Notes#3 - Lecture 3 notes PDF

Preview

Summary

Lecture 3 notes...

Description

Notes on Linear System Theory (ESE 500) Michael A. Carchidi September 24, 2015 Chapter 3 - State Equation Solutions

Linear

The following notes are written by Dr. Michael A. Carchidi for the Course (ESE 500) taught through the Department of Electrical

System Theory

and Systems Engineering at the University of Pennsylvania. These notes are based on the text entitled: by Wilson J. Rugh (2nd edition), and these can be viewed at

Linear System Theory

https://canvas.upenn.edu/ after you log in using your PennKey user name and Password.

1. Introduction The basic questions of existence and uniqueness of solutions are first addressed for the linear state equations unencumbered by inputs and outputs, which means we shall

first consider the equations x˙ (t) = A(t)x(t)

with initial conditions

×

x(t0 ) = x0

(1)

and solve these for x(t) for all t0 ≤ t. The n n matrix function A(t) is assumed to be continuous for all values of t. By definition, a solution to Equation (1) is a

×

continuously differentiable n 1 vector x(t) that satisfies the ordinary differential equation (ODE) x ˙ (t) = A(t)x(t) for all t0 < t, and the initial conditions (IC)

x(t) = x0 for t = t0 . To study the ODE and IC of Equation (1), it is convenient to write it as an integral equation (IE). Toward this end we take the ODE dx(t) = A(t)x(t) dt

and write it as

dx = A(t)x(t)dt.

After integrating, we get

Z

Z

x

A(τ )x(τ )dτ

dx = x0

Z

t

x

with x0

t0

Z

which leads to the IE

x − x0 =

dx = x − x0

t

A(τ )x(τ )dτ t0

Z

or

t

x(t) = x0 +

A(τ )x(τ )dτ.

(2)

t0

Note that this IE includes the initial condition

Z

x(t0 ) = x0

since

t0

A(τ )x(τ )dτ = 0, t0

and it should be clear after differentiating both sides of Equation (2) that

d d d x(t) = x0 + dt dt dt Thus we see that any

x(t)

Z

t

A(τ )x(τ )dτ = 0 + A(t)x(t). t0

that satisfies the ODE and IC of Equation (1) also

satisfies the IE of Equation (2) and conversely, which makes the ODE and IC of Equation (1) equivalent to the IE of Equation (2). 2. Existence of Solutions for the System

Given t0 ,

x˙ (t) = A(t)x(t)

x0 and an arbitrary time T > 0, we will use Equation (2) to construct n × 1 vector functions {xk (t)}∞ k=0 , defined on the interval t0 ≤ t ≤

a sequence of

t0 + T , that can be interpreted as a sequence of ”approximate” solutions to the IE of Equation (2). After proving that this sequence converges uniformly and

absolutely on the interval t0 ≤ t ≤ t0 + T , we shall then show that this limit function is continuously differentiable and satisfies the IE of Equation (2) and hence the ODE and IC of Equation (1). This will settle the question of existence of a solution by actually developing a method for constructing a solution to the ODE and IC of Equation (1).

2

To construct a sequence of approximating functions to the IE of Equation (2), we start with the initial condition x(t0 ) = x0 and define the starting function x0 (t) as the constant function

x0 (t) = x0 for all t0

≤ t ≤ t0 + T , and

x1 (t) = x0 + to get

x1 (t)

from

x0 (t).

Then we set

x2 (t) = x0 + to get

x2 (t)

from

Z

then we set

x1 (t), and

t

A(σ1 )x0 (σ1 )dσ1 t0

Z

Z

t

A(σ2 )x1 (σ2 )dσ2 t0

in general, we set

xk (t) = x0 +

t

A(σk )xk−1 (σk )dσk

(3)

t0

to get

xk (t)

from

xk−1 (t)

for all

k = 1, 2, 3, .....

To determine conditions for which the sequence

{xk (t)}∞ k=0

converges to a

solution of the IE of Equation (2), and hence the ODE and IC of Equation (1), we must first take a moment to discuss the idea of the Euclidean norm of an m × n matrix.

The Euclidean Norm of an m × n Matrix : A Mathematical Tangent Given the real

n×1

vector

⎡

⎢ ⎢ ⎢ x=⎢ ⎢ ⎣

x1 x2 x3 .. .

xn

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

vuuX t

we had already defined the Euclidean norm of

||x|| =

√ xT x =

x

as

n

i=1

3

xi2.

(4a)

Using this, we may now define the Euclidean norm of an

⎡

a11 a12 a13 ⎢ a21 a22 a23 ⎢ ⎢ A = ⎢ a31 a32 a33 ⎢ .. .. ... ⎣ . . am1 am2 am3

as

A|| = max

||

all

··· ···

..

...

.

amn

···

Ax||

||

x6=0 ||x||

p(Ax) (Ax)

which is the same as

a1n a2n a3n

···

A|| = max all

x6=0

√ xT x

matrix

⎥ ⎥ ⎥ ⎥ ⎥ ⎦

,

T

||

m×n ⎤

(4b)

r (Ax) (Ax) T

= max all

xT x

x6=0

.

We remind the student that a norm defined on a real vector space valued function from is denoted by

V

to the real numbers

v∈V

R,

i.e.,

f :V →R

V

is a real-

whose value at

f(v) = ||v|| ∈ R,

and which has the following four properties:

v|| ≥ 0

N1

||

N2

||

N3

||

N4

v|| = 0

for all

v ∈ V,

if and only if

v = 0,

αv|| = |α|||v|| for any scalar α ∈ R and any vector v ∈ V , ||u + v|| ≤ ||u|| + ||v|| for all vectors u, v ∈ V .

Note that a norm on

V

defines a metric on

V

and

via

d(u, v) = ||u − v|| where

u − v = u + (−v), || • || on V .

and this metric is called the metric induced on

V

by

the norm

Going back to Equation (4b), we note that since all x 6= 0, we may write

1 Ax|| || Ax|| = = ||x|| ||x||

||

°° 1 °° x ||

||

4

x||

||

is a positive scalar for

°° °° x °° Ax ° ° = °°A x °° = ||

||

Au||

||

where

x u= ||x||

satisfies

u||

||

°° x °° =° ° x °° = ||

||

1 ||x|| = 1. ||x||

Thus we may write Equation (4b) as

A|| = max

||

all

||u||=1

p(Au) (Au) = T

Au|| = max

||

all

||u||=1

Example #1: The Euclidean Norm of a 2 Consider the

2×2

¸∙ u ¸

Then

Au = with

2 5 2 2

u12 + u22 = 1, and

1

u2

hence

Au|| =

||

=

A|| = 2max 2

||

u1 +u2 =1

(u1 , u2 )

∙

p(2u

and so

Since the point

2

matrix

A=

∙

×

1

∙

2 5 2 2

all

||u||=1

.

¸

for any

1

p(2u

1

+ 5u2 )2 + (2u1 + 2u2 )2 .

+ 5u2 )2 + (2u1 + 2u2 )2

is a maximum is the same in which

(2u1 + 5u2 )2 + (2u1 + 2u2 )2 is a maximum, we may

u=

+ 5u2 )2 + (2u1 + 2u2 )2

in which

p(2u

√ uT AT Au.

Matrix

¸

2u1 + 5u2 2u1 + 2u2

max

first determine the maximum of

f(u1 , u2 ) = (2u1 + 5u2 )2 + (2u1 + 2u2 )2

5

∙

u1 u2

¸

(4c)

subject to the condition that

u21 + u22 = 1,

and then take its square root at the

end of the calculation. Of course, computing

∂f(u1 , u2 ) ∂u1

and setting equal to zero is not valid since

2 constraint u1

u22

∂f(u1 , u2 ) ∂u2

and

u1

and

u2

are dependent through the

= 1.

To get around this we may use the method of Lagrange

Using the method of

Lagrange multipliers, we write the constraint as u21 + u22 −

+

multipliers.

1 = 0 and

consider the function

L(u1 , u2 , λ) = f(u1 , u 2 ) + λ(u21 + u22 − 1) or

L(u1 , u2 , λ) = (2u1 + 5u2 )2 + (2u1 + 2u2 )2 + λ(u21 + u22 − 1) which, through the extra multiplier variable and

λ

µ,

allows one to treat all of

u1 , u2

as independent variables and hence we may now set

∂L = (16 + 2λ)u1 + 28u2 = 0 ∂u1 and

∂L = 28u1 + (58 + 2λ)u2 = 0 ∂u2

and

The

∂L = u21 + u22 − 1 = 0. ∂λ

first two equations lead to ∙

16 + 2λ 28 28 58 + 2λ

¸∙

u1 u2

while the third equation leads to the constraint the solution

u1 = u2 = 0 is

¸ =

∙

0 0

¸ .

u21 + u22 = 1, and since u21 + u22 = 1,

unacceptable, and thus we must require that

∙

16 + 2λ 28 28 58 + 2λ 6

¸−1

does not exist, because otherwise, we would get

∙

u1 u2

¸ =

∙

16 + 2λ 28 28 58 + 2λ

¸−1 ∙

0 0

¸ =

∙

0 0

¸ .

From our knowledge of linear algebra, we know that the nonexistence of

∙

16 + 2λ 28 28 58 + 2λ

¸−1

is equivalent to requiring that

det

∙

¸

16 + 2λ 28 28 58 + 2λ

= 0.

This leads to the quadratic equation

4λ2 + 148λ + 144 = 0 = 4(λ + 1)(λ + 36) resulting in

For

λ = −1

or

λ = −36.

λ = −1, the expression ∙ ¸∙ ¸ ∙ ¸ 16 + 2λ 28 u1 0 = 28 58 + 2λ u2 0

becomes

∙

or

yielding

16 − 2 28 28 58 − 2 ∙

u1 = −2u2 , and

14 28 28 56

¸∙

¸∙ u1 u2

coupling this with

1 u2 = ± √ 5

and hence

u1 u2

¸ =

¸ = ∙

0 0

∙

0 0

¸

¸

u21 + u22 = 1,

we

2 u1 = −2u2 = ∓ √ , 5

and this gives

f(u1 , u2 ) = (2u1 + 5u2 )2 + (2u1 + 2u2 )2

7

find that 5u22 = 1, or

µ

or

2 1 f ∓√ , ± √ 5 5

¶ µ =

2 ∓2 √ 5

µ

or

1 ± 5√ 5

1 2 f ±√ , ±√ 5 5

For

λ = −36,

the expression

becomes

∙

16 + 2λ 28 28 58 + 2λ

∙

16 − 72 28 28 58 − 72

or

∙

yielding

u2 = 2u1 , and

−56 28 28 −14

¸∙

coupling this with

1 u1 = ±√ 5

¶2 µ

2 + ∓2√ 5

¶

¶2

= 1.

¸∙

u1 u2

¸∙

u1 u2

u1 u2

1 ± 2√ 5

¸ =

∙

0 0

=

∙

0 0

¸

¸

=

∙

0 0

u21 + u22 = 1,

¸ ¸

¸

we

find that 5u21 = 1, or

2 u2 = 2u1 = ± √ , 5

and

and this gives

f(u1 , u2 ) = (2u1 + 5u2 )2 + (2u1 + 2u2 )2

µ

or

2 1 f ±√ , ± √ 5 5 or

Thus we

find that

¶ µ =

µ

1 ±2 √ 5

2 ± 5√ 5

¶2 µ

¶

+

u22

2 ± 2√ 5

¶2

1 2 = 36 = fmax . f ±√ , ± √ 5 5 √ √ ||A|| = fmax = 36 = 6.

Note that we may also solve this by setting

2 that u1

1 + ±2√ 5

= 1 is

fi

u1 = cos(θ)

and

u2 = sin(θ)

automatically satis ed. Then

f(u1 , u2 ) = (2u1 + 5u2 )2 + (2u1 + 2u2 )2 = (2 cos(θ) + 5 sin(θ))2 + (2 cos(θ) + 2 sin(θ))2 8

so

which reduces to

f(θ) = 8 cos2 (θ) + 28 cos(θ) sin(θ) + 29 sin 2 (θ) or

f(θ) = 8 + 14 sin(2θ) + 21 or

f(θ) =

µ 1 − cos(2θ) ¶ 2

21 37 + 14 sin(2θ) − cos(2θ) 2 2

Then

f 0 (θ) = 28 cos(2θ) + 21 sin(2θ) = 0 leads to

tan(2θ) = −

4 3

sin(2θ)

or

which says that

sin(2θ) = ∓

4 5

3 cos(2θ) = ± . 5

and

Using the “upper signs” for these gives

µ

4 37 + 14 − f (θ) = 2 5

¶

U

µ ¶

3 21 + − 5 2

and using the “lower signs” for these gives

µ ¶

4 37 + 14 + f (θ) = 5 2 L

showing that

fmax = 36 and

A|| =

hence ||

21 − 2

√ fmax

=1

µ 3¶

= 36 5 √ = 36 = 6. ¥ −

It should be noted that the Euclidean norm of a matrix

A

fi

satis es properties

N1 - N4 for all norms so that:

A|| ≥ 0 for all A ∈ Rm×n , m × n matrices

N1 ||

A|| = 0 if

N2 ||

and only if

αA|| = |α|||A||

N3 ||

where

Rm×n

is the vector space of all real

A = 0,

for any scalar

9

α∈R

and any matrix

A ∈ Rm×n ,

and

N4

A + B|| ≤ ||A|| + ||B||

||

for all matrices

A, B ∈ Rm×n .

It also satisfies the properties

Ax|| ≤ ||A||||x||

||

and if

A

is

m×n

with entries

,

(4f)

AB|| ≤ ||A||||B||

||

aij ,

max |aij | ≤ ||A|| ≤ (i,j)

√ mn max |aij |.

(4g)

(i,j)

Q is a symmetric n × n matrix (so QT = Q) then all of its eigenvalues are real and Q can be diagonalized by an orthogonal matrix P whose columns are simply the corresponding eigenvectors of Q. This says that if λ1 , λ2 , λ3 , ..., λn (with possible repetition) are the eigenvalues of Q, then setting ⎡ ⎤ ⎡ ⎤ λ1 0 0 · · · 0 v11 v21 v31 · · · vn1 ⎢ 0 λ2 0 · · · 0 ⎥ ⎢ v12 v22 v32 · · · vn2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ D = ⎢ 0 0 λ3 · · · 0 ⎥ and P = ⎢ v13 v23 v33 · · · vn3 ⎥ ⎢ .. .. .. . . ⎥ ⎢ .. .. ⎥ . .. ... ... ⎣ . ⎣ .. . . . . . ⎦ . ⎦ 0 0 0 · · · λn v1n v2n v3n · · · vnn We also know from matrix algebra that if

that

with

⎡

⎢ ⎢ ⎢ Q⎢ ⎢ ⎣

we have

vk1 vk2 vk3 .. .

vkn

⎤

⎡

⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = λk ⎢ ⎥ ⎢ ⎦ ⎣

PT = P−1 ,

and

vk1 vk2 vk3 .. .

vkn

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎡ and

PT QP = D.

⎢ ⎢ ⎢ ⎢ ⎢ ⎣

vk1 vk2 vk3 .. .

vkn

⎤T ⎡ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎢ ⎢ ⎢ ⎢ ⎢ ⎣

vk1 vk2 vk3 .. .

vkn

⎤

⎥ ⎥ ⎥ ⎥ = 1, ⎥ ⎦

An Important Inequality Using the fact that

λmax

PT = P−1

are smallest and largest

PT QP = D, let us show that eigenvalues of Q, respectively, then and

λmin xT x ≤ xT Qx ≤ λmax xT x 10

if

λmin

and

(4h)

n × 1 vectors x. Toward this end, P QP = D, we have Q = PDP and so for all real

we note that since

P = P−1 and T

T

T

T

T

T

T

T

T

T

T

T

T

x Qx = x (PDP )x = x PDP x = (x P)D(P x) = (P x) D(P x) or T

T

x Qx = u Du

T

u = P x.

with

But

X

⎡

⎢ ⎢ ⎢ D=⎢ ⎢ ⎣

n

T

u Du =

λi u2i

since

i=1

and so we

λ1 0 0 0 λ2 0 0 0 λ3

···

0 0 0

. . .

.. .

.. .

..

.. .

0

0

0

···

find that

X

··· ···

.

λn

⎡

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

and

⎢ ⎢ ⎢ u=⎢ ⎢ ⎣

u1 u2 u3 . . .

un

n

T

x Qx =

λi ui2.

i=1 But clearly

X

λmin ≤ λi ≤ λmax

for all

n

i=1 or

n

λmin u2i

≤

i=1

and so

≤

i=1

n

u2i ≤ u2i

i=1

λmax ui2

X X n

λi ui2 ≤ λmax

i=1

u2i .

i=1

n

n

λmin

X

X

and so

n

λi ui2

i=1

X X X n

λmin

X

i = 1, 2, 3, ..., n

T

≤ x Qx ≤ λmax

u2i .

i=1

But since

n

u2i = u u = (P x) (P x) = x PP x = x (PP )x = x (I)x = x x T

T

T

T

T

T

T

T

i=1 we see that T

T

T

λmin x x ≤ x Qx ≤ λmax x x 11

T

T

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

and the proof is complete. Note also that if if xmax satisfies Qxmax = λmaxxmax , then

xmin

satisfies

Qxmin = λmin xmin

and