Systems Analysis and Control

Matthew M. Peet
Arizona State University

Lecture 4: The Laplace Transform (and Friends)

Introduction
In this Lecture, you will learn:
Simple ways to use state-space.
• How to find the output given an input.
• Linearity

Linear Systems.
• The Fourier Series
I Representing signals as the sum of sinusoids.
I Representing systems using response to sinusoids.
• The Fourier Transform
• The Laplace Transform
I Representing signals in the frequency domain.

I Representing systems using response to sinusoids.

How to represent a system using a Transfer Function.

• How to find the output given an input.

M. Peet Lecture 4: Control Systems 2 / 23

Recall the state-space form
Find the output given the input

State-Space:
ẋ = Ax(t) + Bu(t) Input Output
u State-Space y
y(t) = Cx(t) + Du(t) x(0) = 0 System

Basic Question: Given an input function, u(t), what is the output?

Solution: Solve the differential Equation.
Example: The equation

ẋ(t) = ax(t), x(0) = x0

has solution
x(t) = eat x0 ,
But we are interested in Matrices!!!
Not a rule, but sometimes... If it works for scalars, it also works for matrices.

M. Peet Lecture 4: Control Systems 3 / 23

The Solution to State-Space
State-Space:
ẋ = Ax(t) + Bu(t)
y(t) = Cx(t) + Du(t) x(0) = 0

The equation
ẋ(t) = Ax(t), x(0) = x0

Solution:
x(t) = eAt x0

The Matrix exponential is defined by the series expansion

1 1 1
eAt = I + (At) + (At)2 + (At)3 + · · · + (At)k + · · ·
2 6 k!
Don’t Worry! You will never have to calculate a matrix exponential by hand.
The important part is that
d At
ẋ(t) = e x0 = AeAt x0 = Ax(t), and x(0) = e0 x0 = x0
dt
What happens when we add an input instead of an initial condition?
M. Peet Lecture 4: Control Systems 4 / 23
Find the output given the input
State-Space:
ẋ = Ax(t) + Bu(t)
y(t) = Cx(t) + Du(t) x(0) = 0

The equation
ẋ(t) = Ax(t) + Bu(t), x(0) = 0

Solution:Z
t
y(t) = CeA(t−s) Bu(s)ds + Du(t)
0

Proof.
Check the solution:
Z t
x(t) = eA(t−s) Bu(s)ds
0
Z t
0
ẋ(t) = e Bu(t) + A eA(t−s) Bu(s)ds = Bu(t) + Ax(t)
0
M. Peet Lecture 4: Control Systems 5 / 23
Calculating the Output
Numerical Example, u(t) = sin(t)
State-Space:
input

ẋ = −x(t) + u(t) 1

0.8

y(t) = x(t) − .5u(t) x(0) = 0 0.6

0.4

A = −1; B = 1; C = 1; D = −.5 0.2

u(t)
0

−0.2

Solution:Z −0.4

t −0.6

CeA(t−s) Bu(s)ds + Du(t)

−0.8

y(t) = −1
0 2 4 6 8 10
0 t

Z t
−t 1
=e es sin(s)ds − sin(t)
0 2 0.6
output

1 −t s
1
e (sin s − cos s)|t0 − sin(t)
0.4
= e
2 2 0.2

1 −t t
1 0

= e e (sin t − cos t) + 1 − sin(t)

y(t)
2 2 −0.2

1 −t
−0.4

= e − cos t −0.6

2 −0.8
0 2 4 6 8 10
t

M. Peet Lecture 4: Control Systems 6 / 23

Calculating Complicated Outputs
Linearity

Problem: Finding the integral of complicated functions is tough.

Solution: Use the linearity of the system.

A system G is a MAP from input functions to

output functions. e.g., G : u 7→ y Input Output
u y=Gu
Z t G
y(t) = CeA(t−s) Bu(s)ds + Du(t)
0

Definition 1.
A SYSTEM is Linear if for any scalars, α and β, and any inputs u1 and u2 ,

G(αu1 + βu2 ) = αGu1 + βGu2

State-space systems are LINEAR.

M. Peet Lecture 4: Control Systems 7 / 23

Properties of Linearity

Properties: Multiplication
• y = Gu implies αy = G(αu).
• larger input implies a larger output.
Properties: Addition
• y1 = Gu1 and y2 = Gu2 implies G(u1 + u2 ) = y1 + y2 .
• Can calculate complicated outputs by addition.
I Step 1: Decompose input into simple parts: u(t) = u1 (t) + u2 (t)

I Step 2: Calculate simple outputs: y1 and y2 .

I Step 3: Calculate the complicated output: y(t) = y1 (t) + y2 (t)

Example: Using previous system,

ẋ = −x(t) + u(t)
y(t) = x(t) − .5u(t) x(0) = 0

Let u(t) = 2 sin(t) − cos(t).

Question: Can we compute the output without recalculating the integral???

M. Peet Lecture 4: Control Systems 8 / 23

Example of Linearity
Recall our simple state-space system:
A = −1; B = 1; C = 1; D = −.5
• We found the that input u1 (t) = sin(t) produces output
y1 (t) = 1
2 (e−t − cos t).

Now if u2 (t) = cos(nt), we have

Z t
y2 (t) = CeA(t−s) Bu2 (s)ds + Du2 (t)
0
Z t
1
= e−t es cos(ns)ds − cos(nt)
0 2
Z t
1  1
= e−t e(1+ın)s + e(1−ın)s ds − cos(nt)
2 0 2
 t
1 −t 1 (1+ın)s 1 (1−ın)s 1
= e e + e − cos(nt)
2 (1 + ın) (1 − ın) 0 2
−e−t + cos(nt) + n sin(nt) 1
= − cos(nt)
1 + n2 2

M. Peet Lecture 4: Control Systems 9 / 23

Example of Linearity

For our simple state-space system:

• We found the that input u1 (t) = sin(t) produces output
y1 (t) = 1
2 (e−t − cos t).

Now if u2 (t) = cos(t), we have

1
sin t − e−t


y2 (t) =
2
So for the composite input, u(t) = 2u1 (t) − u2 (t), we find
y(t) = 2y1 (t) − y2 (t)

1
= e−t − cos t − sin t − e−t


2
3 −t 1
= e − cos t − sin t
2 2

M. Peet Lecture 4: Control Systems 10 / 23

Frequency Domain version 1: Fourier Series
We calculated the response to a complicated signal

u(t) = 2 sin(t) − cos(t)

by adding the responses to the simple sinusoids:

u1 (t) = sin(t) and u2 (t) = cos(t)

So that y(t) = y1 (t) + y2 (t)

Question: What about more complicated
signals???
Claim: We can recreate ANY signal u(t) by adding
up sinusoids.

Conclusion: Since we know the output when the input is sinusoid, we can
calculate the solution for ANY input signal.

M. Peet Lecture 4: Control Systems 11 / 23

Frequency Domain version 1: Fourier Series
Question: How do we express our input using sines and
cosines???
Answer: The Fourier Series.

Theorem 2 (Fourier Series).

Any input signal, u on the time interval [−π, π] is the
combination of sines and cosines:
∞
a0 X
u(t) = + [an sin(nt) + bn cos(nt)]
2 n=1

Z π Z π
1 1
an = u(s) sin(ns)ds bn = u(s) cos(ns)ds
π −π π −π

M. Peet Lecture 4: Control Systems 12 / 23

Frequency Domain version 1: The Fourier Series

Assume we know the output produced by sinusoids:

• y1,n (t) is the output from u1,n (t) = sin(nt)
• y2,n (t) is the output from u2,n (t) = cos(nt)
Then for the input
∞
a0 X
u(t) = + [an sin(nt) + bn cos(nt)]
2 n=1

The output is
∞
b0 X
y(t) = y2,0 (t) + [an y1,n (t) + bn y2,n (t)]
2 n=1

Conclusion: The functions yn (t) make it very easy to calculate the output for
any input.
Problem: We need an infinite number of an and yn (t).

M. Peet Lecture 4: Control Systems 13 / 23

Frequency Domain version 2: The Fourier Transform
An alternative to the Fourier Series is the Fourier Transform.
Instead of a Sum of sinusoids, we express the signal as an Integral of sinusoids:
Z ∞ Z ∞
u(t) = a(ω) sin(ωt)dω + b(ω) cos(ωt)dω
0 0
−ıωt
e ıωt
−e eıωt +e−ıωt
Alternatively, (using sin(ωt) = 2i and cos(ωt) = 2 ):
Z ∞
u(t) = û(ıω)eıωt dω
−∞

Here û(ıω) is the Fourier Transform of the signal u(t) and is computed as
follows:
Theorem 3.
The Fourier Transform of the signal u can be found as
Z ∞
û(ıω) = e−ıωt u(t)dt
−∞

M. Peet Lecture 4: Control Systems 14 / 23

Inverse Fourier Transform

Recall: û is the Fourier Transform of u if

Z ∞
û(ıω) = e−ıωt u(t)dt
−∞

In order to recover u from û, we can use the following:

Theorem 4 (Inverse Fourier Transform).

Given û, the Fourier Transform of u, we can find u as
Z ∞
u(t) = eıωt û(ıω)dω
−∞

M. Peet Lecture 4: Control Systems 15 / 23

The Fourier Transform
Example: Step Function

Consider the input signal:

(
0 t<0
u(t) =
1 t≥0

The Fourier transform is

Z ∞
û(ıω) = e−ıωt dt
0 10

1  −ıωt ∞ 1 9

= e 0
= 8

−ıω −ıω 7

magnitude of u(w)
6

0
−10 −5 0 5 10
Frequency(w)

M. Peet Lecture 4: Control Systems 16 / 23

The Fourier Transform
Example
Consider the input signal:

u(t) = eıat

The Fourier transform is

(
∞ ω=a
û(ıω) =
6 a
0 ω=
= δ(ω − a)

magnitude
δ is the Dirac Delta function.

a frequency
We’ll do many more examples next lecture.
M. Peet Lecture 4: Control Systems 17 / 23
The Fourier Transform
The Fourier transform can be applied to samples of music to understand the
frequency composition.

Application: The Graphic Equalizer:

M. Peet Lecture 4: Control Systems 18 / 23

The Fourier Transform
Application

Figure: Equal Loudness Test

Audio Compression
• Sound is communicated in compression waves
• Humans can only hear sound in the range 20Hz-20kHz
• Music signals are usually simpler in frequency content than in time.
• e.g. The .mp3 standard.
M. Peet Lecture 4: Control Systems 19 / 23
Transfer Functions
Suppose we know the output of a system for every possible input of the form
e−ıωt .
• Input uω (t) = e−ıωt produces output g(ω, t)

Then for any input with Fourier Transform û(ıω), we can calculate the output y
as Z ∞
y(t) = û(ıω)g(ω, t)dω
−∞

Summary:
• Let û(ıω) be the Fourier Transform of the input, u(t).
• Let ŷ(ıω) be the Fourier Transform of the outut, y(t).
• Then
ŷ(ıω) = Ĝ(ıω)û(ıω)
for some Transfer Function, Ĝ
Problem: Sinusoids are NOT REAL! Every signal has to START and STOP...
Next: We consider the Laplace Transform.

M. Peet Lecture 4: Control Systems 20 / 23

Causality

Definition 5.
In a Causal System, a change in input at a later time u(t + τ ) cannot affect the
present output y(t).

Every physical system in the universe is causal.

Causal: Non-Causal
ẋ(t) = x(t) ẋ(t) = x(t + 1)
Non-Causal Systems are usually artificial.
• Noise filtering
I Averages window of future and past.
I Playback must be delayed due to causality
Conclusion: We only consider causal systems in this class.
M. Peet Lecture 4: Control Systems 21 / 23
The Laplace Transform
Transfer Functions

For causal systems, we can use the Laplace Transform:

Definition 6.
The Laplace Transform of the signal u can be found as
Z ∞
û(s) = e−st u(t)dt
0

Similar to the Fourier Transform, we have the following property.

Theorem 7.
For any causal, bounded, linear, time-invariant system, there exists a function
Ĝ(s) such that
ŷ(s) = Ĝ(s)û(s)
where ŷ(s) is the Laplace Transform of the output.
The function, Ĝ is called the Transfer Function of the system.

M. Peet Lecture 4: Control Systems 22 / 23

Summary

What have we learned today?

Simple ways to use state-space.

• How to find the output given an input.
• Linearity

Linear Systems.
• The Fourier Series
I Representing signals as the sum of sinusoids.
I Representing systems using response to sinusoids.
• The Fourier Transform
• The Laplace Transform
I Representing signals in the frequency domain.

I Representing systems using response to sinusoids.

Next Lecture: Calculating the Laplace Transform

M. Peet Lecture 4: Control Systems 23 / 23