Chuong 1 Fourier

Chng 1: Gii tch Fourier

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GII TCH FOURIER

Cui th k 18 nh ton hc, nh vt l ng thi l k s ngi Php tn Jean Baptiste Joseph Fourier c khm ph k l. Trong mt kt qu nghin cu ca mnh v phng trnh o hm ring m t s truyn nhit ca vt th, Fourier khng nh rng mi hm s u c th biu din di dng tng ca chui v hn cc hm lng gic.

Ban u khng nh ca Fourier khng c cc nh ton hc cng thi tin tng v ch n. Tuy nhin khng lu sau cc nh khoa hc nh gi cao kh nng ng dng v lnh vc ng dng rng ln ca tng ny. Pht hin ny ca Fourier c xp hng top ten v thnh tu ton hc trong mi thi i, trong danh sch ny cn c khm ph ca Newton v php tnh vi tch phn, ca Riemann v hnh hc vi phn, v 70 nm sau c l thuyt tng i ca Einstein. Gii tch Fourier l mt thnh phn khng th thiu ca ton hc ng dng hin i, n c ng dng rng ri trong ton l thuyt, vt l, k thut. Chng hn, x l tn hiu hin i bao gm audio, ting ni, hnh nh, video, d liu a chn, truyn sng v tuyn, v.v u c t c s trn gii tch Fourier v nhng dng khc ca n. Nhiu cng ngh tin tin hin i bao gm truyn hnh, CD v DVD m nhc, phim video, ha my tnh, x l nh, phn tch v lu tr du vn tay theo cch ny hay cch khc u c s dng nhng dng khc nhau ca l thuyt Fourier.

V mt l thuyt ngi ta c th phn tch cc tn hiu m thanh pht ra t cc nhc c nh: piano, violin, kn trumpet, kn oboe, trng . thnh chui Fourier tm ra cc tn s c bn (tone, overtone, ). V mt ng dng, l thuyt Fourier cn l mt cng c hiu qu ca m nhc in t hin i; mt nhc c in t c th c thit k sao cho c th t hp cc tng sin v cosin thun ty pht ra cc m thanh k diu ca nhc c. Nh vy, c hai cch t nhin v nhn to m nhc in t u da vo cc nguyn l tng qut ca Fourier.

tng ban u ca Fourier phn tch mt hm s tun hon thnh tng ca mt chui cc hm lng gic c m rng thnh biu din mt vc t ca khng gian Hilbert theo h trc chun y . V vy nu c mt h trc chun th ta c mt cch khai trin Fourier.

Trong chng ny ta xt nhng vn chnh ca gii tch Fourier

Khng gian Hilbert Chui Fourier v php bin i Fourier hu hn Php bin i Fourier Php bin i Fourier ri rc v php bin i Fourier nhanh. Php bin i Fourier hu hn c pht trin trn tng ca khai trin hm s tun

hon thnh chui Fourier, trong mi hm s hon ton c xc nh bi cc h s Fourier ca n v ngc li. C ba dng ca chui Fourier: dng cu phng (cng thc 1.24, 1.28),


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dng cc (cng thc 1.36) v dng phc (cng thc 1.37, 1.41, 1.42). Phn 1 ca mc ny s trnh by ba dng ca chui Fourier, cc cng thc lin h gia chng v km theo li nhn xt nn s dng dng no trong mi trng hp c th. Trng hp hm khng tun hon php bin i Fourier ri rc c thay bng php bin i Fourier, php bin i ngc duy nht c xy dng da vo cng thc tch phn Fourier.

Khi cc hm s biu din cho cc tn hiu th bin i Fourier ca chng c gi l biu din ph. Tn hiu tun hon s c ph ri rc, cn tn hiu khng tun hon s c ph lin tc. i s ca hm tn hiu l thi gian cn i s ca bin i Fourier ca n l tn s, v vy php bin i Fourier cn c gi l php bin i bin min thi gian v min tn s.

Trong thc t ta thng phi tnh ton gi tr s ca cc tn hiu c ri rc ho bng cch chn mu ti mt s hu hn cc thi im, khi ph tng ng cng nhn c ti mt s hu hn cc tn s bng php bin i Fourier ri rc. Ngoi ra thc hin nhanh php bin i Fourier ri rc, ngi ta s dng cc thut ton bin i Fourier nhanh.

Hng ng dng vo vin thng: Phn tch ph, phn tch truyn dn tn hiu, ghp knh v tuyn, ghp knh quang, nh gi cht lng WDM...

1.1. KHNG GIAN HILBERT

Khi nim khng gian Hilbert l s m rng ca khi nim khng gian Euclide, l khng gian vc t hu hn chiu vi tch v hng. Khng gian Euclide c trang b trong chng trnh ton i cng bc i hc.

1.1.1. Tch v hng

Khi nim tch v hng ca hai vc t ca khng gian vc t bt k c khi qut t

tch v hng cos( ; )uv u v u v=GG G G JG G . Trong khng gian vc t tch v hng ca hai vc t n

1 2( , ,..., )nx x x x= , 1 2( , ,..., )ny y y y= c nh ngha nh sau:

1 1 2 2, n nx y x y x y x y= + + +" . (1.1) Tch v hng gi mt vai tr rt quan trng, v l mt khi nim c ng dng rng ri trong ton hc, c hc, vt l Bit tch v hng ca mi cp vc t th c th suy ra di ca vc t (bnh phng di ca vc t bng tch v hng ca vc t y vi chnh n) v gc gia hai vc t (cosin ca gc ny bng tch v hng ca hai vc t chia cho tch ca hai di ca chng). Thnh th trong khi nim tch v hng bao hm kh nng o di, o gc, v t i n nhng khi nim quan trng khc nh tnh trc giao, hnh chiu thng

Khi nim tch v hng c m rng i vi khng gian vc t bt k nh sau:


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Mt dng song tuyn tnh i xng xc nh dng ca khng gian vc t c gi l mt tch v hng ca khng gian vc t .

Nh vy tch v hng ,u v ca hai vc t u , v trong khng gian vc t H c cc tnh cht ct yu sau:

1) , ,u v v u= 2) 1 2 1 2, ,u u v u v u v+ = + , 3) ,u v u v = , vi mi s thc 4) ,u u > 0 nu u v 0 ,u u 0= nu u = 0 . Nu H l khng gian vc t trn trng s phc th iu kin 1) c thay bng

, ,u v v u= , trong ,v u l s phc lin hp ca s phc ,v u . Mt khng gian vc t vi tch v hng c gi l khng gian tin Hilbert.

Vi mi vc t ta nh ngha v k hiu chun hay mun ca vc t v H v qua biu thc

,v v= v . (1.2) Nu 1v = th v c gi l vc t n v. C th kim chng c

1) 0v v 0v = khi v ch khi v = 0 . 2) Vi mi : | |v v = . 3) u v u v+ + .

nh ngha 1.1: Dy cc vc t { } 1n nu = hi t v vc t nu u lim 0nn u u = , ta k hiu , vy lim nn

u u =

lim 0, : ;nnu u N n N u u = > < n (1.3)

Dy cc vc t { } c gi l dy c bn nu 1n nu = ,lim 0n mn m u u = , vy { } 1n nu = l dy c bn khi v ch khi 0, : , ; n mN n m N u u > < .

C th chng minh c rng mi dy hi t l dy c bn,tuy nhin iu ngc li cha chc ng.


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Khng gian tin Hilbert tha mn iu kin mi dy c bn u hi t c gi l khng gian Hilbert (y l tnh cht y ca khng gian Hilbert).

V d 1.1: Ngi ta chng minh c khng gian cc dy bnh phng hi t

20

0( ) : | |n n n

nl

==

2 = < (1.4) vi tch v hng xc nh nh sau

0( ); ( )n n n

n

=n = (1.5)

l mt khng gian Hilbert.

Khng gian cc hm bnh phng kh tch trn on [ ];a b (theo ngha tch phn Lebesgue)

[ ] [ ]{ }2 ; ;( ) : | ( ) |a b a bL x t x t dt2= < (1.6) vi tch v hng xc nh nh sau

[ ];( ); ( ) ( ) ( )a bx t y t x t y t= (1.7) cng l mt khng gian Hilbert.

Ch rng i vi cc hm lin tc hoc lin tc tng khc th tch phn Lebesgue trng vi tch phn theo ngha thng thng.

Hi t trong khng gian v (cng thc 1.7) c gi l hi t bnh phng

trung bnh.

2l [ ]2 ;a bL

1.1.2. Bt ng thc Cauchy-Schwarz

nh l 1.1: Bt ng thc Cauchy-Schwarz

Vi mi , lun c ,u v H

,u v u v (1.8) ng thc xy ra khi v ch khi ph thuc tuyn tnh. ,u v

Chng minh: Nu mt trong hai vc t bng th c hai v ca bt ng thc trn u bng , do bt ng thc nghim ng.

0 0

Gi s , vi mi ta c: v 0 t , 0u tv u tv+ + .


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Mt khc 22( ) , 2 , 2F t u tv u tv t v t v u u= + + = + + l mt tam thc bc hai i vi v lun lun khng m. V vy t 2 2 2' ,F v u v u 0 = . T suy ra bt ng thc Cauchy-Schwarz.

Khi ph thuc th u (hoc v,u v kv= ku= ): 2, ,u v kv v k v kv v u v= = = = .

Ngc li nu ,u v u v= th ' 0F = . Do tn ti sao cho 0t 0 0, 0u t v u t v u t v0+ + = = . nh l c chng minh.

p dng bt ng thc Cauchy-Schwarz vo khng gian vi tch v hng (1.1) ta c bt ng thc Bunnhiacopsky:

n

( ) ( )( )2 2 2 21 1 1 1... ... ...n n n n2x y x y x x y y+ + + + + + (1.9) ng thc xy ra khi v ch khi 1 1, ..., n nx ty x ty= = .

H qu:

1) Nu dy cc vc t { } hi t v vc t th 1n nu = u lim , ,nn u v u v = ng vi mi . v2) Nu dy { } 1n nu = hi t v u v { } 1n nv = hi t v th v ,lim , ,n mn m u v u v = .

Chng minh: 1) 0 , , ,n n nu v u v u u v u u v = 0 khi n .

2) Hai dy { } 1n nu = v { } hi t do chn, v vy tn ti sao cho 1n nv = C nu C , nv C vi mi n .

0 , , , , , ,n m n m m n m mu v u v u u v u v v u u v u v v = + +

( ) 0n m m n mu u v u v v C u u v v + + khi , . n m 1.1.3. H trc chun, trc chun ho Gram-Schmidt

nh ngha 1.2: Hai vc t gi l trc giao nhau, k hiu u,u v H v , nu , 0u v = . H cc vc t { }1,..., ,...nS v v= ca H c gi l h trc giao nu hai vc t bt k ca

h u trc giao nhau. S


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H trc giao cc vc t n v c gi l h trc chun.

Vy h cc vc t { }1,..., ,...nS e e= l h trc chun khi tha mn iu kin ,i j ije e = trong 10ij

i ji j= =

nu

nu l k hiu Kronecker (1.10)

V d 1.2: Trong khng gian vc t [20;2L ] cc hm bnh phng kh tch vi tch v hng xc nh bi cng thc (1.7), h cc hm s sau l mt h trc giao

{ }1, cos ; sin ; 1, 2, ...nt nt n = (1.11) Tht vy

2 2

0 0

cos sin 0 ;ntdt ntdt n

= = (1.12) 2

0

cos sin 0 ; ,nt mtdt n m

= (1.13) 2 2

0 0

cos cos sin sin 0 ;nt mtdt nt mtdt n m

= =

(1.14)

2 22 2

0 0

cos sin ; 0ntdt ntdt n

= = (1.15) nh l 1.2: Mi h trc chun l h c lp tuyn tnh.

Chng minh: Gi s h { }1,..., ,...nS v v= trc chun, khi nu 1 1 ... m mv v + + = 0 th 1 1 ... , 0i m mv v v = + + =i m vi mi 1,...,i = . Do c lp tuyn tnh. S

nh l c chng minh.

nh l 1.3: Gi s { }1,..., ,...nS u u= l mt h cc vc t c lp tuyn tnh ca khng gian Hilbert H . Khi ta c th tm c h trc chun { }1' ,..., ,...nS e e= sao cho

{ } { }1 1span ,..., span ,..., ;k ke e u u= vi mi 1,2,...k = . Chng minh: Ta xy dng h trc chun 'S theo cc bc quy np sau y m c gi l qu trnh trc chun ho Gram-Schmidt.

) : V h c lp nn 1k = S 1u 0 . t 111

ueu

= .


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) : Xt 2k = 2 2 1 1,e u e e= + 2u , ta c 2e 0 (v nu 2e = 0 th , iu ny

tri vi gi thit h c lp). t

2u ke= 1S 22

2

eee

= , h { }1 2,e e trc chun v

{ } { }1 2 1 2span , span ,e e u u= . ) Gi s xy dng c n 1k . Ngha l tn ti { }1,..., ke e 1 trc chun sao cho { } { }1 1 1span ,..., span ,...,ke e u u = 1k . Tng t trn ta xt

1

1,

k

k k i ii

e u e e

=ku= + (1.16)

ta cng c ke 0 ( v nu ke = 0 th l t hp tuyn tnh ca , do l t hp tuyn tnh ca , iu ny mu thun vi gi thit h c lp). t

ku 1,..., ke e 111,..., ku u S

kkk

eee

= (1.17)

th . Vy h ; 1,..., 1k ie e i k = { }1,..., ke e trc chun v { } { } { }1 1 1 1span ,..., span ,..., , span ,..., ,k k ke e e e e u u u= = 1k k .

V d 1.3: Trong xt h 3 vc t c lp: 3 1 (1,1,1)u = , 2 ( 1,1,1)u = , . Hy trc chun ho h

3 (1,2,1)u ={ }1 2 3, ,S u u u=

Bc 1: 1 3u = 111

1 1 1, ,3 3 3

ueu

= = .

Bc 2: 2 2 1 1 21 1 1 1 4 2 2, , , ( 1,1,1) , ,

3 3 33 3 3 3e u e e u = + = + =

22 ( 2,1,1)3

e = 2 2 1 1, ,6 6 6e = .

Bc 3: 3 3 1 1 3 2 2, ,e u e e u e e= + 3u

4 1 1 1 1 2 1 1 1 1, , , , (1, 2,1) 0, ,2 23 3 3 3 6 6 6 6

= + =

31 (0,1, 1)2

e = 3 1 10, ,2 2e = .

{ }1 2 3, ,e e e l h vc t trc chun ho ca h { }1 2 3, ,u u u .


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V d 1.4: Xt h gm ba hm s , , ca khng gian c th cho trong

hnh 1.1 1( )s t 2( )s t 3( )s t [20;TL ]

3( )s t

t / 3T 0 T

1

t0 / 3T

1

1( )s t

0 2 / 3T

1

2 ( )s t

t

Hnh 1.1: th ba hm , , 1( )s t 2( )s t 3( )s t

Ba hm s , , c lp tuyn tnh, trc chun ha Gram-Schmidt ba hm s

ny ta c ba hm , , xc nh nh sau: 1( )s t 2( )s t 3( )s t

1( )e t 2( )e t 3( )e t

( )21 1 10

( ), ( ) ( )3

T Ts t s t s t dt= = 1 13 3 / 0 / 3( ) ( ) 0 T te t s tT = =

nu

nu n

Tgc li

2 1 2 10

( ), ( ) ( ) ( )3

T Ts t e t s t e t dt= =

2 2 11 / 3 2

( ) ( ) ( )03

T t TTe t s t e t = =

nu

nu ngc li/ 3

Vy 23 / / 3 2 / 3( )0

T T t Te t =

nu

nu ngc li

Tng t 33 / / 3( )0

T T te t T =

nu 2

nu ngc li

H trc chun , , c th 1( )e t 2( )e t 3( )e t

3 / T

3( )e t

t 23T 0 T

3 / T

0

2( )e t

t 3T 2

3T

t

1( )e t

0 3T

3 / T

Hnh 1.2: th h trc chun e t , e , e t 1( ) 2 ( )t 3( )


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1.1.4. H trc chun y , chui Fourier

nh l 1.4: Gi s l mt h trc chun ca khng gian Hilbert , vi mi { } 1n ne = H u H ta c:

1) Nu th 1

n nn

u e

== ,n nu e = .

Ta gi ,n u e = n l h s Fourier ca i vi v chui gi l chui

Fourier ca theo h { } . u ne

1n n

ne

=

u 1n ne=

2) 221| |n

nu

= (bt ng thc Bessel).

3) Chui hi t v 1

n nn

e

=

1n n n

nu e

=

e vi mi . n

Chng minh: 1) 1 1 1

, , lim , lim ,n n

m n n m k k m k k mn nn k ku e e e e e e e

= = =

m= = = = . 2) Vi mi : n 2

1 1 1 1, ,

n n n n

k k k k k k k kk k k k

u u u e u e e u e= = = =

= = + +

1 1 1 1, ,

n n n n


e e e u e= = = =

= +

1 1 1 1, ,

n n n n


u e e u e u e= = = =

+ + 2

1 1 1 1 1, ,

n n n n n

k k k k k k k k kk k k k k

e e u e u e= = = = =

= + | | . Vy 22

1| |n

nu

= .

3) T 1) v 2) ta c : vi mi , vi mi : n m n 2, 0m m m

k k k k kk n k n k n

e e= = = = khi

, v v khng gian Hilbert y nn chui Fourier n 1

n nn

e

= hi t.

Vi mi : n1 1

, , , , , limm

n k k n n k k n n k kmk ke u e e u e e e u e e

= =

= = 1

,k=

1

, lim , ,m

n n k k nm ke u e e =

= = 0n = . nh l c chng minh.


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nh ngha 1.3: H trc chun { } 1n ne = ca khng gian Hilbert H c gi l h trc chun y khi ch c vc t 0 mi trc giao vi tt c cc phn t ca h, ngha l:

, nu e = 0 vi mi 1, 2,...n = th u = 0 (1.18) V d 1.5:

1) H cc hm

1 1 1, cos ; sin ; 1, 2, ...2

nt nt n = (1.19)

l mt h trc chun y ca khng gian Hilbert [ ]20;2L . 2) H cc vc t 2ne l , 1, 2, ...n =

{ } , trong 1n ne = 1 (1,0,0,....)e = , 2 (0,1,0,....)e = , (1.20) l mt h trc chun y ca khng gian Hilbert . 2l

nh l 1.5: Gi s { } 1n ne = l mt h trc chun ca khng gian Hilbert , H ,n u e = n l h s Fourier u i vi . Cc mnh sau y tng ng: H ne

1) l mt h trc chun y { } 1n ne =2) Vi mi : u H

1n n

nu e

==

3) Vi mi : ,u v H1

, n nn

u v

== trong ,n v e = n l h s Fourier ca i

vi .

v

ne

4) Vi mi : u H 221

nn

u

== .

Chng minh: 1) 2): Theo kt qu 3) ca nh l 1.4 ta c 1

n n nn

u e

=

e vi mi , vy

theo nh ngha ca h trc chun y (cng thc 1.17): .

n

1 1n n n n

n nu e u

= = = = 0 e

2) 3): 1 1 1 1

, , lim , limn n

k k m m k k m mn nk m k mu v e e e e

= = = =

= =


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1 1 1 1

lim , limn n k

k k m m k k k kn nk m k ke e

= = = =

= = = .

3) 4): Cho u ta c v= 221

, nn

u u u

== = .

4) 1): Gi s ,n nu e = = 0 vi mi 1, 2,...n = th 221

0nn

u

== = , do . Vy h

y .

u = 0

{ } 1n ne =nh l c chng minh.

nh l 1.6: (RieszFischer). Cho { } 1n ne = l mt h trc chun y ca khng gian Hilbert H . Nu dy s { } tha mn iu kin 1n n=

2

1n

n

= < (1.21)

Th s c mt vc t duy nht u nhn cc s H n lm h s Fourier v

1n n

nu e

== , 22

1n

nu

== (1.22)

Chng minh: Vi mi : m n 2,m m m

k k k k kk n k n k n

e e= = = = , iu kin (1.19) ko theo

2 0m

kk n=

khi n , v v khng gian Hilbert y nn chui 1

n nn

e

= hi t.

t , ta c 1

n nn

u

== e

1 1 1, , lim , lim ,

m m

n k k n k k n k k nm mk k ku e e e e e e e

= = =

= = = = n , vy nhn lm h s Fourier v v h u n { } 1n ne = y nn ta cng c (1.21). Ngoi ra nu c vc

t nhn cc s lm h s Fourier th v n , n n nu v e 0 = = vi mi , do n u v = 0 . Vy vc t u nhn cc s lm h s Fourier l duy nht. H n

nh l c chng minh.


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1.2 CHUI FOURIER

1.2.1 Khai trin Fourier ca hm tun hon chu k 2 Trong khng gian cc hm bnh phng kh tch trn on [20;2L ] [ ]0, 2 tch v hng

xc nh theo cng thc (1.7) v h trc chun (1.19) ta c chui Fourier ca hm ( )x t l mt chui lng gic v hn c dng

(01

( ) cos sin2 n nn

a )x t a nt b=

+ + nt (1.23) trong

2 2 2

00 0 0

1 1( ) ; ( ) cos ; ( )sin ; 1, 2, ...n na x t dt a x t ntdt b x t ntdt n

= = = = (1.24) H s 1

2 ca s hng th nht xut pht t s thun li trong vic tnh ton sau ny.

Theo nh l 1.5 chui Fourier (01

cos sin2 n nn

a a nt b n

=+ + )t ca hm ( )x t vi cc h s

tha mn (1.23) hi t v ( )x t theo ngha bnh phng trung bnh (1.3). Tuy nhin cha chc hi t theo im, chnh v vy ngi ta dng k hiu thay cho du =. Cc cu hi c t ra mt cch t nhin:

(i) Khi no chui lng gic v hn (1.23) hi t?

(ii) Loi hm ( )x t no c th biu din thnh tng ca chui Fourier? Ngha l c th thay du = thay cho du .

nh l 1.7(nh l Dirichlet): Nu hm ( )x t tun hon chu k 2 , n iu tng khc v b chn (gi l iu kin Dirichlet), th chui Fourier hi t v du trong cng thc (1.23) c thay bng du = .

Ti cc im gin on ta k hiu

( 0) ( 0)( )2

x t x tx t + + = (1.25)

trong ln lt l gii hn phi v gii hn tri ca ( 0), ( 0)x t x t+ ( )x t ti t . V d 1.6: Xt hm s ( )x t t= , ; tun hon chu k t < < 2 . V ( )x t l hm l nn cc h s Fourier c th tnh nh sau

01 0a tdt

= = , 1 cos 0na t ntdt

= = ,


15

12

00

1 2 2 cos sin 2sin sin ( 1)nnt nt ntb t ntdt t ntdt

n nn

+

= = = + = . Do chui Fourier tng ng

1

1

sin sin 2 sin 3 sin 42 ( 1) 2 sin2 3 4

n

n

nt t t tt tn

+=

= + "+ (1.26) p dng nh l 1.7 ta c

1

1

sin2 ( 1)0

n

n

t tnttn

+=

< < = = nu

nu

Thay 2

t = v chia hai v cho 2 ta c

1 1 1 114 3 5 7 9 = + + "

V d 1.7: Xt hm s ( )x t t= , ; tun hon chu k t < < 2 . V ( )x t l hm chn nn cc h s Fourier c th tnh nh sau

1 sin 0nb t ntdt

= = ; 0 0

1 2a t dt tdt

= = = ,

20 20

0 22 2 sin coscos 4 2 1n t

n kt nt nta t ntdt

n kn nn

=

= = = + = 0

= +

nu

nu.

Do chui Fourier tng ng

21

4 cos 4 cos3 cos5 cos 7cos2 2 9 25 49n

nt t t tt tn

= = + + + "+ (1.27)

Thay ta c 0t =2

20

1 1 1 118 9 25 49 (2 1)n n

= = + + + + = +" .

V d 1.8: Xt hm bc nhy tun hon chu k 2 xc nh nh sau 1 0

( )0 0

tt

t< < = <


16

0

2 2 11 1( )sin sin0 2

nn k

b t ntdt ntdt nn k

= += = = = nu

nu.

Chui Fourier tng ng 1 2 sin 3 sin 5 sin 7( ) sin2 3 5 7

t t tt t + + + + + "

Hnh 1.3: th ca hm bc nhy tun hon

p dng nh l 1.7 ta c cng thc

0 (2 1)1 2 sin 3 sin 5 sin 7sin 1 2 (2 1)2 3 5 7

1/ 2

k tt t tt k

t k

2kt k

+ < < + + + + + = < < + = "

nu

nu

nu

Cc th sau tng ng l th ca tng ring ln lt c 3, 5 v 10 s hng ca chui Fourier ca hm bc nhy tun hon.

Hnh 1.4: th cc tng ring ca chui Fourier ca hm bc nhy tun

T cc th trn ta nhn thy rng mc d hm gc gin on nhng cc tng ring ca chui Fourier tng ng l cc hm lin tc hi t, mc d chm chp. Tuy nhin gn v tr gin on ca hm th th ca cc tng ring Fourier vt qu v tr khong 9%. Vng vt qu v tr ny cng nh khi s cc s hng ca tng ring Fourier tng ln, nhng ln ca n khng thay i. iu ny gii thch tnh cht khng hi t u ca chui Fourier. Hin tng ny ln


17

u tin c nh vt l Josiah Gibbs (ngi M) pht hin v ngy nay ngi ta gi l hin tng Gibbs.

1.2.2 Khai trin Fourier ca hm tun hon chu k lT 20 = Trng hp hm tun hon vi chu k bt k, ta c th i bin a v chu k 2 v

p dng cc kt qu mc trn.

Gi s ( )x t l mt hm tun hon chu k . t 2l ( ) ly t x t= th tun hon chu k . Nu

( )y t

2 ( )x t tha mn iu kin Dirichlet th cng tha mn iu kin Dirichlet, do c th khai trin thnh chui Fourier.

( )y t

( )01

( ) cos sin2 n nn

ay t a nt b nt

== + +

trong v tri ca ng thc trn c quy c nh (1.25). ( )y t

Thay bin s ta c

01

( ) cos sin2 n nn

a n nx t y t a t b tl l

= = = + + l

(1.28)

Cc h s Fourier c tnh theo cng thc sau:

2 2 2

00 0 0

1 1 1( ) ; ( ) cos ; ( )sin ; 1, 2, ...l l l

n nn na x t dt a x t tdt b x t tdt n

l l l l l = = = = (1.29)

V d 1.9: Xt hm s ( )x t t= , ; tun hon chu k . V 1 t <


18

2 2

01 1( ) ; ( ) cos ;

l c l c

nc c

na x t dt a x t tdtll l

+ + = = 21 ( )sin ; 1, 2, ...

l c

nc

nb x t tdt nl l

+ c= = cng thc c tnh i xng ngi ta thng chn c l= :

01 1( ) ; ( ) cos ;

l l

nl l

na x t dt a x t tdtll l

= = 1 ( )sin ; 1, 2, ...l

nl

nt tdt nl l

= =b x (1.30) 2. Nu l hm l tun hon chu k th )(tx 2l ( ) cos nx t

lt l hm l v ( )sin nx t

l t l hm

chn, do cc h s Fourier (1.24) tha mn

00

20; ( )sin ; 1, 2, ...l

n nna a b x t tdt n

l l= = = = (1.31)

3. Nu l hm chn tun hon chu k th )(tx 2l ( ) cos nx tl

t l hm chn v ( )sin nx t tl l

hm l, do cc h s Fourier (1.29) tha mn

00 0

2 20; ( ) ; ( ) cos ; 1, 2, ...l l

n nnb a x t dt a x t tdt n

l l l= = = = (1.32)

4. Nu l hm xc nh, b chn v n iu tng khc trong khong . Ta c th m rng thnh hm tun hon chu k

)(tx ( , )a b2l b a= . Do c th khai trin thnh chui

Fourier, cc h s Fourier c tnh nh sau )(tx

02 2( ) ; ( ) cos ;

b b

na a

na x t dt a x tb a b a b a

= = 2 tdt

2 2( )sin ; 1, 2, ...b

na

nb x t tdt nb a b a

= =)

(1.33)

5. Nu l hm xc nh, b chn v n iu tng khc trong khong . Khi ta c

th m rng thnh hm chn hoc hm l tun hon chu k . Nu m rng thnh hm chn th cc h s Fourier c tnh theo cng thc (1.32) v nu m rng thnh hm l th cc h s Fourier c tnh theo cng thc (1.31).

)(tx (0, l2l

1.2.3 Dng cc ca chui Fourier (Polar Fourier Series)

T cng thc (1.28) nu ta t

2 200 ;2 n naA A a nb= = + (1.34)

v gc xc nh bi , 0 2n n <


19

n ncos , sinn nn n

a bA A

= = (1.35)

th cng thc (1.28) c th vit li

00

1 1( ) cos sin cos

2 n n nn n

a n n nx t a t b t A A tl l l

= = = + + = + n (1.36)

Cng thc (1.28) c gi l chui Fourier dng cu phng (Quadrature Fourier Series). Cng thc (1.36) c gi l chui Fourier dng cc ca . )(tx

1.2.4 Dng phc ca chui Fourier (Complex Fourier Series)

Thay cng thc Euler

cos2

i ie e + = , sin2

i ie ei

=

vo (1.23) ta c

( ) nt nt nt nt0 01 1

( ) cos sin2 2 2

i i i i

n n n nn n

a a e e e ex t a nt b nt a bi

= =

+ = + + = + + 2

nt nt012 2 2

i in n n n

n

a a ib a ibe e =

+ = + + Vy ta c th vit chui Fourier di dng phc

nt 2 22 1 0 1 2( )

i it it itn

nx t c e c e c e c c e c e

== = + + + + + " "it +

2/ 2

(1.37)

trong cc h s Fourier phc xc nh nh sau nc

hoc 0 0 / 2

( ) /( )

n n n

n n n

c ac a ibc a ib

== = +

0 02

( )n n n

n n n

a ca c cb i c c

== +=

(1.38)

Mt khc, tng t (1.7) ta c tch v hng cc hm phc

2

0

; ( ) ( )x y x t y t

= dt Vi tch v hng ny h cc hm { }mti

me

= l mt h trc giao, ngha l tha mn

2

nt

0

20

i imt n me e dtn m

== nu

nu (1.39)


20

V vy cc h s Fourier phc (1.38) c th tnh trc tip

nt1 ( )2

inc x t e

= dt hoc

2nt1 ( ) ,

2

ci

nc

c x t e dt+

c= (1.40) V d 1.10: Xt hm bc nhy tun hon v d 1.8

nt nt

0

1 021 1( ) 0 0

2 21

i in

n

c t e dt e dt n n

nin

== = =

nu

nu chn

nu l

Vy, hm bc nhy n v c khai trin Fourier

(2 1)1( )2 2

m it

m

i etm

+

=

1 + . V d 1.11: Tm khai trin Fourier ca hm m tun hon ( ) atx t e= .

( )nt ( )

0

1 12 2 2 (

a in tat i a in t

nt

ec e e dt e dta in

) =

= = = ( ) ( ) ( )

2 2( )sh( 1) ( 1)

2 ( ) 2 ( ) 2 ( ) ( )

a in t a in a in a an n

t

e e e e e a ina in a in a in a n

=

+= = = +a .

Vy hm c chui Fourier tng ng

nt2 2

sh ( 1) ( )nat in

a a ine ea n

= +

+ . Hm tun hon chu k c khai trin Fourier dng phc 0 2T = l

( )ni tl

nn

x t c e

= , 21 ( ) ,2

nc l i tl

nc

c x t e dtl

+ c= (1.41)

Nu k hiu 00

1fT

= l tn s c bn ca hm tun hon chu k th cng thc (1.41) c biu din

0T

, 02( ) i n f tnn

x t c e

= 02 21 ( ) ,2

c li n f t

nc

c x t e dtl

+ c= (1.42) Nhn xt 1.2: Cng thc (1.34)-(1.38) cho thy dng cc, dng phc v dng cu phng ca chui Fourier l hon ton tng ng, ngha l t dng ny ta c th biu din duy nht qua dng kia v ngc li. Vy th dng no c ng dng tt nht. Cu tr li ph thuc vo tng


21

trng hp c th. Nu bi ton thin v gii tch th s dng dng phc s thun li hn v vic tnh cc h s d hn. Tuy nhin khi o cc hm dng sng c thc hin trong phng th

nghim th dng cc s thun tin hn, v cc thit b o lng nh vn k, my phn tch ph s c c bin v pha. Dng cc kt qu th nghim o c cc nh k thut c th v cc

vch ph mt pha l cc on thng ng vi mi gi tr bin ti tn s

nc

nA 00

nnf nf

T= = .

1.2.5 ng thc Parseval

nh l 1.8: i vi mi hm ( )x t tun hon chu k 0 2T l= tho mn iu kin Dirichlet s xy ra ng thc Parseval

0

22

0

1 ( )c T

nnc

x t dt cT

+

== (1.43)

Chng minh:

0 0 0

2

0 0 0

1 1 1( ) ( ) ( )m nc T c T c T i t i tl l

m nm nc c c

x t dt x t x t dt c e c e dtT T T

+ + +

= =

= =

0

2

0 ,

1m nc T i t i tl l

m n nm n nc

c c e dt cT

+

= == = .

1.2.6 o hm v tch phn ca chui Fourier

i vi chui hm hi t, mt vn t nhin t ra l: khi ly o hm hoc ly tch phn ca tng s hng ca chui ta c chui mi, chui mi ny c hi t v o hm hoc tch phn ca hm tng ca chui ban u khng? Trng hp chui ly tha th cu tr li l khng nh. Vi tng ny ngi ta thng tm nghim ca phng trnh vi phn di dng chui ly tha nu nghim ca phng trnh khng phi l hm s cp.

S hi t ca chui Fourier tinh t hn v vy i hi phi thn trng khi p dng phng php ly o hm hoc tch phn theo cc s hng. Tuy nhin, trong nhiu tnh hung c hai php ton ny em li nhng kt qu th v v cung cp mt cng c hu ch xy dng chui Fourier ca cc hm tng i phc tp.

1.2.6.1 Tch phn ca chui Fourier

Ta thy rng nguyn hm lun mn hn hm gc, v vy c th tin on rng s khng gp kh khn g khi ly tch phn ca chui Fourier. Tuy nhin c mt tr ngi l nguyn hm ca mt hm tun hon cha chc l hm tun hon. Chng hn hm hng 1 l mt hm tun hon nhng c nguyn hm, c th x , khng tun hon. V nguyn hm ca hm sin , hm l hm v hm , do nguyn hm ca tt c cc hm tun hon khc trong chui

coscos sin


22

Fourier cng l hm tun hon. V vy ch c s hng hng 02a c th gy nn kh khn khi ly

tch phn ca chui Fourier.

B 1.1: Gi s ( )x t l hm tun hon chu k 2 , khi tch phn l hm

tun hon chu k 2 khi v ch khi

0

( ) ( )t

y t x u du= ( ) 0x t dt

= (c gi tr trung bnh bng 0).

nh l 1.9: Nu ( )x t l hm lin tc tng khc, tun hon chu k 2 v c gi tr trung bnh bng 0 th c th ly tch phn tng s hng ca chui Fourier ca ( )x t nhn c chui Fourier ca nguyn hm

10

( ) ( ) cos sint

n n

n

b ay t x u du m nt ntn n

= = + + ,

1 ( )2

m y t

= dt (1.44)

V d 1.12: Hm l tun hon chu k 2 v ( )x t t= , do c gi tr trung bnh bng 0. Theo v d 1.6 ta c chui Fourier

1

1

sin2 ( 1)nn

nttn

=

Ly tch phn tng s hng ca chui Fourier ta c

2 2 1 2

21

( 1) cos 2 cos3 cos 42 cos 2 cos2 6 6 4 9 16

n

n

t tnt tn

= = + "

t t +

V 2 21

2 2 6t dx

= .

Nu ( )x t c gi tr trung bnh khc 0, chui Fourier tng ng c s hng 0 0a

( )01

( ) cos sin2 n nn

ax t a nt b

=+ + nt

Trong trng hp ny kt qu ca ly tch phn s l

0

10

( ) ( ) cos sin2

tn n

n

a b ay t x u du t m nt ntn n

= = + + + ;

1 ( )2

m y t

= dt (1.45)

Ch rng v phi ca chui (1.45) khng phi l chui Fourier. Ta c th vit li di dng khai trin chui Fourier ca hm tun hon chu k 2 nh sau


23

0

1( ) cos sin

2n n

n

a b ay t t m nt ntn n

= + + (1.46)

1.2.6.2 o hm ca chui Fourier

Php tnh o hm ngc vi php ly tch phn. o hm c th lm cho hm xu hn. V vy khi s dng phng php ly o hm ca chui Fourier ( )x t chng ta cn phi ch n s tha mn iu kin Dirichlet ca '( )x t . i hi ny c tha mn nu ( )x t kh vi lin tc tng khc n cp 2.

nh l 1.10: Nu ( )x t l hm tun hon chu k 2 v kh vi lin tc tng khc n cp 2 th c th ly o hm tng s hng ca chui Fourier ca ( )x t nhn c chui Fourier ca o hm

(1.47) [1

( ) '( ) cos sinn nn

t x t nb nt na nt

= = ]

V d 1.13: Nu o hm chui Fourier ca hm ( )x t t= (v d 1.7) ta c 4 sin 3 sin 5 sin 7'( ) sin

3 5 7t t tx t t + + + + " .

Mt khc o hm ca ( )x t = t ta c hm du 1 0

sign1 0

td tt

tdt>= =


24

K s ngi Anh Oliver Heaviside l ngi u tin s dng hm delta trong cc ng dng thc t ca mnh, mc d cc nh ton hc l thuyt cng thi cho rng l ngh in r. Ba mi nm sau, nh Vt l l thuyt ni ting Paul Dirac s dng hm delta trong l thuyt c hc lng t ca mnh, nh cui cng cc nh l thuyt chp nhn hm delta. Nm 1944 nh ton hc Php Laurent Schwartz cui cng xy dng c l thuyt phn b kt hp vi hm suy rng iu ny gii thch c s tn ti ca hm delta.

C hai cch khc nhau xy dng hm delta:

Cch th nht xem hm delta l gii hn ca dy hm trn theo ngha bnh thng. Cch th hai xem hm delta nh l mt phim hm tuyn tnh ca khng gian hm thch

hp.

C hai u quan trng v ng quan tm. Tuy nhin cch th nht s d dng tip thu hn, v vy ta ch xt phng php ny.

Phng php gii hn xem hm delta 0( )t t l gii hn ca dy hm kh vi c gi

tr ngy cng tp trung ti v c tch phn lun bng 1.

( )ng t

0t t=

Chng hn xt dy hm 2 2( ) (1 )nng tn t

= + tha mn hai iu kin

0 0lim ( )

0nnt

g tt= =

nu

nu (1.50)

1( ) arctan 1n tg t dt n t

=

= = (1.51)

Hnh 1.5: th cc hm ( )ng t


25

V vy, mt cch hnh thc ta ng nht gii hn ca dy hm l hm delta tp trung ti

gc .

( )ng t0t =

0lim ( ) ( ) ( )nng t t t = = . (1.52)

Hnh 1.5 cho thy cc hm c gi tr ngy cng tp trung ti gc ( )ng t 0t = . Cn ch rng c nhiu cch chn cc hm c gii hn l hm delta. ( )ng t

Trng hp hm delta c gi tr tp trung ti bt k c th nhn c t hm

bng cch tnh tin 0( )t t 0t

( )t

0 0( ) ( )t t t t = . (1.53) V vy, c th xem l gii hn ca dy hm

0( )t t

l ( )0 2 0( ) ( ) 1 ( )nn ng t g t t n t t= = + 2 (1.54) o hm v tch phn ca hm delta

T cng thc (1.48)-(1.49) ta c

Vi mi hm lin tc ( )x t :

0

( ) 0( ) ( )

0

l

vx v v

t x t dtl< = =


26

1 0lim ( ) ( ) 1/ 2 0

0 0nn

tf t t t

t

>= = =


27

Cng thc o hm (1.57) tng ng

6'( ) '( ) ( 1)5

x t y t t= + , 1 1

( ) 2 15

ty t

t t

nu

nu

th hm ( )x t th hm '( )x t

Hnh 1.7: th ca ( )x t v o hm '( )x t

V d 1.15: Xt hm s 20

( ) 1 0 1

2 1t

t t

x t t t

e t


28

Tch phn ca hm bc nhy 0(t t ) gin on l hm dc lin tc 0( )t t

0 0( ) ( )t t t t = ; 0 0 0 000

( ) ( ) ( )0

t

t ta

t t t t at dt u t u t t

a t t > > = = = < =


29

Hm phn b c xc nh t hm khi lng xc sut theo cng thc

{ }( ) ( )k

X X kx x

F x P X x p x

= = th ca hm phn b ( ) l c dng bc thang lin tc phi ti cc bc nhy. XF x

S dng cng thc (1.55) v (1.59) ta c th vit li

( ) ( ) ( ) ( )k k

x

X X k X kx x x

F x p x p x t x dt

= = k V vy ta c th xem hm mt ca bin ngu nhin ri rc l

( ) ( ) ( )k

X X k kx

f x p x x x= . Khai trin Fourier ca hm delta

p dng cng thc (1.56) tnh h s Fourier ta c

-

1 1 1( ) cos cos 0na t ntdt n

= = = , -

1 1( )sin sin 0 0nb t ntdt n

= = = (1.62)

Vy hm delta c khai trin Fourier

(1 1( ) cos cos 2 cos32

t t t + + + + )t " (1.63)

Thay ik ik

cos2

t te ekt+= (cng thc Euler) vo (1.63) ta c

( )ik 2 21 1( ) 12 2t it it it ikt e e e e e =

= + + + + + " t +" (1.64) Cng c th nhn c cng thc khai trin trn bng cch tnh trc tip cc h s theo cng thc (1.40)

ik 0

-

1 1( )2 2

t ikkc t e dt e

= = = 12 .

Tng ring ca chui Fourier

ik12

nt

nk n

s e=

= l tng ca s hng ca cp s nhn c s hng u tin l v cng bi , do : 2 1n + in te ite


30

1 1(2 1) ( 1) in 2 2

in/ 2 / 2

1sin1 1 1 1 1 2

12 2 2 21 1 sin2

i n t i n ti n t i n t t

tn it it it it

n te e e e es e

e e e e t

+ + + +

+ = = = =

Hnh 1.10: th cc tng ring ca chui Fourier hm Delta

1.3 PHP BIN I FOURIER HU HN

Mi hm tun hon c xc nh duy nht bi cc h s Fourier ca n v ngc li (cng thc 1.23, 1.24, 1.28, 1.29), iu ny c suy ra t tnh cht trc giao ca h 1.11, 1.39.

Tng t ta c th chng minh c h cc hm phc tun hon { }2i nf ne = l mt h trc chun trn on [ ] 0, 1

12 2

0

10

i nf i mf n mn m

e e df ==

nu

nu. (1.65)

Da vo h trc chun ny ta nh ngha php bin i Fourier hu hn ca cc tn hiu ri rc nh sau.

1.3.1 nh ngha php bin i Fourier hu hn

nh ngha 1.4: Bin i Fourier hu hn ca dy tn hiu ri rc { } =nnx )( l


31

l { } 2( ) ( ) ( ) i nfn

X f x n x n e =

= = F (1.66) nu chui v phi hi t.

Cng thc bin i ngc

l{ } l110

( ) ( ) ( ) i nf2x n X f X f e= = F df (1.67) V d 1.17: Tm bin i Fourier hu hn ca tn hiu ri rc )(rect)( nnx N= , N l 1 s t nhin.

=

li ngcnunu

01

)(rect10 Nn

nN

Gii: l21

2 22

0

1( ) ( )1

i NfNi nf i nf

i fn n

eX f x n e ee

= =

= = =

( 1)sinsin

i Nf i Nf i Nfi N f

i f i f i fe e e Ne f

fe e e

= = .

Nhn xt 1.3:

1. Trong cng thc bin i Fourier 1.66, 1.67 i s c k hiu cho tn s. C ti liu khng biu din bin i Fourier qua min tn s m qua min tn s gc nh sau

ff

l { }( ) ( ) ( ) i nn

X x n x n e =

= = F , l{ } l210

1( ) ( ) ( )2

i nx n X X e d = = F (1.68)

Hai cch biu din ny tng ng vi nhau qua php i bin s 2 f = . 2. Mt iu kin tn hiu ri rc { } =nnx )( tn ti bin i Fourier hu hn l


32

1.3.2 Cc tnh cht ca php bin i Fourier hu hn

Tng t php bin i Laplace, php bin i Fourier hu hn c cc tnh cht sau:

1. Tuyn tnh:

{ } { } { }( ) ( ) ( ) ( )x n y n x n y n + = +F F F (1.69) Chng minh: { } ( ) 2( ) ( ) ( ) ( ) i nf

nx n y n x n y n e

=

+ = +F

{ } {2 2( ) ( ) ( ) ( )i nf i nfn n

}x n e y n e x n y n = =

= + = + F F . 2. Tr:

l { } { } l020( ) ( ) ( ) ( )i n fX f x n x n n e X = =F F f( )

x n

. (1.70)

Chng minh: { } 0 02 220 0 0( ) ( ) ( )i n f i n n fi nfn n

x n n x n n e e x n n e = =

= = F

. { }0 02 22( ) ( )i n f i n fi nfn

e x n e e =

= = F3. Dch chuyn nh:

l { } { } l02 0( ) ( ) ( ) ( )i nfX f x n e x n X f f= =F F . (1.71) 4. iu ch:

{ } l l0 02 2 00 ( ) (( )cos(2 ) ( ) 2 2i nf i nf

0)X f f X f fe ex n nf x n + ++ = =

F F . (1.72)

5. Lin hp phc: l { } 2( ) ( ) ( ) i nfn

X f x n x n e =

= = F

{ } l2 2( ) ( ) ( ) ( )i nf i nfn n

x n x n e x n e X = =

= = = F f (1.73) Do nu thc th )(nx l l( ) ( )X f X f= .

6. Bin s o: l { } 2( ) ( ) ( ) i nfn

X f x n x n e =

= = F


33

{ } l2 2 ( )( )( ) ( ) ( ) (i nf i n fn n

)x n x n e x n e X = =

= = = F f (1.74) 7. Tch chp:

{ } { } { }( ) ( ) ( ) ( )x n y n x n y n = F F F (1.75) Chng minh: Ta c ( ) ( ) ( ) ( ) ( )

kz n x n y n x k y n k

== =

l 2 2( ) ( ) ( ) ( ) ( )i nf i kf i n k fn k n k

Z f x k y n k e x k e y n k e = = = =

= = 2 ( )

)

l l2 ( ) 2( ) ( ) ( ) (i n k f i kfk n

y n k e x k e X f Y f = =

= = 8. Tch chp nh:

{ } { } { }( ) ( ) ( ) ( )x n y n x n y n = F F F (1.76) Chng minh: { } 2( ) ( ) ( ) ( ) i nf

nx n y n x n y n e

=

= F . Theo 2.71 ta c: 1

2 2 ( )

0

( ) ( ) ( ) ( )i nf i m n u i nfn n m

x n y n e x m e du y n e = = =

= 2

12 ( ) 2

0

( ) ( )i m n u i nfn m

x m e y n e du = =

= l l1 2 2 ( )

0

( ) ( ) ( ) ( )i mu i n f um n

x m e y n e du X f Y f = =

= = . 9. Bin i ca hm tng quan

= =m

yx nmymxnr )()()(, { } l l, ( ) ( ) ( )x yr n X f Y f=F (1.77) nh l Weiner-Khinchin: { } l 2, ( ) ( )x xr n X f=F . (1.78) Nu thc, )(),( nynx =

=myx nmymxnr )()()(, { } l l, ( ) ( ) ( )x yr n X f Y f= F .


34

Chng minh: , ( ) ( ) ( ) ( ) ( )x ym

r n x m y m n x n y n

== = { } l l, ( ) ( ) ( )x yr n X f Y f=F .

Hoc ta c th chng minh trc tip nh sau:

{ } 2, ( ) ( ) ( ) i nfx yn m

r n x m y m n e = =

= F 2 ( ) 2( ) ( ) i n m f i mf

m nx m y m n e e

= =

=

l l2 ( ) 2( ) ( ) ( ) ( )i m n f i mfm n

x m y m n e e X f Y = =

= = f . 10. o hm nh:

l { } { } l( )( ) ( ) ( )2i d X fX f x n nx n

df= = F F (1.79)

Chng minh: { } l22 1 (( ) ( ) ( )2 2

i nfi nf

n n

de i d X fnx n nx n e x ni df

= =

= = = F )df 11. ng thc Parseval:

l l1

0

( ) ( ) ( ) ( )n

x n y n X f Y f df

== ; l

1 22

0

( ) ( )n

x n X f

== df . (1.80)

Chng minh:

l l1 12 20 0

( ) ( ) ( ) ( ) ( ) ( )i nf i nfn n n

x n y n y n X f e df X f y n e df = = =

= =

l l l1 1

2 ( )

0 0

( ) ( ) ( ) ( )i n fn

X f y n e df X f Y f df =

= = .

Ta c th tng kt trong bng sau.

Tnh cht Tn hiu ri rc { } =nnx )( Bin i Fourier l { }( ) ( )X f x= F nTuyn tnh ( ) ( )x n y n + l l( ) ( )X f Y f + Tr 0( )x n n l02 ( )i n fe X f


35

Dch chuyn nh 02 ( )i nfe x n l 0( )X f f

iu ch 0( )cos(2 )x n nf l l

0 0( ) (2

X f f X f f + + )

Lin hp phc ( )x n l( )X f Bin s o ( )x n l( )X f Tch chp ( ) ( )x n y n l l( ) ( )X f Y f Tch chp nh ( ) ( )x n y n l l( ) ( )X f Y f

Hm tng quan , ( ) ( ) ( )x ym

r n x m y m n

== l l( ) ( )X f Y f

o hm nh ( )nx n l( )

2i d X f

df

1.4 PHP BIN I FOURIER

Khi u chui Fourier c xy dng vi mc ch gii quyt cc bi ton tng ng vi cc hm s xc nh trong min b chn hoc hm tun hon. gii quyt cc bi ton c cc hm s xc nh trn ton b tp s thc t < < ngi ta m rng mt cch t nhin phng php chui Fourier, iu ny a n php bin i Fourier. Php bin i Fourier l mt cng c mnh m v ng vai tr ct yu trong nhiu min ng dng nh: Gii phng trnh vi phn, phng trnh o hm ring, x l tn hiu, l thuyt iu khin v trong nhiu lnh vc khc ca ton l thuyt cng nh ton ng dng. i vi cc nh ton hc php bin i Fourier l c bn hn php bin i Laplace.

C s ca php bin i Fourier l cng thc tch phn Fourier, cng thc ny c c bng cch xt chui Fourier trong khong kh ln ty , sau cho khong ny tin n v cng.

1.4.1 Cng thc tch phn Fourier

nh l 1.11: Nu hm kh tch tuyt i trn ton b trc thc ()(tx


36

Chng minh: V hm tha mn iu kin Dirichlet trn ton b trc thc nn vi mi ta c th khai trin thnh chui Fourier trong khong

)(tx0>l );( ll (xem nhn xt v cng thc

1.28).

0

1( ) cos sin

2 n nn

a n nx t a t bl l

=t = + +

Cc h s Fourier c tnh theo cng thc sau:

01 1 1( ) ; ( )cos ; ( )sin ; 1, 2, ...

l l l

n nl l l

n na x u du a x u udu b x u udu nl l l l l

= = = =

1

1 1( ) ( ) ( ) cos cos sin sin2

l l

nl l

n n n nx t x u du x u u t u tl l l l l l

= du = + +

1

1 1( ) ( ) ( )cos ( )2

l l

nl l

nx t x u du x u u tl l l

= = + du (1.82)

V nn khi cho


37

V hm cosin l hm chn v sin l hm l nn t cng thc (1.81) ta cng c:

1 1( ) ( )cos ( ) ( )cos ( )

2 2x t d x u u t du d x u t u

= = du

( ) ( )1 1( ) cos ( ) sin ( ) ( )2 2

i t ud x u t u i t u du d x u e du

= + = Vy

1( ) ( )

2i u i tx t x u e du

= e d (1.86) (1.86) c gi l cng thc tch phn Fourier phc.

Nhn xt 2.4:

1. Cc cng thc trn s dng quy c (1.25) ti nhng im khng lin tc.

2. Nu l hm chn th cng thc (1.86) tr thnh )(tx

0 0

2( ) cos ( )cosx t td x u udu

= . (1.87) 3. Nu l hm l th )(tx

0 0

2( ) sin ( )sinx t td x u udu

= . (1.88) 4. Cc cng thc tch phn Fourier, nh l 1.11 c pht biu v chng minh cho trng hp l hm thc. Tuy nhin do tnh cht tuyn tnh ca tch phn nn cc kt qu trn

vn cn ng cho trng hp hm phc bin thc kh tch tuyt i c phn thc, phn o tha mn iu kin Dirichlet.

)(tx)(tx

5. i bin 2 2f d = = df v thay vo cng thc (1.88) ta c

2 ( ) 2 2( ) ( ) ( )i f u t i fu i ftx t df x u e du x u e du e df

= = (1.89) 1.4.2 Php bin i Fourier

nh ngha 1.5: Bin i Fourier (vit tt l FT) ca hm kh tch tuyt i trn trc thc v tha mn iu kin Dirichlet l

)(tx


38

l { } 2( ) ( ) ( ) ,i ftX f x t x t e dt f

= = F (1.90) Trong k thut, nu l hm dng sng (waveform) theo thi gian )(tx t th c

gi l ph hai pha ca (two - sided spectrum), cn tham s ch tn s, c n v l Hz.

l( )X f)(tx f

T cng thc tch phn Fourier (1.89) ta c cng thc bin i ngc

l{ } l1 2( ) ( ) ( ) i ftx t X f X f e df

= = F (1.91) Hm nh qua php bin i Fourier c th vit di dng cc l( )X f

l l ( )( ) ( ) i fX f X f e = (1.92) trong

l l l( ) ( ) ( )X f X f X f= , l( ) ( )f X f = (1.93) c gi dng bin - pha ca php bin i.

Cp l( ), ( )x t X f c gi l cp bin i Fourier. 1.4.3 Tnh cht php bin i Fourier

Tng t cc tnh cht (1.69)-(1.80) ca php bin i Fourier hu hn, php bin i Fourier c cc tnh cht c tng kt trong bng sau:

Tnh cht Hm )(tx Bin i Fourier l( )X f1. Tuyn tnh 1 2( ) ( )x t x t + l l1 2( ) ( )X f X f +

2. ng dng )(atx l ( )1 /| |

X f aa

3. Lin hp )(tx l( )X f 4. i ngu l( )X t )( fx 5. Tr )( dTtx l2 ( )di Te X f 6. ng dng tnh

tin ( )x at b+ l

2

| |

bi fae fX

a a


39

7. Dch chuyn nh 02 ( )i f te x t l 0( )X f f

8. iu ch 0( )cos 2x t f t l l0 01 1( ) (2 2X f f X f f + + )

9. o hm nn

dttxd )(

( ) l2 (ni f X f )

10. Tch phn

t

duux )( l l1 1( ) (0) ( )2 2

X f X fi f

+

11. o hm nh )(txt n l( )

2

n n

ni d X f

df

12. Tch chp 1 2 1 2( ) ( ) ( ) ( )x t x t x u x t u du

= l l1 2( ) ( )X f X f

13. Tch )()( 21 txtx l l1 2( ) ( )X f X f

Hm trong tnh cht 10. l hm delta Dirac (xem mc 1.2.7). T nh ngha bin i Fourier (1.90) ta nhn thy rng nu l hm thc chn th bin

i Fourier ca n cng l hm thc chn. Kt hp vi tnh cht i ngu 4. ta c th chuyn i

vai tr ca v cho nhau, ngha l

)(tx

)(tx l( )X fl { } l{ }( ) ( ) ( ) ( )X f x t X t x f= =F F (1.94)

1.4.3 nh l Parseval v nh l nng lng Rayleigh

Nu l hai hm bnh phng kh tch (gi l hm kiu nng lng) th ta c

ng thc Parseval )(),( 21 txtx

l l1 21 2( ) ( ) ( ) ( )x t x t dt X f X f df

= (1.95)

Khi )()()( 21 txtxtx == ta c nh l nng lng Rayleigh

l 22 1( ) ( )x t dt X f df

= (1.96)

Nh vy c th chuyn tnh nng lng trong min thi gian bng tnh nng lng trong min tn s.


40

Cng thc (1.96) c th chng minh bng cch s dng cng thc tch phn Fourier nh sau:

l 221 2 1( ) ( ) ( ) ( ) i ftx t x t dt x t X f e df dt

=

l 22 1( ) ( ) i ftX f x t e dt df

=

l l1 2( ) ( )X f X f df

= .

1.4.4 Bin i Fourier ca cc hm c bit

V d 1.18: Bin i Fourier ca xung ch nht hay hnh hp c di 2a

(1.97) 1 | |

( )0 | | ,a

t at

t a a >

nu

nu 0

l 22

2

0

0

1

( ) i ft

i f

a ai fta

aa

e

f

ff e dt

=

= =

nu

nu

0

sin(2 ) 0

1 fa f ff

=

=

nu

nu

t

0

sin( ) 0

1sinc( )

tt t

tt

=

=

nu

nu (1.98)

Ta c

{ }( ) 2 sinc(2 )a t a af =F . (1.99) Php bin i Fourier ngc cho php khi phc li gi tr ca xung ch nht (xem

cng thc (1.91))

( )a t


41

2| |

sin(2 ) | || |

11/ 20

i ftt a

a f t af

t ae df

=

nu

nu

nu

(1.100)

Tch phn thc v phn o (1.100) ta c:

0

| |cos (2 )sin(2 ) | |

| |

/ 2/ 4

0

t aft a f t a

ft a

df

=

nu

nu

nu

;

sin (2 )sin(2 ) 0ft a ff

df

= .

Khi ta c xung hnh vung: 1a =1 | |

( )0 | |

tt

t11

nu

nu v { }( ) 2sinc(2 )t f =F .

p dng cng thc (1.94) ta cng c

{ }2sinc(2 ) ( )t f= F .

Hnh 1.11: th ca ( )t v l( )f

Cng thc (1.100) khi phc gi tr ca ( )t da vo tch phn v hn ca l( )f

l{ } l1 2( ) ( ) ( ) i ftt f f e df

= = F Tuy nhin khi tnh ton s ngi ta ch tnh tch phn trong khong hu hn.

Hai th sau y tng ng l th ca v . l25

2

25

( )i fte f

df dfl50

2

50

( )i fte f


42

Hnh 1.12

V d 1.19: Xung tam gic n v

(1.101) = >

= < >nu

nu. (1.104)

C bin i Fourier


43

l ( 2 )20 0

1( )2 2

i f tt i ft

rt

eX f e e dti f i f

+ =

= = = + + S dng cng thc bin i Fourier ngc ta c

2 01/ 2 0

20 0

ti ft e te df ti f

t

>= = + nu

nu 0 (1.105)

C bin i Fourier

l00 ( 2 )

2 1( )2 2

i f tt i ft

lt

eX f e e dti f i f

=

= = = . Xung dng m chn, hai pha

( ) ; 0tex t e= > (1.106)

R rng ( ) ( ) ( )e l rx t x t x t= + , do bin i Fourier s l l l l

2 21 1 2( ) ( ) ( )2 2 4

e l rX f X f X fi f i f 2f

= + = + = + + (1.107)

p dng bin i Fourier ngc ta c cng thc tnh tch phn

2

2 2 2 2 2 22 2 cos

4 4

i ftt e fe df

f f

= = + + 2 t df . Xung dng m l ( ) (sgn ) tox t t e

= 0

( ) ( ) ( )0 ; 0

t

o r l t

e tx t x t x t

e t

>= = < >nu

nu (1.108)

l l l2 2

1 1 4( ) ( ) ( )2 2 4

o r lfX f X f X f i

i f i f 2f= = = + + (1.109)

S dng cng thc bin i Fourier ngc ta c


44

2

2 2 2 2 2 2sin 2(sgn ) 4 4

4 4

i ftt fe f ftt e i df df

f f

= = + + (1.110) C th nghim li c

( )1( ) (sgn ) t eodx tx t t e

dt= = ,

do l l 2 21 4( ) ( 2 ) ( )

4o e

fX f i f X f i 2f= = + .

Mt khc, v hm xung dng m l gin on ti 0t = vi bc nhy bng 2, do o hm ca n cha hm delta v tha mn

( ) 2 ( ) ( ) 2 ( )to edx t e t x t

dt= + = + t

Cng thc ny cng thng nht vi kt qu ca bin i Fourier

l l2 2 22 2 2 2 2 28 2( 2 ) ( ) 2 2 ( ) ( )

4 4o e

fi f X f f X ff f

= = = + + .

V d 1.21: Xt hm hu t

2 21( )x t

t c= + , 0c >

p dng cng thc (1.96) v (1.106)-(1.107) ta c

2 2 22 , 04

fet

= > + F .

2 2

2 2 2 2 2 22

4 2 ( / 2 )

i ft i ftf e ee dt

t t

= = + + dt l 2 2

2 2( )i ft

c feX f dt ect c

= =+ Vy

22 2

1 ,c fe cct c

0= > + F (1.111)

V d 1.22: Bin i Fourier ca Hm delta Dirac ( )t tp trung gi tr ti . 0t =T cng thc (1.48)-(1.49) ta c


45

00

0( )

tt

t

= =

vi

vi v ( ) 1t dt

= (1.112)

1. ( ) ( ) (0)x t t dt x

= vi mi hm ( )x t lin tc ti 0.

2. . (1.113) { } { }2 1( ) ( ) 1 ( ) 1i ft i ftt t e dt t e

= = = = F F 2 df

df

3. Nu gi thit l hm chn th ( )t

2( ) ( ) i ftt t e

= = . (1.114) 4. p dng tnh ng dng ca bin i Fourier ta c

1( ) ( )at ta

= . (1.115)

5. i bin s ly tch phn ta c

0 0 0( ) ( ) ( ) ( ) ( )0x t t x t t t dt x t

= = (1.116)

vi mi hm ( )x t lin tc ti . 0t

Mt kt qu th v nhn c t xung dng m chn (1.106) l nu ly gii hn khi ta nhn c hm hng ng nht bng 1. 0

Mt khc ly gii hn ca bin i Fourier (1.107) ca xung dng m chn ta c

2 2 20

0 02lim04

fff = = +

nu

nu (1.117)

Gii hn ny gip ta nh n nh ngha ca hm delta

2 2 2 20( ) lim lim

(1 ) ( )nntn t t

= = + + , trong 1n = .

V d 1.23: Hm bc nhy

(1.118) 00

1( ) ( )

0

ttt

t

>

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