SPFirst-L06EV

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Copyright © Monash University 2009

Signal

ProcessingFirst

Lecture 6

Fourier Analysis

And

Fourier Synthesis

1




READING ASSIGNMENTS• This Lecture:

– Fourier Series

in

Ch

3,

Sects

3‐4,

3‐5 &

3‐6

• Other Reading:

– Next Lecture:

Sampling

2




LECTURE OBJECTIVES

• Work with the Fourier Series Integral

• ANALYSIS via Fourier Series

– For PERIODIC signals:

x(t+T 0) = x(t)

– Later: spectrum from the Fourier Series

3

0(2 / )( )

j k T t

k

k

x t a e

0

0

00

(2 / )1 ( )T

j k T t

k T a x t e dt




SPECTRUM DIAGRAM• Recall Complex Amplitude vs. Freq

4

0 100 250 –100 –250f (in Hz)

3/7 je3/7 je

2/4 je

2/4 je

10 k k a X 21

X k Ak e j k

* *12 k k X a

N

k

t f jk

t f jk

k k e X e X X t x1

2

212

21

0)(

k a{ *k a0a




Harmonic Signal

5

PERIOD/FREQUENCY of each COMPLEX EXPONENTIAL:

t f k j

k

k eat x 02)(

00 0 0

0 0

2 1 12 or wherek

k

T kf k T T T kf k f

Signals of the following for are periodic of period T 0




Fourier Series Synthesis

6

k jk k

N

k k k

e A X

t kf A At x

1 00

)2cos()(

k jk k k e A X a

21

21

COMPLEX

AMPLITUDE

t f k j

k k eat x

02

)(




Harmonic Signal (3 Freqs)

3/15/20127

T = 0.1

a3 a5

a1




Example )3(sin)(

3

t t x

t jt jt jt j e j

e j

e j

e j

t x 9339

88

3

8

3

8)(

8




In this case, analysis

just requires picking

off the coeffic ients.

t jt jt jt j e j

e j

e j

e j

t x 9339

88

3

8

3

8)(

3k 1k 1k 3k

k a

9

Example )3(sin)(

3

t t x




SYNTHESIS vs. ANALYSIS• SYNTHESIS

– Easy – Given (k,Ak,k) create

x(t)

• Synthesis can be HARD – Synthesize Speech so that it

sounds good

• ANALYSIS

– Hard – Given x(t), extract

( k,Ak,k)

– How many?

– Need algorithm for

computer

10




STRATEGY: x(t) ak

• ANALYSIS

– Get representation

from

the

signal

– Works for PERIODIC Signals

• Fourier

Series – Answer is: an INTEGRAL over one period

11

0

0

0

0

)(1T

dt et xa t k jT k




INTEGRAL Property of exp(j)

• INTEGRATE over ONE PERIOD

12

00

0 0

0

0

(2 / ) (2 / )0

0 0

20

(2 / )

00

2

( 1)2

0

T T j T mt j T mt

j m

T j T mt

T e dt e

j m

T e

j m

e dt T

0m

0

0

2

T

0m




ORTHOGONALITY of exp(j)

• PRODUCT of exp(+j ) and exp(‐ j )

13

k

k dt ee

T

T

kt T jt T j

1

01 0

00

0

)/2()/2(

0

0

0

0

))(/2(

0

1

T

t k T j dt eT




Isolate One FS Coefficient0

0 0

0 0 0

0 0

0 0

0 0 0

0 0

0

0

(2 / )

(2 / ) (2 / ) (2 / )1 1

0 0

(2 / ) (2 / ) (2 / )1 1

0 0

(2 / )1

( )

( )

( )

( )

j T k t

k

k

T T

j T t j T k t j T t

k T T

k

T T j T t j T k t j T t

k T T

k

j T l t l T

x t a e

a x t e dt a e e dt

x t e dt a e e dt a

a x t e d

0

0

T

t k forexcept

zeroisIntegral




SQUARE WAVE EXAMPLE

15

0 –.02 .02 0.04

1

t

x(t)

.01

sec.04.0for

0

01)(

0

0021

02

1

T

T t T

T t t x

Duty cycle 50%




FS for a SQUARE WAVE {ak}

16

)0()(1 0

0

0

)/2(

0

k dt et xT a

T

kt T jk

02.

0

)04./2(

)04./2(04.1

02.

0

)04./2(104.

1 kt j

k j

kt jk edt ea

k je

k j

k k j

2

)1(1)1(

)2(

1 )(




Special care for the DC Coefficient: a0

17

)0()(1 0

0

0

)/2(

0

k dt et x

T a

T

kt T jk

)Area(1

)(1

000

0

0

T dt t xT a

T

21

02.

0

0 )002(.04.11

04.1 dt a




Fourier Coefficients ak

18

ak is a function of k

Complex Amplitude for k-th Harmonic

This one doesn’t depend on the period, T0

which only decides the fundamental frequency

0

,4,20

,3,11

2

)1(1

21 k

k

k k j

k ja

k

k




Spectrum from Fourier Series

19

0

,4,20

,3,1

21 k

k

k k

j

ak

)25(2)04.0/(20




FS: Rectified Sine Wave {ak

}

20

)1()(1 0

0

0

)/2(

0

k dt et x

T a

T

kt T jk

2/

0))1)(/2((2

)1)(/2(2/

0))1)(/2((2

)1)(/2(

2/

0

)1)(/2(

21

2/

0

)1)(/2(

21

2/

0

)/2()/2()/2(

1

2/

0

)/2(21

0

00

00

00

0

0

0

0

0

0

0

0

0

00

0

0

0

00

2

)sin(

T

k T jT j

t k T jT

k T jT j

t k T j

T

t k T j

T j

T

t k T j

T j

T

kt T jt T jt T j

T

T kt T j

T T k

ee

dt edt e

dt e jee

dt et a

Half-Wave Rectified Sine




1 1,4 4

j ja a

FS: Rectified Sine Wave {ak }

21

0 00 0

0 0 0 0

0 0 0 0

/2 /2(2 / )( 1) (2 / )( 1)

2 ( (2 / )( 1)) 2 ( (2 / )( 1))0 0

(2 / )( 1) /2 (2 / )( 1) /21 14 ( 1) 4 ( 1)

( 1) ( 1)1 14 ( 1) 4 ( 1)

1 ( 1)

4 (

1 1

1 1

T T j T k t j T k t

k j T j T k j T j T k

j T k T j T k T

k k

j k j k k k

k k

k

e ea

e e

e e

2

2

1)

1

2 ( 1)

0 odd ( 1)

( 1) 1 ? 1 even

k

k

k

k k

Try yourselves




Fourier Series Integral• HOW do you determine ak from x(t) ?

22

00 /1 FrequencylFundamenta T f

component)(DC)(

)(

0

0

0

0

0

0

10

0

)/2(1

T

T

T t k T j

T k

dt t xa

dt et xa

realis)(when*t xaa k k



Copyright © Monash University 2009 23

Fourier Series Synthesis• HOW do you APPROXIMATE x(t) ?

• Use FINITE number of coefficients

0

0

0

0

)/2(1 )(T

t k T j

T k dt et xa

realis)(when*t xaa k k

t f k j N

N k

k eat x 02)(




3/15/2012

24

Fourier Series Synthesis



Copyright © Monash University 2009 25

Synthesis: 1st & 3rd Harmonics

))75(2cos(3

2))25(2cos(

2

2

1)(

22

t t t y




Synthesis: up to 7th Harmonic

26

)350sin(7

2)250sin(

5

2)150sin(

3

2)50cos(

2

2

1)(

2 t t t t t y



Copyright © Monash University 2009 3/15/2012

27

Fourier Synthesis

)3sin(3

2)sin(

2

2

1)( 00 t t t x N






3/15/2012

29

fseriesdemo GUI




Vector space interpretation

Scalar product: , cos( )v w v w

a

b

, cos( ) cos( )

, cos( ) sin( )2

a v i v i c

b v i v j c

v ai b j

c

i

j

,i j

Orthonormal basis i.e |i|=| j|=1 and <i,j> = 0

θ




Hilbert space interpretation

Scalar product:

( ), ( )v x t w y t

0

2 0( ) ( ), ( )

1

k j t T

k k l

k

k lt e t t

k l

Orthonormal basis:

Periodic signals correspond to vectors in a vector space where

0

*

0 0

1( ), ( ) ( ) ( )

T

x t y t x t y t dt T

02 2

0 0

1( ) ( )

T

x t x t dt T

Norm:




Hilbert space interpretation

i.e. the signal is decomposed along the orthonormal basisto get the spectrum components ak

The Fourier series becomes:

( ) ( ) ( ), ( )k k k k

k

x t a t a x t t

Signal energy: Px = ||x(t)||2 = square length of the vector = …

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SPFirst-L06EV