Lect 12 Dynamic Solution

8/3/2019 Lect 12 Dynamic Solution

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Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Lecture 12: Dynamic Solution in State Space

MAE3406/TRC3600

Aerospace Control/Modeling and Control

Dr Chao Chen

Monash University

2010

1



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Motivating Problems



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

What have we learned?

Laplace transform.

Model a system in frequency domain, by means of Laplace transform.

Model a system in time domain (state space).

3



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Objectives

Solving homogeneous state space equations.

Solving nonhomogeneous state space equations. Introduction of pole-placement method.



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example: Inverted Pendulum

l

State variables: x1 = y, x2 = θ , x3 = ˙ y, x4 = θ

These four equations can be put in state-space form

(u = f ):

x = Ax + Bu, or

˙ x1

˙ x2

˙ x3

˙ x4

=

0 0 1 0

0 0 0 1

0 −mg/ M 0 0

0 ( M +m)g/ Ml 0 0

x1

x2

x3

x4

+

0

0

1/ M

−1/ Ml

u



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion


l

Assume M = 1kg, m = 1kg, g = 10m2 /s, and l = 1m.

The state-space equations become into

˙ x1

˙ x2˙ x3

˙ x4

=

0 0 1 0

0 0 0 1

0 −10 0 0

0 20 0 0

x1

x2 x3

x4

+

0

0

1

−1

u

We are interested in the behaviour of this inverted

pendulum with certain inputs and initial conditions.



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion


l


˙ x1

˙ x2

˙ x3

˙ x4

=

0 0 1 0

0 0 0 1

0 −10 0 0

0 20 0 0

x1

x2

x3

x4

+

0

0

1

−1

u

A trivial example is no input with zero initial conditions,

i.e.,

x1 = x2 = x3 = x4 = 0



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion


l


˙ x1

˙ x2

˙ x3

˙ x4

=

0 0 1 0

0 0 0 1

0 −10 0 0

0 20 0 0

x1

x2

x3

x4

+

0

0

1

−1

u

How about a tiny deviation in y, i.e.,

x1 = 0.01, x2 = x3 = x4 = 0

How about a tiny deviation in θ , i.e.,

x2 = 0.01, x1 = x3 = x4 = 0



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Homogeneous and Time-Invariant System

x = Ax

no forcing,

no changing of system with respect to time.



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Homogeneous Differential Equation

An ordinary differential equation is given by

˙ x = ax

The solution is given by

x(t ) = eat x(0)

A vector-matrix differential equation

x = Ax

It can be proved that the solution is given by

x(t ) = eAt x(0)

where

eAt ≡∞

∑k =0

Ak t k

k !

10



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

State Transition Matrix

x(t ) = eAt x(0)

The state transition matrix (STM) is defined as

ΦΦΦ(t , t 0)≡ eA(t −t 0)

If t 0 = 0, then

ΦΦΦ(t , t 0) = ΦΦΦ(t )≡ eAt

Properties of the STM ΦΦΦ(t )

1. ΦΦΦ(t ,t ) =

I (No change in time = no state change)

2. ΦΦΦ−1(t ) = ΦΦΦ(−t ) (Reverse time = reverse state change)

3. ΦΦΦ(t 1 + t 2) = ΦΦΦ(t 1)ΦΦΦ(t 2)4. [ΦΦΦ(t )]n =ΦΦΦ(nt )5. ΦΦΦ(t 2− t 1)ΦΦΦ(t 1− t 0) = ΦΦΦ(t 2− t 0) = ΦΦΦ(t 1− t 0)ΦΦΦ(t 2− t 1)

(Reversible)

11



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Computing State Transition Matrix

ΦΦΦ(t )≡ eAt ≡∞

∑k =0

Ak t k

k !

Cayley-Hamilton Theorem: Consider n×n matrix A and

its characteristic equation given by

|λ I−A|= λ n +a1λ

n−1 + . . . +an−1λ +an = 0

Then,

An +a1An−1 + . . . +an−1A +anI = 0

Theorem

eAt = α 0I +α 1A + . . . +α n−1An−1

eλ it = α 0 +α 1λ i + . . . +α n−1λ n−1i

where λ i are the eigenvalues of the matrix A.12



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Computing State Transition Matrix

x(t ) = eAt x(0) = ΦΦΦ(t )x(0)

Step 1: Find all eigenvalues of A.

Step 2: write down a set of equations with eigenvalues,

eλ 1t = α 0 +α 1λ 1 + . . .+α n−1λ n−11

eλ 2t

= α 0 +α 1λ 2 + . . .+α n−1λ

n−1

2

... =...

eλ nt = α 0 +α 1λ n + . . .+α n−1λ n−1n

Step 3: solve these equations for α i.

Step 4: compute the state transition matrix by

eAt = α 0I +α 1A + . . . +α n−1An−1

13



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

Motivating

Examples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

l

˙ x1

˙ x2

˙ x3

˙ x4

=

0 0 1 0

0 0 0 1

0 −10 0 0

0 20 0 0

x1

x2

x3

x4

,

x1(0) x2(0) x3(0) x4(0)

=

0

0.01

0

0

x(t ) = eAt x(0) Step 1: find all eigenvalues,

λ 1 = 0, λ 2 = 0, λ 3 = 4.47, λ 4 =−4.47

14



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example Step 2: write down a set of equations with eigenvalues,

eλ 1t = α 0 +α 1λ 1 + . . .+α 3λ 31

eλ 2t = α 0 +α 1λ 2 + . . .+α 3λ 32

eλ 3t = α 0 +α 1λ 3 + . . .+α 3λ 33

eλ 4t = α 0 +α 1λ 4 + . . .+α 3λ 34

with λ 1 = 0, λ 2 = 0, λ 3 = 4.47, λ 4 =−4.47.

Here we encounter a special case with two identicaleigenvalues. In order to deal with it, we need to buildanother equation, which can be done by

deλ t

d λ =

d α 0 +α 1λ + . . .+α n−1λ n−1

d λ

⇒

teλ t = α 1 + 2α 2λ + . . . + (n−1)α n−1λ n−2

This is a general rule for identical eigenvalues. Times of

differentiation is equal to the repeat number.15

L t 12



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

Therefore, we setup a set of equations as follows,

eλ 1t = α 0 +α 1λ 1 + . . .+α 3λ 31

teλ 2t = α 1 + 2α 2λ 2 + 3α 3λ 2

eλ 3t = α 0 +α 1λ 3 + . . .+α 3λ 33

eλ 4t = α 0 +α 1λ 4 + . . .+α 3λ 34

with λ 1 =

0,

λ 2 =

0,

λ 3 =

4.47

,λ

4 =−4.47.

Step 3: solving these equations yields

α 0 = 1

α 1 = t

α 2 =−0.05 + 0.025e−4.47t + 0.025e4.47t

α 3 =−0.05t −0.005e−4.47t + 0.005e4.47t

Step 4: compute the state transition matrix by

ΦΦΦ(t ) = eAt = α 0I +α 1A + . . . +α n−1An−1

16

L t 12



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

Step 5: write down the solutions as

x(t ) = ΦΦΦ(t )x(0)

where x(0) = 0 0.01 0 0T

.

The solutions are

x1 = 0.005−0.0025e−4.47t −0.0025e4.47t

x2 = 0.005e−4.47t + 0.005e4.47t

x3 = 0.011e−4.47t −0.011e4.47t

x4 = −0.022e−4.47t + 0.022e4.47t

17

Lecture 12:



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

l

18

Lecture 12:

E l



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

l

Question: how can we stabilize this system? Using input force f .

How?

Let’s choose some inputs for test.

˙ x1˙ x2

˙ x3

˙ x4

=

0 0 1 00 0 0 1

0 −10 0 0

0 20 0 0

x1

x2

x3

x4

+

00

1

−1

u

with x(0) = 0 0.01 0 0T

.

1

Lecture 12:

I h d Ti I i S



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Inhomogeneous and Time-Invariant System

x = Ax + Bu

Rewrite the state equations as

x−Ax = Bu

Multiply both sides by e−At :

e−At (x−Ax) = e−At Bu(t )

Considering a function of e−At x

d

dt e−At x = e−At Bu(t )

Time integration from t 0 to t :

e−At x(t )− e−At 0 x(t 0) =

t t 0

e−Aτ Bu(τ )d τ

0

Lecture 12:

I h d Ti I i t S t



Lecture 12:

Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Inhomogeneous and Time-Invariant System

x = Ax + Bu

Rearranging:

x(t ) = eA(t −t 0)x(t 0) +

t t 0

eA(t −τ )Bu(τ )d τ

So

x(t ) = ΦΦΦ(t , t 0)x(t 0) +

t t 0


In case of t 0

= 0, we have

x(t ) = ΦΦΦ(t )x(0) +

t 0


1

Lecture 12:

E l



Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

l

˙ x1

˙ x2

˙ x3

˙ x4

=

0 0 1 0

0 0 0 1

0 −10 0 0

0 20 0 0

x1

x2

x3

x4

+

0

0

1

−1

u

with x(0) =

0 0.01 0 0T

.

Given an input u = f = 2N, find out the state response of

the system.

2

Lecture 12:

E ample



Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

The solution is

x(t ) = ΦΦΦ(t )x(0) +

t 0


where

A =

0 0 1 0

0 0 0 1

0 −10 0 0

0 20 0 0

, B =

0

0

1

−1

, x(0) =

0

0.01

0

0

, u = 2

Further ΦΦΦ(t ) has been computed.

3

Lecture 12:

Example



Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

The solutions are given by

x1(t ) = 0.005−0.0025e−4.47t

−0.0025e4.47t

+0.02520t 2e2(2.23t )−2e2(2.23t ) + 1 + e4(2.23t )

e2(2.23t )

x2(t ) = 0.005e−4.47t + 0.005e4.47t −0.05−2e2(2.23t ) + 1 + e4(2.23t )

e2(2.23t )

4

Lecture 12:

Example



Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

How about another input f = u = 0.95sin(t )N.

The solution is

x(t ) = ΦΦΦ(t )x(0) +

t 0


5

Lecture 12:

D i S l i iExample



Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

l

˙ x1

˙ x2

˙ x3

˙ x4

=

0 0 1 0

0 0 0 1

0 −10 0 0

0 20 0 0

x1

x2

x3

x4

+

0

0

1

−1

u

with x(0) =

0 0.01 0 0T

.

Given an input of u = f = 2N, the balance is maintainedfor about 0.3s

Given an input of u = f = 0.95sin(t )N, the balance is

maintained for about 1.2s.

Is there an input to stabilize the system constantly?

6

Lecture 12:

D i S l ti iState Feedback



Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

State Feedback

x = Ax + Bu

We can choose the control signal to be

u =−Kx

The original system is now converted to a homogeneous

system (a CL system):

x = (A−BK)x = A2x

The characteristic equation of the CL system is given by

|sI−A2|= 0

If all the CL poles are in the left-half s plane, the system is

stable.

We can choose K properly to enforce the CL poles in the

left. How?7

Lecture 12:

Dynamic Solution inExample



Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

l

˙ x1

˙ x2

˙ x3

˙ x4

=

0 0 1 0

0 0 0 1

0 −10 0 00 20 0 0

x1

x2

x3

x4

+

0

0

1−1

u, x(0) =

0

0.01

00

Assume u =−Kx =−(K 1 x1 +K 2 x2 +K 3 x3 +K 4 x4)

The CL system is hence given by

˙ x1

˙ x2

˙ x3

˙ x4

=

0 0 1 0

0 0 0 1

−K 1 −10−K 2 −K 3 −K 4K 1 20 +K 2 K 3 K 4

x1

x2

x3

x4

8

Lecture 12:




Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example The CL characteristic equation is then given by

|sI−A2|= s4 + (K 3−K 4)s3 + (K 1−K 2−20)s2−10K 3s−10K 1 = 0

We can create a CL characteristic equation with all poleson the left-half s plane, i.e.,

(s+ 3)(s+ 4)(s+ 5)(s+ 6) = s4 + 18s3 + 119s2 + 342s+ 360 = 0

Comparing these two characteristic equations yields a setof equations on K i,

−K 4 +K 3−18 = 0

−139−K 2 +K 1 = 0

−10K 3−342 = 0

−10K 1−360 = 0

Solving the above equations yields

K 1 =−36, K 2 =−175, K 3 =−171/5, K 4 =−261/5

Lecture 12:




Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

The state-feedback system is given by

˙ x1

˙ x2

˙ x3

˙ x4

=

0 0 1 0

0 0 0 1

36 165 171/5 261/5

−36 −155 −171/5 −261/5

x1

x2

x3

x4

, x(0) =

0

0.01

0

0

Solving it yields

x1(t ) = 1.38e−4t t −0.415e−3t −0.13e−5t + 0.55e−6t

x2(t ) = −3.68e−4t t + 1.63e−3t −1.15e−5t −0.46e−6t

30

Lecture 12:




Dynamic Solution in

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Example

l

31

Lecture 12:

Dynamic Solution inMotivating Problems



y

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

Motivating Problems

32

Lecture 12:

Dynamic Solution inSummary



y

State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

y

Solving homogeneous state space equations.

Solving nonhomogeneous state space equations. Introduction of pole-placement method.

33

Lecture 12:

Dynamic Solution inWhat is Next?



State Space

Chao Chen

Outline

MotivatingExamples

Solving

Homogeneous State

Equations

Solving

Nonhomogeneous

State Equations

Stabilization

Conclusion

We will not discuss the state-space technology further in

this unit.

More advanced topics and systematic studies in state

space will be in MEC4418, Control Systems.

We will study the technologies in classical control

engineering in the rest of this unit.

34

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Lect 12 Dynamic Solution

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