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
AOE 5204
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
AOE 5204
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
AOE 5204
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
AOE 5204
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
AOE 5204
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
AOE 5204
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
AOE 5204
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
AOE 5204
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
AOE 5204
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
AOE 5204
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
AOE 5204
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)
AOE 5204
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.
AOE 5204
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?
AOE 5204
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.
AOE 5204
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
AOE 5204
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
AOE 5204
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
AOE 5204
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)]
AOE 5204
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
AOE 5204
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
AOE 5204
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?
AOE 5204
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
AOE 5204
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
AOE 5204
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