linear systems analysis - virginia techcdhall/courses/aoe5204/linears… · · 2006-09-27aoe 5204...

AOE 5204

Linear Systems AnalysisLast time:

Axisymmetric, torque-free rigid bodyLinear equations, complete analytical solution

Asymmetric, torque-free rigid bodyNonlinear equations, analytical solution for angular velocities; linearized about principal axis spin

AOE 5204

Rigid Body Dynamics Equations of MotionThe complete set of coupled

translational and rotational equa-

tions of motion for a rigid body:

ω = I−1h

v =1

mp

f = f(r,v,R,ω)

g = g(r,v,R,ω)

r =1

mp− ω×r

p = −ω×p+ f˙q = Q(q)ω

h = −ω×h+ g

The differential equations are

coupled through ω, as well as

through the dependence of force

and moment on position, velocity,

attitude, and angular velocity

The force and moment can also

include controls; for example,

the torque on a submersible ve-

hicle can include environmental

torques due to viscosity, buoy-

ancy, and gravity, as well as con-

trol torques due to actuators such

as momentum wheels and fins

The equations are also nonlinear

AOE 5204

Control of Desired MotionsMany dynamics and control prob-

lems involve relatively simple,

steady motions

Examples

• straight and level flight of anairplane

• Earth-pointing attitude of a

satellite in a circular orbit

• straight and level motion of asubmersible vehicle

• straight motion of a surface ma-rine vessel

The nonlinear equations of mo-

tion are applicable

Disturbances affect the motion

(i.e., wind gusts, solar radiation

pressure, ocean currents, waves)

Controls are used to counter the

effects of the disturbances

Linearization about the desired

motion simplifies system analyis

and control synthesis

AOE 5204

Linear Systems Analysis and Control

•Linearization of the equations of motion•Stability analysis•Modal analysis•PID and LQR control synthesis•Controllability and observability•Stabilizability and detectability

AOE 5204

General Nonlinear System

The function f is sometimes

called the vector field

Often f will not depend on t, and

in that case the system is called

autonomous, or time-invariant

Furthermore, y usually does not

depend on u; if y does depend on

u, then the system is said to have

a feedforward connection between

the input and the output

Generally, the equations describ-

ing dynamics and control prob-

lems can be developed in the

form:

x = f(x,u, t)

y = g(x,u, t)

where x ∈ Rn, y ∈ Rm, and u ∈Rp

The vector x is the state vector,

the vector u is the input vector,

and the vector y is the output

vector

AOE 5204

Linearization

θ

Equations of motion for a pendu-

lum:

θ +g

lsin θ = 0

There is no input, but clearly

θ∗ = 0 and θ∗ = π are two equlib-

rium solutions

Using physical reasoning, we also

see that θ∗ = 0 is stable, whereas

θ∗ = π is unstable

Exercise

Suppose that there is some con-

stant input, u∗, and a constant

state vector, x∗, such that

x = f(x∗,u∗, t) = 0

This combination of input and

state is called an equilibrium,

since x = 0 implies that the state

remains constant

Dynamics and control analysts

are often interested in the stabil-

ity of equilibrium motions, and in

controlling motions near equilib-

rium motions

AOE 5204

Linearization (2)Define small perturbations away from the equilibrium state and

control:

x = x∗ + δx

u = u∗ + δu

Recall that for a scalar function f(x), the Taylor series is

f(x) = f(x∗ + δx)

= f(x∗) + f 0(x∗)δx+1

2!f 00(x∗)δx2 + · · ·

For the vector functions of vector arguments here, the Taylor series

is expressed similarly:

f(x) = f(x∗ + δx) = f(x∗) +∂f

∂x(x∗)δx+ · · ·

where the · · · represent the higher order terms in the Taylor series.

AOE 5204

Linearization (3)The vector field depends on x and u

x = f(x,u, t) = f(x∗ + δx,u∗ + δu, t)

Apply the Taylor series expansion to f(x,u, t) as

f(x,u, t) = f(x∗,u∗, t) +∂f

∂x(x∗,u∗, t)δx+

∂f

∂u(x∗,u∗, t)δu+ · · ·

By definition, the term f(x∗,u∗, t) = 0, so that the differential

equation is approximated as

˙δx =∂f


∂f

∂u(x∗,u∗, t)δu

The terms ∂f∂x (x

∗,u∗, t) and ∂f∂u(x

∗,u∗, t) are n× n and n× p ma-trices, respectively

AOE 5204

Linearization (4)

The first matrix in the linearized equation

˙δx =∂f


∂f


is

∂f

∂x(x∗,u∗, t) =

⎡⎢⎣∂f1∂x1

· · · ∂f1∂xn

... · · ·...

∂fn∂x1

· · · ∂fn∂xn

⎤⎥⎦¯¯(x∗,u∗)

≡ A(t)

where the notation |(x∗,u∗) denotes that the terms in the matrixare evaluated at (x,u) = (x∗,u∗)

This matrix, A(t) is the plant matrix, or simply The Plant

If A(t) does not depend on t, then the system is a time-invariant,

or constant-coefficient system

AOE 5204

Linearization (5)

The second matrix in the linearized equation

˙δx =∂f


∂f


is

∂f

∂u(x∗,u∗, t) =

⎡⎢⎢⎣∂f1∂u1

· · · ∂f1∂up

... · · ·...

∂fn∂u1

· · · ∂fn∂up

⎤⎥⎥⎦¯¯(x∗,u∗)

≡ B(t)

This matrix, B(t) is the input matrix

Using these matrix definitions, the linear equation can be written

˙δx = A(t)δx+B(t)δu

AOE 5204

Linearization (6)

The linearized equation

˙δx =∂f


∂f


is abbreviated as˙δx = A(t)δx+B(t)δu

Frequently, the “δ” and the time-dependence are understood, and

the equation is written

x = Ax+Bu

This is a standard form for the linear state-space differential equa-

tion

One also sees

x = Fx+Gu

AOE 5204

Linearization (7)The “output” equation is handled similarly

y = g(x,u, t)

y∗ + δy = g(x∗ + δx,u∗ + δu, t)

δy =∂g


∂g


δy = C δx+D δu

y = Cx+Du

Thus, the general nonlinear system

x = f(x,u, t)

y = g(x,u, t)

is approximated by

x = Ax+Bu

y = Cx+Du

AOE 5204

Linearization ExampleConsider the motion of a spinning rigid body with a constant

“environmental” torque. The desired motion of spinning body is

ω∗ = [Ω1 Ω2 Ω3]. We want to linearize the equations of motion

about the desired motion.

Expressed in a principal frame, the environmental torque is ge.

The “control” torque, g∗, required to maintain the desired motion

is easily computed using Euler’s equations:

ω = −I−1ω×Iω + I−1ge + I−1g = f(x,u)0 = −I−1ω∗×Iω∗ + I−1ge + I−1g∗ = f(x∗,u∗)g∗ = ω∗×Iω∗ − ge

Given the desired motion, ω∗ and the environmental torque ge, we

can easily compute g∗, which is clearly constant

In this problem, x = ω, u = g, x∗ = ω∗, and u∗ = g∗

AOE 5204

Linearization Example (2)Rosetta stone: x = ω, u = g, x∗ = ω∗, and u∗ = g∗

Equations in scalar form:

ω1 =I2 − I3I1

ω2ω3 +ge1I1+g1I1= f1(x,u)

ω2 =I3 − I1I2


ω3 =I1 − I2I3


The matrix, ∂f∂x (x∗,u∗, t), is

∂f

∂x(x∗,u∗, t) =

⎡⎢⎣ 0 I2−I3I1

Ω3I2−I3I1

Ω2I3−I1I2

Ω3 0 I3−I1I2

Ω1I1−I2I3

Ω2I1−I2I3

Ω1 0

⎤⎥⎦ ≡ AClearly A depends only on the inertias and the desired steady

motion ω∗.

AOE 5204

Linearization Example (3)Rosetta stone: x = ω, u = g, x∗ = ω∗, and u∗ = g∗

Equations of motion:

ω1 =I2 − I3I1


ω2 =I3 − I1I2


ω3 =I1 − I2I3


The matrix ∂f∂u (x

∗,u∗, t) is

∂f

∂u(x∗,u∗, t) =

⎡⎣ 1I1

0 0

0 1I2

0

0 0 1I3

⎤⎦ ≡ BAs with A, the matrix B is constant and depends only on the

inertias. Note that B = I−1

AOE 5204

Linearization Example (4)Thus, the linearized equations of motion are

˙δω = A δω +B δg or x = Ax+Bu

where

A =

⎡⎢⎣ 0 I2−I3I1

Ω3I2−I3I1

Ω2I3−I1I2

Ω3 0 I3−I1I2

Ω1I1−I2I3

Ω2I1−I2I3

Ω1 0

⎤⎥⎦ and B =

⎡⎣ 1I1

0 0

0 1I2

0

0 0 1I3

⎤⎦Note that δω = ω−ω∗, and δg = g−g∗. A typical control problemis to determine the control δg that will maintain the desired motion

in the presence of initial condition errors and other disturbances.

A recommended exercise is to choose values of I, ge, and ω∗, and

then compare the results of integrating the nonlinear differential

equations with the results of integrating the linear equations.

AOE 5204

Stability AnalysisThe linearization process serves

to shift the origin to the equilib-

rium of interest

Before applying a control, we

should determine whether the

origin is stable in the absence of

any control (i.e., u = 0)

x = Ax

We further assume that A is con-

stant, so that we have a time-

invariant system

Stability analysis is based on the

eigenvalues of A

Consider the special case where

A is diagonal

The differential equation decou-

ples into the n first-order differ-

ential equations

x1 = A11x1

xj = Ajjxj

xn = Annxn

and each solution is immediately

integrable as

xj(t) = eAjjtxj(0)

Exercise: Solve for xj(t)

AOE 5204

Stability Analysis (2)If A is diagonal

xj(t) = eAjjtxj(0)

If Ajj < 0, then limt→∞ xj(t) = 0

If Ajj > 0, then limt→∞ xj(t) =

∞

Thus, Ajj < 0⇒ stability,

And Ajj > 0⇒ instability

If all states are stable, then the

system is stable.

If any of the states are unstable,

then the system is unstable.

The diagonal elements of a diago-

nal matrix are the eigenvalues of

the matrix, typically denoted

λj(A), j = 1, ..., n

For a diagonal matrix,

λj < 0∀ j ⇒ stability

λj > 0 for any j ⇒ instability

AOE 5204

Stability Analysis (3)More generally,A is not diagonal,

and its eigenvalues are either real,

or complex conjugate pairs

Denote the real and imaginary

parts of the eigenvalues by

Re λj and Im λj

The stability condition can then

be written as

Re λj < 0∀ j ⇒ stability

Re λj > 0 for any j ⇒ instabil-

ity

If any eigenvalue has Re λj = 0,

then the linear stability analysis

is inconclusive

We will develop the eigenvalue

decomposition next time

AOE 5204

Linearization ReviewThe general nonlinear system of equations is

x = f(x,u, t)

y = g(x,u, t)

where the state is x ∈ Rn, the input is u ∈ Rp, and the output isy ∈ Rm

Equilibrium motions, (x∗,u∗) satisfy

x = f(x∗,u∗, t) = 0

Linearization about (x∗,u∗), and simplifying notation with δx→ x

gives the linear system

x = Ax+Bu

y = Cx+Du

AOE 5204

State Feedback Control PreviewGeneral form for linear system is

x = Ax+Bu

y = Cx+Du

A standard control application is to let the input depend on the

state:

u = −Kx

With this form for the input, the state dynamics equation becomes

x = Ax−BKx= [A−BK]x= Ax

As we continue with linear stability analysis, keep in mind that the

analysis of x = Ax is equally applicable to x = Ax

AOE 5204

Stability AnalysisThe linearization process serves

to shift the origin to the equilib-

rium of interest

Before applying a control, we

should determine whether the

origin is stable in the absence of

any control (i.e., u = 0)

x = Ax

We further assume that A is con-

stant, so that we have a time-

invariant system

Stability analysis is based on the

eigenvalues of A

Consider the special case where

A is diagonal

The differential equation decou-

ples into the n first-order differ-

ential equations

x1 = A11x1

xj = Ajjxj

xn = Annxn

and each solution is immediately

integrable as

xj(t) = eAjjtxj(0)

Exercise: Solve for xj(t)

AOE 5204

Stability Analysis (2)If A is diagonal

xj(t) = eAjjtxj(0)

If Ajj < 0, then limt→∞ xj(t) = 0

If Ajj > 0, then limt→∞ xj(t) =

∞

Thus, Ajj < 0⇒ stability,

And Ajj > 0⇒ instability

If all states are stable, then the

system is stable.

If any of the states are unstable,

then the system is unstable.

The diagonal elements of a diago-

nal matrix are the eigenvalues of

the matrix, typically denoted

λj(A), j = 1, ..., n

For a diagonal matrix,

λj < 0∀ j ⇒ stability

λj > 0 for any j ⇒ instability

AOE 5204

Stability Analysis (3)More generally,A is not diagonal,

and its eigenvalues are either real,

or complex conjugate pairs

Denote the real and imaginary

parts of the eigenvalues by

Re λj and Im λj

The stability condition can then

be written as

Re λj < 0∀ j ⇒ stability

Re λj > 0 for any j ⇒ instabil-

ity

If any eigenvalue has Re λj = 0,

then the linear stability analysis

is inconclusive

We will develop the eigenvalue

decomposition next time

AOE 5204

Eigenvalue Decomposition of AWork with the zero-input state dynamics equation

x = Ax

Assume solutions of the form

x(t) = eλte, with constants λ ∈ C, e ∈ Cn

Substitute into the differential equation, and obtain

λeλte = Aeλte

The scalar, eλt can never be zero, so

λe = Ae

[λ1−A] e = 0

det [λ1−A] = 0

p(λ) = 0 characteristic polynomial

AOE 5204

Eigenvalue Decomposition of A (2) Working with the zero-input state dynamics equation

x = Ax, x(t) = eλte

Leads to

det [λ1−A] = 0

p(λ) = 0 characteristic polynomial

Example:

A =

∙1 2−1 6

¸⇒ [λ1−A] =

∙λ− 1 −21 λ− 6

¸det [λ1−A] = 0 ⇒ p(λ) = (λ− 1)(λ− 6) + 2 = 0λ2 − 7λ+ 8 = 0 ⇒ λ = 5.5616, 1.4384

Matlab: eig([1 2; -1 6]) and roots([1 -7 8]) both give

this result.

AOE 5204

Eigenvalue Decomposition of A (3) Generally, there are n eigenvalues (roots of the characteristic poly-

nomial), and they are distinct

There can be repeated eigenvalues

Eigenvalues can be real or complex; if complex, they always come

in complex conjugate pairs

If the eigenvalues are distinct, then there are n linearly independent

eigenvectors, ej , one associated with each eigenvalue, λj

Linear independence means that no eigenvector can be written as

a linear combination of the remaining eigenvectors

With n linearly independent eigenvectors, any vector x ∈ Rn canbe written as a linear combination of the eigenvectors:

x =nXj=1

cjej

AOE 5204

Eigenvalue Decomposition of A (4) Each eigenvalue-eigenvector pair satisfies

Aej = λjej

We can rewrite as (Exercise)

A[e1 · · · en] = [e1 · · · en] diag[λ1 · · ·λn]⇒ AE = EΛ

Since the eigenvectors are linearly independent, the matrix E is

non-singular, so

A = EΛE−1

Thus, we have factored the matrix A into the product of three

matrices, one of which is diagonal (for distinct eigenvalues)

This factorization is called the eigenvalue decomposition; there are

other matrix decompositions

AOE 5204

Eigenvalue Decomposition of A (5) We are still assuming distinct eigenvalues, in which case

A = EΛE−1

where E is the eigenvector matrix and Λ is the diagonal eigenvalue

matrix

Recall the assumed form of the solution to x = Ax:

xj(t) = eλjtej

Since the eigenvectors are linearly independent, any solution can

be written as a linear combination of these n solutions:

x(t) =nXj=1

cjeλjtej

where the constants cj depend on the initial conditions

AOE 5204

Eigenvalue Decomposition of A (5)

The general solution to x = Ax is

x(t) =nXj=1

cjeλjtej

where the constants cj depend on the initial conditions

We can rewrite the summation as (Exercise):

x(t) = E diag£eλ1t · · · eλjt · · · eλnt

¤c = EeΛtc = EeΛtE−1x(0)

Note the similarity in the eigenvalue decomposition of A and its

application to the solution:

A = EΛE−1

x(t) = EeΛtE−1x(0)

AOE 5204

Eigenvalue Decomposition of A (6)

For constant A with distinct eigenvalues, we have

A = EΛE−1

x(t) = EeΛtE−1x(0)

Where Λ is the diagonal eigenvalue matrix, E is the eigenvector

matrix, and x(0) is the initial state

Premultiply the solution by E−1 to obtain

E−1x(t) = eΛtE−1x(0)

Notice that E−1x appears on both sides. Define z(t) = E−1x(t),

and obtain

z(t) = eΛtz(0) z = Λz

Thus, for the constant-coefficient case, ifA has distinct eigenvalues,

we can transform the system to a decoupled system

AOE 5204

Eigenvalue Decomposition Example

A =

⎡⎣ −2 −1 1−0.5 −2.5 0.50.5 −0.5 −1.5

⎤⎦ Matlab: [E,Lambda]=eig(A)

Λ =

⎡⎣ −3 0 00 −1 00 0 −2

⎤⎦E =

⎡⎣ 0.7071 −0.7071 00.7071 0 0.7071

0 −0.7071 0.7071

⎤⎦Matlab check: E*Lambda*inv(E) should = A

Initial conditions

x(0) = [2 − 3 1]T

z(0) = E−1[2 − 3 1]T Matlab: z0 = E\x0z(0) = [−1.4142 − 4.2426 − 2.8284]T

AOE 5204

Eigenvalue Decomposition Example (2)

Started with

A =

⎡⎣ −2 −1 1−0.5 −2.5 0.50.5 −0.5 −1.5

⎤⎦Found the eigensystem. Transformed initial conditions from x(0)

to z(0). Now we can write the solution as

z(t) = z1(0)e−3t

⎡⎣ 100

⎤⎦+ z2(0)e−1t⎡⎣ 010

⎤⎦+ z3(0)e−2t⎡⎣ 001

⎤⎦or

x(t) = c1e−3t

⎡⎣ 0.70710.70710

⎤⎦+c2e−1t⎡⎣ −0.70710−0.7071

⎤⎦+c3e−2t⎡⎣ 00.70710.7071

⎤⎦

AOE 5204

Complex λ Eigenvalue Example

A =

⎡⎣ −1 0 00 3 −20 5 −3

⎤⎦ Λ =

⎡⎣ i 0 00 −i 00 0 −1

⎤⎦E =

⎡⎣ 0 0 10.5071 + 0.1690i 0.5071− 0.1690i 0

0.8452 0.8452 0

⎤⎦Λ is diagonal, but it is also complex. We can create a block-diagonal

matrix and block-diagonal eigenvalue matrix as follows

Eb = [Re e1 Im e1 e3]

=

⎡⎣ 0 0 10.5071 0.1690 00.8452 0 0

⎤⎦Λb = E−1b AEb

=

⎡⎣ 0 1 0−1 0 00 0 −1

⎤⎦

AOE 5204

Complex λ Eigenvalue Example (2)System is block-diagonalized. Not quite as good as diagonal, but

better than nothing.

Eb =

⎡⎣ 0 0 10.5071 0.1690 00.8452 0 0

⎤⎦Λb =

⎡⎣ 0 1 0−1 0 00 0 −1

⎤⎦x(t) = Ebe

ΛbtE−1b x(0)

What is eΛbt? It turns out to be

eΛbt =

⎡⎣ cos t sin t 0− sin t cos t 0

0 0 e−t

⎤⎦

AOE 5204

Complex λ Eigenvalue Example (3)

x(t) = EbeΛbtE−1b x(0)

where

eΛbt =

⎡⎣ cos t sin t 0− sin t cos t 0

0 0 e−t

⎤⎦

z(t) = E−1b x(t)

z(t) = eΛbtz(0)

z(t) = z1(0)

⎡⎣ cos t− sin t

0

⎤⎦+ z2(0)⎡⎣ sin tcos t0

⎤⎦+ z3(0)⎡⎣ 0

0e−t

⎤⎦+The block-diagonalization procedure works for complex eigenvalues

of the form σ + ωi, and makes use of Euler’s identity:

eσ+ωi = eσeωieσ (cosω + i sinω)

AOE 5204


Review of linearization about equilibrium motions

Diagonal systems

Eigenvalue decomposition

Diagonalization and block-diagonalization

AOE 5204

Eigenvalue Decomposition of AIf A ∈ Rn×n has distinct eigenvalues, then we can factor A into

the product of three matrices:

A = EΛE−1

where Λ is the diagonal matrix of eigenvalues, and E is the matrix

of linearly independent eigenvectors.

In general, Λ ∈ Cn×n, and E ∈ Cn×n

With the eigenvalue decomposition, we can write

eAt = EeΛtE−1

The advantage of this formulation of the matrix exponential is that

eΛt = diag£eλ1t eλjt eλnt

¤T

AOE 5204

Zero-Input SolutionThe solution to the differential equation

x = Ax

is

x(t) = eAtx(0) = EeΛtE−1x(0)

Decoupled system dynamics is obtained by transforming from x to

z

E−1x(t) = eΛtE−1x(0)

z(t) = eΛtz(0)

If Λ and E are complex, then a real-valued partial decoupling is

possible by defining Eb and Λb; e.g.

Eb = [Ree1 Ime1 e3] , Λb = E−1b AEb

AOE 5204

Block Diagonalization ProcedureIf A has complex eigenvalues and eigenvectors, they always come

in complex conjugate pairs: λj,j+1 = σ ± iω, ej,j+1 = Reej ±iImej For each c.c. pair of eigenvalues and eigenvectors, replaceeigenvectors (j and j + 1 columns of E) with Reej and Imej;e.g.

Eb = [Ree1 Ime1 e3]

Then form the block-diagonal eigenvalue matrix using

Λb = E−1b AEb

The 2× 2 blocks of Λb associated with the complex conjugate pairwill always be in the form ∙

σ ω−ω σ

¸Diagonal blocks associated with real eigenvalues are unaffected

AOE 5204

Eigenvalue Recognition: Pop QuizWhat are the eigenvalues of the following matrices?

(a)

∙1 00 −3

¸(b)

∙1 02 −3

¸(c)

∙1 40 −3

¸

(d)

⎡⎣ 1 0 00 −3 00 0 7

⎤⎦ (e)

⎡⎣ 1 0 011 −3 03 1 7

⎤⎦ (f)

⎡⎣ 1 11 30 −3 10 0 7

⎤⎦What are the roots of the following characteristic polynomials?

(g) λ2 + 3λ+ 2

(h) λ(λ2 + 3λ+ 2)

(i) (λ+ 2)(λ+ 1)

(j) λ(λ+ 2)(λ+ 1)

(k) (λ+ 7)(λ+ 2)(λ+ 1)

(l) (λ+ 7)(λ+ 2)(λ+ 1)(λ− 3)(λ− 9)

AOE 5204

Matrix Exponential and Matrix FunctionsThe matrix exponential eAt is defined so that its Taylor series is in

the same form as that for the scalar exponential

eat = 1 + at+1

2!a2t2 +

1

3!a3t3 + · · · 1

n!antn + · · ·

eAt = 1+At+1

2!A2t2 +

1

3!A3t3 + · · · 1

n!Antn + · · ·

The Cayley-Hamilton Theorem: A matrix A satisfies its own char-

acteristic polynomial:

p(λ) = λn + an−1λn−1 + · · ·+ a1λ+ a0 = 0

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

Example application: Multiply p(A) by A−1 and solve for A−1:

A−1 = − 1a0

£An−1 + an−1A

n−2 + · · ·+ a11¤

AOE 5204

Matrix Exponential ExamplesTaylor series for scalar and matrix exponentials:

eat = 1 + at+1

2!a2t2 +

1

3!a3t3 + · · · 1

n!antn + · · ·

eAt = 1+At+1

2!A2t2 +

1

3!A3t3 + · · · 1

n!Antn + · · ·

Compute the exponential of a 2× 2 diagonal matrix.

A =

∙a1 00 a2

¸eAt = 1+At+

1

2!A2t2 +

1

3!A3t3 + · · · 1

n!Antn + · · ·

=

∙1 00 1

¸+

∙a1 00 a2

¸t+

1

2!

∙a1 00 a2

¸2t2 +

1

3!

∙a1 00 a2

¸3t3 + · · ·

=

∙1 + a1t+

12!a

21t2 + 1

3!a31t3 + · · · 0

0 1 + a2t+12!a

22t2 + 1

3!a32t3 + · · ·

¸=

∙ea1t 00 ea2t

¸

AOE 5204

Matrix Exponential Examples (2)Compute the exponential of a 2×2 block-diagonal matrix,A =

∙0 ω−ω 0

¸

eAt = 1+At+1

2!A2t2 +

1

3!A3t3 + · · · 1

n!Antn + · · ·

=

∙1 00 1

¸+

∙0 ω−ω 0

¸t+

1

2!

∙0 ω−ω 0

¸2t2 +

1

3!

∙0 ω−ω 0

¸3t3 + · · ·

=

∙1 00 1

¸+

∙0 ω−ω 0

¸t+

1

2!

∙−ω2 00 −ω2

¸t2 +

1

3!

∙0 −ω3ω3 0

¸t3 + · · ·

=

∙1− 1

2!ω2t2 + · · · ωt− 1

3!ω3t3 + · · ·

−ωt+ 13!ω

3t3 + · · · 1− 12!ω

2t2 + · · ·

¸=

∙cosωt sinωt− sinωt cosωt

¸

AOE 5204

Matrix Exponential Examples (3)Consider complex conjugate eigenvalues λ1,2 = σ ± iω

The 2× 2 block of Λb is

A =

∙σ ω−ω σ

¸Instead of Taylor series, apply eAt = EeΛtE−1

The eigenvector matrix and its inverse are

E =

∙1 1i −i

¸and E−1 =

∙1/2 −i/21/2 i/2

¸Note that these eigenvectors are not normalized, nor is

normalization of eigenvectors required

The exponential term is

eΛt =

∙e(σ+iω)t 00 e(σ−iω)t

¸

AOE 5204

Matrix Exponential Examples (4)Continuing with c.c. eigenvalues λ1,2 = σ ± iω

The diagonalized matrix exponential term is

eΛt =

∙e(σ+iω)t 00 e(σ−iω)t

¸Recall that e(σ+iω)t = eσt(cosωt+ i sinωt)

Carrying out the required matrix multiplication leads to

eAt = exp

∙σ ω−ω σ

¸t

=

∙1 1i −i

¸ ∙eσt(cosωt+ i sinωt) 0

0 eσt(cosωt− i sinωt)t

¸×∙

1/2 −i/21/2 i/2

¸=

∙cosωt sinωt− sinωt cosωt

¸eσt

AOE 5204

Matrix Exponential Examples (5)An example with 5 eigenvalues: one real, λ1; two pure imaginary,

λ2,3 = ±iω1; and two complex conjugates, λ4,5 = σ ± iω2

Λb =

⎡⎢⎢⎢⎢⎣λ1 0 0 0 00 0 ω1 0 00 −ω1 0 0 00 0 0 σ ω20 0 0 −ω2 σ

⎤⎥⎥⎥⎥⎦

eΛbt =

⎡⎢⎢⎢⎢⎣eλ1t 0 0 0 00 cosω1t sinω1t 0 00 − sinω1t cosω1t 0 00 0 0 eσt cosω2t eσt sinω2t0 0 0 −eσt sinω2t eσt cosω2t

⎤⎥⎥⎥⎥⎦Any matrix A that has these five eigenvalues will have this block

diagonal structure

AOE 5204

Cayley-HamiltonThe Cayley-Hamilton Theorem: A matrix A satisfies its own char-

acteristic polynomial:

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

Example application: Multiply p(A) by A−1 and solve for A−1:

A−1 = − 1a0

£An−1 + an−1A

n−2 + · · ·+ a11¤

A 2× 2 example: A =

∙1 2−1 6

¸, p(λ) = λ2 − 7λ+ 8 = 0

A−1 = −18

½∙1 2−1 6

¸− 7∙1 00 1

¸¾=

∙0.7500 −0.25000.1250 0.1250

¸Exercise: check this calculation with Matlab, and try the procedure

on some matrices with larger n

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 Ree2 Ime2] 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


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


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.

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


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


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