ECE504: Lecture 4 - spinlab - Worcester Polytechnic Institute

ECE504: Lecture 4

ECE504: Lecture 4

D. Richard Brown III

Worcester Polytechnic Institute

23-Sep-2008

Worcester Polytechnic Institute D. Richard Brown III 23-Sep-2008 1 / 28

ECE504: Lecture 4

Lecture 4 Major Topics

We are now starting Part II of ECE504: Quantitative and

qualitative analysis of systems

mathematical description → results about behavior of system

Today:

1. Solution of state equations for discrete-time systems

2. Solution of state equations for continuous-time systems

3. Some necessary linear algebra (and calculus review)

4. Examples

You should be reading Chen Chapter 4 now. You should also readChen 3.2-3.3 to learn about “basis”, “linear independence”, andsolutions to linear algebraic equations like Ax = y.


ECE504: Lecture 4

Linear State-Space Description of Discrete-Time Systems

x[k + 1] = A[k]x[k] + B[k]u[k]

y[k] = C[k]x[k] + D[k]u[k]

We assume a general model with p inputs, q outputs, and n states.

Given an initial time k0 ∈ Z, an initial state x[k0] ∈ Rn, how does the

state evolve for k = k0 + 1, k0 + 2, . . . ?


ECE504: Lecture 4

Solution to State Equation

Following our induction, for all k ≥ k0, we can write

x[k] = Φ[k, k0]x[k0] +

k−1∑

ℓ=k0

Φ[k, ℓ + 1]B[ℓ]u[ℓ]

where Φ is an n × n matrix valued function with two time arguments:

Φ[k, j] =

undefined k < j

In k = j

A[k − 1]A[k − 2] · · ·A[j] k > j

Remarks:◮ In is the n × n identity matrix.◮ The order of the product A[k − 1]A[k − 2] · · ·A[j] is important

because matrices don’t usually commute.◮ The matrix function Φ : Z

2 7→ Rn×n is called the state transition

matrix (STM) corresponding to A[k].


ECE504: Lecture 4

Zero-Input Response

Recall that linear systems have the nice property that we canseparately analyze the zero-input response and the zero-stateresponse.

Zero-input response: Given u[k] = 0 for all k ≥ k0, we can write

x[k] = Φ[k, k0]x[k0]

The state transition matrix Φ[k, k0] describes how the state at timek0 evolves to the state at time k ≥ k0 (in the absence of an input).

x[k0]

x[k]

Φ[k, k0]

If the STM Φ[k, k0] is invertible, then Φ−1[k, k0] = Φ[k0, k]. But

there is no guarantee that it is invertible. This operation is one-way.Worcester Polytechnic Institute D. Richard Brown III 23-Sep-2008 5 / 28

ECE504: Lecture 4

Zero-State Response

Zero-state response: Given x[k0] = 0, we can write

x[k] =

k−1∑

ℓ=k0

Φ[k, ℓ + 1]B[ℓ]u[ℓ]

In this case, we have to compute several state transition matrices:Φ[k, k0 + 1],Φ[k, k0 + 2], . . . ,Φ[k, k].

This looks like it might require a lot of computation as k gets larger.Fortunately, there are some nice properties of the state transitionmatrix that can ease the computational burden...


ECE504: Lecture 4

Some Basic Properties of the State Transition Matrix

1. Φ[j, j] = In for all j ∈ Z.

2. Φ[k + 1, j] = A[k]Φ[k, j] for all k ≥ j.

3. If ℓ ≤ j ≤ k, then Φ[k, ℓ] = Φ[k, j]Φ[j, ℓ].

This last property is called the “semigroup” property. It intuitivelysays that the transition from x[ℓ] to x[k] is the same as the transitionfrom x[ℓ] to x[j] followed by the transition from x[j] to x[k].

x[ℓ]x[j]

x[k]

Φ[j, ℓ]

Φ[k, j]


ECE504: Lecture 4

Special Case: A[k] ≡ A for all k ≥ k0

When A[k] ≡ A for all k ≥ k0, the product

A[k − 1]A[k − 2] · · ·A[j] = AA · · ·A

How many A’s are involved in this product?

Hence, when A[k] ≡ A for all k ≥ k0, the state transition matrix can bewritten as

Φ[k, j] =

undefined k < j

In k = j

Ak−j k > j.

In this case, the solution to the DT state-update difference equation is

x[k] = Ak−k0x[k0] +

k−1∑

ℓ=k0

Ak−ℓ−1B[ℓ]u[ℓ]

for all k ≥ k0.Worcester Polytechnic Institute D. Richard Brown III 23-Sep-2008 8 / 28

ECE504: Lecture 4

Discrete-Time Output Solution

For all k ≥ k0, we can just plug our solution to the state equation into ourstate-space output equation to get

y[k] = C[k]Φ[k, k0]x[k0]︸︷︷︸

zero-input response

+ C[k]k−1∑

ℓ=k0

Φ[k, ℓ + 1]B[ℓ]u[ℓ] + D[k]u[k]

︸︷︷︸

zero-state response

If the system is time-invariant, then we can write

y[k] = CAk−k0x[k0]︸︷︷︸

zero-input response

+ C

k−1∑

ℓ=k0

Ak−ℓ−1Bu[ℓ] + Du[k]

︸︷︷︸

zero-state response


ECE504: Lecture 4

Remarks on Discrete-Time State-Space Solutions

For causal, linear, lumped discrete-time systems with p input terminals, q

output terminals, and n states, we have shown that, given x[k0] and u[k]for all k ≥ k0, there exists a unique solution to the discrete-timestate-update difference equation:

x[k] = Φ[k, k0]x[k0] +k−1∑

ℓ=k0

Φ[k, ℓ + 1]B[ℓ]u[ℓ]

for all k ≥ k0 with Φ[k, j] as defined earlier.

This also implies that, given x[k0] and u[k] for all k ≥ k0, there exists aunique solution to the discrete-time output equation.


ECE504: Lecture 4

Discrete-Time State-Space Example


ECE504: Lecture 4

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 4

Theorem Remarks

◮ Note that this theorem claims two things:

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

◮ Our strategy to prove the theorem:

1. We will first show that, given two solutions to thestate-update equation, they must be identical. Thisestablishes uniqueness.

2. We will then establish existence constructively by giving asolution and showing that it satisfies the state-updateequation.

Before doing any of this, however, we are going to need to learnsome more linear algebra (and a calculus refresher)...


ECE504: Lecture 4

Euclidean Norm of a Vector

Definition

For x ∈ Rn, the Euclidean norm of x is given as

‖x‖ :=(x2

1 + · · · + x2n

)1/2.

The Euclidean norm of vectors in R1, R

2, or R3 is just your normal notion

of distance/length.

Some useful facts (easy to show from the definition):

◮ ‖x‖2 = x⊤x.

◮ ‖αx‖ = |α|‖x‖ for any α in R.

◮ ‖x + y‖ ≤ ‖x‖ + ‖y‖ for any x ∈ Rn and any y ∈ R

n. This is oftencalled the triangle inequality.


ECE504: Lecture 4

Induced Euclidean Norm of a Matrix

Definition

For A ∈ Rn×n, the induced Euclidean norm of the matrix A is given as

‖A‖ := maxx∈Rn and ‖x‖=1

‖Ax‖.

◮ The set of vectors x where ‖x‖ = 1 is a unit-sphere in Rn.

◮ The induced Euclidean norm of A is the maximum value of ‖Ax‖ asx ranges over all points on this unit-sphere.

◮ Intuitively, ‖A‖ gives a measure of how much A can magnify thelength (Euclidean norm) of a vector in R

n.

Some useful facts (not too hard to show from the definition):◮ ‖A + B‖ ≤ ‖A‖ + ‖B‖ for any A ∈ R

n×n and any B ∈ Rn×n.

◮ ‖AB‖ ≤ ‖A‖‖B‖ for any A ∈ Rn×n and any B ∈ R

n×n.◮ ‖Ax‖ ≤ ‖A‖‖x‖ for any A ∈ R

n×n and any x ∈ Rn.


ECE504: Lecture 4

Schwarz Inequality

Theorem

Given x ∈ Rn and y ∈ R

n, then

|x⊤y| = |x1y1 + · · · + xnyn| ≤ ‖x‖‖y‖

Proof:


ECE504: Lecture 4

Leibniz’ rule

Theorem

If f(t, τ) is continuous and all of the necessary derivatives exist, then

d

dt

∫ w(t)

v(t)f(t, τ) dτ = w(t)f(t, w(t)) − v(t)f(t, v(t)) +

∫ w(t)

v(t)

d

dtf(t, τ) dτ

The proof can be found in most calculus textbooks.

Two particularly useful special cases are

d

dt

∫ t

af(τ) dτ = f(t)

d

dt

∫ a

tf(τ) dτ = −f(t)

where a is not a function of t.Worcester Polytechnic Institute D. Richard Brown III 23-Sep-2008 17 / 28

ECE504: Lecture 4

Back to the Theorem: Uniqueness Proof

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

We first want to show that any solution x(t) to this state-updatedifferential equation must be unique.

To show this, suppose we had two solutions to the state-update differentialequation, x1(t) and x2(t) for t ∈ [s1, t1], both of which satisfy the initialcondition x1(t0) = x2(t0) = x(t0). Let’s prove that x1(t) must beidentical to x2(t)...


ECE504: Lecture 4

Theorem: Existence Proof Warmup

We now know that, if a solution to the state-update DE exists, it must beunique. We now need to show that a solution always exists.

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

sa(τ) dτ

}

What is φ(s, s)?

What is ddtφ(t, s)?


ECE504: Lecture 4

Theorem: Existence Proof Warmup

Note that φ(t, s) = exp{∫ t

s a(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 solution is valid, we should 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 4

Theorem: Existence Proof

For the general (non-scalar) 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)

To complete the existence proof, we need to:

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

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

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


ECE504: Lecture 4

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

1. Computation of Φ(t, s) is almost always difficult.2. The Peano-Baker series is only one way to compute Φ(t, s). Other

(perhaps better?) ways:◮ Directly solve the matrix state-update differential equations (not always

possible)◮ Fundamental matrix method (see Chen 4.5)◮ Other methods...

3. Question: Is it possible that different methods for computing theSTM will lead to different Φ(t, s)?

4. 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.

5. 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.


ECE504: Lecture 4

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

sAA · · ·A︸︷︷︸

k−fold product

dτk · · · dτ1

=

∞∑

k=0

Ak

∫ t

s

∫ τ1

s· · ·

∫ τk−1

sdτk · · · dτ1

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

◮ What is M0(t, s)?





ECE504: Lecture 4


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

Suppose, for x ∈ C, we have

f(x) =

∞∑

k=0

xk

k!

Math trivia question: What is f(x)?


ECE504: Lecture 4

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 4


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

Φ(t, s) = exp {(t − s)A}

Then the solution to the 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 4

Contrast/Comparison Between CT and DT Solutions

Similarities

◮ Results have same “look”.

◮ Both have state transition matrices with same intuitive 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 4

Conclusions

◮ Solution to LTI or LTV discrete-time state-spacedifference equations (existence and uniqueness)

◮ Solution to LTI or LTV continuous-time state-spacedifferential equations (existence and uniqueness)

◮ Special case: time-invariant A matrix

◮ LTI discrete-time systems: Ak−j

◮ LTI continuous-time systems: exp{(t − τ)A}


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