Fourier series - City University of Hong Konghcso/ee31112_fs.pdf · 2010-05-25 · H. C. So Page 1...

H. C. So Semester A, 2001

Fourier seriesRevisiting what you have learned in “Advanced Mathematical Analysis”

Let )(xf be a periodic function of period π2 and is integrable over aperiod. )(xf can be represented by a trigonometric series,

!+++++=

∑ ++=∞

=

)2sin()2cos(sincos

))sin()cos(( )(

22110

10

xbxaxbxaa

nxbnxaaxfn

nn

where the coefficients are computed as

∫π= π

π− dxxfa )(21

0

∫π= π

π− dxnxxfan )cos()(1 and ∫π= π

π− dxnxxfbn )sin()(1

!,3,2,1=n


Consider a continuous-time signal )(tx periodic with period 0>T : )()( Ttxtx += , for all t

T : fundamental period Tπ=ω 20 : fundamental frequency (in radians/second)

)(tx can also be represented as complex exponential function

!! ++++++=

∑=

ωωω−−

ω−−

ω∞

−∞=tjtjtjtj

tjk

kk

eaeaaeaea

eatx

)2(210

)(1

)2(2

0000

0)(

As )(tx can be complex-valued, all ka ’s are generally complex 0a : DC component tjea )(

10ω±

± : fundamental (first harmonic) components tjea )2(

20ω±

± : second harmonic components … : third harmonic components


Example 1Consider a periodic signal )(tx , which is of the form

∑−=

=3

3

2)(k

tjkk eatx π

where 10 =a , 4111 == −aa , 2122 == −aa , 3133 == −aa .

(1) The fundamental frequency π=ω 20 or 1=T .(2) )(tx contains only (the first) three harmonic components,

function real a )6cos(32)4cos()2cos(

211

][31][

21][

411

31

21

411

41

21

31)(

)6(6)4(4)2(2

642)2()4()6(

→π+π+π+=

++++++=

++++++=

π−ππ−ππ−π

ππππ−π−π−

ttt

eeeeee

eeeeeetx

tjtjtjtjtjtj

tjtjtjtjtjtj


DCcomponent

1st harmoniccomponent

3rd harmoniccomponent

2ndharmonic


As in the above example, for a real periodic signal )(tx , its complexconjugate equals itself, i.e. )()( txtx =∗ .

Since the conjugate of tjke 0ω is tjke 0ω− , we have

⇒ ∗

−= kk aa or kk aa −∗ = , thus

∑+=∑ ++=∞

=

ω∞

=

ω−∗ω

10

10 }Re{2][)( 000

k

tjkk

k

tjkk

tjkk eaaeaeaatx

Writing ka as polar form, i.e., kjkk eAa θ= , gives

∑ θ+ω+=∞

=100 )cos(2)(

kkk tkAatx

∑==∑=

∑=

∞

−∞=

ω∞

−∞=

ω−∗∞

−∞=

ω∗

k

tjkk

k

tjkk

k

tjkk eatxeaeatx 000 )(conj)(


Alternatively, write kkk jCBa += ,

∑ ω−ω+=∞

=1000 )]sin()cos([2)(

kkk tkCtkBatx

To summarize, for a real periodic signal, the following three forms ofFourier series representation are equivalent:

tjk

kk ea 0ω

∞

−∞=∑

∑ θ+ω+∞

=100 )cos(2

kkk tkAa

∑ ω−ω+∞

=1000 )]sin()cos([2

kkk tkCtkBa

summation of complexexponentials

summation of cosinefunctions of non-zero phase

summation of cosine and sinefunctions with zero phase


Computing Fourier Series Coefficients

Given tjk

kkeatx 0)( ω∞

−∞=∑= ,

tnkj

kk

tjntjk

kk

tjn eaeeaetx 0000 )()( ω−∞

−∞=

ω−ω∞

−∞=

ω− ∑=⋅∑=⋅

Integrating both sides from 0 to T ,

∫ ∑ ∫=∑=∫∞

−∞=

ω−ω−∞

−∞=

ω− T

k

T tnkjk

tnkj

kk

T tjn dteadteadtetx 0 0)()(

0 ][)( 000

Knowing that

≠=

=∫ ω−

nknkT

dteT tnkj,0,

0)( 0 ,

∫= ω−T tjnn dtetx

Ta 0

0)(1, for !,2,1,0 ±±=n


Example 2

Find the Fourier series coefficients of )sin()( 0ttx ω= .Solution: By Euler’s relation,

jeettx

tjtj

2)sin()(

00

0

ω−ω −=ω=

Thus 00 =a , )2(11 ja = , )2(11 ja −=− , and 0=ka for !,3,2 ±±=n

Example 3Find the Fourier series coefficients of

( )42cos)cos(2)sin(1)( 000 π+ω+ω+ω+= ttttx .Solution:

tjjtjjtjtj eeeeej

ej

tx 0000 2)4

(2421

21)

211()

211(1)( ω−

π−ωπ

ω−ω ++−+++=


Thus,

10 =a , ja2111 −= , ja

2111 +=−

)1(42

2 ja += , )1(42

2 ja −=−0=ka for !,4,3 ±±=n

( ) ( )22 }Im{}Re{ kkk aaa +=

=∠ −

}Re{}Im{tan}{ 1

k

kk a

aa

Amplitude

Phase


Example 4Consider the following periodic square wave

Find its Fourier series representation.

Solution:The signal is periodic with period T . Also, it is an even signal.

Over the specific period from 2T− to 2T , the signal is defined as,

<<−=

otherwise,0,1

)( 11 TtTtx


It is easy to show that

∫=∫= + ω−ω− Ttt

tjkT tjkk dtetx

Tdtetx

Ta 0

000 )(1)(1

0

Thus we can do the integration over ]2,2[ TT−

∫=∫= −ω−

−ω− 1

100 1)(1 2

2TT

tjkTT

tjkk dte

Tdtetx

Ta

For 0=k ,

TTdt

Ta T

T1

0211 1

1=⋅= ∫− ,

For 0≠k ,

πω=

ω−=∫=

−

ω−−

ω−

kTke

Tjkdte

Ta

T

T

tjkTT

tjkk

)sin(11 10

0

1

1

011

0


T=4T1

T=8T1

T=16T1


Convergence Problems of Fourier series

The problems:

1) The integral ∫ −T tjk dtetxT 0

0)(1 ω may not converge, i.e. ∞→ka

2) Even if all ka ’s are finite, the summation tjk

kkea 0ω∑

∞

−∞= may not

be equal to the original signal )(tx .

Virtually all periodic signals arising in engineering do have a Fourierseries representation, without convergence problems.

tjk

kkeatx 0)( ω∑

∞

−∞== ∫ −= T tjk

k dtetxT

a0

0)(1 ω


Dirichlet Conditions of ConvergenceFor a periodic signal )(tx to have converged Fourier series,1) )(tx must be absolutely integrable over any period, i.e.

∫+

∞<Tt

tdttx0

0)( for any 0t

2) )(tx has a finite number ofmaxima & minima over anyperiod

3) )(tx has a finite number ofdiscontinuous over any period

∫+

∞→Tt

tdttx0

0)( unbounded

times ofoscillation

infinite number ofdiscontinuities


Convergence at Discontinuities

1) If a periodic signal has no discontinuities, its Fourier seriesrepresentation! converges, and! equals the original signal at every value of t .

2) If the signal has a finite number of discontinuities in eachperiod, its Fourier series representation! equals the original signal everywhere except at the

discontinuities! converges to the midpoint of )(tx at each discontinuity


Example 5

Expanding )(tx ,

tjk

kkeatx 0)( ω∞

−∞=∑= , where

πω=

kTkak

)sin( 10 and 10 =a

Let )(txN denote the Fourier series truncated at the N th harmonics,i.e.

tjkN

NkkN eatx 0)( ω

−=∑=

)(txN becomes a good approximation of )(tx if N is large enough.

<<−=

otherwise,0,1

)( 11 TtTtx


Properties of Fourier series

Notations: The relation that ka ’s are Fourier series coefficients of )(tx is

represented by katx →←FS)( .

Assume katx →←FS)( and kbty →←FS)( and )(tx and )(ty have

the same fundamental period T (or 0

2ωπ

).

Signal FS coefficientsLinearity )()( tBytAx + kk BbAa +Time shifting )( 0ttx − 00tj

kea ω−

Frequency shifting )(0 txe tjMω Mka −


Signal FS coefficientsConjugation )(* tx ∗

−kaTime reversal )( tx − ka−Time scaling )( tx α

period changed to αTka

Periodic convolution ∫ ττ−τT dtyx )()( kkbTaMultiplication )()( tytx

∑∞

−∞=−

llklba

Differentiation

dttdx )( kajk 0ω

Integration ∫ ∞−t dttx )(

kajk 0

1ω


Example 6Derive the Fourier series representation of the following signal.

Solution:

Define )(tg over a period,

≤<−≤<

=42212021

)(tt

tg

Let ktg ρ →←FS)( , for ,0=k 0])21(

21[

41 4

2200 =∫ −+∫=ρ dtdt

242

4

2

42

2

0

)2/sin()21(

41

21

41 π−

π−

π−

ππ=∫ −+∫=ρ jk

tjktjkk e

kkdtedte , 0≠k

Period T=4


Alternatively, we can use properties of Fourier series to solve it.Recalling the signal )(tx in Example 4:

)(tg can be expressed as 21)1()( −−= txtg , with 4=T & 11 =T

By the property of time-shift,0)1( ω− →←− jk

keatx FS ⇒ 2)1( π− →←− kjkeatx FS

Subtracting the DC offset of 21 from 0a ,

=

≠ππ

=

=−≠=ρ

π−π−

0,0

0,)2sin(

0,210, 2

0

2

k

kekk

kakea kjkj

kk

,)sin( 10πω=

kTkak

2/1/2 10 == TTa


Note that when we chose the interval [0,4):

∫

−+∫

=

π−

π− 4

2

22

0

221

41

21

41 dtedtea

tjktjkk

−

π−=

⋅π

−⋅−⋅π

−⋅=

π−

π−

π−

π−

4

2

2

2

0

2

4

2

2

2

0

2

41

2/1

81

2/1

81

tjktjk

tjktjk

eejk

ejk

ejk

[ ])(14

1 2 π−π−π− −−−π

−= jkkjjk eeejk


[ ] keejk

kjjk integerfor112

1 2 =−π

−= π−π− "

On the other hand, if we choose [-2,2)

∫

+∫

−=

π−

−

π− 2

0

20

2

221

41

21

41 dtedtea

tjktjkk

[ ]

2/

2/

2/2/2/

)2/sin(

)2/sin(22

12

1

π−

π−

ππ−π−

⋅ππ=

π−⋅⋅π

−=

−⋅⋅π

−=

jk

jk

jkjkjk

ekk

kjejk

eeejk


[ ][ ] 2/

2

0

2

0

2

2

2

0

2

0

2

2

)2/sin(12

1

)1()1(4

1

41

2/1

81

2/1

81

π−π−

π−π−

π−

−

π−

π−

−

π−

⋅ππ=−

π=

−−−π

−=

−π

=

⋅π

−⋅+⋅π

−⋅−=

jkjk

jkjk

tjktjk

tjktjk

ekke

jk

eejk

eejk

ejk

ejk

You can choose intervals [3,7), [4,8), etc. Make use of12 =πkje you can get the same answer.


Fourier series of even & odd signals

Real even signalsBy the property of time reversal, we have katx − →←− FS)( .

)(tx is even ⇒ )()( txtx =− ⇒ katx − →←FS)()(tx is real ⇒ kk aa =∗− ⇒ ∗

− →← katx FS)( Therefore, we have ∗= kk aa for all ∞<<∞− k . For a real even signal

1) }{ ka are real: ∗= kk aa2) }{ ka are even: kk aa −=

)cos(2)()( 01

01

000 tkaaeeaatx

kk

tjktjk

kk ω∑+=+∑+=

∞

=

ω−ω∞

=


Real odd signalsSimilar to the above derivation, we have ∗−= kk aa . For a real odd signal

1) }{ ka are purely imaginary: ∗−= kk aa2) }{ ka are odd: kk aa −−=3) ;00 =a

Let kk jCa = , where kC is real and kk CC −−=

)sin(2)()( 011

00 tkCeejCtxk

ktjktjk

kk ω∑−=−∑=

∞

=

ω−ω∞

=


Parseval’s Theorem

Recall tjk

kkeatx 0)( ω∞

−∞=∑=

The total average power over a period equals the sum of the average powers in all of the harmonic components.

∑∫∞

−∞==

kkT

adttxT

22)(1

220

1kT

tjkk adtea

T=∫ ω

Average powerover a period

Sum of squaredmagnitudes of allharmonics

Average power ofthe kth harmonic


Proof:

∑=

∫∑∑=

∑∫

∑=∫

∞

−∞=

−

ω−∞

−∞=

∞

−∞=

∞

−∞=

ω

−

∞

−∞=

ω

−

mm

T

T

tnmj

nnm

m

n

tjnn

T

T m

tjmm

T

T

a

dteaaT

dteaeaT

dttxT

2

2/

2/

)(*

*2/

2/

2/

2/

2

||

1

1|)(|1

0

00

Example 7Suppose we want to use )(tgN to approximate )(tg of Example 6where )(tgN denotes the Fourier series truncated at the N th

harmonics, i.e. tjkN

NkkN etg 0)( ω

−=∑ρ= . What is the minimum value of

N so that )(tgN contains at least 90% power of )(tg ?


Solution:

Average power of )(tg = 41)

21(

41)(1 24

0

2

0=∫=∫ dtdttg

T

T

Try 1=N , we have

2026.02)2/(sin2 22

22

12

121

1≈

π=

ππ=ρ+ρ=ρ∑ −

−=k

k

Try 2=N , 22

2

22

22

2222

2

2)(sin222π=

ππ+

π=ρ+ρ+

π=ρ∑ −

−=k

koddis,0 kk =ρ⇒

Try 3=N , we have

419.0

920

)3()2/3(sin222

22

2

22

32

3223

3×>

π=

ππ+

π=ρ+ρ+

π=ρ∑ −

−=k

k

Hence the minimum value of N is 3.


Discrete-time Fourier Series! Fourier Series is a frequency analysis tool for continuous-time

periodic signals while Discrete-Time Fourier Series (DTFS) isused for analyzing discrete-time periodic signals

! In fact, DTFS can be derived from the Fourier Series (I can showyou if you are interested on this)

! Similar to continuous-time case, a discrete-time signal ][nx isperiodic with fundamental period N if

][][ Nnxnx +=where N is the smallest integer for which the equation holds. Thefundamental frequency is defined as N/20 π=ω .

! However, notice that sampling a continuous-time periodic signaldoes not necessarily give a discrete-time periodic sequence.


Example 8

Consider a continuous-time periodic signal )5.0cos()( ttx = .Sampling )(tx with a sampling period of 1s gives

}),1cos(),5.0cos(),0cos(),5.0cos({]}[{ !! −=nx⇒ ][nx is not periodic

Consider another continuous-time periodic signal )5.0cos()( tty π= .Sampling )(ty with a sampling period of 1s yields

},0,1,0,1,0{}),cos(),5.0cos(),0cos(),5.0cos({]}[{

!!

!!

=πππ−=ny

⇒ ][ny is periodic and the fundamental period is 2=N

We can use DTFS to analyze ][ny but not ][nx . In fact, we can usediscrete-time Fourier transform (DTFT), which will be discussedlater, to analyze ][nx .


Fourier Series & LTI Systems

The output of a continuous-time periodic signal )(tx to a LTI systemwith impulse response )(th is given by

∫ ττ−τ=∗= ∞∞− dthxthtxty )()()()()(

where )(ty is also periodic with the same fundamental frequency as)(tx . That is, if

tjk

kk

tjk

kk ebtyeatx 00 )()( ω∞

−∞=

ω∞

−∞=∑=⇒∑=

⇒ only phases and magnitudes are changed


Proof:∫ ττ−τ=∫ ττ−τ=∗= ∞∞−

∞∞− dtxhdthxthtxty )()()()()()()(

Consider the k th component of )(tx , i.e., tjkk ea 0ω . The system

output due to this component is:

∫ τ⋅τ⋅⋅=

∫ τ⋅τ=∞∞−

τω−ω

∞∞−

τ−ω

dehae

deahtyjk

ktjk

tjkkk

00

0

)(

)()( )(

where we can see that

∫ τ⋅τ⋅= ∞∞−

τω− dehab jkkk

0)(

Combining all harmonic components, we have

tjk

kk

kk ebtyty 0)()( ω∞

−∞=

∞

−∞=∑=∑=


Example 9Given a LTI system with impulse response )()( tueth t−= and aninput signal )2cos(5.0)( ttx π= . Find the system output )(ty .

Solution:

∞π+−τπ−∞π+τ−π

∞ πττ−π−∞ πτ−τ−π

∞ τ−π−τ−πτ−∞ τ−

∞∞−

τ−∞∞−

π+−⋅+

π+−⋅=

∫ τ+∫ τ=

∫ τ+=∫ ττ−π=

∫ ττ−π⋅τ=∫ ττ−τ=

0)21(2

0)21(2

022

022

0)(2)(2

0

)21(1

41

)21(1

41

41

41

)(41))(2cos(

21

))(2cos(21)()()()(

ej

eej

e

deeedeee

deeedte

dtuedtxhty

tt

jtjt

tjtj


( )

)2sin()41(2

2)2cos()41(2

1

)2sin(2)2cos()41(2

12121

21)2sin()2cos(Re

21

21)2sin()2cos(Re

21

21Re

42

212141

212141)(

22

2

2

*2222

tt

tt

jj

jtjt

jtjt

je

je

je

je

jety

t

tttt

ππ+π+π

π+=

ππ+ππ+

=

π−π−⋅

π+π+π=

π+π+π=

π+=

π+

+π+

=

π−

+π+

=

π

πππ−π


Express )(ty as a function of )2cos( tπ only, we get

( )( ))2(tan2cos

4121

)2sin())2(sin(tan)2cos())2(cos(tan412

1

)2sin(41

2)2cos(41

1412

1)(

12

112

222

π−ππ+

=

ππ+πππ+

=

π

π+π+π

π+π+=

−

−−

t

tt

ttty

Comparing )(tx ( )2cos(5.0 tπ ) and )(ty , we see that they are of thesame frequency but their phases and magnitudes are different.

Nevertheless, we notice that computing the convolution is tediouseven for simple signal and system. We will solve this problem againwith the use of Fourier transform, which is shown to be easier.

Fourier series - City University of Hong Konghcso/ee31112_fs.pdf · 2010-05-25 · H. C. So Page 1...

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Transcript of Fourier series - City University of Hong Konghcso/ee31112_fs.pdf · 2010-05-25 · H. C. So Page 1...