AUTO_L7_2012 Control Lecture Note

7/27/2019 AUTO_L7_2012 Control Lecture Note

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AUTO-1029 Lecture 6 Slide 1

AUTO-1029: AUTOMOTIVESYSTEMS & CONTROL

Prof Pavel M. TRIVAILO, PhD(Professor of Aerospace Engineering)

E-mail: [email protected]

SAMME, RMIT University, Melbourne, Australia

Lecturer: Prof P.M.Trivailo c2012 SAMME, RMIT


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Lecture 7:

INTRODUCTION INTO OPTIMAL CONTROL

LINEAR QUADRATIC REGULATOR (LQR) LQR Control of 2-DOF Systems using SIMULINK

(diagram method)



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Lecture 7:FEEDBACK? PROBLEMS?

ANNOUNCEMENT: A-2

Assignment is due on Friday, 21-Sept-2012, 18:00

Groups (with up to 4 Members)

Registration of your Groups is due today!



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LINEAR QUADRATIC

REGULATOR(LQR)



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HINTS: LQR

Let us inspect a cost function J for a linearised n-

DOF system with state vector x = [q q]T

COST: J=

0

(xTQx + uTRu + 2xTNu) dt

Firstly, let us determine the dimensions of the ma-

trices in the cost function matrix expression:

for n-DOF: x2n1; un1; Q2n2n; Rnn; N2nnLecturer: Prof P.M.Trivailo c2012 SAMME, RMIT


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These matrices will be used in the lqr(SYS,Q,R,N)or lqr(A,B,Q,R,...) MATLAB commands.

Note, that in these commands, there are peculiar

requirements to the matrices, for example:the [R] matrix must be symmetric positive definite

with as many columns as [B]. Also, [Q N;N R] must

be positive definite (you can use EIG to check pos-

itivity).



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Let us then distinquish a special case (we will call it

Case-V), for which

[Q] =

[k]nn [0]nn[0]nn [0]nn

; [R] [0]nn; [N] = [0]2nn;

Please observe that for this case the cost is approxi-

mately equal to the doubled POTENTIAL ENERGY

of the system V:

J 2qT[k]q = 2V



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Therefore, Case-V will correspond to the control

strategy, where the main consideration is given to

the minimisation of the potential energy of the

system.

Similarly, let us distinquish a case (we will call it

Case-T), for which

[Q] =

[0]nn [0]nn

[0]nn [m]nn

; [R] [0]nn; [N] = [0]2nn



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Prove that for this case the cost is approximately

equal to the doubled KINETIC ENERGY of the

system T:

J 2qT[m]q = 2T

Therefore, Case-T will correspond to the control

strategy, where the main consideration is given to

the minimisation of the kinetic energy of the sys-

tem.



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Combining two previous cases, we can introduce

case (we will call it Case-E), for which

[Q] =

[k]nn [0]nn

[0]nn [m]nn

; [R] [0]nn; [N] = [0]2nn

Please observe that for this case the cost is approx-

imately equal to the doubled TOTAL ENERGY

of the system E = V+ T:

J 2 qT[k]q + 2 qT[m]q = 2V+ 2T = 2 E



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Therefore, the last Case-E will correspond to the

control strategy, where the main consideration is

given to the minimisation of the total energy of

the system.

We can also introduce some other control strtate-

gies, including minimisation of the control efforts

(Case-u) : [Q] [0]2n2n; [R] = [diag(1)]nn; [N] =[0]2nn OR a mixture of the abovementioned Cases.



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All of these control strategies can be easily imple-mented in MATLAB and/or SIMULINK. You may

employ the following steps:

Enter the system matrices [m], [k], [c].

Calculate the state-space matrices [A], [B], [C]

and [D].



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Describe (represent) the system in MATLAB, us-

ing the ss MATLAB comand:

>> SYS = ss(A,B,C,D)

Based on the requirements to your system, in-troduce matrices [Q], [R] and [N], as discussed

before on Slides 5-11.

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Calculate the optimal control gain matrix K using

the MATLAB command

>> [K,S,e] = lqr(SYS,Q,R,N)

In the Project, we do not need to use calculated

matrices [S] a nd [e]. However, the matrix [K]

will be very important. The values in this matrix,gains, will be used to calculate the optimal control

feedback as u = Kx.



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Create a SIMULINK closed-loop model for yoursystem

Simulate your system in SIMULINK

Plot results of the simulation in MATLAB



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LQR EXAMPLE: Test 2-DOF System

Parameters: m1 = 2; m3 = 3 kg

k1 = 600; k2 = 280 N/mADDED DAMPING: c1 = 3.8; c2 = 10 Ns/m.



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EXCITATION EXTERNAL FORCES

0 5 10 15

0

1

2

3

4

5

Q1

[N]

0 5 10 151

0.5

0

0.5

1

Q

2[N]

time [s]



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LQR: SIMULINK DIAGRAM



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TRICK: option for GAIN



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LQR: RESULTS for displacements q1 and q2

0 5 10 150

2

4

Q1,2

[N]

Q1

Q2

0 5 10 155

0

5

10

15x 10

3

q1

[m]

0 5 10 150.01

0

0.01

0.02

q2

[m]

time [s]



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LQR: RESULTS for velocities q1 and q2

0 5 10 150

2

4

Q1,2

[N]

Q1

Q2

0 5 10 150.1

0.05

0

0.05

0.1

q1

[m/s]

0 5 10 150.1

0.05

0

0.05

0.1

q2

[m/s]

time [s]



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COMARISON: NO CONTROL vs LQR

0 5 10 15

5

0

5

10

15x 10

3

q1

[m]

NO CONTROL

0 5 10 15

5

0

5

10

15x 10

3

LQR CONTROL

0 5 10 15

5

0

5

10

15

x 103

q2[m]

time [s]0 5 10 15

5

0

5

10

15

x 103

q2[m]

time [s]



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INSPECT MATLAB FILE

My detailed MATLAB/SIMULINK example file is

provided. In order to re-produce simulation you can

run one single file twoDOFprojectALL.m:

>> twoDOFprojectALL

Note that the SIMULINK model twoDOFprojectLQR.m

(shown on Slide 18) is run from inside MATLAB

using the following command:

>> sim(twoDOFprojectLQR);

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