1

Signals and Systems Lecture 7

•Convergence of CTFS•Properties of CTFS

2

Chapter 3 Fourier Series

§3.4 Convergence （收敛） of the Fourier Series

1. Approximation( 近似性 )

tjkk

N

NkN eatx 0ˆ

txtxte NN ——Error

1 NEN

dtetxT

aa tjk

Tkk0

1ˆ 2

dtteE NTN

2

EN 最小

0 N NE

dtteteT NN

If T dttx 2|)(| , then the series is convergent. ( xN(t) x(t) )

tjkk

k

eatx 0

0

00

1 jk tk Ta x t e dt

T

3

ka

Chapter 3 Fourier Series

2. Dirichlet Conditions:

Condition 1 dttx

T

dtetxT

a tjk

Tk0

1

dttx

T T

1

1T , 1t0 , /1 ttx

4

Chapter 3 Fourier Series

Condition 2.

In any finite interval , is of bounded variation. tx

1T , 1t0 , /2sin ttx

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Chapter 3 Fourier Series

Condition 3.

In any finite interval , there are only a finite number

of discontinuities.

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The Dirichlet Conditions (cont.)

• Dirichlet conditions are met for most of the signals we will encounter in the real world. Then– The Fourier series = x(t) at points where x(t) is continuous– The Fourier series = “midpoint” at points of discontinuity

• Still, convergence has some interesting characteristics:

• As N → ∞, exhibits Gibbs’ phenomenon at points of discontinuity.

N

Nk

tjkN eatx 0)(

)(txN

Chapter 3 Fourier Series

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Chapter 3 Fourier Series

Gibbs Phenomenon:

Figure 3.9

Any continuity:

xN(t1) x(t1)

Vicinity of discontinuity:

ripples

peak amplitude does not seem to decrease

Discontinuity:

overshoot 9%

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Chapter 3 Fourier Series

§3.5 Properties of Continuous-Time Fourier Series

§3.5.1 Linearity

k kx t a y t b FS FS

k k kz t Ax t By t c Aa Bb FS

§3.5.2 Time Shifting

kx t aFS

0 00

jk tkx t t a e FS

x(t) and y(t) may have the same period T.

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Chapter 3 Fourier Series

§3.5.3 Time Reversal

kx t aFS kx t a FS

x t x t k ka a

x t x t k ka a

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§3.5.6 Conjugation and Conjugate Symmetry（共轭及共轭对称性）

kk atxatx FSFS

Chapter 3 Fourier Series

kk aa txtx kk aa

or

txtx k ka a k ka a

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Example (more symmetry properties )3.42 (P262)

kk atxatx FSFS

Chapter 3 Fourier Series

real even tx ka real even

real odd tx kaPurely imaginary

odd

[x(t)real ] }{ txEv }Re{ ka

[x(t)real ] }{ txOd }Im{ kaj

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Chapter 3 Fourier Series

§3.5.4 Time Scaling

tjkk

k

eatx 0

0jk a t

kk

x at a e

0kak 0akak

§3.5.5 Multiplication （相乘）

kk btyatx FSFS

tytx FSk m k m

m

h a b

Convolution Sum

The Fourier series representation has changed!

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Average Powerof kth harmonic

Average Power of tx

Chapter 3 Fourier Series

§3.5.7 Parseval’s Relation （帕兹瓦尔关系式）

221k

kT

adttxT

§3.5.8 Differential Property

katx FS

knn ajktx 0

FS

kajktx 0FS

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Example (Proof Multiplication and Parseral’s Relation )3.46 (P264)

Chapter 3 Fourier Series

k

tjkkectytxtz 0)(

dtetytxT

dtetzT

cT

tjk

T

tjkk

0

0

0

0 )(11

00

dtetyeaT T

tjk

n

tjnk

0

00 )(1

0

nT

tnkjn dtety

Ta

0

0)(

0

)(1

nnknba

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Example ( Continue )

Chapter 3 Fourier Series

k

tjkkectxtxtz 0)(*

nnnk

nnnkk aaaac **

k

tjk

nnnk eaatx 0*2

dteaaT

dttxT T

n

tjn

knnkT

0

0

0

*

0

2

0

11

dteeaT

ak

T

tjktknj

nnkn

0

00)(

0

* 1

dtetxT

ak

T

tjkk

0

0)(1

0

*

k

kk

kk aaa2*

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Chapter 3 Fourier Series

Example 3.6 tg

-2 -1 0 1 2 t

2

1

2

1

Figure of Example 3.5

2/ t T 0

t 1

1

1

T

Ttx

1 tx

-T -T/2 –T1 0 T1 T/2 T t

Based on Property of linear and time-shifting, we may get

g(t)=x(t-1)-1/2

dk=bk+ck

2

jk

kk eab

02/1

00

kfor

kforck

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Chapter 3 Fourier Series

1 kctx FS

-4 -2 0 2 4 t

Example 3.7

Figure of Example 3.6-2 -1 0 1 2 t

2

1

2

1 ketg FS

dt

tdgtx

)(

kk ejkc 0Differentiation Property

2

20

T

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Chapter 3 Fourier Series

Example 3.9

21

210)( tjtj eaeaatx

According to Fact 3

2

20

T

According to Fact 2

According to Fact 1 *11 aa

Synthesis Equation

Symmetry Property

}Re{2

)(2

10

2*1

210

tj

tjtj

eaa

eaeaatx

So

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Chapter 3 Fourier Series

According to Fact 4 Time-ReversalTime-Shifting

kkFS abtxty )()(

Example 3.9 Continue

kjk

kFS aebtxty 0)1()(

So, ck=a-k => z(t)=x(-t)

=x(-(t-1))

bk=e-jkπ/2ck => y(t)=z(t-1)

Because bk is odd, b0=0, b1=-b-1

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Chapter 3 Fourier Series

According to Fact 5

2

1)(

4

1

)1(4

1)(

4

1

2

1

2

14

2

4

2

4

2

bbdtty

dttxdttx

Example 3.9 Continue

Parseval’s Relation

So, b1=-b-1 =±j/2

Because bk=e-jkπ/2a-k, so, a0=0, a1=b-1ejπ/2=jb-1

then, x(t)=±cos(πt/2)

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Chapter 3 Fourier Series

kk btyatxFSFS

kkk BbAactBytAxtz FS

00

FS

0tjk

keattx

kk aa txtx kk aa

or

txtx k ka a k ka a

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Readlist

• Signals and Systems:– 3.6~3.7

• Question: – Calculation of DTFS.

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Problem Set

• 3.5 P251

• 3.8 P252 Reference Example 3.9(P210)

• 3.40 P261 Reference Table3.1(P206)