response methods, do the following:
a. Find the value of K to yield 25% overshoot for a step input.
b. Repeat Part a using MATLAB.

Question 11.CS.1: Our ongoing antenna azimuth position control system serves n...

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