ece504: lecture 5ece504: lecture 5 lecture 5 major topics we are still in part ii of ece504:...

ECE504: Lecture 5

ECE504: Lecture 5

D. Richard Brown III

Worcester Polytechnic Institute

30-Sep-2008

Worcester Polytechnic Institute D. Richard Brown III 30-Sep-2008 1 / 43

ECE504: Lecture 5

Lecture 5 Major Topics

We are still in Part II of ECE504: Quantitative and qualitative

analysis of systems

mathematical description → results about behavior of system

Today:

1. Existence and uniqueness of solutions to state equations forcontinuous-time systems

2. Linear algebraic tools that we are going to need for analysis ofAk and exp{At}.

◮ Subspaces◮ Nullspace and range◮ Rank◮ Matrix invertibility

You should be reading Chen Chapter 4 now (and referring back toChapter 3 for the necessary linear algebra).


ECE504: Lecture 5

Continuous-Time Linear Systems

x(t) = A(t)x(t) + B(t)u(t) (1)

y(t) = C(t)x(t) + D(t)u(t) (2)

Theorem

For any t0 ∈ R, any x(t0) ∈ Rn, and any u(t) ∈ R

p for all t ≥ t0, thereexists a unique solution x(t) for all t ∈ R to the state-update differentialequation (1). It is given as

x(t) = Φ(t, t0)x(t0) +

∫ t

t0

Φ(t, τ)B(τ)u(τ) dτ t ∈ R

where Φ(t, s) : R2 7→ R

n×n is the unique function satisfying

d

dtΦ(t, s) = A(t)Φ(t, s) with Φ(s, s) = In.


ECE504: Lecture 5

Theorem Remarks

◮ Note that this theorem claims two things:

1. A solution to the state-update equation always exists.2. The solution is unique.

◮ Why is this important?◮ Not every differential equation has a solution, e.g.

x(t) =1

twith x(0) = 5

◮ Not every differential equation has a unique solution

x(t) = 3(x(t))2/3 with x(0) = 0

◮ We have already established uniqueness for the vector/matrix stateupdate differential equation x(t) = A(t)x(t) + B(t)u(t) with initialcondition x(t0).

◮ We still need to establish existence.


ECE504: Lecture 5

Theorem: Existence Proof Warmup #1

To develop some intuition, let’s first assume that everything is scalar,i.e. p = q = n = 1. Our state update equation becomes

x(t) = a(t)x(t) + b(t)u(t)

Let

φ(t, s) := exp

{∫ t

s

a(τ) dτ

}

What is φ(s, s)?

What is ddt

φ(t, s)?


ECE504: Lecture 5


Note that φ(t, s) = exp{∫ t

sa(τ) dτ

}

always exists and satisfies its own

differential equation:

d

dtφ(t, s) = a(t)φ(t, s) with φ(s, s) = 1.

Now lets try the following solution to the scalar state-update differentialequation with initial state condition x(t0):

x(t) = φ(t, t0)x(t0) +

∫ t

t0

φ(t, τ)b(τ)u(τ) dτ ∀t ∈ R

To see that this is indeed a solution, we need to confirm two things:

1. Does our solution satisfy the initial condition requirement of thescalar state-update DE?

2. Does our solution really solve the scalar state-update DE?


ECE504: Lecture 5


To develop additional intuition, let’s now assume that everything istime-invariant, i.e. A(t) ≡ A and B(t) ≡ B. Our state update equationbecomes

x(t) = Ax(t) + Bu(t)

Let

Φ(t, s) :=

∞∑

k=0

Ak 1

k!(t − s)k

What is Φ(s, s)?

What is ddt

Φ(t, s)?


ECE504: Lecture 5


Note that Φ(t, s) =∑

∞

k=0 Ak 1k!

(t − s)k exists for any A ∈ Rn×n, t ∈ R,

and s ∈ R. Moreover, Φ(t, s) satisfies its own differential equation:

d

dtΦ(t, s) = AΦ(t, s) with Φ(s, s) = In.

Now lets try the following solution to the scalar state-update differentialequation with initial state condition x(t0):

x(t) = Φ(t, t0)x(t0) +

∫ t

t0

Φ(t, τ)B(τ)u(τ) dτ ∀t ∈ R

To see that this is indeed a solution, we need to confirm two things:

1. Does our solution satisfy the initial condition requirement of thescalar state-update DE?

2. Does our solution really solve the scalar state-update DE?


ECE504: Lecture 5

Theorem: Existence Proof for General Case

For the general (non-scalar, time-varying) case, we propose the solution

x(t) = Φ(t, t0)x(t0) +

∫ t

t0

Φ(t, τ)B(τ)u(τ) dτ (3)

where the state transition matrix satisfies the matrix differential equation

d

dtΦ(t, s) = A(t)Φ(t, s) with Φ(s, s) = In. (4)

Note that (4) is consistent with our two warmup cases.

To complete the existence proof, we need to:

1. Show that (3) with Φ(t, s) defined according to (4) satisfies theinitial condition requirement of the state-update DE.

2. Show that (3) with Φ(t, s) defined according to (4) is indeed asolution to the state-update DE.

3. Show that there always exists a solution to the matrix DE (4).


ECE504: Lecture 5

Theorem: Existence Proof for General Case: Part 1

Show that (3) with Φ(t, s) defined according to (4) satisfies the initialcondition requirement of the state-update DE.


ECE504: Lecture 5


Show that (3) with Φ(t, s) defined according to (4) is indeed a solution tothe state-update DE.


ECE504: Lecture 5


Show that there always exists a solution to the matrix DE (4).


ECE504: Lecture 5

Peano-Baker Series Example

A(t) =

[0 0t 0

]


ECE504: Lecture 5

Fundamental Matrix Method

While the Peano-Baker series establishes existence (and thus concludes theproof of the existence and uniqueness theorem), it is sometimes easier tofind Φ(t, s) via the “fundamental matrix method” (Chen section 4.5).

Basic idea:

1. Consider the the continuous time DE with x(t) ∈ Rn

x(t) = A(t)x(t) (5)

2. Choose n different initial conditions x1(t0), . . . ,xn(t0). These n

initial condition vectors must be linearly independent.

3. These n different initial conditions lead to n different solutions to (5).Call these solutions x1(t), . . . ,xn(t) and put them into a matrixX(t) = [x1(t), . . . ,xn(t)] ∈ R

n×n.

4. Note that X(t) = A(t)X(t). The quantity X(t) is called afundamental matrix of (5). Is the fundamental matrix unique?


ECE504: Lecture 5

Fundamental Matrix Method

Let X(t) be any fundamental matrix of (5). Note that X(t) is invertiblefor all t (see Chen p. 107). The state transition matrix Φ(t, s) can then becomputed as

Φ(t, s) = X(t)X−1(s).

Check:

Φ(s, s) =

d

dtΦ(t, s) =


ECE504: Lecture 5

Fundamental Matrix Example

A(t) =

[0 0t 0

]


ECE504: Lecture 5

Remarks on the CT State-Transition Matrix Φ(t, s)

1. There are many ways to compute Φ(t, s). Some are easier thanothers, but computing Φ(t, s) is almost always difficult.

2. Do different methods for computing Φ(t, s) lead to differentsolutions?

3. Unlike the DT-STM Φ[k, j], the CT-STM Φ(t, s) is defined for any(t, s) ∈ R

2. This means that we can specify an initial state x(t0) andcompute the system response at times prior to t0.

4. It is easy to show that Φ(t, s) possesses the semi-group property, i.e.

Φ(t, τ) = Φ(t, s)Φ(s, τ)

for any (t, τ, s) ∈ R3 from the fundamental matrix formulation:

Φ(t, τ) = Φ(t, s)Φ(s, τ) = X(t)X−1(s)X(s)X−1(τ) = X(t)X−1(τ)


ECE504: Lecture 5

Important Special Case: A(t) ≡ A

When A(t) ≡ A, the state-transition matrix Peano-Baker series becomes

Φ(t, s) =∞∑

k=0

Mk(t, s)

=∞∑

k=0

∫ t

s

∫ τ1

s

· · ·

∫ τk−1

s

AA · · ·A︸︷︷︸

k−fold product

dτk · · · dτ1

=

∞∑

k=0

Ak

∫ t

s

∫ τ1

s

· · ·

∫ τk−1

s

dτk · · · dτ1

To compute Mk(t, s), let’s look at k = 0, 1, 2, . . . to see the pattern...


ECE504: Lecture 5


By induction, we can show that

Mk(t, s) = Ak 1

k!(t − s)k

hence

Φ(t, s) =

∞∑

k=0

Ak 1

k!(t − s)k

which is consistent with our earlier result (warmup #2).

Suppose, for x ∈ C, we have

f(x) =

∞∑

k=0

xk

k!

What is f(x)?Worcester Polytechnic Institute D. Richard Brown III 30-Sep-2008 19 / 43

ECE504: Lecture 5

Matrix Exponential

Definition (Matrix Exponential)

Given W ∈ Cn×n, the matrix exponential is defined as

exp(W ) =∞∑

k=0

W k

k!

Note that the matrix exponential is not performed element-by-element, i.e.

exp

([w11 w12

w21 w22

])

6=

[ew11 ew12

ew21 ew22

]

Matlab has a special function (expm) that computes matrix exponentials.Calling exp(W) will not give the same results as expm(W).


ECE504: Lecture 5


Putting it all together, when A(t) ≡ A, we can say that

Φ(t, s) =

∞∑

k=0

Ak 1

k!(t − s)k = exp {(t − s)A}

Then the solution to the LTI continuous-time state-update DE is

x(t) = exp {(t − t0)A}x(t0) +

∫ t

t0

exp {(t − τ)A}B(τ)u(τ) dτ

and the output equation is

y(t) = C(t) exp {(t − t0)A}x(t0) + C(t)

Z

t

t0

exp {(t − τ )A}B(τ )u(τ ) dτ + D(t)u(t)


ECE504: Lecture 5

Contrast/Comparison Between CT and DT Solutions

Similarities

◮ CT and DT solutions have same “look”.

◮ CT and DT solutions have state transition matrices with sameintuitive properties, e.g. semigroup.

Differences

◮ In DT systems, x[k] is only defined for k ≥ k0 because the DT-STMΦ[k, k0] is only defined for k ≥ k0.

◮ In CT systems, x(t) is only defined for all t ∈ R because the CT-STMΦ(t, t0) is defined for all (t, t0) ∈ R

2.

◮ We didn’t prove this, but the CT-STM Φ(t, t0) is always invertible.This is not true of the DT-STM Φ[k, k0].


ECE504: Lecture 5

What We Know

◮ We know how to solve discrete-time LTV and LTI systems.“Solve” means “write an analytical expression for x[k] and y[k]given A[k], B[k], C[k], D[k], and x[k0]”.

◮ We know that solutions must exist and must be unique.

◮ We know how to solve continuous-time LTV and LTI systems.“Solve” means “write an analytical expression for x(t) and y(t)given A(t), B(t), C(t), D(t), and x(t0)”.

◮ We know that solutions must exist and must be unique.

◮ We also know two ways to compute the state transition matrix.

◮ We know some of the properties of state transition matrices.

◮ We know differences between the DT-STM and the CT-STM.


ECE504: Lecture 5

Where We Are Heading

Our focus is going to shift primarily to LTI systems for a little while.

Recall that, when A is not a function of time, the state transitionmatrices become

Φ[k, j] = Ak−j (discrete time)

Φ(t, s) = exp{(t − s)A} (continuous time)

We would like to be able to better analyze these matrix functions inorder to, for example, efficiently compute Ak−j.

We are going to need to learn some more linear algebra first...


ECE504: Lecture 5

Sets and Subspaces

Let A and B be sets.

◮ A ⊂ B means that all elements of the set A are also in the set B.

◮ x ∈ A to mean that x is an element of the set A.

◮ A ⊂ B and x ∈ A implies that x ∈ B.

Definition

S ⊂ Rn is a subspace if and only if S is closed under addition and scalar

multiplication, i.e.

x ∈ S and y ∈ S ⇒ x + y ∈ S

and

x ∈ S and α ∈ R ⇒ αx ∈ S.

Note that subspaces must always include the zero vector.


ECE504: Lecture 5

Spanning Set of a Subspace

Definition

A spanning set for the subspace S ⊂ Rn is a set of vectors s1, . . . sp, each

in S, such that every element of S can be expressed as a linearcombination of the vectors s1, . . . sp, i.e.

x ∈ S ⇒ there exists α1, . . . , αp such that x = α1s1 + · · · + αpsp

where αi ∈ R for i = 1, . . . , p.

Example: Suppose S is the xy plane in R3. Which of the following are

spanning sets?

8

<

:

2

4

1

0

0

3

5

9

=

;

or

8

<

:

2

4

1

0

0

3

5 ,

2

4

0

1

0

3

5

9

=

;

or

8

<

:

2

4

1

0

0

3

5 ,

2

4

0

1

0

3

5 ,

2

4

0

0

1

3

5

9

=

;

or

8

<

:

2

4

1

0

0

3

5 ,

2

4

1

1

0

3

5

9

=

;

or

8

<

:

2

4

1

0

0

3

5 ,

2

4

1

1

0

3

5 ,

2

4

0

1

0

3

5

9

=

;


ECE504: Lecture 5

Some Facts About Subspaces, Spanning Sets, and Bases

1. Every S ⊂ Rn possesses a linearly independent spanning set. Such a

set is called a basis for S. This basis is not unique, of course.

2. The number of vectors in any basis for S is the same. This number iscalled the dimension of S. We use the notation dim(S) to denotethe dimension of a subspace.

3. If S is a subspace of Rn, then dim(S) ≤ n with equality if and only if

S = Rn.

4. Any spanning set for S contains at least dim(S) vectors.

5. Any set with elements from S containing more than dim(S) vectors islinearly dependent.

6. A basis is a minimally-sized spanning set of S.

7. A basis is a maximally-sized linear independent set of vectors in S.


ECE504: Lecture 5

Nullspace and Range

Given W ∈ Rm×n (not necessarily square), there are two important

subspaces related to this matrix.

Definition

The nullspace of W is defined as the set of all x ∈ Rn such that

Wx = 0. We denote this subspace of Rn as null(W ).

Definition

The range of W is defined as the set of all y ∈ Rm such that there exists

an x satisfying Wx = y. We denote this subspace of Rm as range(W ).

The range is also sometimes called the “column space” because it is thesubspace generated by linear combinations of the columns of W .

Note that both subspaces always include the zero vector.


ECE504: Lecture 5

Nullspace

The matrix W ∈ Rm×n maps vectors from R

n to Rm. The nullspace of

W is a subspace of Rn.

null

Rn R

m


ECE504: Lecture 5

Range

The matrix W ∈ Rm×n maps vectors from R

n to Rm. The range of W is

a subspace of Rm.

range

Rn R

m


ECE504: Lecture 5

Nullspace and Range Examples

Suppose

W =

1 11 11 1

(6)

W =

[1 1 11 1 1

]

(7)

W =

[1 10 1

]

(8)

What is the nullspace and range of W in each case?


ECE504: Lecture 5

Existence and Uniqueness of Solutions to Ax = b

For any A ∈ Rm×n and b ∈ Rm, a solution to Ax = b

exists if and only if b ∈ range(A).

For any A ∈ Rm×n and b ∈ Rm, a solution to Ax = b isunique if and only if dim(null(A)) = 0.


ECE504: Lecture 5

Gaussian Elimination and Echelon Form

◮ GE is an algorithm for reducing a matrix to echelon form.◮ Once you have a matrix in echelon form, you can easily determine its

range and the dimension of its nullspace.◮ This allows you to easily answer questions about the existence and

uniqueness of solutions to Ax = b.

1. Form “augmented matrix” U = [A | b] ∈ Rm×n+1.

2. Notation U (k, :) is the kth row of U and U(k, j) is the k, jth

element of U .3. Force U(2, 1) = 0 by forming an appropriate combination of other

rows and subtracting this combination from U(2, :).4. Force U(3, 1) = U (3, 2) = 0 using the same technique.5. Keep doing this until you have an upper triangular matrix.6. You can now solve the last row since it has only one unknown.7. Back substitute your answer and solve the second last row.8. Keep doing this until you solve the top row.Worcester Polytechnic Institute D. Richard Brown III 30-Sep-2008 33 / 43

ECE504: Lecture 5

Gaussian Elimination and Echelon Form Examples


ECE504: Lecture 5

Using the Echelon Form to Determine a Basis for range(A)

The pivot columns of A for a basis for the range of A. Note that theechelon form matrix tells you which columns to pick from A. Example:

A =

1 2 −1 3 0−1 −2 2 −2 −11 2 0 4 00 0 2 2 −1

and b =

1167

(9)

After reduction to echelon form of U = [A | b], we have

1 2 −1 3 0 10 0 1 1 −1 20 0 0 0 1 30 0 0 0 0 0

(10)

The pivot columns here are the first, third, and fifth. Continued...


ECE504: Lecture 5

Using the Echelon Form to Determine a Basis for range(A)

Hence a basis for the range of A is

1−110

,

−1202

,

0−10−1

You should be able to verify that these vectors are linearly independent.

The range of A is the subspace formed by all linear combinations of thesevectors. Solutions to Ax = b exist only when b ∈ range(A).


ECE504: Lecture 5

Using the Echelon Form to Determine dim(null(A))

The dimension of the nullspace of A (also called the nullity) is simply thenumber of non-pivot columns in the echelon form.

You can use the echelon form to also find a basis for null(A) (see anygood linear algebra textbook for the details). In our example, a basis forthe nullspace is

−21000

,

−40−110

You should be able to verify that these vectors are linearly independentand that Ax = 0 if x is any linear combination of these basis vectors.

Most importantly, solutions to Ax = b are unique only whendim(null(A)) = 0.


ECE504: Lecture 5

Rank

Definition

The rank of W is defined as the dimension of the range of W , i.e.

rank(W ) := dim(range(W )).

Some useful facts:

◮ For W ∈ Rm×n, 0 ≤ rank(W ) ≤ min{m,n}.

◮ rank(W ) is equal to the number of pivot columns in the echelonform of W .

◮ Since dim(null(W )) is equal to the number of non-pivot columns inthe echelon form, rank + nullity must equal n.

◮ 0 ≤ rank(UW ) ≤ min{rank(U ), rank(W )}. In other words, matrixmultiplication can only decrease rank.


ECE504: Lecture 5

Matrix Transpose

Definition

Given

W =

w11 w12 . . . w1n

w21 w22 . . . w2n

......

...wm1 wm2 . . . wmn

∈ R

m×n

the transpose of W is given as

W⊤ =

w11 w21 . . . wm1

w12 w22 . . . wm2

......

...w1n w2n . . . wmn

∈ R

n×m.


ECE504: Lecture 5

An Important Property of the Matrix Transpose

For any A ∈ Rm×p and B ∈ R

p×n, the product C = AB is an m × n

real-valued matrix. The transpose of C is

C⊤ = (AB)⊤ = B⊤A⊤ ∈ Rn×m

Note that the order of the matrix product has been changed by thetranspose. Do the matrix dimensions agree?


ECE504: Lecture 5

Invertibility of Square Matrices

Definition

Given W ∈ Rn×n, we say that W is invertible if there exists V ∈ R

n×n

such that V W = WV = In. The quantity V is called the matrix inversefor W and we use the notation: V = W−1.

The matrix inverse does not always exist, but when it does, it is unique.

Fact: If W is invertible, then W⊤ is also invertible. To see this, just usewhat you know about the matrix inverse and the matrix transpose

WW−1 = In

(WW−1)⊤ = I⊤

n

(W−1)⊤W⊤ = In

and, by the definition, (W−1)⊤ = (W⊤)−1.


ECE504: Lecture 5

Invertibility of Square Matrices: Equivalences

The following statements are equivalent:

1. W is invertible.

2. The only x ∈ Rn satisfying Wx = 0 is x = 0.

3. For every b ∈ Rn, there exists a unique x ∈ R

n solving Wx = b.

4. The echelon form of W has no rows composed of all zeros.

5. det(W ) 6= 0.

6. rank(W ) = n.

Proofs...


ECE504: Lecture 5

Conclusions

1. Existence and uniqueness of solutions to CT and DT

systems.

2. Proofs were constructive: you can now “solve” thesesystems.

3. Linear algebra tools to lay foundation for analysis ofAk and exp (A):

◮ Subspaces◮ Nullspace and range◮ Rank◮ Matrix invertibility equivalences


ece504: lecture 5ece504: lecture 5 lecture 5 major topics we are still in part ii of ece504:...

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