Frequency Domain Modelling

8/18/2019 Frequency Domain Modelling

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AMME 3500/9501 :

System Dynamics and Control

Frequency Domain Modelling

Dr. Ian R. Manchester

A/Prof Ian R. Manchester Amme 3500 : Introduction


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Slide 3

Overview

• Linear Time-Invariant (LTI) Models

– Linearisation

– The Principle of Superposition

• Laplace transform

• Transfer Functions



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Slide 4Dr. Ian R. Manchester Amme 3500 : Introduction

Mathematical Modelling

• A mathematical model is one or more equations

that describe the relationship between the system

variables

• These models are usually derived from basic

physical principles, often with some parameters

that need to be determined experimentally


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Slide 5

Linearity

• The net response to a sum of inputs is

the sum of the output responses of each

input considered in isolation

• Mathematically: = ()

• Linear: 1 + 2 = 1) + (2



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Dynamic Models

• A linear differential equation for a single-input, single-

output system has the form

Plant

u(t) y(t)

• (an-1, …, a0, bm, … b0) are the system’s parameters

• An LTI system has parameters that are time-invariant

• n is the order of the system

1

1 0 01

( ) ( ) ( )( ) ( )

n n m

n mn n m

d y t d y t d u t a a y t b b u t

dt dt dt


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A Familiar Mechanical Example

• Spring damper system

• Mass M is suspended by

a spring and damper

• Force f is applied

• How do we predict the

motion of this system?M

f(t)

y(t)


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Translation Mechanical Elements

• Force-velocity,

force-

displacement, and

impedancetranslational

relationships

for springs,

viscous dampers,and mass


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Rotational Mechanical Elements

• Torque-angular

velocity, torque-

angular

displacement,and impedance

rotational

relationships for

springs, viscousdampers, and

inertia


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Free Body Diagram

• Draw an FBD

• Insert forces acting on

body

• Sum forces

d K yM

f(t)

y(t)

Ky(t ) K d y(t )

F =myå

my = f (t )- Ky(t )- K d y(t )my+ K d y(t )+ Ky(t ) = f (t )


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Slide 11

Nonlinear Systems

• Most real systems are nonlinear:

– A spring has only so much travel available

– Wind resistance is a quadratic function of

velocity

– Trigonometric terms show up for rotational

joints



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Slide 12

Linearisation (Taylor Series)

• A local approximation to a nonlinear

function can be found by linearisation:

≈ 0 +

− 0 + ℎ. . .



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Slide 13

Pendulum Example

• + sin =



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Slide 14

Hydraulic Pump Example



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Impulse Response

• vid


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Slide 16

Impulse Response

Dr. Ian R. Manchester Amme 3500 : Introduction


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Slide 17

Examples

• Response of a car suspension system to a pothole

• Flexing “modes” of an aircraft wing

• Or a high-precision industrial robot arm

• The response of blood glucose concentration, insulin

production, etc after eating a meal



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The Unit Impulse

• An impulse is an infinitely short pulse at t =0

• Any signal can be thought of as

the summation (integral) of

many impulses at different points

in time

• By the principal of

superposition, if we can find the

response of the system to one

impulse, we will be able to find

the response to an arbitrary

input

( ) ( ) ( ) f t d f t

f(t)

t


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Slide 19

Impulse Response

• Suppose the input consisted of just three

impulses at times t = 0, 1, 2 of size 7, 8, 9.

• What is the value of y(t ) at time t = 5?


y = f (u1 + u2 + u3)

= f (u1)+ f (u2)+ f (u3)

= 7h(5)+ 8h(4)+ 9h(3)


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Convolution Integral

• The output is the sum (integral) of each impulse

response of the system to each individual impulse of

the input , delayed by the appropriate time

• We call this convolution, and it is written as

y(t ) = h(t )u(t -t )d t 0

¥

ò

y(t ) = h(t )*u(t )


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Example : Solving in time

• Assume we have a system described by the

following differential equation

• The impulse response for this system is

( ) kt h t e

˙ y(t )+ ky(t ) = u(t )


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• If we want to find the response of the system

to a sinusoidal input, sin(wt )

Example : Solving in time

( ) ( ) ( )

sin( ( ))k

y t h u t d

e t d

w


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Cascaded systems

• Now what happens if

we have multiple

components in thesystem?

h (t)u(t) y1(t)

h1(t)y(t)

1

1 1

1

( ) ( ) ( )

( ) ( ) ( )

( ) ( ( ) * ( )) * ( )

y t u t h d

y t y t h d

y t u t h t h t


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Step Response

• Often we are interested inthe response of a system to astep change, e.g. the change

of reference signal


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Slide 25

Paradox?

• How can we have a step response for a

system defined by derivatives of the input?


1

1 0 01

( ) ( ) ( )( ) ( )

n n m

n mn n m

d y t d y t d u t a a y t b b u t

dt dt dt


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The input signal as an Exponential

• This provides us with a rich basis function for

describing functions

• Through Euler ’s formula, we find

• Fourier analysis tells us that this is sufficient for

representing any signal

( )

(cos sin )

st j t

t j t

t

e e

e e

e t j t

w

w

w w


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Slide 27

The “Frequency Domain”

• We can also consider the response of a system

to sinusoidal inputs

• Fourier theory tells us that all signals can be

decomposed into a sums of sinusoids

• Our “basis signals” are complex exponentials:


u(t ) = e st = e(s + jw )t


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Laplace Transform

• The Laplace transform is defined as:

• The inverse transform is:

F ( s ) = f (t )e - st dt 0

¥

ò

1( ) ( )

2

j st

j f t F s e ds

j

Where

s =s + jw


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Slide 29

Laplace Transform of a Step


F ( s ) = f (t )e - st dt 0

¥

ò

f (t )=1

Step input

Laplace Trans.

If s > 0


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Table of Laplace Transforms

• What about that nastyintegral in the Laplaceoperation?

• We normally use tablesof Laplace transformsrather than solving the

preceding equationsdirectly

• This greatly simplifiesthe transformation

process


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Laplace Transform Theorems


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How does this help us?

• Convolution in time is

equivalent to

multiplication in theLaplace domain

( ) ( ) * ( )

( ) ( ) ( )

y t u t h t

Y s U s H s

H(s)U(s) Y1(s)

H1(s)Y(s)

1( ) ( ) ( ) ( )Y s U s H s H s


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Comparing Solution Methods

• Starting with an impulse response, h(t), and an

input, u(t), find y(t)

u(t), h(t)

U(s), H(s)

convolution

y(t)

L L -1

Multiplication,

algebraic manipulation

Y(s)


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Transfer Function

• The transfer function H(s) of a system isdefined as the ratio of the Laplace transforms

output and input with zero initial conditions

• It is also Laplace transform of the impulse

response ( ) ( ) * ( )

( ) ( ) ( )

y t u t h t

Y s U s H s

H ( s) = Y ( s)U ( s)


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Example : Transfer Function

• Recall the system we examined earlier

• To find the transfer function for this system,we perform the following steps

sY ( s) - y(0)+ kY ( s) =U ( s)

Y ( s)( s+k )=U ( s)

H ( s) =Y ( s)

U ( s)=

1

s+ k

( ) kt h t eor

˙ y(t )+ ky(t ) = u(t )


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Slide 36Dr Ian R Manchester Amme 3500 : Introduction

Conclusions

The “Linear Time Invariant” abstraction

allows us to completely understand system

response by looking at certain basic responses

(impulse, step, frequency)

The Laplace transform (of signals) and transfer

functions (of systems) are a very convenient

representations for analysis

Frequency Domain Modelling

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