ECE504: Lecture 2 - spinlab.wpi.edu · ECE504: Lecture 2 Preliminary Deﬁnition: Relaxed Systems Deﬁnition A system is said to be “relaxed” at time t = t0 if the output y(t)

ECE504: Lecture 2

ECE504: Lecture 2

D. Richard Brown III

Worcester Polytechnic Institute

09-Sep-2008

Worcester Polytechnic Institute D. Richard Brown III 09-Sep-2008 1 / 30

ECE504: Lecture 2

Lecture 2 Major Topics

We are still in Part I of ECE504: Mathematical

description of systems

model → mathematical description

You should be reading Chen chapters 2-3 now.

1. Advantages and disadvantages of differentmathematical descriptions

2. Transfer functions review

3. Relationships between mathematical descriptions


ECE504: Lecture 2

Preliminary Definition: Relaxed Systems

Definition

A system is said to be “relaxed” at time t = t0 if theoutput y(t) for all t ≥ t0 is excited exclusively by the

input u(t) for t ≥ t0.


ECE504: Lecture 2

Input-Output Description: Capabilities and Limitations

Example:

ay(t) +by(t)

cy(t)= du(t) + eu(t)

+ Can describe memoryless, lumped, or distributed systems.

+ Can describe causal or non-causal systems.

+ Can describe linear or non-linear systems.

+ Can describe time-invariant or time-varying systems.

+ Can describe relaxed or non-relaxed systems (non-zero initialconditions).

— No explicit access to internal behavior of systems, e.g. doesn’tdirectly to apply to systems like “sharks and sardines”.

— Difficult to analyze directly (differential equations).


ECE504: Lecture 2

State-Space Description: Capabilities and Limitations

Example:

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

y(t) = Cx(t) + Du(t)

— Can’t describe distributed systems. Only memoryless or lumpedsystems.

— Can’t describe non-causal systems. Only causal systems.+ Can describe linear or non-linear systems (the example is linear).+ Can describe time-invariant or time-varying systems.+ Can describe relaxed or non-relaxed systems (non-zero initial

conditions).+ Explicit description of internal system behavior, e.g. we can

determine the stability of signals internal to a system.+ Abundance of analysis techniques. Linear state-space descriptions

are analyzed with linear algebra, not calculus.Worcester Polytechnic Institute D. Richard Brown III 09-Sep-2008 5 / 30

ECE504: Lecture 2

Transfer Function Description: Capabilities and Limitations

Example:

g(s) =as2 + bs + c

ds3 + 1

+ Can describe memoryless, lumped, and some distributed systems.

— Can’t describe non-causal systems. Only causal systems.

— Can’t describe non-linear systems. Only linear systems.

— Can’t describe time-varying systems. Only time-invariantsystems.

— No explicit access to internal behavior of systems.

— Can’t describe systems with non-zero initial conditions. Implicitlyassumes that system is relaxed.

+ Abundance of analysis techniques. Systems are usually analyzedwith basic algebra, not calculus.


ECE504: Lecture 2

Impulse Response Description: Capabilities and Limitations

◮ Recall that the impulse response of a system is the output of thesystem given an input u(t) = δ(t) and relaxed initial conditions.

◮ Example:

g(t) =

{

βe−αt t ≥ 0

0 t < 0

◮ What are the capabilities and limitations of the impulse-responsedescription?


ECE504: Lecture 2

Impulse Response Matrix

Suppose you have a linear system with p inputs and q outputs. Ratherthan a simple impulse response function, you now need an impulseresponse matrix:

G(t, τ) =

g11(t, τ) . . . g1p(t, τ)...

...gq1(t, τ) . . . gqp(t, τ)

where gkℓ(t, τ) is the response at the kth output from an impulse at theℓth input at time τ . The vector output can be computed as

y(t) =

∫

∞

−∞

G(t, τ)u(τ) dτ

◮ Sanity check: What are the dimensions of everything here?

◮ How do we integrate the vector G(t, τ)u(τ)?


ECE504: Lecture 2

The One-Sided Laplace Transform

◮ Suppose f (t) : R+ 7→ Rq×p is a matrix valued function of t ≥ 0, i.e.

f(t) =

f11(t) . . . f1p(t)...

...fq1(t) . . . fqp(t)

◮ Define the Laplace transform of f (t)

f (s) =

∫

∞

0e−stf (t) dt

◮ The integral of a matrix is done element by element.◮ The notation here is consistent with your textbook:

f (s) = L [f (t)]

f (t) = L−1[

f (s)]

f (t) ↔ f (s)


ECE504: Lecture 2

Convergence of the One-Sided Laplace Transform

f (s) =

∫

∞

0e−stf (t) dt

Let Λkℓ ⊆ R be the set of all σ ∈ R such that∫

∞

0 |fkℓ(t)|e−σt dt < ∞.

If

Λ =

q⋂

k=1

p⋂

ℓ=1

Λkℓ = ∅

then the one-sided Laplace transform doesn’t exist. Otherwise, leta = min Λ. The set of complex numbers such that {s ∈ C : Re [s] > a} iscalled the “region of absolute convergence” of the Laplace transform ofthe matrix valued function f(t).


ECE504: Lecture 2

The Inverse Laplace Transform

f(t) =1

2πj

∫ σ+j∞

σ−j∞

estf(s) ds

where j :=√−1 and σ is any point in the region of absolute convergence

of f(s).

◮ This is complex integration on a path in the complex plane.

◮ In general, this integral is usually not easy to compute.

◮ Whenever possible, use tables instead.


ECE504: Lecture 2

Continuous-Time Transfer Function

Definition

Given a causal, linear, time-invariant system with p input terminals, q

output terminals, and relaxed initial conditions at time t = 0

x(0) = 0u(t), t ≥ 0

}

→ y(t), t ≥ 0

then the transfer function matrix is defined as

g(s) :=

g11(s) . . . g1p(s)...

...gq1(s) . . . gqp(s)

where gkℓ(s) :=

yk(s)

uℓ(s)

for k = 1, . . . , q and ℓ = 1, . . . , p.


ECE504: Lecture 2

Continuous-Time Transfer Function: Remarks

◮ By the relaxed assumption, the transfer function describes thezero-state response of the system.

◮ Since the transfer function must be the same for any input/outputcombination, most textbooks (including Chen) define it as theLaplace transform of the impulse response of the system.

◮ What is the Laplace transform of δ(t)?◮ Hence, given a causal, linear, time-invariant system with p input

terminals, q output terminals, and relaxed initial conditions

x(0) = 0uℓ(t) = δ(t) and um(t) = 0 for all m 6= ℓ, t ≥ 0

}

→ y(t), t ≥ 0

then

g1ℓ(s)...

gqℓ(s)

=

y1(s)uℓ(s)

...yq(s)uℓ(s)

=

y1(s)...

yq(s)


ECE504: Lecture 2

Rational Transfer Functions and Degree

Theorem (stated as a fact, Chen p.14)

If a continuous-time, linear, time-invariant system is lumped, each transfer

function in the transfer function matrix is a rational function of s.

The proof for this theorem can be found in some of the other textbooksmentioned in Lecture 1. In any case, this result means that

gkℓ(s) =N(s)

D(s)=

bmsm + bm−1sm−1 + · · · + b1s + b0

ansn + an−1sn−1 + · · · + a1s + a0

where N(s) and D(s) are polynomials of s.

Definition

The degree of a polynomial in s is the highest power of s in thepolynomial.

Example: deg(0 · s4 + 2 · s3 + 6) = .Worcester Polytechnic Institute D. Richard Brown III 09-Sep-2008 14 / 30

ECE504: Lecture 2

The Three Most Important Laplace Transforms: #1

eat ↔ 1

s − a

◮ Easy to show directly by integrating according to the definition.◮ Application: For a continuous-time, linear, time-invariant, lumped

system, we have

g(s) =bmsm + bm−1s

m−1 + · · · + b1s + b0

ansn + an−1sn−1 + · · · + a1s + a0

Compute partial fraction expansion ...

g(s) =c1

s − d1+ · · · + cn

s − dn

Then what can we say about g(t)?◮ Note that this is slightly more complicated if the roots of the

denominator are repeated.Worcester Polytechnic Institute D. Richard Brown III 09-Sep-2008 15 / 30

ECE504: Lecture 2


Notation: g(n)(t) := dgn(t)dtn

.

g(n)(t) ↔ sng(s) − sn−1g(0) − sn−2g(0) − · · · − g(n−1)(0)

◮ n = 1 case is especially useful: g(t) ↔ sg(s) − g(0).

◮ General relationship can be shown inductively using the definition andintegration by parts.


ECE504: Lecture 2


Application #1: Given the input-output differential equation description ofa continuous-time, linear, time-invariant, lumped system, we can easilycompute the transfer function.

any(n)(t) + an−1y(n−1)(t) + · · · + a1y(t) + a0y(t) =

bmu(m)(t) + bm−1u(m−1)(t) + · · · + b1u(t) + b0u(t)


ECE504: Lecture 2


Application #2: Given the state space description of a continuous-time,linear, time-invariant, lumped system, we can easily compute the transferfunction.




ECE504: Lecture 2


Notation (convolution assuming a causal relaxed system):

f(t) ∗ g(t) :=

∫ t

0f(t − τ)g(τ) dτ =

∫ t

0f(τ)g(t − τ) dτ.

It isn’t too hard to show that

f(t) ∗ g(t) ↔ f(s)g(s)

◮ Application #1: By definition of the transfer function of a linear,time-invariant, causal system, y(s) = g(s)u(s). This implies that

y(t) =

∫ t

0g(t − τ)u(τ) dτ =

∫ t

0g(τ)u(t − τ) dτ.

◮ Application #2: When u(t) = δ(t), you can show that y(t) = g(t).Hence g(t) = L−1 [g(s)] is the impulse response of the system.


ECE504: Lecture 2

What We Know: Moving Between System Descriptions

impulse

response

description

transfer

function

description

state-space

description

input-output

di�erential

equation

we will show

these soon

◮ You’ve seen transfer functions in an undergraduate course (hopefully).◮ A transfer function (matrix) describes the zero-state response of a

causal LTI system (relaxed initial conditions at time t0).◮ Transfer functions can be used to represent some distributed systems,

but these systems can’t be represented with a state-space description.


ECE504: Lecture 2

Moving Between SS and I/O Descriptions

For now, we will focus on the linear, time-invariant case.

Theorem

Given a continuous-time input-output differential equation



a state-space description of this systems exists if and only if m ≤ n < ∞.

◮ Note that this theorem states “if and only if”. What does this mean?

◮ How can we prove this theorem?


ECE504: Lecture 2

An Easy Way to Go From an LTI I/O to SS Description

To prove the “if” part of the theorem, we are going to show that given acontinuous-time, linear, time-invariant, lumped system with input-ouputdifferential equation



we can always write a linear, time-invarant state-space description



of the system.


ECE504: Lecture 2


First derive a state dynamic equation...


ECE504: Lecture 2


Now derive the output equation...

◮ Case I: m < n < ∞.

◮ Case II: m = n < ∞.

Hence, we have proved the “if” part of the theorem.


ECE504: Lecture 2

Going from an LTI SS to I/O Description

To prove the “only if” part of the theorem, we are going to show that alinear, time-invaraint state-space description



can always be written as an I/O differential equation



with m ≤ n < ∞.

To do this, however, we are going to need to learn a bit of linear algebra:◮ The determinant of a square matrix.◮ The adjoint of a square matrix.◮ The matrix inverse.◮ The matrix inverse in terms of the determinant and the adjoint.Worcester Polytechnic Institute D. Richard Brown III 09-Sep-2008 25 / 30

ECE504: Lecture 2

The Determinant of a Square Matrix

Given W ∈ Rn×n, let M ij be the (n − 1) × (n − 1) square matrix formed

by deleting the ith row and the jth column of W .

Definition

Given W ∈ Rn×n, the determinant is defined recursively as

det[W ] =n

∑

i=1

wij(−1)i+jdet[M ij]

for any j ∈ {1, . . . , n} and where the determinant of any scalar x ∈ R issimply det[x] = x.

Remark: cij := (−1)i+jdet[M ij] is called the ijth cofactor of W .

Examples...


ECE504: Lecture 2

The Adjoint of a Square Matrix

Definition

The adjoint J ∈ Rn×n of the matrix W ∈ R

n×n is defined as

J = adj(W ) =

c11 . . . c1n

......

cn1 . . . cnn

⊤

where cij the ijth cofactor of W .

Remarks:

◮ Note the transpose.

◮ The adjoint of any scalar x ∈ R is simply adj[x] = 1.

Examples...


ECE504: Lecture 2

The Matrix Inverse in Terms of the Det. and the Adjoint

Theorem (Hoffman and Kunze, Linear Algebra, 2nd Edition, p.160)

Let A be an n × n matrix. When A is invertible, the unique inverse for A

is

A−1 =adj(A)

det(A)

Remarks:

◮ See any decent linear algebra textbook for the proof.

◮ Note that A is invertible if and only if det(A) 6= 0.

◮ This is not how matrix inverses are actually computed in programslike Matlab and Octave (there are more computationally efficientways to get the same answer). Nevertheless, we can use this result inour proof of the “only if” part.


ECE504: Lecture 2

Going from an LTI SS to I/O Description

Back to the “only if” part of the proof. Recall that we can go from alinear, time-invariant state-space description to a transfer function bycomputing

g(s) = C(sI − A)−1B + D =N(s)

D(s)

Using what we now know about matrix inverses, we can write

C(sI − A)−1B + D =Cadj(sI − A)B

det(sI − A)+ D =

N(s)

D(s)+ D

◮ What can you say about the degree of D(s)?

◮ What can you say about the degree of N(s)? Recall that theelements of C and B are constants (not functions of s).

◮ But what about D?Worcester Polytechnic Institute D. Richard Brown III 09-Sep-2008 29 / 30

ECE504: Lecture 2

Conclusions

◮ Capabilities and limitations of different mathematicaldescriptions of systems.

◮ Continuous time transfer function review.

◮ Linear, time-invariant state-space → transfer function

relationship

◮ Linear algebra: identity matrix, matrix inverse,determinant, adjoint.

◮ Linear, time-invariant state-space ↔ I/O differentialequation relationship.


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ECE504: Lecture 2 - spinlab.wpi.edu · ECE504: Lecture 2 Preliminary Deﬁnition: Relaxed Systems Deﬁnition A system is said to be “relaxed” at time t = t0 if the output y(t)