Fourier Series (for Periodic Signals)

7/31/2019 Fourier Series (for Periodic Signals)

1/21

EECE 301

Signals & Systems

Prof. Mark Fowler

Note Set #13

C-T Signals: Fourier Series (for Periodic Signals) Reading Assignment: Section 3.2 & 3.3 of Kamen and Heck


2/21

Ch. 1 Intro

C-T Signal Model

Functions on Real Line

D-T Signal Model

Functions on Integers

System Properties

LTICausal

Etc

Ch. 2 Diff EqsC-T System Model

Differential Equations

D-T Signal ModelDifference Equations

Zero-State Response

Zero-Input Response

Characteristic Eq.

Ch. 2 Convolution

C-T System Model

Convolution Integral

D-T Signal Model

Convolution Sum

Ch. 3: CT FourierSignalModels

Fourier Series

Periodic Signals

Fourier Transform (CTFT)

Non-Periodic Signals

New System Model

New Signal

Models

Ch. 5: CT FourierSystem Models

Frequency Response

Based on Fourier Transform

New System Model

Ch. 4: DT Fourier

SignalModels

DTFT

(for Hand Analysis)DFT & FFT

(for Computer Analysis)

New Signal

Model

Powerful

Analysis Tool

Ch. 6 & 8: LaplaceModels for CT

Signals & Systems

Transfer Function

New System Model

Ch. 7: Z Trans.

Models for DT

Signals & Systems

Transfer Function

New System

Model

Ch. 5: DT Fourier

System Models

Freq. Response for DT

Based on DTFT

New System Model

Course Flow DiagramThe arrows here show conceptual flow between ideas. Note the parallel structure between

the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis).


3/21

3.2 & 3.3 Fourier Series

In the last set of notes we looked at building signals using:

=

=N

Nk

tjk

kectx0)(

We saw that these build periodic signals.

Q: Can we getany periodic signal this way?

A: No! There are some periodic signals that need an infinite number of terms:

=

++=N

k

kk tkAAtx1

00 )cos()(

=

=k

tjk

kectx0)(

Fourier Series

(Complex Exp. Form)

kare integers

=

++=1

00 )cos()( k kk tkAAtx

kare integers

Fourier Series

(Trig. Form)

These are two

different

forms of the

same tool!!

There is a 3rd

form that

well see later.

Sect. 3.3

There is a related

Form in Sect. 3.2


4/21

Q: Does this now let us get any periodic signal?

A: No! Although Fourier thought so!

Dirichlet showed that there are some that cant bewritten in terms of a FS!

But those will never show up in practice!

See top of

p. 155

So what??!! Here is what!!

We can now break virtually any periodic signal into a sum of simple things

and we already understand how these simple things travel through an LTI system!

So, instead of:

)(th)(tx )()()( thtxty =

We breakx(t) into its FS components and find how each component goes

through. (See chapter 5)


5/21

To do this kind of convolution-evading analysis we need to be able to solve the

following:

Given time-domainsignal modelx(t)

Find the FScoefficients {ck}

Time-domain model Frequency-domain model

Converting time-domain signal model into

a frequency-domain singal model


6/21

Q: How do we find the (Exp. Form) Fourier Series Coefficients?

A: Use this formula (it can be proved but we wont do that!)

+ =

Tt

t

tjk

k dtetxT

c0

0

0)(1 Slightly different

than book

It uses t0 = 0.

Integrate over

any complete

period

where: T= fundamental period ofx(t) (in seconds)

0= fundamental frequency ofx(t) (in rad/second)

= 2/T

t0 = any time point (you pickt0 to ease calculations)

k

all integers

Comment: Note that for k= 0 this gives

+

=Tt

tdttx

Tc

0

0

)(10


7/21

Summarizing rules for converting between the

Time-Domain Model & the Exponential Form FS Model

Analysis Synthesis

+ =

Tt

t

tjk

k etxT

c0

0

0)(1

=

=k

tjk

kectx0)(

Use signal to figure out

the FS Coefficients

Eat food and figure out recipe

Use FS Coefficients to Build

the Signal

Read recipe and cook food

Time-Domain Model: The Periodic Signal Itself

Frequency-Domain Model: The FS Coefficients

There are similar equations for finding the FS coefficients for

the other equivalent forms But we wont worry about them

because once you have the ckyou can get all the others easily


8/21

K

,3,2,121

21

00

=

=

=

=

keAc

eAc

Ac

k

k

j

kk

j

kk

( )

( )

K,3,2,1

2

1

21

00

=

+=

=

=

k

jbac

jbac

ac

kkk

kkk

Exponential Form

=

=k

tjk

kectx0)(

Trig Form: Amplitude & Phase

=

++=1

00 )cos()(k

kk tkAAtx

Trig Form: Sine-Cosine

[ ]

=

++=1

000 )sin()cos()(k

kk tkbtkaatx { }

{ }K

K

,3,2,1,Im2

,3,2,1,Re2

00

==

==

=

kcb

kca

ca

kk

kk

K,3,2,1

2

00

=

=

=

=

k

c

cA

cA

kk

kk

=

+=

=

k

kk

kkk

a

b

baA

aA

1

22

00

tan)sin(

)cos(

00

kkk

kkk

Ab

Aa

ca

=

=

=

Three (Equivalent) Forms of FS and Their Relationships


9/21

Example of Using FS Analysis

In electronics you have seen (or will see) how to use diodes and an RC filter circuit

to create a DC power supply:

60Hz Sine

wave with

around110V RMS

msT 67.1660

11 ==

x(t) R C

60 Hz

t

ms67.16

x(t)

T = T/2 = 8.33ms

t

1. This signal goes into the RC

filter what comes out?

(Assume zero ICs)2. How do we choose the desired

RCvalues?


10/21

t )(th)(tx ?)(

x(t)=tyA

Later (Ch. 5) we will use it to analyze whaty(t) looks like

The equation for the FS coefficients is: =

Ttjk

k dtetxT

c0

0)(1

T

20 =

Choose t0 = 0 here to make things easier:

t

x(t)A

t0 t0+T

This kind of choice would

make things harder: t

x(t)

A

t0t0+T


11/21

Now what is the equation forx(t) over t [0,T]?

TttT

Atx

= 0sin)(

So using this we get:

=

=

T tT

jk

Ttjk

k

dtetT

AT

dtetxT

c

0

2

0

sin1

)(1

0

T

20 =

Determined bylooking at the plot

t

x(t)A

t0 t0+T

So now we just have to evaluate this integral as a function ofk

jk2

1


12/21

we do a Change of Variables. There are three steps:

1. Identify the new variable and sub it into the integrand

2. Determine its impact on the differential

3. Determine its impact on the limits of integration

=

T tT

jk

k dtetT

AT

c0

2

sin1

To evaluate the integral:

dtT

d

=

dT

dt=Step 2:

when t = 0

when t= T

Step 3: 00 ==T

== TT

tT

=Step 1: 2)sin( jke

( )

( )

=

=

=

0

2

0

2

0

2

sin

sin

1

sin1

deA

d

T

eAT

dtetTATc

jk

jk

T t

T

jk

k


13/21

use your favorite Table of Integrals (a short one is available on the

course web site):

+

=22

)]cos()sin([

)sin( ba

bxbbxaedxbxe

axax

We get our case with: a = -j2k b = 1

So

0

2

2

41

)]cos()sin(2[

=

k

kjeAc

kj

k

So [ ]

0

2

2 )cos()41(

kj

k ek

A

c

=

Recall: sin(0) = sin() = 0 So the sin term above goes away

(Finesse the problem dont use brute force!)

( )=

0

2sin deA

cjk

kSo to evaluate the integral given by:

S


14/21

So[ ])0cos()cos(

)41(

022

2

kjkj

k eek

Ac

=

=1 =1=-1 =1

=-2So

)41(

22

k

Ack

=

Notes: 1. This does not depend on T

2. ckis proportional toA

So: 1. If you change the input sine wave's

frequency the ckdoes not change

2. If you, say, doubleA youll double ck

Things you would

never know if youcant work

arbitrary cases


15/21

Re

Im

ckk0

0

2

0

0

=

=

c

A

c

0

0)14(

22

=

=

kc

kk

A

c

k

k

Because c0 is real and > 0 Because ckis real and < 0

Now find the magnitude and phase of the FS coefficients:

)41(

22

k

A

ck =

Re

Im

c0

+

-


16/21

So the two-sided spectrum after the rectifier:

A2

3

2A

0 02 03 04 05002030405

0 02 03 04 05002030405

kc

kc

The spectrum before the rectifier (input is a single sinusoid):

2

0

2

0

2

0

2

0

kc~

kc~

Note: Therectifier

creates

frequencies

because it

isnon-

linear

Now you can find the Trigonometric form of FS


17/21

The rectified sine wave has Trig. Form FS:

=

+

+=1

02)cos(

)14(

42)(

k

tkk

AAtx

Ak kA0

Now you can find the Trigonometric form of FS

Once you have the ckfor the Exp. Form, Eulers formula gives the Trig Form as:

=

++=1

00 )cos(2)(k

kk ctkcctx

So the input to the RC circuit

consists of a superposition of

sinusoids and you know

from circuits class how to:1. Handle a superposition of

inputs to an RC circuit

2. Determine how a single

sinusoid of a given frequencygoes through an RC circuit

A2

3

4A

0 02 03 04 05

The one-sided spectrum is:

0 02 03 04 05

MagnitudeSpectrum

Phase

Spectrum

Ak

k

Preliminary to Parsevals Theorem (Not in book)


18/21

Preliminary to Parseval s Theorem (Not in book)

Imagine that signalx(t) is a voltage.

Ifx(t) drops across resistanceR, the instantaneous power is R

tx

tp

)()(

2

=

Sometimes we dont know whatR is there so we normalize this by ignoring theR

value: )()( 2 txtpN =

Energy in one period ++ ==

0

0

0

0

)()( 2tTt

tT

tdttxtdE

Recall: power = energy per unit time

(1 W = 1 J/s)dt

tdEtp

)()( = dttxtdE )()( 2=

differential increment

of energy

Once we have a specificR we can always un-normalize viaR

tpN )(

(In Signals & Systems we will drop theNsubscript)

The Total Energy

= dttx )(2

= for a periodic signal


19/21

+=

0

0 )(

1 2tT

tdttx

TP

Average power over one period =T

PeriodOneinEnergy

Recall: power = energy per unit time

For periodic signals we use the average power as measure of the sizeof a signal.

The Average Power of practical periodic signals is finite and non-zero.

(Recall that the total energy of a periodic signal is infinite.)

P l Th


20/21

Parsevals Theorem

We just saw how to compute the average power of a periodic signal if we are

given its time-domain model:

+

=0

0

)(1 2tT

tdttx

TP

Q: Can we compute the average power from the frequency domain modelA: Parsevals Theorem says Yes!

{ } ,...2,1,0, =kck

=

=k

kcP2

Another way to view Parsevals theorem is this equality:

=

+=

k

k

Tt

tcdttx

T

220

0

)(1

Parsevals theorem gives this equation

as an alternate way to compute the average power of a periodic signal whose

complex exponential FS coefficients are given by ck

I t ti P l Th


21/21

=

+

=k

k

Tt

t cdttxT

220

0 )(

1

sum of squares in

time-domain model

sum of squares in

freq.-domain model

=

)(2 tx = power at time t(includeseffects of all frequencies)

=2

kcpower at frequency k0

(includes effects of all times)

Interpreting Parsevals Theorem

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Fourier Series (for Periodic Signals)