EE360 Control systems: Optional Compensation assignment for low grade students

Instructions:

1. Submit this assignment before classes end for spring 2014 semester.2. Quiz will be conducted for about 30 minutes in your last class based on this

assignment.

Your Score:

3 points for submission before Final Exam [Submit a good report]3 points for QuizBased on your performance in this assignment/quiz, your score will be compensated/adjusted inthe final exam.

Question-1: Performance comparison of Lead and Lag compensator

Consider the following unity feedback system

Given specifications: (Each part carries 0.3 points)A. Design a lead compensator C(s) using Bode Plot technique so that the closed-loop system

satisfies the following performance specifications:i. The steady-state error to a unit-ramp reference input is less than 0.01.

ii. The damping ratio of the dominant closed-loop poles is no less than 0.4 or the OS% nomore than 25% (Transient performance specification).

B. Obtain the unit step response of the resulting closed-loop system. Does the closed-loop systemsatisfy the above OS% requirement? Use MATLAB to show this.

C. Show Bode plots of uncompensated and Compensated System.D. Show on Bode Plot the values of Phase Margin and Gain Margin.E. Implement the Lead compensator using OP-AMP circuit and find the values of resistors and

capacitors.F. Now implement the Lag controller using Root Locus technique for the same specifications given

in part-A, and plot the resulting Bode diagram of compensated system.G. Obtain the unit step response of the resulting closed-loop system with Lag controller in the loop.

Does the closed-loop system satisfy the above OS% requirement? Use MATLAB.H. Plot step responses from Part-B and Part-G on a single figure and conclude which controller is

better in terms of performance.I. Implement the Lag compensator using OP-AMP circuit and find the values of resistors and

capacitors. PlotJ. Cascade lead+lag compensator with the given system G(s) and plot the resulting closed loop step

response to see if the specifications are met.

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Lead and PD Controllers (II)

Lead compensation: D(s)=K(Ts+1)/(Ts+1), <1 Lead compensator is a high-pass filter (app.PD control) It is used whenever substantial improvement in damping

is required The maximum phase contribution is

max

max

max

sin1sin1

1

TExample:

sysD=tf([10 1],[1 1])bode(sysD)

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Example: PD Controller for Antenna Control (I) Design the low frequency gain K with respect to the steady-

state error specification: Tracking error to a ramp input of slope 0.01rad/sec to be less than 0.01rad

Ramp Input (R(s) = 1/s^2): (MM5)

Antenna system case..... K=1

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Example: PD Controller for Antenna Control (II) PD controller: D(s)=10s+1

sysd=tf(1,[10.0 1.0 0])*tf([10.0 1.0],1); syscl2=feedback(sysd,1); step(syscl2)

Check whether this controller is ok?

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Example: Intuitive Lead Design Design the low frequency gain K with respect to the steady-

state error specification Antenna system case..... K=1 Lead controller: D(s)=(10s+1)/(s+1)

sysd=tf(1,[10.0 1.0 0])*tf([10.0 1.0],[1 1]); syscl3=feedback(sysd,1); step(syscl3)

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Example: Comparison of PD and Lead Design bode(syscl2,syscl3); grid

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Goals for this lecture (MM10) An illustrative example

Frequency response analysis Frequency response design

Lead and lag compensators What’s a lead/lag compensator? Their frequency features

A systematical procedure for lead compensator design

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Lead Compensator Design Procedure (I)

Step 1: Design the low frequency gain K with respect to the staedy-state error specification

Antenna system case..... K=1

Step Input (R(s) = 1/s):

Ramp Input (R(s) = 1/s^2):

Parabolic Input (R(s) = 1/s^3):

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Lead Compensator Design Procedure (II)

Step 2: Determine the needed phase lead Original system PM:

sys=tf(1,[10 1 0]), margin(sys)... PM=18 at 0.308 Expected PM:

Expected overshoot limit (16%)

Dampling ratio ξ ≥ 0.5

Expected PM ≈ 100*0.5=50 degree Directly needed phase lead: 50-18=32 degree Expected phase lead: 32+ (7~10) degree

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Lead Compensator Design Procedure (III)

Lead compensation: D(s)=K(Ts+1)/(Ts+1), <1

Step 3: Determine coefficient α

Step 4: Determine coefficient T

2174.040sin140sin1

sin1sin1

max

max

622.492.0

2)2/(

11

max

n

T max

max

max

sin1sin1

1

T

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Step 5: Draw the compensated frequency response, check PMsysD=tf([4.622 1],[1.0137 1]); sysC=sys*sysD; margin(sysC);

step(feedback(sysC,1))

Lead Compensator Design Procedure (IV)

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Check the poles & zeros of the closed-loop...syscl=feedback(sysC,1); pzmap(syscl);

Compare with a standard 2nd-order system...Sys2=tf(0.3099, conv([1 0.384-0.403i], [1 0.384+0.403i])) ;

step(feedback(sysC,1),Sys2)

Why this Lead Compensator doesn’t work? (I)

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Lead Compensator Design Procedure (V)

Step 6: Iterate on the design until all specifications are metsysD=tf([9.7457 1],[1.0911 1]) ; sysC=sys*sysD; margin(sysC);

sysCL=feedback(sysC,1); step(sysCL)sysD=tf([9 1],[1 1]); sysCL=feedback(sys*sysD,1); step(sysCL)

112.053sin153sin1

sin1sin1

max

max

10911.117457.9)(

7457.992.0

3)5.1/(

11

max

sssD

Tn

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