1 fourier analysis

8/18/2019 1 Fourier Analysis

1/60

Discrete Time Periodic Signals

A discrete time signal x[n] is periodic with period N if and only if

][][ N n xn x +=for all n .

Definition:

N

][n x

n

Meaning: a periodic signal keeps repeating itself forever


2/60

!xample: a Sin"soid

( )[ ] 2cos 0.2 0.9 x n nπ π = +#onsider the Sin"soid:

$t is periodic with period since 10= N

( )

( ) ][29.02.0cos2

9.0)10(2.0cos2]10[

n xn

nn x

=++=++=+

π π π

π π

for all n.


3/60

%eneral Periodic Sin"soid

+= α π n

N

k An x 2cos][

#onsider a Sin"soid of the form:

$t is periodic with period N since

][22cos

)(2cos][

n xk n N

k A

N n N

k A N n x

=

×++=

++=+

π α π

α π

for all n.

with k & N integers'


4/60

( )π π 1.03.0cos5][ −= nn x#onsider the sin"soid:

$t is periodic with period since 20= N

( )( ) ][231.03.0cos5

1.0)20(3.0cos5]20[n xn

nn x=×+−=

−+=+π π π

π π

for all n.

(e can write it as:

−= π π 1.0203

2cos5][ nn x

!xample of Periodic Sin"soid


5/60

n N

k j

Aen x

=π 2

][

#onsider a #omplex !xponential of the form:

for all n.

$t is periodic with period N since

Periodic #omplex !xponentials

][

][

2

2

)(2

n xe Ae

Ae N n x

jk n

N k j

N n N

k j

=×=

=+

+

π

π

π

1=


6/60

n je jn x π 1.0)21(][ +=

#onsider the #omplex !xponential:

(e can write it as

!xample of a Periodic #omplex !xponential

n j

e jn x

+= 201

2

)21(][π

and it is periodic with period N = )*'


7/60

%oal:(e want to write all discrete time periodic signals in terms

of a common set of +reference signals,'

-eference .rames

$t is like the pro/lem of representing a vector in a

reference frame defined /y• an origin +*,

• reference vectors

x

,..., 21 ee

x

0

1e

2eReference

Frame


8/60

-eference .rames in the Plane and in Space

.or example in the plane we need two reference vectors

x

0

1e

2e

21,ee

Reference

Frame

0 while in space we need three reference vectors 321 ,, eee

0

1e2e

ReferenceFrame

x3e


9/60

A -eference .rame in the Plane

$f the reference vectors have "nit length and they areperpendic"lar 1orthogonal 2 to each other& then it is very simple:

2211 eaea x +=0

11ea

22ea

(here pro3ection of along

pro3ection of along

1a2a 2e

1e x x

The plane is a ) dimensional space'


10/60

A -eference .rame in the Space

$f the reference vectors have "nit length and they areperpendic"lar 1orthogonal 2 to each other& then it is very simple:

332211 eaeaea x ++=0

11ea22ea

(here pro3ection of along

pro3ection of along pro3ection of along

1a

2

a2e

1e x

x

The +space, is a 4 dimensional space'

3a x 3e

33ea


11/60

!xample: where am I 5

6

!0

1e

2e

x

m300

m200

Point + x” is 4**m !ast and )**m 6orth of point +*,'


12/60

-eference .rames for Signals

(e want to expand a generic signal into the s"m of reference

signals'

The reference signals can /e& for example& sin"soids or complex

exponentials

n

][n x

=

∑

reference signals


13/60

7ack to Periodic Signals

A periodic signal x [n] with period N can /e expanded in terms of N

complex exponentials

1,...,0 ,][2

−== N k enen

N

k j

k

π

as

∑−

=

=1

0

][][ N

k

k k nean x


14/60

A Simple !xample

Take the periodic signal x [n] shown /elow:

n0

1

2

6otice that it is periodic with period N=)'

Then the reference signals are

nn j

nn j

ene

ene

)1(][

11][

2

12

1

2

02

0

−==

===

π

π

(e can easily verify that 1try to /elieve2:

nn

nenen x

)1(5.015.1

][5.0][5.1][ 10

−×−×=

+=

for all n.


15/60

Another Simple !xample

Take another periodic signal x [n] with the same period (N 8)2:

n0

3.0−

3.1

Then the reference signals are the same0

2 20

12

21

[ ] 1 1

[ ] ( 1)

j n n

j nn

e n e

e n e

π

π

= = =

= = −(e can easily verify that 1again try to /elieve2:

nn

nenen x

)1(8.015.0

][8.0][5.0][ 10

−×+×=+=

for all n.

Same reference signals& 3"st different coefficients


16/60

9rthogonal -eference Signals

6otice that& given any N, the reference signals are all orthogonal

to each other& in the sense

=

≠=∑

−

= k m N

k mnene

N

n

mk if

if 0][][

1

0

*

∑∑∑

−

=−

−−−

=

−−

= −

−

=

==

1

0 2

)(221

0

21

0

*

1

1

][][

N

n N

k m j

k m jn

N

k m j N

n

n N

k m j N

nmk

e

e

eenene π

π π π

Since

/y the geometric s"m


17/60

0 apply it to the signal representation 0

k

N

m

N

n

mk m

N

n

k

n x

N

m

mm

N

n

k

Nanenea

neneanen x

=

=

=

∑ ∑

∑ ∑∑

−

=

−

=

−

=

−

=

−

=

1

0

1

0

*

1

0

*

][

1

0

1

0

*

][][

][][][][

and we can comp"te the coefficients' #all thenk Nak X =][

1,...,0 ,][][1

0

2

−=== ∑−

=

− N k en x Nak X

N

n

kn N

j

k

π


18/60

Discrete .o"rier Series

1,...,0 ,][][1

0

2

−== ∑−

=

− N k en xk X

N

n

kn N

j π

%iven a periodic signal x [n] with period N we define theDiscrete .o"rier Series 1D.S2 as

Since x [n] is periodic& we can s"m over any period' The general

definition of Discrete .o"rier Series 1D.S2 is

{ } 1,...,0 ,][][][1 20

0

−=== ∑−+=

− N k en xn x DFS k X

N n

nn

kn N

j π

for any 0n


19/60

$nverse Discrete .o"rier Series

{ } ∑−=

==1

0

2

][1][][ N

k

kn N

j

ek X N

k X IDFS n xπ

The inverse operation is called $nverse Discrete .o"rier Series

1$D.S2& defined as


20/60

-evisit the Simple !xample

-ecall the periodic signal x [n] shown /elow& with period N=):

n0

1

2

1,0,)1(21)1](1[]0[][][1

0

2

2

=−×+=−+== ∑=− k x xen xk X k k

n

nk j

π

Then 1]1[,3]0[ −== X X Therefore we can write the se"ence as

{ }

n

k

kn j

ek X k X IDFS n x

)1(5.05.1

][2

1][][

1

0

2

2

−×−=

== ∑=

π


21/60

!xample of Discrete .o"rier Series

#onsider this periodic signal

The period is N=;*' (e comp"te the Discrete .o"rier Series

25

102 2

9 4 210 10

100 0

1 if 1, 2,...,9[ ] [ ]

1

5 if 0

j k

j kn j kn j k

n n

ek k x n e e

e

k

π

π π

π

−

− −−

= =

−

== = = −

=

∑ ∑

][n x

n010

1


22/60

0 now plot the val"es 0

0 2 4 6 8 100

5magnitude

0 2 4 6 8 10-2

0

2phase (rad)

k

k

|][| k X

][k X


23/60

!xample of D.S

#omp"te the D.S of the periodic signal

)5.0cos(2][ nn x π =

#omp"te a few val"es of the se"ence

,...0]3[,2]2[,0]1[,2]0[ ==== x x x xand we see the period is N=)' Then

k

n

kn j

x xen xk X )1(]1[]0[][][1

0

2

2

−×+==

∑=−

π

which yields

2]1[]0[ == X X


24/60

Signals of .inite 1ie )*&***

samples?sec2' Then we have

( ) ( ) pointsdata 60010201030 33 =×××= − N


25/60

Series !xpansion of .inite Data

(e want to determine a series expansion of a data set of length N.

@ery easy: 3"st look at the data as one period of a periodic se"ence

with period N and "se the D.S:

n

1− N 0


26/60

Discrete .o"rier Transform 1D.T2

%iven a finite interval of a data set of length N, we define the

Discrete .o"rier Transform 1D.T2 with the same expression as theDiscrete .o"rier Series 1D.S2:

{ }

21

0[ ] [ ] [ ] , 0,..., 1

N j kn N

n X k DFT x n x n e k N

π − −

== = = −∑And its inverse

{ }21

0

1[ ] [ ] [ ] , 0,..., 1

N j kn N

n

x n IDFT X k X k e n N N

π −

=

= = = −∑


27/60

Signals of .inite 1ie )*&***

samples?sec2' Then we have

( ) ( ) pointsdata 60010201030 33 =×××= − N


28/60

Series !xpansion of .inite Data

(e want to determine a series expansion of a data set of length N.

@ery easy: 3"st look at the data as one period of a periodic se"ence

with period N and "se the D.S:

n

1− N 0


29/60

Discrete .o"rier Transform 1D.T2

%iven a finite of a data set of length N we define the Discrete .o"rier

Transform 1D.T2 with the same expression as the Discrete .o"rierSeries 1D.S2:

{ }

21

0[ ] [ ] [ ] , 0,..., 1

N j kn N

n X k DFT x n x n e k N

π − −

== = = −∑and its inverse

{ }21

0

1[ ] [ ] [ ] , 0,..., 1

N j kn N

n

x n IDFT X k X k e n N N

π −

=

= = = −∑


30/60

!xample of Discrete .o"rier Transform

#onsider this signal

The length is N=;*' (e comp"te the Discrete .o"rier Transform

25

102 29 4

210 10

100 0

1 if 1, 2,...,9[ ] [ ]

1

5 if 0

j k

j kn j kn j k

n n

ek k x n e e

e

k

π

π π

π

−

− − −

= =

−

== = = −

=

∑ ∑

][n x

n0 9

1


31/60

0 now plot the val"es 0

0 2 4 6 8 100

5magnitude

0 2 4 6 8 10-2

0

2phase (rad)

k

k

|][| k X

][k X


32/60

D.T of a #omplex !xponential

#onsider a complex exponential of fre"ency rad '0

ω

,][ 0n j

Aen x ω = +∞


33/60

-ecall Magnit"de& .re"ency and Phase

0π ω π − < ≤ +)' (e represented it in terms of magnitude and phase:

( )rad ω 0ω

0ω

magnitude

phase

|| A

A∠

-ecall the following:

;' (e ass"me the fre"ency to /e in the interval

π −

π − π

π

( )rad ω


34/60

#omp"te the D.T0

{ }

00

21

0

221 1

0 0

[ ] [ ] [ ]

, 0,..., 1

N j kn N

n

N N j k n j kn j n N N

n n

X k DFT x n x n e

Ae e Ae k N

π

π π ω

ω

− −

=

− − − −− ÷

= =

= =

= = = −

∑

∑ ∑

6otice that it has a general form:

0

2[ ] N X k A W k

N

π ω

= × − ÷

1

0

1( )

1

j N N j n

N jn

eW e

e

ω

ω

ω ω

−−−

−=

−= =

−∑

where 1"se the geometric series2


35/60

See its general form:

1( 1)/2

0

sin2

( )

sin2

N j n j N

N

n

N

W e eω ω

ω

ω

ω

−− − −

=

÷ = =

÷

∑


36/60

0 since:

1

0

/2 / 2 /2 / 2

/ 2 /2 /2 / 2

/ 2 / 2 /2( 1)/2

/ 2 / 2 / 2

1( )

1

sin2

sin

2

j N N j n

N jn

j N j N j N j N

j j j j

j N j N j N j N

j j j

eW e

e

e e e e

e e e e

N

e e ee

e e e

ω

ω

ω

ω ω ω ω

ω ω ω ω

ω ω ω

ω

ω ω ω

ω

ω

ω

−− −−

=

− − −

− − −

− −− −

− −

−= =−

−=

− ÷ − = = ÷ ÷− ÷

∑


37/60

0 and plot the magnit"de

-3 -2 -1 0 1 2 30

2

4

6

8

10

12

( ) N W ω

ω

π π −

N

2

N

π 2

N

π −


38/60

!xample

#onsider the se"ence

0.3[ ] , 0,...,31 j n x n e nπ = =

$n this case 32,3.00 == N π ω

Then its D.T /ecomes

( ) 23232

[ ] 0.3 , 0,...,31k

X k W k π ω

ω π =

= − =


39/60

''' first plot this 0

0 1 2 3 4 5 60

10

20

30

40

( )π ω 3.032 −W

ω

2π

32= N

π ω 3.00 =

d th th l t f it D.T


40/60

0 and then see the plot of its D.T

0 5 10 15 20 25 300

5

10

15

20

25

30

35

( ) 2[ ] 0.3 , 0,..., 1

N k N

X k W k N π ω

ω π

== − = −

k The max corresponds to fre"ency π π π ω 3.0312.032/25 ≅=×=

S ! l i M tl /


41/60

Same !xample in Matla/

%enerate the data:

BB n8*:4;C

BBx8exp13*'4pin2C

#omp"te the D.T 1"se the +.ast, .o"rier Transform& ..T2:

BB E8fft1x2C

Plot its magnit"de:

BB plot1a/s1E22

0 and o/tain the plot we saw in the previo"s slide'

S ! l i M tl /


42/60


%enerate the data:

BB n8*:4;C

BBx8exp13*'4pin2C


BB E8fft1x2C

Plot its magnit"de:

BB plot1a/s1E22


S ! l 1 d t i t 2


43/60

Same !xample 1more data points2


0.3[ ] , 0,..., 255 j n x n e nπ = =

$n this case 0 0.3 , 256 N ω π = =BB n8*:)FFC

BBx8exp13*'4pin2C

BB E8fft1x2C

BB plot1a/s1E22

See the plot …

d it it d l t


44/60

0 and its magnit"de plot

0 50 100 150 200 250 3000

50

100

150

200

k

| [ ] | X k

(h t d it 5


45/60

(hat does it mean5

The max corresponds to fre"ency

38 2 / 256 0.2969 0.3ω π π π= × = ≅

A peak at index means that yo" have a fre"ency0k

0 50 100 150 200 250 3000

50

100

150

200

k

| [ ] | X k

0 38k =

( )0 0 2 /k N ω π ;

! l


46/60

!xample

Go" take the ..T of a signal and yo" get this magnit"de:

0 50 100 150 200 250 3000

200

400

600

800

1000

1200

|][| k X

k 271 =k 2 81k =

There are two peaks corresponding

to two fre"encies:π

π π ω

π π π

ω

6328.0256

281

2

2109.0256

227

2

22

11

===

===

N k

N k

D.T f Si id


47/60

D.T of a Sin"soid

#onsider a sin"soid with fre"ency rad '0

ω

0[ ] cos( ), x n A nω α = + +∞


48/60

Sin"soid 8 s"m of two exponentials

-ecall that a sin"soid is the s"m of two complex

exponentials n j jn j j ee A

ee A

n x 0022

][ ω α ω α −−+=

( )rad ω 0ω

0ω

magnitude

phase

/ 2 A

α

π −

π − π

π

( )rad ω

0ω −

0

ω −

/ 2 A

α −

Hse of positive fre"encies


49/60

Hse of positive fre"encies

0 0[ ]2 2

j n j n j j A A X k DFT e e DFT e eω ω α α −− = +

Then the D.T of a sin"soid has two components

0 /"t we have seen that the fre"encies we comp"te

are positive' Therefore we replace the last exponential

as follows:

0 0(2 )[ ]2 2

j n j n j j A A X k DFT e e DFT e eω π ω α α −− = +

-epresent a sin"soid with positive fre


50/60

-epresent a sin"soid with positive fre'

Then the D.T of a sin"soid has two components

( )rad ω 0ω

0ω

magnitude

phase

/ 2 A

α

π

π

( )rad ω 2π

2π 0

2π ω −

/ 2 A

02π ω −

α −

0 0(2 )[ ]2 2

j n j n j j A A X k DFT e e DFT e eω π ω α α −− = +

!xample


51/60

!xample


[ ] 2cos(0.3 ), 0,...,31 x n n nπ = =

$n this case 32,3.00 == N π ω

Then its D.T /ecomes

( ) ( ) 232 3232

[ ] 0.3 1.7 , 0,...,31k

X k W W k π ω

ω π ω π =

= − + − =


52/60

''' first plot this 0

0 1 2 3 4 5 60

5

10

15

20

( ) ( )32 321

0.3 1.72

W W ω π ω π − + −

/ 2 32 / 2 N =

π ω 3.00 = 0 1.7ω π =

/ 2 32 / 2 N =

ω

2π

and then see the plot of its D.T


53/60

0 and then see the plot of its D.T

k The first max corresponds to fre"ency π π ω 3.032/25 ≅×=

( ) ( )32 32 21

[ ] 0.3 1.7 , 0,..., 12 k

N

X k W W k N π

ω

ω π ω π

== − + − = −

0 5 10 15 20 25 30 350

5

10

15

20

his is N! a fre"uency

Symmetry


54/60

Symmetry

$f the signal is real& then its D.T has a symmetry: ][n x

*][][ k N X k X −=

$n other words:

][][

|][||][|

k N X k X

k N X k X

−−∠=∠−=

Then the second half of the spectr"m is red"ndant 1it does not

contain new information2

7ack to the !xample:


55/60

7ack to the !xample:

0 5 10 15 20 25 30 350

5

10

15

20

$f the signal is real we 3"st need the first half of the spectr"m&

since the second half is red"ndant'

Plot half the spectr"m


56/60

Plot half the spectr"m

$f the signal is real we 3"st need the first half of the spectr"m&

since the second half is red"ndant'

0 5 10 150

5

10

15

20



57/60


%enerate the data:

BB n8*:4;C

BBx8cos1*'4pin2C


BB E8fft1x2C

Plot its magnit"de:

BB plot1a/s1E22




58/60



[ ] cos(0.3 ), 0,..., 255 x n n nπ = =

$n this case 0 0.3 , 256 N ω π = =BB n8*:)FFC

BBx8cos1*'4pin2C

BB E8fft1x2C

BB plot1a/s1E22

See the plot …

and its magnit"de plot


59/60

0 and its magnit"de plot

k

| [ ] | X k

0 50 100 150 200 2500

20

40

60

80

100

0 38k = 0 218 N k − =

The first max corresponds to fre"ency 0 38 2 / 256 0.3ω π π = × ≅

!xample


60/60

!xample

Go" take the ..T of a signal and yo" get this magnit"de:

|][| k X

k

There are two peaks corresponding

to two fre"encies:π

π π ω

π π π

ω

6328.0256

281

2

2109.0256

227

2

22

11

===

===

N k

N k

0 50 100 150 200 250 3000

50

100

150

271 =k 2 81k =

1 fourier analysis

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