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EL2620 Nonlinear Control
Lecture 9
• Nonlinear control design based on high-gain control
Lecture 9
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Today’s Goal
You should be able to analyze and design
• High-gain control systems
• Sliding mode controllers
Lecture 9
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History of the Feedback Amplifier
New York–San Francisco communication link 1914.
High signal amplification with low distortion was needed.
r
f (·)
f (·)
−
f (·)
y
k
Feedback amplifiers were the solution!
Black, Bode, and Nyquist at Bell Labs 1920–1950.
Lecture 9
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Linearization Through High Gain Feedback
α2 e f (e)
r
−
f (·)
y
α1 e
e
K
f (e)
α1 ≤
≤ α2
e
choose K
⇒
α1
α2
r ≤ y ≤
r
1 + α1 K
1 + α2 K
≫ 1/α1 , yields
1
y≈ r
K
Lecture 9
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A Word of Caution
Nyquist: high loop-gain may induce oscillations (due to dynamics)!
Lecture 9
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Inverting Nonlinearities
Compensation of static nonlinearity through inversion:
Controller
−
F (s)
fˆ−1 (·)
f (·)
G(s)
Should be combined with feedback as in the figure!
Lecture 9
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Remark: How to Obtain f −1 from f using
Feedback
f (u)
v
−
k
s
u
u
f (·)
u̇ = k v − f (u)
If k
> 0 large and df /du > 0, then u̇ → 0 and
0 = k v − f (u)
⇔
f (u) = v
⇔
Lecture 9
u = f −1 (v)
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Example—Linearization of Static Nonlinearity
1
r
e
−
K
u
f (·)
y
0.8
y(r)
0.6
0.4
f (u)
0.2
0
0
Linearization of f (u)
The case K
Lecture 9
0.2
0.4
0.6
0.8
1
= u2 through feedback.
= 100 is shown in the plot: y(r) ≈ r.
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The Sensitivity Function S
= (1 + GF )−1
r
The closed-loop system is
−
G
Gcl =
1 + GF
G
y
F
Small perturbations dG in G gives
dGcl
1
=
dG
(1 + GF )2
⇒
1
dG
dG
dGcl
=
=S
Gcl
1 + GF G
G
S is the closed-loop sensitivity to open-loop perturbations.
Lecture 9
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Distortion Reduction via Feedback
The feedback reduces distortion in each link.
Several links give distortion-free high gain.
−
f (·)
K
Lecture 9
−
f (·)
K
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Example—Distortion Reduction
r
Let G = 1000,
distortion dG/G
Choose K
−
= 0.1
= 0.1
⇒
G
y
K
S = (1 + GK)−1 ≈ 0.01. Then
dG
dGcl
=S
≈ 0.001
Gcl
G
100 feedback amplifiers in series give total amplification
Gtot = (Gcl )100 ≈ 10100
and total distortion
dGtot
= (1 + 10−3 )100 − 1 ≈ 0.1
Gtot
Lecture 9
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Transcontinental Communication Revolution
The feedback amplifier was patented by Black 1937.
Lecture 9
Year
Channels
Loss (dB)
No amp’s
1914
1
60
3–6
1923
1–4
150–400
6–20
1938
16
1000
40
1941
480
30000
600
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Sensitivity and the Circle Criterion
r
−
−1
r
GF
y
f (·)
GF (iω)
Consider a circle C
:= {z ∈ C : |z + 1| = r}, r ∈ (0, 1).
GF (iω) stays outside C if
|1 + GF (iω)| > r
⇔
|S(iω)| ≤ r−1
1
f (y)
1
Then, the Circle Criterion gives stability if
≤
≤
1+r
y
1−r
Lecture 9
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Small Sensitivity Allows Large Uncertainty
If |S(iω)| is small, we can choose r large (close to one).
This corresponds to a large sector for f (·).
Hence, |S(iω)| small implies low sensitivity to nonlinearities.
k2 y f (y)
1
k1 =
1+r
1
k2 =
1−r
Lecture 9
k1 y
y
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On–Off Control
On–off control is the simplest control strategy.
Common in temperature control, level control etc.
r
e
−
u
G(s)
y
The relay corresponds to infinite high gain at the switching point
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Example: Stabilizing control of inverted
pendulum
Model with x1
= θ − α, α = π/2 and x2 = θ̇
ẋ1 = x2
g
k0
1
ẋ2 = − sin(x1 + α) − x2 + 2 u
l
m
ml
Thus,
g
k0
1
f (x) = − sin(x1 + α) − x2 , g(x) =
l
m
ml2
We choose the sliding manifold
σ(x) = x2 + ax1 = 0
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Control design
The control law
u(x) = β(x) − Ksign(σ)
where
f (x) + ax2
= gml sin(x1 + α) + k0 ml2 x2 − ml2 ax2
β(x) = −
g(x)
and we choose
K = k1 + k2 σ 2
Lecture 9
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0.15
0.1
x
1
0.05
0
x
2
-0.05
-0.1
0
2
4
6
8
10
time [s]
Lecture 9
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0.1
0.08
0.06
0.04
x2
0.02
0
-0.02
-0.04
-0.06
-0.08
0
0.02
0.04
0.06
x
0.08
0.1
0.12
0.14
1
Lecture 9
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65
60
Control Input
55
50
45
40
35
0
1
2
3
4
5
time [s]
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Design of Sliding Mode Controller
Idea: Design a control law that forces the state to σ(x) = 0. Choose
σ(x) such that the sliding mode tends to the origin. Assume

  
f1 (x) + g1 (x)u
x1

 
x1
d 

 x2  
 = f (x) + g(x)u
 ..  = 
..
dt  .  

.
xn−1
xn
T
T
Choose σ(x) = p x with p = p1
stable polynomial Then the control law
...
pn the coefficients of a
pT f (x)
µ
u=− T
− T
sgn σ(x),
p g(x) p g(x)
where µ > 0 is a design parameter, will make the sliding mode and
the equilibrium globally asymptotically stable.
Lecture 9
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Closed-Loop Stability
Consider V (x)
= σ 2 (x)/2 with σ(x) = pT x. Then,
T
T
T
T
V̇ = σ (x)σ̇(x) = x p p f (x) + p g(x)u
With the chosen control law, we get
V̇ = −µσ(x) sgn σ(x) < 0
so x tend to σ(x)
= 0.
0 = σ(x) = p1 x1 + · · · + pn−1 xn−1 + pn xn
(0)
(1)
+
p
x
+
·
·
·
+
p
x
= p1 x(n−1)
n n
n−1 n
n
where x(k) denote time derivative. Now p corresponds to a stable
differential equation, and xn → 0 exponentially as t → ∞ . The
state relations xk−1 = ẋk now give x → 0 exponentially as t → ∞.
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Example—Sliding Mode Controller
Design state-feedback controller for
Choose p1 s + p2
given by
1 0
1
ẋ =
x+
u
1 0
0
y= 0 1 x
= s + 1 so that σ(x) = x1 + x2 . The controller is
µ
pT Ax
− T sgn σ(x)
u=− T
p B
p B
= 2x1 − µ sgn(x1 + x2 )
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Phase Portrait
Simulation with µ
= 0.5. Note the sliding surface σ(x) = x1 + x2 .
x2
1
0
−1
−2
Lecture 9
−1
0
1
2
x1
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Time Plots
x1
Initial condition
x(0) = 1.5 0
T
x2
.
Simulation agrees well with
time to switch
σ0
=3
ts =
µ
and sliding dynamics
ẏ = −y
Lecture 9
u
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The Sliding Mode Controller is Robust
Assume that only a model ẋ = fb(x) + g
b(x)u of the true system
ẋ = f (x) + g(x)u is known. Still, however,
T
T
b
p g
p (f gb − f g )p
− µ T sgn σ(x) < 0
V̇ = σ(x)
T
p gb
p gb
if sgn(pT g)
T
T
= sgn(pT gb) and µ > 0 is sufficiently large.
The closed-loop system is thus robust against model errors!
(High gain control with stable open loop zeros)
Lecture 9
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Comments on Sliding Mode Control
• Efficient handling of model uncertainties
• Often impossible to implement infinite fast switching
• Smooth version through low pass filter or boundary layer
• Applications in robotics and vehicle control
• Compare pulse-width modulated control signals
Lecture 9
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Today’s Goal
You should be able to analyze and design
• High-gain control systems
• Sliding mode controllers
Lecture 9
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Next Lecture
• Lyapunov design methods
• Exact feedback linearization
Lecture 9
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