PS#4 - Nonlinear Systems and Control Spring 2011

PS#4 - Nonlinear Systems and Control Spring 2011
Problem 1
The magnetic suspension system is modeled by
˙x1 = x2
˙x2 = g − k
m x2
L0ax
−
2 3
2m(a + x1) 2
˙x3 =
1
−Rx3 +
L(x1)
L0ax2x3
+ u
(a + x1) 2
where x1 = y is the elevation, x2 = ˙y, x3 = I is the current, and u = V is the input voltage. Use the
following numerical data: m = 0.1kg, k = 0.001N/m/sec, g = 9.81m/sec2 , a = 0.05m, L0 = 0.01H,
.
L1 = 0.02H, R = 1Ω, and L(x1) = L1 + L0
1+ x 1
a
1. Find the steady-state values Iss and Vss of I and V , respectively, which are needed to balance the
ball at a given position y = r.
2. Is the equilibrium point obtained above stable?
3. Linearize the system and design a state feedback control law to stabilize the ball at y = 0.05m.
Solution:
1. Set ˙x1 = ˙x2 = ˙x3 = 0. Then, we have
0 = x2
0 = g − k
m x2
L0ax
−
2 3
2m(a + x1) 2
0 =
1
−Rx3 +
L(x1)
L0ax2x3
+ u
(a + x1) 2
Now, set x1 = yss = r, x3 = Iss, and u = Vss. Then, (choosing the positive sign)
Iss =
2mg(a + r) 2
,
L0a
Vss = RIss = Iss
2. We linearize the system around the equilibrium point xss = [r, 0, Iss] T to obtain
where c1 to c5 are positive constants given by
A = ∂f
∂x |xss
⎡
⎤
0 1 0
= ⎣c1
−c2 −c3⎦
0 c4 −c5
c1 = L0aI 2 ss
m(a + r) 3 , c2 = k
m , c3 = L0aIss
m(a + r) 2 , c4 =
Now, the charactersitic equation is given by
λ 3 + (c2 + c5)λ 2 + (c2c5 + c3c4 − c1)λ − c1c5 = 0
L0aIss
L(r)(a + r) 2 , c5 = R
L(r)
which is clearly unstable (using Routh-Hurwitz criterion) due to the sign of the last element being
negative.
1

3. The linearized system is given by
˙xδ = Axδ + Buδ
where xδ = x − xss and uδ = u − uss, with A as in part 2 and B = [0 0 L(r)] T . It can easily be
verified that the pair (A, B) is controllable, and hence we can design a state feedback gain K,
such that A − BK has eigenvalues at −2 and −1 ± j. The control input is given by
where Vss = Iss = 6.26.
Problem 2
Let
Aa =
u = − −256.2 −65.5 3.07 (x − xss) + Vss
An×n 0n×p
Cp×n 0p×p
, Ba =
(n+p)×(n+p)
where A, B, and C satisfy the rank condition
A
rank
C
B
= n + p.
0
Bn×p
0p×p
Show that if (A, B) are controllable, then (Aa, Ba) are also controllable.
Solution:
First note that since (A, B) are controllable then
rank (C(A, B)) . = rank B|AB| · · · |A n−1 B = n
(n+p)×p
Now, let us look at the controllability matrix of C(Aa, Ba):
rank (C(Aa, Ba)) = rank Ba|AaBa| · · · |A n+p−1
a Ba
B | AB | A2B | · · · | An+p−1B = rank
0 | CB | CAB | · · · | CAn+p−2
B
A B
= rank .
C 0 Ip×p | 0p×p | 0p×p | · · · | 0p×p
0n×p | B | AB | · · · | A n+p−2 B
A
Since rank
C
B
0n×p
= n + p, and rank
0
| B | AB | · · · | An+p−2 Ip×p | 0p×p | 0p×p | · · · |
B
= n + p,
0p×p
it follows that rank (C(Aa, Ba)) = n + p, and therefore the augmented system is controllable.
Problem 3
A simplified model of the low-frequency motion of a ship is given by
τ ¨ ψ + ˙ ψ = kσ
where ψ is the heading angle of the ship and σ is rudder angle, viewed as the control input. The time
constant τ and the gain k depend on the forward speed of the ship v, according to the expressions
τ = τ0 v0
v
v and k = k0 v0 , where τ0, k0, and v0 are constants.
1. Assuming a constant forward speed, design a state feedback integral controller so that ψ tracks a
desired angle ψr.
2. Use gain scheduling to compensate for varying forward speed.
2

Solution:
1. Take x1 = ψ − ψr, x2 = ˙ ψ, and u = σ, then the state-space model is given by
˙x1 0 1
=
˙x2 0 − 1
x1 0k
+ u
τ
x2 τ
A
B
As such, we would like to regulate the system to ψr, using integral control. First, it is easy to
check that the pair (A, B) is controllable and that for C = [1 0]
A
rank
C
B
= 2 + 1 = 3
0
Note that k = 0 and τ = 0. Therefore, taking ˙η = ψ − ψr results in the following augmented
system
˙x = Ax + Bu
˙η = Cx
Since the system above is controllable (see the previous exercise), we can design
u = − k1 k2
k3
x
η
that stabilizes the overall system to the desired equilibrium. The closed-loop system matrix is
given by
A
C
0 B k1
−
0 0
k2
⎡
0
k3 = ⎣−
1 0
kk1
τ
1 − τ (1 + kk2) − kk3
1 0
⎤
⎦
τ
0
which has the following characteristic polynomial
λ 3 + 1
τ (1 + kk2)λ 2 + kk1 kk3
λ +
τ τ
All the roots of the characteristic polynomial lie in the left half plane iff the following conditions
are satisfied
1
τ (1 + kk2)
kk1 kk3 1
kk1
> 0, > 0, > 0, (1 + kk2) · −
τ τ τ τ
kk3
> 0
τ
which give the desired stabilizing control gains k1, k2, and k3.
2. Whenever the forward speed is varying, the resulting system matrices depend on v, i.e., A(v) and
B(v), because τ = τ(v) and k = k(v). We can design control gains k1(v) k2(v) k3(v) such
that the over all closed-loop system matrix
A(v) 0
C 0
is Hurwitz for a given velocity v.
−
= 0
B(v) k1(v) k2(v) k3(v)
0
3

Problem 4
Consider the following scalar (x ∈ R 1 ) system
˙x = −ax + bu (1)
The values of a and b are unknown, however we assume that b > 0, i.e., the sign of b is known. Now
the objective is to make system track the following known reference model
˙xm = −amxm + bmur
where am > 0 and the reference input ur is known. Let the controller be
u = θ1ur − θ2x (3)
where the parameters θ1 and θ2 are the ones that we would like to adjust adaptively online. Define the
error between the system and the reference model
1. MIT rule design
e = x − xm
• Substitute the value of u in (3) into equation (1), and derive the transfer function from x to
ur (under zero initial condition).
• Using the transfer function derived above and the definition of e in (4), derive the following
quantities: ∂e ∂e and . Then, use these quantities to form the adaptation equations obtained
∂θ1 ∂θ2
via the MIT rule, i.e., minimize J (θ) = 1
2e2 . Can you implement the MIT controller that
you derived? Why?
• Use the following nominal approximation a + bθ 0 2 ≈ am in your answers for part 2. Can you
implement the resulting adaptation laws now? What is the final MIT rule controller?
2. Lyapunov-based design
Consider the following Lyapunov function candidate
V (e, θ1, θ2) = 1
e
2
2 + 1
bγ (bθ2 + a − am) 2 + 1
bγ (bθ1 − bm) 2
Design the update equations of θ1 and θ2 so that we obtain a stable system, i.e., ˙ V ≤ 0. Use the
following equation
˙e = −ame − (bθ2 + a − am)x + (bθ1 − bm)ur
3. Simulate both controllers that you got in parts 1 and 2 for the following parameters: a = 1,
b = 0.5, am = 2, and bm = 2. Take ur to be a square-wave with amplitude 1, and γ = 1.
Solution:
1. MIT rule design: It is reasonable to change the design parameters θ1 and θ2 in the direction of
the opposite gradient of J , i.e., ˙ θ = −γe ∂e
∂θ , for γ > 0.
• The closed-loop plant equation is
bθ1
˙x = (a + bθ2)x + bθ1ur ⇒ x =
s + a + bθ2
4
ur
(2)
(4)

• As such, we have that
∂e
∂θ1
b
=
ur,
s + a + bθ2
∂e
∂θ2
b
= −
2θ1 (s + a + bθ2) 2 ur
b
= − x
s + a + bθ2
Since we do not know a and b, the partial derivatives above cannot be used in the MIT-based
adaptation given by
˙θ = −γe ∂e
∂θ
• Using the approximation a + bθ0 2 ≈ am, we obtain the following adaptation rules instead
˙θ1 = −γ ′
˙θ2 =
am
ur e
s + am
−γ ′
− am
x e
s + am
where γ ′ = γ b . The sign of b is crucial for this scheme to work!
am
2. Take the derivative of V and substitute the given expression for ˙e to obtain
˙V = −ame 2 + 1
γ (bθ2
+ a − am) ˙θ2 − γxe + 1
γ (bθ1
− bm) ˙θ1 + γure
Then pick the adaptation parameters in order to obtain ˙ V ≤ 0, i.e.,
˙θ1 = −γure
˙θ2 = γxe
This guarantees that (e, θ1, θ2) is bounded. With some more work, one can show that e → 0 as
t → ∞, but not θ1 and θ2.
3. For the simulation results run the file sol4adaptive.mdl.
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