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Transcript of Chuong 1 Fourier
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Chng 1: Gii tch Fourier
3
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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Chng 1: Gii tch Fourier
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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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Chng 1: Gii tch Fourier
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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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Chng 1: Gii tch Fourier
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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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Chng 1: Gii tch Fourier
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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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Chng 1: Gii tch Fourier
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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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Chng 1: Gii tch Fourier
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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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Chng 1: Gii tch Fourier
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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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Chng 1: Gii tch Fourier
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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
k k k k k k k kk k k k
e e e u e= = = =
= +
1 1 1 1, ,
n n n n
k k k k k k k kk k k k
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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Chng 1: Gii tch Fourier
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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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Chng 1: Gii tch Fourier
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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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Chng 1: Gii tch Fourier
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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
= = ,
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Chng 1: Gii tch Fourier
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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< < = <
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Chng 1: Gii tch Fourier
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
-
Chng 1: Gii tch Fourier
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 <
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Chng 1: Gii tch Fourier
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 <
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Chng 1: Gii tch Fourier
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)
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Chng 1: Gii tch Fourier
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
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Chng 1: Gii tch Fourier
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
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Chng 1: Gii tch Fourier
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
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Chng 1: Gii tch Fourier
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>= =
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Chng 1: Gii tch Fourier
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
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Chng 1: Gii tch Fourier
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< = =
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Chng 1: Gii tch Fourier
26
1 0lim ( ) ( ) 1/ 2 0
0 0nn
tf t t t
t
>= = =
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Chng 1: Gii tch Fourier
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
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Chng 1: Gii tch Fourier
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 > > = = = < =
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Chng 1: Gii tch Fourier
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
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Chng 1: Gii tch Fourier
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
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Chng 1: Gii tch Fourier
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
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Chng 1: Gii tch Fourier
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
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Chng 1: Gii tch Fourier
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 .
-
Chng 1: Gii tch Fourier
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
-
Chng 1: Gii tch Fourier
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
-
Chng 1: Gii tch Fourier
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
-
Chng 1: Gii tch Fourier
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
-
Chng 1: Gii tch Fourier
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
-
Chng 1: Gii tch Fourier
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.
-
Chng 1: Gii tch Fourier
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
-
Chng 1: Gii tch Fourier
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
-
Chng 1: Gii tch Fourier
42
Hnh 1.12
V d 1.19: Xung tam gic n v
(1.101) = >
= < >nu
nu. (1.104)
C bin i Fourier
-
Chng 1: Gii tch 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
-
Chng 1: Gii tch Fourier
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
-
Chng 1: Gii tch Fourier
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
>