Control Systems - Chibum Lee...Chibum Lee -Seoultech Advanced Control Theory Ex. 6 2 Observation...

transcript

Control Systems

Design of State Feedback

Control

chibum@seoultech.ac.kr

Advanced Control TheoryChibum Lee -Seoultech

Outline

Design of State feedback control

• Dominant pole design

• Symmetric root locus (linear quadratic regulation)

Selection of closed-loop poles

Now, in principle, we can place the closed-loop poles (CLPs) ‘arbitrarily’

Is it practical to make a truck behave like a sports car?

The further we try to shift the open-loop poles (OLPs), the greater the control effort required

• expense?

• actuator saturation?

• component strength?

If an OLP is nearly cancelled by an open-loop zero (OLZ), that mode will be almost uncontrollable

• will require large control gain (and effort) to shift

Ex. How zero location can affect the control law

A specific thermal system is in OCF with a zero at 𝑠 = 𝑧0.

𝑥1 𝑥2=

−7 1−12 0

𝑥1𝑥2

+1−𝑧0

𝑦 = 1 0 𝑥

(a) Find a state feedback gain necessary for placing the poles of the

system at the roots of 𝑠2 + 2𝜁𝜔𝑛𝑠 + 𝜔𝑛2

(b) Repeat the computation with MATLAB with 𝑧0 = 2, 𝜁 = 0.5 and 𝜔𝑛 =

2rad/sec

• Sol) The closed-loop characteristic equation: det(𝑠𝐼 − 𝐴 − 𝐵𝐾2)=0

𝑠2 + 7 + 𝐾1 − 𝑧0𝐾2 𝑠 + 12 − 𝐾2 7𝑧0 + 12 − 𝐾1𝑧0=0

Equating the desired characteristic equation

7 + 𝐾1 − 𝑧0𝐾2=2𝜁𝜔𝑛

12 − 𝐾2 7𝑧0 + 12 − 𝐾1𝑧0 = 𝜔𝑛2

The solutions are

𝐾1 =𝑧0 14𝜁𝜔𝑛 − 37𝜔𝑛

2 + 12(2𝜁𝜔𝑛 − 7)

(𝑧0 + 3)(𝑧0 + 4)

𝐾2=𝑧0 7−2𝜁𝜔𝑛 +12−𝜔𝑛

(𝑧0+3)(𝑧0+4)

(b) MATLAB

Ao=[-7 1; -12 0];

Bo=[1;-zo];

pc=roots([1 2 4])

K=place(Ao,Bo,pc)

If zo=-2.99 (closed to open-loop pole) then

K=[-3.80 0.60]

K=[2052.5 -688.1]

Open-loop Poles @ -3, -4

2 Observation from the example

• 1st. The gains grow as the zero 𝑧0 approaches open loop poles

(either - 3 or -4), the values where this system loses

controllability. As controllability is almost lost, the control gains

become very large.

• 2nd. Both 𝐾1 and 𝐾2 grow as the desired closed-loop bandwidth

given by 𝜔𝑛 is increased. From this we conclude that

To move the poles a long way requires large gains

The system has to work harder and harder to achieve control as controllability slips away.

Basic approaches to design state feedback control

Dominant pole design• try to make CL system ‘like’ a simple 2nd-order system

Symmetric root locus LQR• select CLPs with an explicit trade-off between control effort and performance

Dominant pole design-selection of CL poles

Seek to make CL system behave like a simple second-

order system:

Then choose ‘dominant’ pair of CLPs based on well-

known relationships between 𝜁 and 𝜔𝑛 and performance

measures such as:

• rise time

• overshoot,

• settling time, etc.

Choose other CLPs to have well-damped responses

𝐺(𝑠)=𝜔𝑛2

𝑠2+2𝜁𝜔𝑛𝑠+𝜔𝑛2

Ex. Tape Drive – Dominant 2nd order poles

A simplified sketch of

a computer tape drive system

Design the tape servomotor by the dominant second-order poles

method to have no more than 5% overshoot and a rise time of no

more than 4 sec. Keep the peak tension as low as possible.

𝐴 =

0 2 0 0 0−0.1 −0.35 0.1 0.1 0.750 0 0 2 00.4 0.4 −0.4 −1.4 00 −0.03 0 0 −1

, 𝐵 =

𝐶 = 0.5 0 0.5 0 0 ,𝐷 = 0

Solution)

2 dominant poles (complex conjugate poles):

Damping ratio 𝜁 = 0.7 overshoot 5%

Natural frequency 𝜔𝑛 = 1/1.5 rise time 4sec

The other three poles:

Need to be placed far to the left of the dominant pair. “far“ means

the transients due to the fast poles should be over well before the

transients due to the dominant poles. (Assume a factor of 4 in the

respective undamped natural frequencies to be adequate.)

pc = [−0.707 + 0.707 𝑗; −0.707 − 0.707 𝑗; −4; − 4; − 4] / 1.5

Use function acker with A and B to find the control gains

𝐾 = [8.5123 20.3457 − 1.4911 − 7.8821 6.1927].

Symmetric root locus

For a SISO regulator, define a cost function 𝐽 which penalizes

• (generalized) output excursions from the set-point, 𝑦

• control effort, 𝑢

Plant transfer function:

It can be shown that:

• 𝐽 will be minimized by the control law 𝑢 = −𝐾𝑥

• the eigenvalues of the closed-loop system are the left half-plane roots of

the 2nd degree polynomial equation

𝐽 = 0∞𝜌𝑧2 𝑡 + 𝑢2 𝑡 𝑑𝑡

Linear quadratic regulator

𝒙 = 𝑨𝒙 + 𝑩𝒖𝒛 = 𝑪𝟏𝒙

𝑍(𝑠)

𝑈(𝑠)= 𝐶1 𝑠𝐼 − 𝐴 −1𝐵 =

𝑁(𝑠)

𝐷(𝑠)

𝛼𝑐 𝑠 𝛼𝑐 −𝑠 = 𝐷 𝑠 𝐷 −𝑠 + 𝜌𝑁 𝑠 𝑁(−𝑠)=0

Optimal pole placement for SISO systems

Hence we can see the effects of the output-weighting factor 𝜌 on the closed-loop poles by plotting a root locus for

If branches of the 180º SRL lie on the imaginary axis, plot the 0º locus instead

• Note that 𝑠 and −𝑠 affect in an identical manner. Any root 𝑠0 of the above Eq., there will also be a root at −𝑠0.

The SRL provides a basis for specifying CLPs in a SISO pole placement design

• Increasing 𝜌 places more emphasis on minimizing output excursions, at the expense of control effort and make us get a fast response.

• Decreasing 𝜌 places more emphasis on minimizing input power with a low control effort and make us get a slow response

• It’s a trade-off between performance and control effort!!

𝛼𝑐 𝑠 𝛼𝑐 −𝑠 = 𝐷 𝑠 𝐷 −𝑠 + 𝜌𝑁 𝑠 𝑁(−𝑠)=0

1 + 𝜌𝑁(𝑠)𝑁(−𝑠)

𝐷(𝑠)𝐷(−𝑠)= 0

Ex. Servo speed control

Plot the SRL for the servo control system with 𝑧 = 𝑦.

𝑦 = −𝑎𝑦 + 𝑢

Sol) The Transfer function 𝐺0 𝑠 =1

𝑠+𝑎

The SRL equation

1 + 𝜌𝑁(𝑠)𝑁(−𝑠)

𝐷(𝑠)𝐷(−𝑠)= 1 + 𝜌

(−𝑠 + 𝑎)(𝑠 + 𝑎)= 0

𝑠 = − 𝑎2 + 𝜌

The closed loop poles 𝑠 = − 𝑎2 + 𝜌 minimizes 𝐽 = 0∞𝜌𝑦2 𝑡 + 𝑢2 𝑡 𝑑𝑡

Ex. Satellite system

Plot the SRL for the satellite system with 𝑧 = 𝑦.

𝑥1 𝑥2=

0 10 0

𝑥1𝑥2

+01𝑢

𝑦 = 1 0 𝑥

Sol) The transfer function 𝐺0 𝑠 =1

SRL equation

1 + 𝜌𝑁(𝑠)𝑁(−𝑠)

𝐷(𝑠)𝐷(−𝑠)= 1 + 𝜌

𝑠2 −𝑠 2 = 1 + 𝜌1

𝑠4= 0

Note that the (stable) closed-loop poles

has damping of 𝜁 = 0.707

Ex. Satellite system

Nyquist plot shows excellent stability PM=65°, GM= ∞

General feature of LQR design but may not possible if not full

statefeedback!

LQR design

A general form for optimal design is linear quadratic regulator (LQR)

• If we take 𝑄 = 𝜌𝐶𝑇𝐶 and 𝑅 = 1, the general LQR design becomes SRL

Bryson’s rule: An appropriate choice to obtain acceptable values

𝑥 and 𝑢 is to choose diagonal matrices 𝑄 and 𝑅 such that

𝑄ii = 1/maximum acceptable value of [𝑥𝑖2]

𝑅𝑖𝑖 = 1/maximum acceptable value of [𝑢𝑖2]

Feedback control law that minimizes the value of the cost

𝑢 = −𝐾 𝑥 where 𝐾 = 𝑅−1𝐵𝑇𝑃

𝑃 from solving the continuous time algebraic Riccati equation(ARE)

𝐴𝑇𝑃 + 𝑃𝐴 − 𝑃𝐵𝑅−1 𝐵𝑇𝑃 + 𝑄 = 0

)( dtRuuQxxJ TT

LQR design

The direct solution for the optimal control gain can be get from

MATLAB function

K=lqr(A,B,Q,R)

More generalized form can be studied in optimal control class

Ex. Tape Drive system – LQR design

(a) Find the optimal control for the tape drive using the position 𝑥3 as

the output for the performance index. Let 𝜌 = 1. Compare the results

with that of dominant second order obtained before.

(b) Compare the LQR designs for 𝜌 = 0.1, 1, 10.

(a) The performance index matrix 𝑅 = 1

The state-cost matrix 𝑄 = 𝐶𝑇𝐶 =

0 2 0 0 0−0.1 −0.35 0.1 0.1 0.750 0 0 2 00.4 0.4 −0.4 −1.4 00 −0.03 0 0 −1

where 𝐶 = 0.5 0 0.5 0 0

The gain can be obtained by MATLAB K=lqr(A,B,Q,R)

𝐾 = 0.6526 2.1667 0.3474 0.5976 1.0616

Ex. Tape Drive system – LQR design

Control Systems - Chibum Lee...Chibum Lee -Seoultech Advanced Control Theory Ex. 6 2 Observation...

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