Unformatted text preview: ME 575 Handouts Example of Forced Response
1⎤
⎡0
⎡0 ⎤
x=⎢
x + ⎢ ⎥u
⎥
⎣ −2 −3 ⎦
⎣ 1⎦
find x(t) for u(t) a unit step and x(0) = 0:
t x(t) = e At x(0) + ∫ e A( t −τ ) B u( τ)dτ
0 ME575 Session 27: Controllability School of Mechanical Engineering
Purdue University Slide 1 Example (cont.)
⎡ ⎡e−( t −τ ) − 1 e−2( t −τ ) ⎤ t ⎤
2
⎣
⎦0 ⎥
x(t) = ⎢
t⎥
⎢
⎡ −e−( t −τ ) + e−2( t −τ ) ⎤ ⎥
⎢⎣
⎦0 ⎦
⎣ ME575 Session 27: Controllability School of Mechanical Engineering
Purdue University Slide 2 1 ME 575 Handouts Controllability (Kalman  1960)
Complete State Controllability:
The system
x = A x +Bu
is completely state controllable if a control u exists
that will transfer the state of the system from any
x(to) = xo to any x(t1) = x1 in a finite time t1−to. School of Mechanical Engineering
Purdue University ME575 Session 27: Controllability Slide 3 Example  Inverted Pendulum
d(t) g m = 1 kg θ
=1m u(t) −10 V +10 V vpot M = 10 kg Is it possible to achieve zero position of both the cart
and the rod with only a single control input u?
ME575 Session 27: Controllability School of Mechanical Engineering
Purdue University Slide 4 2 ME 575 Handouts Simpler Example
1⎤
⎡0
⎡0 ⎤
x=⎢
⎥ x + ⎢ 1⎥ u
⎣ −2 −3 ⎦
⎣⎦
Is it completely state controllable?
Transform to diagonal form: ME575 Session 27: Controllability School of Mechanical Engineering
Purdue University Slide 5 Controllability
Note that controllability depends not only on B, but
on the combination of A (through T) and B. Can we
generalize this? YES
The system
x = A x +Bu
is completely state controllable iff the column vectors
of the controllability matrix
n−1 W C = [B A B A B
A B]
span the ndimensional space (i.e., WC has rank n).
2 ME575 Session 27: Controllability School of Mechanical Engineering
Purdue University Slide 6 3 ME 575 Handouts Proof of Controllability Theorem
Assume u is a scalar:
t
x(t) = e At x(0) + ∫ e A( t −τ ) B u( τ)dτ
0 let x(0) = 0, x(t1) = x1
t1 x(t1 ) = x1 = ∫ e A( t1−τ ) B u( τ)dτ but 0 1
e A( t1−τ ) = I + A(t1 − τ) + 2! A (t1 − τ)2 +
2 t1 t1 0 0 x1 = B ∫ u( τ)dτ + A B ∫ (t1 − τ)u( τ)dτ +
+ A B∫
2 ME575 Session 27: Controllability t1 1
02 (t1 − τ)2 u( τ)dτ + School of Mechanical Engineering
Purdue University Slide 7 Controllability Proof (cont.)
If only linearly independent vectors exist, then
linearly A B = α0 B + α1 A B + α 2 A B +
2 +1 A B = α 0 A B + α1 A B + α 2 A B +
2 3 −1 + α −1 A B
+ α −1 A B Thus, AkB is linearly dependent on B, . . ., A −1B,
for all k ≥ .
rank[WC] is the order of the largest nonzero minor
in WC. This works even if WC is nonsquare.
ME575 Session 27: Controllability School of Mechanical Engineering
Purdue University Slide 8 4 ME 575 Handouts Return to Inverted Pendulum
Define state vector: ⎡0
⎢10.79
x=⎢
⎢0
⎢
⎣ −0.98 ME575 Session 27: Controllability 1
0
0
0 x = [θ 0
0
0
0 θ x x]T 0⎤
⎡0⎤
⎢ −0.1⎥
0⎥
⎥x+ ⎢
⎥u
1⎥
⎢0⎥
⎥
⎢
⎥
0⎦
⎣ 0.1 ⎦ School of Mechanical Engineering
Purdue University Slide 9 Check Controllability
Controllability Matrix: ⎡0
⎢ −0.1
WC = ⎢
⎢0
⎢
⎣ 0.1 ME575 Session 27: Controllability −0.1
0
0.1
0 0
−1.08
0
0.1 −1.08 ⎤
0⎥
⎥
0.1 ⎥
⎥
0⎦ School of Mechanical Engineering
Purdue University Slide 10 5 ME 575 Handouts Another Example
⎡ −2
⎢
A=⎢ 0
⎢0
⎣ ME575 Session 27: Controllability 1
−2
0 0⎤
⎡0 ⎤
⎥
0⎥ , B = ⎢b ⎥
⎢⎥
⎥
a⎦
⎢ 1⎥
⎣⎦ School of Mechanical Engineering
Purdue University Slide 11 Output Controllability
The system x = A x +Bu
y = C x +Du
is output controllable if a control u exists that will
transfer the output of the system from any y(to) to
any y(t1) in a finite time t1−to.
This is true iff
rank[C B CAB ME575 Session 27: Controllability 2 CA B School of Mechanical Engineering
Purdue University n−1 CA B D] = m Slide 12 6 ...
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This note was uploaded on 12/09/2011 for the course ME 575 taught by Professor Meckl during the Fall '10 term at Purdue.
 Fall '10
 Meckl
 Mechanical Engineering

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