(c)
ẋ1 = x2
ẋ2 = −2x1 − 2x2 − sat(−x1 − x2 )
Which phase portrait belongs to what system?
Phase plane
Phase plane
Phase plane
2
2
1
0.8
1.5
1.5
0.6
1
1
0.4
0.5
0
x2
0.2
x2
x2
0.5
0
−0.2
−0.5
0
−0.5
−0.4
−1
−1
−0.6
−1.5
−1.5
−2
−2
−0.8
−1.5
−1
−0.5
0
x1
0.5
1
1.5
−1
−1
2
−0.8
−0.6
−0.4
−0.2
0
x1
0.2
0.4
0.6
0.8
−2
1
−1.5
−1
−0.5
0
x1
0.5
1
1.5
2
Figure 2.10: Phase portraits for saturated linear systems in Exercise 3.4
Figure 1: Phase portraits for saturated linear systems in Exercise 3.4.
Exercise 3.4: Saturations constitute a severe restriction for stabilization of system. Figure 1 shows
three phase portraits, each corresponding to one of the following linear systems under saturated
feedback control.
27
(a)
ẋ1 = x2
ẋ2 = x1 + x2 − sat(2x1 + 2x2 )
(b)
ẋ1 = x2
ẋ2 = −x1 + 2x2 − sat(3x2 )
(c)
ẋ1 = x2
ẋ2 = −2x1 − 2x2 − sat(−x1 − x2 )
Which phase portrait belongs to what system?
Answer: What we agreed so far was that system (a) is associated with phase portrait (right). To
show this, note that if |2x1 + 2x2 | ≤ 1 (and, as a result, sat(2x1 + 2x2 ) = 2x1 + 2x2 ), we can rewrite
the system state-space model as
ẋ1 = x2
ẋ2 = −x1 − x2 ,
or, equivalently,
ẋ1
ẋ2
=
|
0
−1
{z
+1
−1
F
x1
x2
.
}
Now, we can see that
det(λI − F ) = det
λ
+1
−1
λ+1
= λ2 + λ + 1,
√
which shows that the eigenvalues of F are equal to (−1 ± i 3)/2 and, hence, system (a) is stable
close to the origin. Particularly, the origin is a stable focus! What happens if |2x1 + 2x2 | > 1? In
this case, we get
ẋ1
x2
=
,
ẋ2
x1 + x2 ± 1
|
{z
}
f (x)
1
for which we can calculate the equilibria (i.e., x∗ such that f (x∗ ) = 0) in
+1
−1
∗
∗
x =
and x =
.
0
0
These equilibria are indeed feasible since |2x∗1 + 2x∗2 | = 2 > 1 for both of them. Now, the dynamic
of the system around these equilibria is governed by
x̃1
x̃˙ 1
0 +1
.
=
x̃2
+1 +1
x̃˙ 2
{z
}
|
F
where
x̃1
x̃2
x1
x2
λ
−1
=
− x∗ .
For this new system, we can see that
det(λI − F ) = det
= λ2 − λ − 1,
−1
λ−1
√
which shows that the eigenvalues of F are equal to (−1 ± 5)/2 and, hence, both these equilibria
are saddle points! This is perfectly what phase portrait (right) describes.
X
Now, we are ready to analyze system (c). Yes, I am going to consider the last system first. Note
that −sat(−x1 − x2 ) = sat(x1 + x2 ) and, as a result,
ẋ1 = x2
ẋ2 = −2x1 − 2x2 + sat(x1 + x2 ).
Now, if |x1 + x2 | ≤ 1 (and, as a result, sat(x1 + x2 ) = x1 + x2 ), we get
ẋ1 = x2
ẋ2 = −x1 − x2 ,
which, with a similar argument as in the last part, shows the origin is a stable focus! If |x1 + x2 | > 1,
we get
ẋ1
x2
.
=
ẋ2
−2x1 − 2x2 ± 1
|
{z
}
f (x)
Finding x∗ such that f (x∗ ) = 0 results in
+1/2
−1/2
x∗ =
and x∗ =
.
0
0
However, both these points are infeasible because |x∗1 + x∗2 | = 1/2 ≤ 1 which implies the saturations
must be linear (and, therefore, ẋ = f (x) is not describing the underlying system)! Hence, system (b)
has a unique equilibrium at the origin which we know is a stable focus. The hand-wavy argument that
we talked about during the class was that the trajectories of this system are actually pointing inward
for large numbers and, hence, the system must be globally stable! However, we want something
rigorous (which makes the problem a bit more fun and a lot more difficult). For any n > 1, we may
define the set Ωn as in Figure 2 (the red shaded area inside the irregular octagon). Now, we are
going to prove that Ωn is an invariant set (i.e., if x(0) ∈ Ωn , then x(t) ∈ Ωn for all t ≥ 0). Why is
this useful? Well, the answer to that is if this set is invariant the trajectories of the system cannot
2
x2
−n
n
S1
−n
0
S2
S6
0
n
n
0
S3
S5
0
−n
S4
n
−n
x1
Figure 2: The set Ωn is defined with the shaded area inside the red irregular octagon and its boundary
∂Ωn is the union of the red lines.
not leave it and, hence, the system cannot be unstable and we can rule out phase portrait (left)!
How do we actually prove that Ωn is invariant? Notice that since the solutions of the systems are
continuous, if they want to leave Ωn , they can only do so at the boundary of Ωn which is denoted
and calculated as
∂Ωn = S1 ∪ S2 ∪ S3 ∪ S4 ∪ S5 ∪ S6 ,
where Si , 1 ≤ i ≤ 6, are defined in Figure 2. Now, let me show that the trajectories of the system (c)
cannot leave the set Ωn using the boundary line shown by S1 . We can mathematically define S1 set
as
x1
2
S1 =
∈ R g(x) = 0, −n ≤ x1 ≤ 0
x2
where g(x) = x2 − n. If we can show that ∇g(x)> f (x) ≤ 0, then the trajectories of the system
cannot get out using this boundary set1 . Evidently, we have
> 0
x2
>
∇g(x) f (x) =
1
−2x1 − 2x2 + sat(x1 + x2 )
= −2(x1 + x2 ) + sat(x1 + x2 )
= −2(x1 + n) + sat(x1 + n)
because x2 = n
= −(x1 + n) + [sat(x1 + n) − (x1 + n)]
≤ −(x1 + n)
≤0
because sat(u) ≤ u, ∀u > 0
because − n ≤ x1 ≤ 0.
With the same line of reasoning (or using the symmetry between S1 and S4 ), we can prove that
the trajectories of the system (c) cannot leave using the boundary S4 . Now, we can show that the
1 I will describe this part in much detail when I am solving Exercise 4.7. Notice that inside the set Ω and alongside
n
the boundary set S1 , and by definition, g(x) ≤ 0. Now, we can see that
d
g(x) = ∇g(x)> ẋ = ∇g(x)> f (x) ≤ 0.
dt
Therefore, along the trajectories of the system g(x) is going to be negative if it starts from a negative value and, as a
result, the trajectories cannot get out of Ωn through this boundary (because they need to make g(x) positive if they
wish to get out).
3
−n
0
−n
n
x
2
S1
S6
S5
0
−n
S2
0
n
n
0
x1
S3
S4
n
−n
Figure 3: The set Θn is defined as the shaded area outside the red irregular octagon and its boundary
∂Θn is the union of the red lines.
trajectories of the system (c) cannot leave through the boundary set S2 . We can mathematically
define S2 set as
x1
S2 =
∈ R2 g 0 (x) = 0, 0 ≤ x1 ≤ n
x2
where g 0 (x) = x1 + x2 − n. Clearly, we get
0
>
∇g (x) f (x) =
1
1
> x2
−2x1 − 2x2 + sat(x1 + x2 )
= x2 − 2(x1 + x2 ) + sat(x1 + x2 )
= n − x1 − 2n + sat(n)
because x1 + x2 = n
= (1 − n) − x1
because sat(n) = 1 for n > 1
≤ −x1
because n > 1
≤0
because 0 ≤ x1 ≤ n.
This shows that the trajectories of the system (c) cannot leave through the boundary set S2 . Now,
using the symmetry between S2 and S5 , we can see that the trajectories of the system (c) cannot
leave using the boundary S5 as well. Let us show that the trajectories of the system (c) cannot leave
through the boundary set S3 . We can mathematically define S3 set as
x1
2 00
S3 =
∈ R g (x) = 0, −n ≤ x2 ≤ 0
x2
where g 00 (x) = x1 − n. Evidently, we have
∇g 00 (x)> f (x) =
= x2
1
0
> x2
−2x1 − 2x2 + sat(x1 + x2 )
≤0
because − n ≤ x2 ≤ 0.
This shows that the trajectories of the system (c) cannot leave through the boundary set S3 . Now,
again, using the symmetry between S3 and S6 , we can prove that the trajectories of the system (c)
4
cannot leave using the boundary S6 as well. Therefore, we managed to show that Ωn is invariant for
all n > 1. This implies that the solutions of the system (c) cannot converge to infinity and, hence,
the phase portrait (center) describes its behavior because in the other one, for large x the system is
unstable.
This finally implies that system (b) can only match phase portrait (left) because all the other
phase planes are reserved for other systems. This actually can be proved independently using the
same line of reasoning (as in the last part). For any n > 1, we can define the set Θn as in Figure 3
(the red shaded area outside the irregular octagon). Now, we can prove that Θn is an invariant set.
Here, I am not going to prove this fact since the calculations are similar to the earlier parts of this
report. Now, recalling the shape of the set Θn in Figure 3, we can immediately see that if the initial
condition is far from the origin, the trajectories of the system cannot converge to the origin and,
therefore, the origin is not globally asymptotically stable.
5