Antenna Control: Cascade Compensation Design

PROBLEM: Given the antenna azimuth position control system block diagram shown on the front endpapers, Configuration 1, use frequency response techniques and design cascade compensation for a closed-loop response of 20% overshoot for a step input, a fivefold improvement in steady-state error over the gain-compensated system operating at 20% overshoot, and a settling time of 3.5 seconds.

Question

Antenna Control: Cascade Compensation Design

PROBLEM: Given the antenna azimuth position control system block diagram shown on the front endpapers, Configuration 1, use frequency response techniques and design cascade compensation for a closed-loop response of 20% overshoot for a step input, a fivefold improvement in steady-state error over the gain-compensated system operating at 20% overshoot, and a settling time of 3.5 seconds.

Accepted Answer

Following the lag-lead design procedure, we first determine the value of gain, K, required to meet the steady-state error requirement.

1. Using Eq. (10.55)

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

with ζ = 0.456, and [latex]T_{s}=3.5[/latex] seconds, the required bandwidth is 3.3 rad/s.

2. From the preceding case study, the gain-compensated system’s open-loop transfer function was, for K = 50.7,

[latex] G(s)H(s)=\frac{6.63K}{s(s+1.71)(s+100)}=\frac{336.14}{s(s+1.71)(s+100)} [/latex]

This function yields [latex]K_{ u } =1.97[/latex]. If K = 254, then [latex]K_{ u } =9.85[/latex], a fivefold improvement.

3. The frequency response curves of Figure 10.61, which are plotted for K = 1, will be used for the solution.

4. Using a second-order approximation, a 20% overshoot requires a phase margin of 48.1°.

5. Select ω = 3 rad/s to be the new phase-margin frequency.

6. The phase angle at the selected phase-margin frequency is -152°. This is a phase margin of 28°. Allowing for a 5° contribution from the lag compensator, the lead compensator must contribute (48.1° - 28° + 5°) = 25.1°.

7. The design of the lag compensator now follows. Choose the lag compensator upper break one decade below the new phase-margin frequency, or 0.3 rad/s. Figure 11.8 says that we can obtain 25.1° phase shift from the lead if β = 0.4 or γ = 1/β = 2.5. Thus, the lower break for the lag is at 1/(γT) = 0.3/2.5 = 0.12 rad/s.

Hence,

[latex]G_{lag}(s)=0.4\frac{(s+0.3)}{(s+0.12)}[/latex]

8. Finally, design the lead compensator. Using Eq. (11.9),

[latex]\omega _{\max } =\frac{1}{T\sqrt{B} }[/latex]

we have

[latex]T=\frac{1}{\omega _{\max }\sqrt{\beta } } =\frac{1}{3\ sqrt {0.4} } =0.527[/latex]

Therefore, the lead compensator lower break frequency is 1/T = 1.9 rad/s, and the upper break frequency is 1/(βT) = 4.75 rad/s. Thus, the lag-lead-compensated forward path is

[latex]G_{lag-lead-comp}(s)=\frac{(6.63)(254)(s+0.3)(s+1.9)}{s(s+1.71)(s+100)(s+0.12)(s+4.75)} [/latex]

9. A plot of the open-loop frequency response for the lag-lead compensated system shows -7 dB at 5.3 rad/s. Thus, the bandwidth meets the design requirements for settling time. A simulation of the compensated system shows a 20% overshoot and a settling time of approximately 3.2 seconds, compared to a 20% overshoot for the
uncompensated system and a settling time of approximately 5 seconds. [latex]K_{ u }[/latex] for the compensated system is 9.85 compared to the uncompensated system value of 1.97.

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 onthe front endpapers (Configuration 3). Using frequency response methods, do the following:

a. Design a lag-lead compensator to yield a 15% overshoot and [latex] K_{ u } =20[/latex].In order to speed up the system, the compensated system’s phase-margin frequency will be set to 4.6 times the phase-margin frequency of the uncompensated system.
b. Repeat Part a using MATLAB.

Question 11.CS.2: Antenna Control: Cascade Compensation Design PROBLEM: Given ...

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