AOE 5204

Linear Systems AnalysisLast time:

Diagonalization and block-diagonalization

Eigenvalue recognition

Cayley-Hamilton Theorem

Matrix Exponential

AOE 5204

Another Cayley-Hamilton ApplicationSince A satisfies its own characteristic polynomial,

p(A) = An + an−1An−1 + · · ·+ a1A+ a01 = 0

every higher order power of A can be written as a linear combina-

tion of {1,A,A2, · · ·An−1}

A useful application of this fact is the expression of eAt:

eAt = 1+At+1

2!A2t2 + · · ·

= α0(t)1+ α1(t)A+ α2(t)A2 + · · ·+ αn−1(t)A

n−1

=n−1Xj=0

αj(t)Aj

where the αj(t) terms are independent functions of time

We can calculate the αj(t) functions, but the most useful applica-

tion of this expression is in the development of the conditions for

controllability and observability

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Modal AnalysisWe have already seen the modal decomposition that arises when

we diagonalize or block-diagonalize a linear system

A more detailed example:

A =

⎡⎣ 4 −6 −12 −1 −14 −4 −4

⎤⎦has eigenvalues (−3.6252, 1.3126 ± i1.9478) (stable or unstable?),and eigenvectors

E =

⎡⎣ 0.2565 −0.8154 −0.81540.1673 −0.2967 + i0.2740 −0.2967− i0.27400.9520 −0.4109− i0.0556 −0.4109 + i0.0556

⎤⎦Block-diagonalize by defining

Eb = [e1 Re{e2} Im{e2}] and Λb = E−1b AEb

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Modal Analysis (2)

A =

⎡⎣ 4 −6 −12 −1 −14 −4 −4

⎤⎦ Λ = diag

⎡⎣ −3.62521.3126 + i1.94781.3126− i1.9478

⎤⎦

E =

⎡⎣ 0.2565 −0.8154 −0.81540.1673 −0.2967 + i0.2740 −0.2967− i0.27400.9520 −0.4109− i0.0556 −0.4109 + i0.0556

⎤⎦

Eb =

⎡⎣ 0.2565 −0.8154 00.1673 −0.2967 0.27400.9520 −0.4109 −0.0556

⎤⎦Λb =

⎡⎣ −3.6252 0 00 1.3126 1.94780 −1.9478 1.3126

⎤⎦We can use x, z, or zb

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Modal Analysis (3)

x = Ax, z = Λz, zb = Λbzb

x is coupled; z is completely decoupled; zb is partially decoupled

With x, one cannot easily see how the three states relate to the

eigenvalues and eigenvectors

With z, the fact that two states are complex-valued makes it diffi-

cult to understand the dynamics

With zb, the coupling between z2 and z3 is natural and easily un-

derstood

Consider the initial condition x(0) = [1 0 0]T

Note that the period of the oscillatory motion is T = 2π/ω, where

ω is the imaginary part of the eigenvalue

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Modal Analysis (4)

0 0.5 1 1.5 2 2.5 3 3.5−40

−20

0

20

40

60

80

100

t

x

x1

x2

x3

Initial state is x(0) = [1 0 0]T . Even though x2(0) = x3(0), all

three states diverge due to the coupling and the instability

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Modal Analysis (5)

0 0.5 1 1.5 2 2.5 3 3.5−100

−80

−60

−40

−20

0

20

40

t

z

z1

z2

z3

Initial state is x(0) = [1 0 0]T ⇒ z(0) = [−0.6897 − 1.4433 −1.1421]T . The state associated with the stable eigenvalue is not

unstable

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Modal Analysis (6)

0 0.5 1 1.5 2 2.5 3 3.5−0.7

−0.6

−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

t

z

z1

z2

z3

Initial state is z(0) = [−0.6897 0 0]T , corresponding to the stable

mode. The other states remain zero

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Modal Analysis (7)

0 0.5 1 1.5 2 2.5 3 3.5−100

−80

−60

−40

−20

0

20

40

t

z

z1

z2

z3

Initial state is z(0) = [0 − 1.4433 − 1.1421]T , corresponding tothe unstable mode. The other state remains zero

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Modal Analysis (8)

Modal analysis is based on block-diagonalizing x = Ax to obtain

zb = Λbzb

Whereas x is coupled, zb is partially decoupled

The real eigenvalues have one-dimensional real modes with expo-

nential (aka first-order) behavior

The complex conjugate eigenvalues have two-dimensional real

modes with exponential+oscillatory (aka second-order) behavior

By decomposing the motion into different modes, we can determine

the behavior of a dynamic system and focus attention on specific

modes of interest

Typically unstable, marginally stable, or barely stable modes are of

the most interest

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Back to the SolutionThe solution to the differential equation x = Ax is x(t) =

eAtx(0) = EeΛtE−1x(0), but what about

x = Ax+Bu ?

Consider the n = p = 1 case

x = ax+ u ⇒ x− ax = u

where a is constant and u is time-varying

One way to solve this first-order ODE is to turn the left-hand side

into a total derivative

Multiply by an unknown function y(t) and determine the form of

y(t) so that the left-hand side is d(xy)/dt

xy − axy = d

dt(xy) = xy + xy

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Back to the Solution (2)

We are seeking y(t) so that

xy − axy = d

dt(xy) = xy + xy

Subtract xy from both sides to obtain

−axy = xy ⇒ −ay = y ⇒ y(t) = e−aty(0)

Choose y(0) = 1, so y(t) = e−at

Now, back to the original differential equation (with y(t) in it):

d

dt(xy) = uy ⇒ d(xy) = uy dtZ (·)(t)

(·)(0)d(xy) =

Z t

0

uy(τ) dτ

x(t)y(t)− x(0)y(0) =

Z t

0

uy(τ) dτ

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Back to the Solution (2)

x(t)y(t)− x(0)y(0) =

Z t

0

uy(τ) dτ

x(t)e−at − x(0) =

Z t

0

ue−aτ dτ

x(t)e−at = x(0) +

Z t

0

ue−aτ dτ

x(t) = eatx(0) + eatZ t

0

ue−aτ dτ

The form of the solution for general n and p is the same as above:

x(t) = eAtx(0) + eAtZ t

0

e−AτBu(τ) dτ

The first term is the zero-input solution, and the second term isthe zero-state solution:

x(t) = xzi(t) + xzs(t)

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Controllability and Observability

The general solution to x = Ax+Bu is

x(t) = eAtx(0) + eAtZ t

0

e−AτBu(τ) dτ

The first term is the zero-input solution, and the second term isthe zero-state solution:

x(t) = xzi(t) + xzs(t)

Controllability. Does a control exist such that xzs(tf ) = xf forany xf and finite tf? If yes, then the system is controllable.

The output equation is y = Cx+Du

Observability. Does an initial state x(0) exist such that the zero-input solution produces identically zero output? If so, then thesystem is unobservable.

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ControllabilityThe zero-state solution to x = Ax+Bu is

xzs(t) = eAtZ t

0

e−AτBu(τ) dτ

Controllability. Does a control exist such that xzs(tf ) = xf forany xf and finite tf? If yes, then the system is controllable.

Keep in mind that x ∈ Rn, u ∈ Rp, A ∈ Rn×n, and B ∈ Rn×p

Also recall that the Cayley-Hamilton Theorem leads to

eAt =n−1Xj=0

αj(t)Aj

where the αj(t) terms are independent functions of time

Remember, the question is: Can we get from zero to anywhere infinite time?

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Controllability (2)Consider the system with

A =

∙1 00 3

¸, B =

∙01

¸Since the system is diagonal, the dynamics are decoupled and thetwo states are governed by

x1 = x1 and x2 = 3x2 + u

The control has no effect on x1; thus x1 is uncontrollable. Thereis no control u that can drive x1 to any non-zero value in finite (oreven infinite) time.

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Controllability (3)One way to add controllability to the problem in the previous ex-ample is to modify the B matrix

Consider the system with

A =

∙1 00 3

¸, B =

∙11

¸Since the system is diagonal, the dynamics are decoupled and thetwo states are governed by

x1 = x1 + u and x2 = 3x2 + u

The control affects both states, and, as we show later, the systemis controllable

Controllability depends on A and B

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Controllability (4)The zero-state solution to x = Ax+Bu is

xzs(t) = eAtZ t

0

e−AτBu(τ) dτ

and we want to determine whether any xzs(t) is possible.

xzs(t) = eAtZ t

0

e−AτBu(τ) dτ

=

Z t

0

eA(t−τ)Bu(τ) dτ

Recall that Cayley-Hamilton leads to

eAt =n−1Xj=0

αj(t)Aj

Substitute this sum into the integral

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Controllability (5)

xzs(t) =n−1Xj=0

Z t

0

αj(t− τ)AjBu(τ) dτ

=n−1Xj=0

AjB

Z t

0

αj(t− τ)u(τ) dτ

=n−1Xj=0

Aj

∙b1

Z t

0

αj(t− τ)u1(τ) dτ+

b2

Z t

0

αj(t− τ)u2(τ) dτ + · · ·+

bp

Z t

0

αj(t− τ)up(τ) dτ¸

The integrals are functions of t:

γjk(t) =

Z t

0

αj(t− τ)uk(τ) dτ, k = 1, · · · , p

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Controllability (6)

xzs(t) =n−1Xj=0

Aj

∙b1

Z t

0

αj(t− τ)u1(τ) dτ+

b2

Z t

0

αj(t− τ)u2(τ) dτ + · · ·+

bp

Z t

0

αj(t− τ)up(τ) dτ¸

The integrals are functions of t:

γjk(t) =

Z t

0

αj(t− τ)uk(τ) dτ, k = 1, · · · , p

The zero-state solution is

xzs(t) =n−1Xj=0

Aj [b1γj1(t) + b2γj2(t) + · · ·+ b1γjp(t)]

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Controllability (7)The zero-state solution is

xzs(t) =n−1Xj=0

Aj [b1γj1(t) + b2γj2(t) + · · ·+ bpγjp(t)]

Suppose that the set of n× 1 vectors©b1,Ab1, · · · ,An−1b1

ªis linearly independent.

Then any state xzs(t) can be written as a linear combination ofthese n vectors, and the system is controllable

One way to state mathematically that the set of vectors is inde-pendent is to state that

rank£b1 Ab1 · · · An−1b1

¤= n

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Controllability (8)The zero-state solution is

xzs(t) =n−1Xj=0

Aj [b1γj1(t) + b2γj2(t) + · · ·+ bpγjp(t)]

If the set of vectors ©b1,Ab1, · · · ,An−1b1

ªis not linearly independent, then it has rank < n

However, there are still p− 1 controls remaining, and perhaps theadditional Ajbk vectors can be used to construct a linearly inde-pendent set of n vectors

One way to state mathematically that the set of vectors providesa set of n linearly independent vectors is

rank£B AB · · · An−1B

¤= n

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Controllability (9)The test for controllability, therefore, is to construct the controlla-bility matrix,

Mc =£B AB · · · An−1B

¤and determine whether it has rank n.

If rank B = p, the reduced controllability matrix can be used:

Mc =£B AB · · · An−pB

¤The controllability test for this case is also

rank Mc = n

What are the dimensions of these two controllability matrices?

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Controllability (10)Determine whether system

A =

∙1 00 3

¸, B =

∙01

¸is controllable.

Note that n = 2 and p = 1.

Mc = [B AB] =

∙01

03

¸The columns are linearly dependent and thus rank Mc = 1

The system is not controllable

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Controllability (11)Determine whether system

A =

∙1 00 3

¸, B =

∙11

¸is controllable.

Note that n = 2 and p = 1.

Mc = [B AB] =

∙11

13

¸The columns are linearly independent and thus rank Mc = 2

The system is controllable

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ObservabilityThe zero-input solution to x = Ax+Bu is

xzi(t) = eAtx(0)

The output equation is

yzi(t) = Cx = CeAtx(0)

Observability. Is there an initial condition, x(0) such thatyzi(t) = 0 for all time? If not, then the system is observable.

Keep in mind that x ∈ Rn, y ∈ Rm, A ∈ Rn×n, and C ∈ Rm×n

Also recall that the Cayley-Hamilton Theorem leads to

eAt =n−1Xj=0

αj(t)Aj

where the αj(t) terms are independent functions of time