lecture #5 chapter 3 - web.njit.edu · lecture #5 chapter 3. bme 310 biomedical computing -...

BME 310 Biomedical Computing -J.Schesser

118

Spectrum Representation

Lecture #5Chapter 3


119

Mathematical Forms of a Decomposed Signal

2 2

1 1

2 2 2

1

Two Sided Spectrum

( ) cos(2 ) { }2 2

*{ }2 2

Using the subsitution ( ) for the second summation to combi

k k k k

k k k

N Nj j f t j j f tk k

o k k k ok k

N Nj f t j f t j f tk k

o kk k N

A Ax t A A f t X e e e e

X XX e e a e

k k

ne into a single summation

1where for 0; for 02 2

kjko o k k

Xa A k a A e k

2

1 1

Single Sided Spectrum

( ) cos(2 ) { }

Where DC component and is the phasor for frequency

k

k

N Nj f t

o k k k o kk k

jo k k

x t A A f t X e X e

X X A e k


120

Periodic Waveforms• A periodic waveform must satisfy this condition:

x(t+To) = x(t)where To is the period of x(t).

– It can be shown that periodical waveforms are made up of sinusoidal functions which are harmonically related frequencies.

– Harmonic frequencies are frequencies which are multiples of each other.

• If a periodic signal, x(t), has a period of To then it has a fundamental frequency fo = 1/ To.

• Furthermore, we say that x(t) has frequencies which are harmonics of fo.– That is, the kth harmonic of x(t) is k fo


121

Periodic Vs Nonperiodic Signals• We say that is nonperiodic

since there is no assumption about theindividual frequencies.

• We say that is periodic since

there is a fundamental period associated withthe signal

• The fundamental frequency fo is the greatest common divisor of fk

N

Nk

tfjk

keatx 2)(

N

Nk

tkfjk

oeatx 2)(


122

Periodic Vs Nonperiodic Signals

-2-1.5

-1-0.5

00.5

11.5

22.5

33.5

0 0.05 0.1

f0 100 200

21

-100-200

21

f0 165 233.33

21

-165-233.33

21

Spectrum of a Periodic Signal Fundamental Frequency = 100Hz

Spectrum of a Nonperiodic Signal No Fundamental Frequency

-4

-3

-2

-1

0

1

2

3

4

0 0.05 0.1


123

Fourier Analysis• We have shown that a signal can be formulated in

terms of a sum of sinusoids which defines its spectrum

• What we like to find are the ak’s so we can completely determine an signal’s spectrum

• If a signal is periodic, (i.e., the fk’s are multiples of fo) then the determination of these the ak’s are easily achieved using Fourier Analysis.

N

Nk

tfjk

keatx 2)(

N

Nk

tkfjk

oeatx 2)(


124

Another Mathematical Form of a Decomposed Signal

Two Sided Spectrum with N

x(t) Ao Ak cos(2 fkt k )k1

X o {Ak

2e jk e j2 fkt

k1

Ak

2e jk e j2 fkt}

akej2 fkt

k

where fk kfo; ak Ao for k 0;ak X k

2 1

2Ake

jk for k 0

ao akej2kfot ak *e j2kfot

k1

ao 2e{akej2kfot}

k1

since s s* 2e{s}

ao 2e{ ak e jk e j2kfot}k1

ao 2 ak cos(2kfot k )k1


125

Summary of Mathematical Forms of a Decomposed Signal

2

1 1

1

Single Sided Spectrum with

( ) cos(2 ) { }

Where DC component and is the phasor for frequency

Two Sided Spectrum with

( ) cos(2 )

k

k

j f to k k k o k

k kj

o k k

o k k kk

N

x t A A f t X e X e

X X A e k

N

x t A A f t

2 2

1

2

2

1

1

{ }2 2

1 where for 0; for 02 2

2 { }

2 cos(2 )

k k k k

k k

o

j j f t j j f tk ko

k

j f t jkk k o k k

k

j kf to k

k

o k o kk

A AX e e e e

Xa e a A k a A e k

a e a e

a a kf t


126

Fourier Analysis and Synthesis

• Given a Fourier Series (note the limits go to -∞ to ∞ )

• Fourier Analysis is the technique to determine the ak’sfrom x(t).

• Fourier Synthesis is the (opposite) technique to determine x(t) from the ak’s.

2

1

( ) cos(2 )

1 1Where ; ; *;2

o

k

j kf tk o k o k

k k

jo o k k k k o

o

x t a e A A f kt

A a a A e a a fT


127

Fourier Analysis• The way to perform Fourier Analysis is by using the

Fourier Series Integral:

• This states to obtain the ak’s1. Take the periodic function x(t) and multiply it by e-j2π/Tokt

2. integrate the product over 1 period, To, of x(t) 3. Divide the result by To

o

o

T ktT

j

o

k dtetxT

a0

)2(

)(1


128

Derivation of the Fourier Integral

• Some background: The Orthogonality Property• This is a property of certain functions• Orthogonality means perpendicular• So if two functions are orthogonal they are

perpendicular as defined by the following:

lkT

dttvtv

lk

o

T

lk

o

if

)(*)(

if 0

0


129

Sinusoids and Orthogonality• Note that sinusoidal and complex exponential

functions are orthogonal.• First recall:

kT

j

e

kT

j

edte

o

kTT

j

T

oo

ktT

jT kt

Tj

oo

o

ooo

21

2 )1

2

2

0

2

1012sin2cos since

0for ;021

21

2 )1

2

22

2

0

2

jkjke

kk

Tj

e

kT

j

e

kT

j

edte

kjo

kj

o

kTT

j

T

oo

ktT

jT ktT

j

oo

o

oo

o

kT

j

e

kT

j

e

kT

j

edte

o

kj

o

kTT

j

T

oo

ktT

jT kt

Tj

oo

o

ooo

21

21

2 )1

22

2

0

2

o

oo

o

T

oo

ktT

jT kt

Tj

kT

j

edte

2 )1

2

0

2

o

o

T ktT

j

dte0

2

)1


130


functions are orthogonal.• First recall:

1012sin2cos since

0for ;021

21

2 )1

2

22

2

0

2

jkjke

kk

Tj

e

kT

j

e

kT

j

edte

kjo

kj

o

kTT

j

T

oo

ktT

jT ktT

j

oo

o

oo

o


131


functions are orthogonal.• For k = 0:

oo

TTT t

TjT kt

Tj

TT

tdtdtedte o

oo

o

o

o

0

|1 000

02

0

2


132

Sinusoids and Orthogonality• Another way

kT

kT

kTTj

kT

kT

kTT

kT

ktTj

kT

ktT

ktdtT

jktdtT

dte

o

o

o

o

o

o

o

o

T

o

o

T

o

o

T

o

T

o

T ktT

j

oo

oooo

2

02cos2cos

2

02sin2sin

2

2cos

2

2sin

2sin2cos )2

00

000

2

0for 0211

200

20cos2cos

20sin2sin

kk

T

jk

T

kT

kjk

T

k

oo

oo

oT

TTTT

o

T

o

T

o

T

o

T ktT

jTtdtdtjdttdt

Tjtdt

Tktdt

Tjktdt

Tdte

k

o

oooooooo

o

000000000

2

|10102sin02cos2sin2cos

0for


133

Sinusoids and Orthogonality

• Now let’s check sinusoids and complex exponential function for orthogonality:

lkT

dtdtj

dtjdte

dtedte

dteedttvtv

o

TT

TTj

T tllT

jT tlkT

j

T ltT

jktT

jT

lk

oo

oo

oo

oo

ooo

o

if

)01(

)0sin0(cos

)(*)(

00

00

0

0

)(2

0

)(2

0

22

0

lk

dte

dteedttvtv

oo

o

oo

o

T tlkT

j

T ltT

jktT

jT

lk

if0

)(*)(

0

)(2

0

22

0


134

Proof of Fourier Series Integral

oo

oo

oo

T tllT

j

l

T tllT

j

l

T tllT

j

l dteadteadtea0

)1(2

10

)(2

0

)1(2

1

olTa

00 11 loll aTaa

k

T tlkT

j

k

o

o dtea0

)(2

dteadtetxo

o

o

o

T

k

tlkT

j

k

T ltT

j

0

)(2

0

2

}{)(

k

tlkT

j

kk

ltT

jktT

j

k

ltT

joooo eaeeaetx

)(2222

)(

k

ktT

j

koeatx2

)(

o

o

T ktT

j

o

k dtetxT

a0

)2(

)(1


135

Fourier Series

• Fourier Analysis:

• Fourier Synthesis:

o

o

T ktT

j

o

k dtetxT

a0

)2(

)(1

k

ktT

j

koeatx2

)(


136

Example• Find the spectrum of: sin3(3πt)

32;

23

23;3

)833(1

)833(1

)2

(1

)3(sin1

2

0

9339

2

0

963369

2

0

333

2

0

3

ooo

ktT

jT tjtjtjtj

o

ktT

jT tjtjtjtjtjtj

o

ktT

jT tjtj

o

ktT

jT

ok

Tf

dtej

eeeeT

dtej

eeeeeeT

dtejee

T

dtetT

a

o

o

o

o

o

o

o

o


137


3/2

0

3)3(3/2

0

3)1(3/2

0

3)1(3/2

0

3)3(

3/2

0

)39()33()33()39(

33/2

0

9339

3/223/2

0

9339

163

169

169

163

)833(

23

)833(

23

)833(

23

dtej

dtej

dtej

dtej

dtj

eeee

dtej

eeee

dtej

eeeea

tkjtkjtkjtkj

tkjtkjtkjtkj

ktjtjtjtjtj

ktjtjtjtjtj

k


138


2/3 2/3(3 )3 2

30 0

2/3 2/3(1 )3 2

10 0

2/3 2/3(1 )3

0 0

3 3 3 2 1 1; for 3;16 16 16 3 8 8 8 8

9 9 9 2 3 3 3 3; for 1;16 16 16 3 8 8 8 8

9 9 9 216 16 16 3

jj k tk

jj k t

j k t

j ja e dt dt k a ej j j j

j je dt dt k a ej j j j

e dt dtj j j

21

2/3 2/3(3 )3 2

30 0

3 3 3 3; for 1;8 8 8 8

3 3 3 2 1 1; for 3;16 16 16 3 8 8 8 8

j

jj k t

j jk a ej

j je dt dt k a ej j j j


139

Fourier Series of a Square Wave

To.5To-.5To-To

1

s(t)

oo

o

TtT

ts

Tt

21for 0

)(210for 1

t


140

Fourier Series of a Square WaveAnalysis

kkkj

k

kj

o

kT

jTkT

j

o

T

o

ktT

j

o

T ktT

j

o

T

T

ktT

j

o

T ktT

j

o

T ktT

j

o

k

jkjke

kkj

kje

kT

j

eeT

kT

j

eT

dteT

dteT

dteT

dtetsT

a

o

o

o

o

ooo

o

o

oo

o

oo

)1(0)1()sin()cos(

0;2

)1(1

21

)2(

1

|)2(

111

0111

)(1

0)2(2

)2(

2/0

)2(2/

0

)2(

2/

)2(2/

0

)2(

0

)2(

2

/2

00 0

1 ( 1) ; 02

1 ( 1) 1 1 ; odd2

1 (1) 0; even2

1 1 1( ) 12

o o

k

k

j

k

T T

o o

a kj k

a e kj k j k k

kj k

a s t dt dtT T


141

Fourier Series of a Square WaveSynthesis

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-0.3

-0.1

0.1

0.3

0.5

0.7

0.9

1.1

1.3

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

2 terms To=4

-0.3

-0.1

0.1

0.3

0.5

0.7

0.9

1.1

1.3

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

3 terms To=458 terms To=4

2 2

01

22

2( )2

( ) 2 { }

1 2 { }2

1 12 { }21 1 22 cos( )2 2

o o

o

o

j kt j ktT T

k kk k

jj kt

T

k odd

j ktT

k odd

k odd o

x t a e a e a e

ee ek

e ek

ktk T

t

t t


142

Spectrum of a Square Wave2

0

1 1 ; odd

0; even12

j

k

k

a e kj k k

a k

a

0

To=4

fo=0.25 Hz

0.25 0.75 1.25-1.25 -0.75 -0.25

k=1

a1=-j/π

|a1|=1/π k=3

a3=-j/3π

|a3|=1/3π

k=5

a5=-j/5π

|a5|=1/5π

k=-1

a1=j/π

|a-1|=1/πk=-3

a3=j/3π

|a-3|=1/3π

k=-5

a-5=j/5π

|a-5|=1/5π

-0.5-1.00 0.5 1.00

k=0

a0=1/2

f


143

Fourier Series of a Sawtooth Wave

To-To-2To

1s(t)

oo TtTtts 0for /)(

2To 3To-3Tot


144

Fourier Series of a Sawtooth WaveAnalysis

01)2sin()2cos( since

}1{32

1)(32

1(}03{32

1[91

])32

1(|32

[91

331

321,

31,

3

331)(

31

2

22

3

0

3230

32

3

0

32

32

32

3

0

323

0

32

jkjke

ekjkj

ekj

dtekj

ekj

t

dteta

ekj

vdtedv

dtdutu

dtetdtetsa

kj

kjkj

ktjktj

ktj

k

ktjktj

ktjtkjk


145

Fourier Series of a Sawtooth WaveAnalysis

2

3 3

00 0

1

1 1 1 1[ {3 1} ( )( {1 1}9 2 3 2 3 2 31 1[ {3 1} 09 2 3

1 1 1 1 ; 03 ( 2 3) 2 2

1 1 1 9( ) ( ) 0.53 3 3 9 21 1 2( ) 2 cos( )2 2 3 2

k

j

aj k j k j k

j k

j e kj k k k

ta f t dt dt

s t ktk


146

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-0.5

-0.3

-0.10.1

0.3

0.5

0.7

0.91.1

1.3

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Fourier Series of a Sawtooth WaveSynthesis

2 terms To=3

3 terms To=358 terms To=3

11 terms To=3

t t

t t


147

Spectrum of a Sawtooth Wave2

0

1 1 ; 02 2

12

j

ka j e kk k

a

0

To=3

fo=0.33 Hz

0.33 1.00 1.66-1.66 -1.00 -0.33

k=1

a1=j/2π

|a1|=1/2π k=2

a3=j/4π

|a3|=1/4π

k=-1

a1=-j/2π

|a-1|=1/2πk=-2

a3=-j/4π

|a-3|=1/4π

-0.66-1.33 0.66 1.33

k=0

a0=1/2

f


148

Time-Frequency Spectrum• Up till now we have seen signals whose frequencies

do not change with time.• However, in reality, signals can produce different

frequencies at different times– Music– Voice– Frequency Modulation

• To plot the spectrum of such signals, we can use a 3-D plot called a spectrogram.– Plots time on the x-axis, frequency on the y-axis and

magnitude on the z-axis (out of the plane of the paper)


149

Example of a Spectrogram• Three sinusoids (0.1 Hz, 0.2 Hz, 0.4Hz) in sequence


150

Frequency Modulation

• We have signals of the type:

where the angle (of the cosine) varies linearly with time and the time derivative of the angle is o, constant

• However, we can general this signal such that the angle varies with time such that its derivative is not constant.

( )( ) { } cos( )oj tox t e Ae A t

)()(

))(cos(}{)( )(

tdtdt

tAAeetx tj

Instantaneous Frequency


151

FM Radio

• Frequency Modulation is the scheme used in FM broadcast.

• For example, if ω(t)=ωc+m(t) where ωc is called the carrier frequency and m(t) contains the “information” (voice or music), then FM broadcast is:

dttmtdttmt cc )(})({)(


152

Homework

• Exercises:– 3.4 – 3.8

• Problems:

3.5 Instead use x(t) 10 20cos(2 (200)t 14 ) 10cos(2 (250)t)

3.8,3.9 Use Matlab for plotting the spectrum; submit your code 0 for 0 t 2.53.12 Instead use x(t) Use Matlab for plotting the spectrum; submit your code 2 for 2.5 t 5

3.13,3.14

lecture #5 chapter 3 - web.njit.edu · lecture #5 chapter 3. bme 310 biomedical computing -...

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