Dynamics and motion control
Lecture 5
Model following control
-servo design
Jan Wikander,Bengt Eriksson
KTH, Machine Design
Mechatronics Lab
e-mail: benke@md.kth.se
KTH
Date: 2009-08-27
File: Q:\md.kth.se\md\mmk\gru\mda\mf2007\arbete\Lectures\FinalVersions\For2009\L5.fm
MMK
Mechatronics Lab
Slide: 1(35)
5.1. Lecture outline
• 1. Introduction
•
2. Transfer function based model following
•
3. Time domain based model following
•
4. An example
•
5. Matlab example for the dc motor
KTH
Date: 2009-08-27
File: Q:\md.kth.se\md\mmk\gru\mda\mf2007\arbete\Lectures\FinalVersions\For2009\L5.fm
MMK
Mechatronics Lab
Slide: 2(35)
5.1.1. The servo problem
•
The servo problem, here with error feedback.
Feed forward
controller
Set point Trajectory
planner
rs
u ff
Model following
control
r
Feedback
controller
u
Controlled Output
system
•
Characterized by frequent changes in set point (e.g in an industrial
robot)
•
Control design mainly to follow (track) the reference (e.g. position or
velocity, or position, velocity, acceleration and jerk).
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Date: 2009-08-27
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Mechatronics Lab
Slide: 3(35)
5.1.2. Feedback control properties
•
The main principle in control engineering
•
Typically model based (but not required to be)
•
Produces control signals after an error has
occurred
•
Disturbance rejection is achieved
•
Effect of process parameter variations is reduced
•
Leads to a closed loop
•
Sensor noise may be amplified and deteriorate performance
•
May lead to instability if designed incorrectly
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Date: 2009-08-27
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Slide: 4(35)
5.1.3. Feed forward control properties
Disturbance
(measured or modelled)
Reference signal
Feed forward
control
Disturbance
Process
Output
•
Model based (but fairly simple models typically helps a lot)
•
Produces control signals before error has occurred
•
Uses measured or modelled disturbance and compensates for it
•
Uses carefully designed reference signals to make the process follow
the references “exactly” and without saturating the control signal.
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Mechatronics Lab
Slide: 5(35)
5.1.4. The model based control concept
Control u ff
Model following
control
Setpoint r s
Output y
Trajectory
planner
S
--R
Reference r
Controlled
system
Control u
0.6
Amplitude
0.4
rs
0.2
r
0
−0.2
0
0.5
1
1.5
Time
2
2.5
3
Two tasks
1.) Design the reference signals in
the trajectory planner, r ( t ) .
2.) Design a model following
controller, u ff ( t ) .
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Date: 2009-08-27
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Mechatronics Lab
Slide: 6(35)
5.1.5. The ultimate goal, Exact model following
Model following
control
uff
Set point
Trajectory
planner
Reference
Controlled Output
system
•
Design the trajectory planner and the model following controller such
that the feed forward input signal uff(t) drives the output to the desired
setpoint
•
Note that the model following control part can be a nonlinear system
•
Two main concepts, Transfer function and Time domain.
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Date: 2009-08-27
File: Q:\md.kth.se\md\mmk\gru\mda\mf2007\arbete\Lectures\FinalVersions\For2009\L5.fm
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Mechatronics Lab
Slide: 7(35)
5.2. Lecture outline
•
1. Introduction
• 2. Transfer function based model following
•
3. Time domain based model following
•
4. An example
•
5. Matlab example for the dc motor
KTH
Date: 2009-08-27
File: Q:\md.kth.se\md\mmk\gru\mda\mf2007\arbete\Lectures\FinalVersions\For2009\L5.fm
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Slide: 8(35)
5.2.1. Model following TF-based
Feed forward
setpoint r s
1/Gp(s)
y
r
+
Gm(s)
Gp(s)
•
Gm(s) is the reference model, i.e it is designed to provide reasonable
references to the controller, and to make Gm ⁄ G p proper.
•
Gp(s) is the transfer function of the process. No feedback controller yet!
•
Ideally the inverse of the process, i.e. 1/Gp(s), would provide exact
model following: 1/Gp(s) * Gp(s) = 1, and the closed loop y ⁄ rs = G m ( s )
•
Clearly, the model following concept is directly dependent on the
accuracy of the process model.
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Date: 2009-08-27
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Slide: 9(35)
5.2.2. The reference model Gm(s)
•
The task of the reference model is to generate smooth references that
the process is able to follow.
•
To implement complete model following, i.e a full process inverse, we
must require that the pole excess dm of the reference model is larger
than or equal to the pole excess d of the process, i.e if
B(s)
G p ( s ) = ----------A(s)
is the pole excess of the process dp
d p = deg ( A ) – deg ( B )
and finally the requirement
dm ≥ dp
Bm ( s )
G m ( s ) = --------------Am ( s )
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d m = deg ( A m ) – deg ( B m )
Mechatronics Lab
Slide: 10(35)
5.2.3. Model following TF-based cont.
•
Since exact model following involves taking the inverse of the process
model, the inverse must be possible to implement. I.e. the inverse must
constitute a stable dynamic system.
•
A stable dynamic system has no right half plane (RHP) poles. This
means for model following that the process must not have any RHP
zeros.
•
Care must be taken when doing the discrete time implementation
because, as we know, a continuous time system with only LHP zeros
may become zeros outside the unit circle when transformed into
discrete time.
This often happens for very low sampling periods.
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Slide: 11(35)
5.2.4. Implementing the TF based servo control
Gm ( s )
--------------Gp ( s )
rs
Gm(s)
u ff
r
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Feedback
controller
MMK
u
Controlled Output
system
Mechatronics Lab
Slide: 12(35)
5.3. Lecture outline
•
1. Introduction
•
2. Transfer function based model following
• 3. Time domain based model following
•
4. An example
•
5. Matlab example for the dc motor
KTH
Date: 2009-08-27
File: Q:\md.kth.se\md\mmk\gru\mda\mf2007\arbete\Lectures\FinalVersions\For2009\L5.fm
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Slide: 13(35)
5.3.1. Model following Time domain (TD)
•
Instead of the combined trajectory and model following transfer function
Gm ( s )
--------------Gp ( s )
are the trajectory and model following implemented separately
and based on trajectories in time.
•
Model following: If the inverse model (from y -> u) can be written as a
function of the output and n-times the derivatives of the output (n:th
order model).
n
⎛ dy d 2 y
d y⎞
u ( t ) = f ⎜ y, ------, --------, …, --------⎟
n
⎝ dt dt 2
dt ⎠
2
•
Trajectory planner: Design
n
dy d y
d y
y, ------, --------, …, -------2
n
dt
dt
dt
based on closed loop
specifications and process limitations, e.g., saturation.
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Date: 2009-08-27
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Slide: 14(35)
5.3.2. The concept, an example
Process model DC-motor with non-linear friction
Gives the feed forward control,
.
··
·
·
u ff ( t ) = Jϕ r ( t ) + dϕ r + F c sgn ( ϕ r ) .
d---dt
··
ϕr
G(s)
J
–1
d
·
ϕr
d---dt
··
·
·
Jϕ = u – dϕ – F c sgn ( ϕ )
sgn()
Fc
u ff
rs
Trajectory
planner
r(t)
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Date: 2009-08-27
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Feedback
Controller
MMK
u fb
ϕ
u
Process
Mechatronics Lab
Slide: 15(35)
5.3.3. Practical implementation
•
We want to avoid to take the derivative of signals.
-> The Trajectory planner must provide the derivatives
G(s)
J
–1
d
rs
Trajectory
planner
··
ϕr
·
ϕr
r = ϕr
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sgn()
Fc
u ff
Feedback
Controller
MMK
u fb
ϕ
u
Process
Mechatronics Lab
Slide: 16(35)
5.3.4. n-times differentiable reference signals
•
The DC-motor is a second order diff. eq. from input to position
- The reference position must be two times differentiable (velocity and acceleration)
- Which is the same as: The acceleration reference must be finite
ex.
··
ϕr
1
0.8
·
ϕr
0.6
0.4
ϕr
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
0
0.5
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1
MMK
1.5
Mechatronics Lab
Slide: 17(35)
5.3.5. How to choose the reference signals
•
Fastest possible positioning (or specific time trajectories)
- Calculate max acceleration and velocity of the process, without saturating the
input to the motor.
- For the DC-motor: If max input torque is ± M max then max acceleration at zero
velocity is a max = ( ± M max ) ⁄ J .
Max velocity at zero acceleration is v max = ( ± M sat ) ⁄ ( d + F c )
a max
1
0.8
Example
reference
trajectories
r pos
v max
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
0
0.5
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Date: 2009-08-27
File: Q:\md.kth.se\md\mmk\gru\mda\mf2007\arbete\Lectures\FinalVersions\For2009\L5.fm
1
1.5
MMK
2
2.5
Mechatronics Lab
Slide: 18(35)
5.3.6. Example: how to calculate the ref. traj.
v = at , zero initial condition!
v max
t 1 = ------------ , the time when maximum velocity is reached using maximum acceleration
a max
t1
2
v
2
1
1 max
y ( t 1 ) = a max t dt = --- a max t 1 = --- ------------ , position reached at time t 1
2
2 a max
0
∫
v max
rs
r s – 2y ( t 1 ) = v max ( t 2 – t 1 ) -> t 2 = ------------ – ------------ + t 1 is the latest time to start braking.
v max a max
80
Example::
60
40
r s = 60,
v max = 55
a max = 70
t2
20
0
t1
−20
−40
−60
−80
0
0.2
0.4
0.6
0.8
1
1.2
1.4
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1.6
1.8
MMK
2
Mechatronics Lab
Slide: 19(35)
5.3.7. General remarks
•
The two concepts of servo control are often combined
•
Some process models are inherently difficult to invert.
- Flexible links: Approximate the flexible model with a stiff model, the inertia and
the friction of the process is still fed forward.
•
You need some margin to the theoretical maximum acceleration
and speeds.
- delays in computer implementation, flexibility.
- Maximum torque is a function of motor speed.
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File: Q:\md.kth.se\md\mmk\gru\mda\mf2007\arbete\Lectures\FinalVersions\For2009\L5.fm
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Slide: 20(35)
5.4. Lecture outline
•
1. Introduction
•
2. Transfer function based model following
•
3. Time domain based model following
• 4. An example
•
5. Matlab example for the dc motor
KTH
Date: 2009-08-27
File: Q:\md.kth.se\md\mmk\gru\mda\mf2007\arbete\Lectures\FinalVersions\For2009\L5.fm
MMK
Mechatronics Lab
Slide: 21(35)
5.4.1. Model following control example I
•
Consider a second order motion control system. The output is position.
The electrical motor (the coil inductance is neglected) is operating
against a linear spring with spring constant K
0
0
1
x· =
K emk ⋅ K m + K m ⋅ u
Kf
-------– ----- – ⟨ -------------------------⟩
J
J
J
•
y = 1 0 ⋅
x1
x2
Or in transfer function form
Km ⁄ J
G p ( s ) = ---------------------------------------------------------Kf
K m ⋅ K emk
2
s + s ⎛ -------------------------⎞ + ----⎝
⎠ J
J
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Slide: 22(35)
5.4.2. Model following control example II
•
Assuming
1
G p ( s ) = ---------------------------------2
2s + 1.5s + 1
•
We get the following principle model following control scheme
r
setpoint rs
Model
following
u ff
Reference
model,Gm(s)
•
1
y
2s2 + 1.5s +1
Task: Design the reference model and the model following controller
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Slide: 23(35)
5.4.3. Implementing the model following
•
The perfect model following is the inverse of the model
2
u ff ( s )
Gm ( s )
2s + 15s + 1
--------------- = -------------- = --------------------------------Gp ( s )
rs ( s )
1
•
Gm ( s ) = 1
(not proper)
Use a reference model with pole excess dm > 2 (version 1)
2
2
G m ( s ) = ω ⁄ ( s 2 + 2ξωs + ω )
2
2
u ff ( s )
( 2s + 15s + 1 )ω
-------------- = --------------------------------------------2
rs ( s )
s 2 + 2ξωs + ω
•
(proper)
Or use a trajectory planner, position, velocity and acceleration
(version 2)
2
u ff ( s ) = ( 2s + 15s + 1 )r s ( s )
··
·
u ff ( t ) = 2r s ( t ) + 15r s ( t ) + r s ( t )
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Slide: 24(35)
5.4.4. Output from the trajectory planner
150
acceleration
100
Amplitude
50
velocity
position
0
−50
−100
−150
•
0
0.5
1
Time
1.5
2
The reference model provides acceleration, velocity and position
references (acceleration step, velocity ramp and a smooth position
reference)
1
r vel = --- r acc
t
1
1
r pos = --- r vel = ---- r acc
2
t
t
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Observe, initial conditions
Mechatronics Lab
Slide: 25(35)
5.4.5. Acceleration feed forward
300
control signal
200
position reference
Amplitude
100
system response
0
−100
−200
−300
•
0
0.5
1
Time
1.5
2
Let’s start out with feed forward to cope better with the acceleration step
u ff = 2r·· .
•
The process follows fairly good initially when the acceleration signal
dominates
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Slide: 26(35)
5.4.6. Acceleration and velocity feed forward
400
control signal, u
300
200
position reference, r
100
system response, y
0
−100
−200
−300
•
0
0.5
1
1.5
2
Next we add velocity feed forward which gives improved following of
the reference.
u ff = 2r·· + 15r·
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Slide: 27(35)
5.4.7. Exact model following
400
control signal
300
Amplitude
200
100
system response
(equal to reference position
0
−100
−200
−300
•
0
0.5
1
Time
1.5
2
Finally we add position feed forward which gives exact model following.
u ff = 2r·· + 15r· + r
•
Note again, that we assume exact knowledge of the model!
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Slide: 28(35)
5.4.8. Adding a zero to the model
Model
inverse
Reference
s+5
y
2s2 + 1.5s +1
•
The process is extended with LHP zero.
•
Exact model following is still possible since the inverse system is stable.
•
The static gain of the process is 5.
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Slide: 29(35)
5.4.9. Feed forward without numerator inverse
control signal
80
60
Amplitude
40
system response
20
0
reference position
−20
−40
−60
•
0
0.5
1
Time
2
Let’s first have look on the response with only changing the gain of the
feed forward controller.
•
1.5
·· + 15r· + r
2r
----------------------------.
u ff =
5
The added dynamics (model zeros) is clearly visible.
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Slide: 30(35)
5.4.10. Exact model following for LHP zero
60
50
system response, y
and reference
40
Amplitude
30
20
control signal, u
10
0
−10
−20
−30
•
0
0.5
1
Time
1.5
2
Exact model following is implemented by including also the inverse of
the numerator.
1
u ff = ( 2r·· + 15r· + r ) ----------s+5
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Slide: 31(35)
5.4.11. A non-minimum phase system
Model
inverse
Reference
model
-s+5
y
2s2 + 1.5s +1
•
Clearly, we have a RHP zero, making exact model following impossible.
That is, the inverse of the process model constitute an unstable dynamic
system.
•
The static process gain is still 5.
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Slide: 32(35)
5.4.12. Approximate model following (RHP zero)
control signal
80
reference position
60
Amplitude
40
20
system response
0
−20
−40
−60
•
0
0.5
1
Time
1.5
2
An approximate model following can be implemented by taking the
inverse of the static portion of the numerator
•
1
u ff = ( 2r·· + 15r· + r ) --- .
5
Observe the typical response starting in the wrong direction.
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Slide: 33(35)
5.4.13. Summary
Control u ff
Model following
control
Setpoint r s
Output y
Trajectory
planner
S
--R
Reference u c
Controlled
system
Control u
Design the trajectory planner based on the performance of the motor.
Design the model following control based on an inverse model of the
process. Excellent for non-linear phenomena.
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File: Q:\md.kth.se\md\mmk\gru\mda\mf2007\arbete\Lectures\FinalVersions\For2009\L5.fm
MMK
Mechatronics Lab
Slide: 34(35)
5.5. Lecture outline
•
1. Introduction
•
2. Transfer function based model following
•
3. Time domain based model following
•
4. An example
• 5. Matlab example for the dc motor
KTH
Date: 2009-08-27
File: Q:\md.kth.se\md\mmk\gru\mda\mf2007\arbete\Lectures\FinalVersions\For2009\L5.fm
MMK
Mechatronics Lab
Slide: 35(35)