Control Systems - Chibum Lee...Chibum Lee -Seoultech Advanced Control Theory Ex. 6 2 Observation...
Transcript of Control Systems - Chibum Lee...Chibum Lee -Seoultech Advanced Control Theory Ex. 6 2 Observation...
Advanced Control TheoryChibum Lee -Seoultech
Outline
Design of State feedback control
• Dominant pole design
• Symmetric root locus (linear quadratic regulation)
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Advanced Control TheoryChibum Lee -Seoultech
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
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Advanced Control TheoryChibum Lee -Seoultech
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
𝑢
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
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Advanced Control TheoryChibum Lee -Seoultech
Ex.
The solutions are
𝐾1 =𝑧0 14𝜁𝜔𝑛 − 37𝜔𝑛
2 + 12(2𝜁𝜔𝑛 − 7)
(𝑧0 + 3)(𝑧0 + 4)
𝐾2=𝑧0 7−2𝜁𝜔𝑛 +12−𝜔𝑛
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(𝑧0+3)(𝑧0+4)
(b) MATLAB
Ao=[-7 1; -12 0];
zo=2;
Bo=[1;-zo];
pc=roots([1 2 4])
K=place(Ao,Bo,pc)
If zo=-2.99 (closed to open-loop pole) then
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K=[-3.80 0.60]
K=[2052.5 -688.1]
Open-loop Poles @ -3, -4
Advanced Control TheoryChibum Lee -Seoultech
Ex.
6
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.
Advanced Control TheoryChibum Lee -Seoultech
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
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Advanced Control TheoryChibum Lee -Seoultech
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
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𝐺(𝑠)=𝜔𝑛2
𝑠2+2𝜁𝜔𝑛𝑠+𝜔𝑛2
Advanced Control TheoryChibum Lee -Seoultech
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.
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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 =
00001
𝐶 = 0.5 0 0.5 0 0 ,𝐷 = 0
Advanced Control TheoryChibum Lee -Seoultech
Ex. Tape Drive – Dominant 2nd order poles
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Solution)
2 dominant poles (complex conjugate poles):
Damping ratio 𝜁 = 0.7 overshoot 5%
Natural frequency 𝜔𝑛 = 1/1.5 rise time 4sec
)(11
tan1
2
1
dd
rt
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eM p
Advanced Control TheoryChibum Lee -Seoultech
Ex. Tape Drive – Dominant 2nd order poles
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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].
Advanced Control TheoryChibum Lee -Seoultech
Ex. Tape Drive – Dominant 2nd order poles
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Advanced Control TheoryChibum Lee -Seoultech
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
With
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
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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
Advanced Control TheoryChibum Lee -Seoultech
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!!
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𝛼𝑐 𝑠 𝛼𝑐 −𝑠 = 𝐷 𝑠 𝐷 −𝑠 + 𝜌𝑁 𝑠 𝑁(−𝑠)=0
1 + 𝜌𝑁(𝑠)𝑁(−𝑠)
𝐷(𝑠)𝐷(−𝑠)= 0
Advanced Control TheoryChibum Lee -Seoultech
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
1
(−𝑠 + 𝑎)(𝑠 + 𝑎)= 0
SRL
𝑠 = − 𝑎2 + 𝜌
The closed loop poles 𝑠 = − 𝑎2 + 𝜌 minimizes 𝐽 = 0∞𝜌𝑦2 𝑡 + 𝑢2 𝑡 𝑑𝑡
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Advanced Control TheoryChibum Lee -Seoultech
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
𝑠2
SRL equation
1 + 𝜌𝑁(𝑠)𝑁(−𝑠)
𝐷(𝑠)𝐷(−𝑠)= 1 + 𝜌
1
𝑠2 −𝑠 2 = 1 + 𝜌1
𝑠4= 0
Note that the (stable) closed-loop poles
has damping of 𝜁 = 0.707
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Advanced Control TheoryChibum Lee -Seoultech
Ex. Satellite system
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Nyquist plot shows excellent stability PM=65°, GM= ∞
General feature of LQR design but may not possible if not full
statefeedback!
Advanced Control TheoryChibum Lee -Seoultech
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
0
)( dtRuuQxxJ TT
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Advanced Control TheoryChibum Lee -Seoultech
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
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Advanced Control TheoryChibum Lee -Seoultech
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.
Sol)
(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
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Advanced Control TheoryChibum Lee -Seoultech
Ex. Tape Drive system – LQR design
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