toan_vienthong

SCH HNG DN HC TP

TON CHUYN NGNH

(Dng cho sinh vin ngnh T-VT h o to i hc t xa)

Lu hnh ni b

H NI - 2006

========== HC VIN CNG NGH BU CHNH VIN THNG

HC VIN CNG NGH BU CHNH VIN THNG

SCH HNG DN HC TP

TON CHUYN NGNH

Bin son : Ts. L B LONG

LI NI U

Tip theo chng trnh ton hc i cng bao gm gii tch 1, 2 v ton i s. Sinh vin chuyn ngnh in t-vin thng cn cn trang b thm cng c ton xc sut thng k v ton k thut.

p ng nhu cu hc tp ca sinh vin chuyn ngnh in t vin thng ca Hc vin, chng ti bin son tp bi ging Ton k thut t nm 2000 theo cng chi tit mn hc ca Hc vin. Qua qu trnh ging dy chng ti thy rng cn hiu chnh v b sung thm cung cp cho sinh vin nhng cng c ton hc tt hn. Trong ln ti bn ln th hai tp bi ging c nng ln thnh gio trnh, ni dung bm st hn na nhng c th ca chuyn ngnh vin thng. Chng hn trong ni dung ca php bin i Fourier chng ti s dng min tn s f thay cho min . Da vo tnh duy nht ca khai trin Laurent chng ti gii thiu php bin i Z biu din cc tn hiu ri rc bng cc hm gii tch. Tuy nhin do c th ca phng thc o to t xa nn chng ti bin son li cho ph hp vi loi hnh o to ny.

Tp gio trnh bao gm 7 chng. Mi chng cha ng cc ni dung thit yu v c coi l cc cng c ton hc c lc, hiu qu cho sinh vin, cho k s i su vo lnh vc vin thng. Ni dung gio trnh p ng y nhng yu cu ca cng chi tit mn hc c Hc vin duyt. Trong tng chng chng ti c gng trnh by mt cch tng quan i n cc khi nim v cc kt qu. Ch chng minh cc nh l i hi nhng cng c va phi khng qu su xa hoc chng minh cc nh l m trong qu trnh chng minh gip ngi c hiu su hn bn cht ca nh l v gip ngi c d dng hn khi vn dng nh l. Cc nh l kh chng minh s c ch dn n cc ti liu tham kho khc. Sau mi kt qu u c v d minh ho. Cui cng tng phn thng c nhng nhn xt bnh lun v vic m rng kt qu hoc kh nng ng dng chng. Tuy nhin chng ti khng i qu su vo cc v d minh ho mang tnh chuyn su v vin thng v s hn ch ca chng ti v lnh vc ny v cng v vt ra khi mc ch ca cun ti liu.

Th t ca tng V d, nh l, nh ngha, c nh s theo tng loi v chng. Chng hn V d 3.2, nh ngha 3.1 l v d th hai v nh ngha u tin ca chng 3 Nu cn tham kho n v d, nh l, nh ngha hay cng thc no th chng ti ch r s th t ca v d, nh l, nh ngha tng ng. Cc cng thc c nh s th t theo tng chng.

H thng cu hi n tp v bi tp ca tng chng c hai loi. Loi trc nghim ng sai nhm kim tra trc tip mc hiu bi ca hc vin cn loi bi tp tng hp gip hc vin vn dng kin thc mt cch su sc hn.

V nhn thc ca chng ti v chuyn ngnh in t Vin thng cn hn ch nn khng trnh khi nhiu thiu st trong vic bin son ti liu ny, cng nh cha a ra ht cc cng c ton hc cn thit cn trang b cho cc cn b nghin cu v chuyn ngnh in t vin thng. Chng ti rt mong s ng gp ca cc nh chuyn mn chng ti hon thin tt hn tp ti liu ny.

Tc gi xin by t li cm n ti PGS.TS. L Trng Vinh, TS T Vn Ban, c bn tho v cho nhng kin phn bin qu gi v c bit ti KS Nguyn Ch Thnh ngi gip ti bin tp hon chnh cun ti liu.

Chng 1: Hm bin s phc

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Cui cng, tc gi xin by t s cm n i vi Ban Gim c Hc vin Cng ngh Bu Chnh Vin Thng, Trung tm o to Bu Chnh Vin Thng 1 v bn b ng nghip khuyn khch, ng vin, to nhiu iu kin thun li chng ti hon thnh tp ti liu ny.

H Ni 5/2006

Tc gi


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CHNG I: HM BIN S PHC

PHN GII THIU

Gii tch phc l mt b phn ca ton hc hin i c nhiu ng dng trong k thut. Nhiu hin tng vt l v t nhin i hi phi s dng s phc mi m t c. Trong chng ny chng ta tm hiu nhng vn c bn ca gii tch phc: Ln cn, gii hn, hm phc lin tc, gii tch, tch phn phc, chui s phc, chui ly tha, chui Laurent nghin cu cc vn ny chng ta thng lin h vi nhng kt qu ta t c i vi hm bin thc. Mi hm bin phc ( ) ( ) ( , ) ( , )w f z f x iy u x y iv x y= = + = + tng ng vi hai hm thc hai bin

( , )u x y , ( , )v x y . Hm phc ( )f z lin tc khi v ch khi ( , )u x y , ( , )v x y lin tc. ( )f z kh vi khi v ch khi ( , )u x y , ( , )v x y c o hm ring cp 1 tha mn iu kin Cauchy-Riemann. Tch phn phc tng ng vi hai tch phn ng loi 2 Mi chui s phc tng ng vi hai chui s thc c s hng tng qut l phn thc v phn o ca s hng tng qut ca chui s phc cho. S hi t hay phn k c xc nh bi s hi t hay phn k ca hai chui s thc ny.

T nhng tnh cht c th ca hm bin phc chng ta c cc cng thc tch phn Cauchy. l cng thc lin h gia gi tr ca hm phc ti mt im vi tch phn dc theo ng cong kn bao quanh im ny. Trn c s cng thc tch phn Cauchy ta c th chng minh c cc kt qu: Mi hm phc gii tch th c o hm mi cp, c th khai trin hm phc gii tch thnh chui Taylor, hm gii tch trong hnh vnh khn c khai trin thnh chui Laurent.

Bng cch tnh thng d ca hm s ti im bt thng c lp ta c th p dng tnh cc tch phn phc v tch phn thc, tnh cc h s trong khai trin Laurent v php bin i Z ngc.

Da vo tnh duy nht ca khai trin Laurent ta c th xy dng php bin i Z.Php bin i Z cho php biu din dy tn hiu s ri rc bng hm gii tch.

hc tt chng ny hc vin cn xem li cc kt qu ca gii tch thc.

NI DUNG

1.1. S PHC

1.1.1. Dng tng qut ca s phc

S phc c dng tng qut z x iy= + , trong ,x y l cc s thc; 12 =i . x l phn thc ca z , k hiu Re z . y l phn o ca z , k hiu Im z .

Khi 0y = th z x= l s thc; khi 0x = th z iy= gi l s thun o. S phc x iy , k hiu z , c gi l s phc lin hp vi s phc z x iy= + .


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Hai s phc 1 1 1z x iy= + v 2 2 2z x iy= + bng nhau khi v ch khi phn thc v phn o ca chng bng nhau.

1 21 1 1 2 2 2 1 2

1 2

, ;x x

z x iy z x iy z zy y== + = + = =

(1.1)

Tp hp tt c cc s phc k hiu .

1.1.2. Cc php ton

Cho hai s phc 1 1 1z x iy= + v 2 2 2z x iy= + , ta nh ngha: a) Php cng: S phc ( ) ( )1 2 1 2z x x i y y= + + + c gi l tng ca hai s phc 1z v

2z , k hiu 1 2z z z= + . b) Php tr: Ta gi s phc z x iy = l s phc i ca z x iy= + . S phc ( ) ( )1 2 1 2 1 2( )z z z x x i y y= + = + c gi l hiu ca hai s phc 1z v 2z ,

k hiu 1 2z z z= . c) Php nhn: Tch ca hai s phc 1z v 2z l s phc c k hiu v nh ngha bi

biu thc:

( )( ) ( ) ( )1 2 1 1 2 2 1 2 1 2 1 2 1 2z z z x iy x iy x x y y i x y y x= = + + = + + . (1.2) d) Php chia: Nghch o ca s phc 0z x iy= + l s phc k hiu 1

z hay 1z , tha

mn iu kin 1 1zz = . Vy nu 1 ' 'z x iy = + th

2 2 2 2' ' 1

' , '' ' 0

xx yy x yx yyx xy x y x y

= = = + = + + . (1.3)

S phc 1 1 2 1 2 1 2 1 21 2 2 2 2 22 2 2 2

x x y y y x x yz z z ix y x y

+ = = ++ + c gi l thng ca hai s phc 1z v

2z , k hiu 12

zzz

= ( 2 0z ).

V d 1.1: Cho z x iy= + , tnh 2 ,z zz . Gii: ( ) ( ) ( )22 2 2 2z x iy x y i xy= + = + , 2 2zz x y= + .

V d 1.2: Tm cc s thc ,x y l nghim ca phng trnh

( )( ) ( )( )5 1 2 3 3 11x y i x i i i+ + + + = . Gii: Khai trin v ng nht phn thc, phn o hai v ta c

2 5 2 3 73,4 5 6 11 5

x yx y

x y+ + = = = + = .


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V d 1.3: Gii h phng trnh 1

2 1z iw

z w i+ = + = + .

Gii: Nhn i vo phng trnh th nht v cng vo phng trnh th hai ta c

( ) ( )( )1 2 21 2 4 32 1 22 5 5

i ii ii z i zi

+ + ++ = + = = =+ ,

( ) 1 3 315 5

i iw i z i + + = = = .

V d 1.4: Gii phng trnh 2 2 5 0z z+ + = . Gii: ( ) ( ) ( ) ( )( )2 2 22 2 5 1 4 1 2 1 2 1 2z z z z i z i z i+ + = + + = + = + + + . Vy phng trnh c hai nghim 1 21 2 , 1 2z i z i= + = .

1.1.3. Biu din hnh hc ca s phc, mt phng phc

Xt mt phng vi h ta trc chun Oxy , c vc t n v trn hai trc tng ng l

iJG

v jJG

. Mi im M trong mt phng ny hon ton c xc nh bi ta ( ; )x y ca n tha

mn OM x i y j= +JJJJG JG JG . S phc z x iy= + cng hon ton c

xc nh bi phn thc x v phn o y ca n. V vy ngi ta ng nht mi im c ta ( ; )x y vi s phc z x iy= + , lc mt phng ny c gi l mt phng phc.

1.1.4. Dng lng gic ca s phc

Trong mt phng vi h ta trc chun

Oxy , nu ta chn OxJJG

lm trc cc th im

( ; )M x y c ta cc ( );r xc nh bi ( ), ,r OM Ox OM= = JJG JJJJG

tha mn cossin

x ry r

= =

Ta k hiu v gi

2 2z r OM x y= = = + (1.4) Argz 2 ,k k= + (1.5)

l m un v argument ca s phc z x iy= + .

xx

My

y

O iJJG

jJJG

r x x

M y

y

O iJJG

jJJG


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Gc ca s phc 0z x iy= + c xc nh theo cng thc sau

+==

22cos

tg

yxx/

y/x (1.6)

Gi tr ca Argz nm gia v c gi l argument chnh, k hiu arg z . Vy arg z < .

T cng thc (1.4) ta c

( )cos sinz x iy r i = + = + (1.7) gi l dng lng gic ca s phc.

S dng khai trin Maclaurin c th chng minh c cng thc Euler

cos sinie i = + (1.8)

Do cos , sin2 2

i i i ie e e ei

+ = = . (1.9)

T (1.7)-(1.8) ta c th vit s phc di dng m

iz z e = (1.10) Cc tnh cht ca s phc

1 11 2 1 2 1 2 1 22 2

; ; z zz z z z z z z zz z

+ = + = = . (1.11)

Re ; Im2 2

z z z zz zi

+ = = . z z z = . (1.12)

1 2 1 21 21 2 1 2arg arg Arg Arg 2

z z z zz z

z z z z k = = = = = +

(1.13)

2zz z= , 21

z

zzzz

z== , 1 1 22

2 2

z z zz z

= . (1.14)

111 2 1 2 1 2 1 22 2

, ,zzz z z z z z z z

z z= = + + . (1.15)

( ) 11 2 1 2 1 22

Arg Arg Arg , Arg Arg Argzz z z z z zz

= + = (1.16)

iyxz +=

zy

zx v yxz + (1.17)


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V d 1.5: a) Tp cc s phc z tha mn 2 3z = tng ng vi tp cc im c khong cch n (2;0)I bng 3, tp hp ny l ng trn tm I bn knh 3.

b) Tp cc s phc z tha mn 2 4z z = + tng ng vi tp cc im cch u (2;0)A v ( 4;0)B l ng trung trc ca on AB c phng trnh 1x = .

1.1.5. Php nng ly tha, cng thc Moivre

Ly tha bc n ca s phc z l s phc n

nz zz z="ln

T cng thc (1.15)-(1.16) ta c cng thc Moivre:

( )cos sin , Arg 2nnz z n i n z k = + = + . (1.18) c bit, khi 1z = ta c

( ) ( )cos sin cos sinni n i n + = + (1.18)' V d 1.6: Tnh ( )101 3i + .

Gii: ( ) 1010 102 2 20 201 3 2 cos sin 2 cos sin3 3 3 3i i i + = + = + 10 10 9 9

2 2 1 32 cos sin 2 2 323 3 2 2

i i i = + = + = + .

1.1.6. Php khai cn

S phc c gi l cn bc n ca z , k hiu n z= , nu zn = . Nu vit di dng lng gic: )sin(cos,)sin(cos +=+= iirz th

+==

+=

==nk

r

kknrz

nn

n2,2

. (1.19)

V Argument ca mt s phc xc nh sai khc mt bi s nguyn ca 2 nn vi mi s phc 0z c ng n cn bc n . Cc cn bc n ny c cng m un l n r , Argument nhn cc gi tr

nk

n+= 2 ng vi 1,...,1,0 = nk , v vy nm trn nh ca n-gic u ni tip

trong ng trn tm O bn knh n r .

V d 1.7: Gii phng trnh 014 =+z Gii: Nghim ca phng trnh l cn bc 4

ca += sincos1 i tng ng l: x

y

0z 1z

2z 3z

O 1

i

4


10

21

4sin

4cos0

iiz +=+= ,

21

01iizz +== ,

2

102

izz == ,

21

03iizz == .

1.1.7. Cc khi nim c bn ca gii tch phc

1.1.7.1. Mt cu phc

Trong 1.1.3 ta c mt biu din hnh hc ca tp cc s phc bng cch ng nht mi s phc iyxz += vi im M c ta );( yx trong mt phng vi h ta Oxy . Mt khc nu ta dng mt cu )(S c cc nam tip xc vi mt phng Oxy ti O, khi mi im z thuc mt phng Oxy s tng ng duy nht vi im l giao im ca tia Pz v mt cu

)(S , P l im cc bc ca )(S .

Vy mi im trn mt phng Oxy c xc nh bi mt im trn mt cu )(S ngoi tr im cc bc P.

Ta gn cho im cc bc ny s phc v cng . Tp hp s phc thm s phc v cng c gi l tp s phc m rng . Nh vy ton b mt cu )(S l mt biu din hnh hc ca tp s phc m rng.

Quy c: ==+== zzzzzz ,,)0(,)0(0

.

1.1.7.2. Ln cn, min

a. Ln cn

Khi nim ln cn ca 0z c nh ngha hon ton tng t vi ln cn trong 2 , l hnh trn c tm ti im ny v bn knh bng .

( ) { }= NzzBN (1.23)

b. im trong, tp m

Gi s E l mt tp cc im ca mt phng phc hoc mt cu phc. im 0z c gi l im trong ca E nu tn ti mt ln cn ca 0z nm hon ton trong E .

Tp ch gm cc im trong c gi l tp m.

z x O y

P )(S


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c. im bin

im 1z , c th thuc hoc khng thuc E , c gi l im bin ca E nu mi ln cn ca 1z u c cha cc im thuc E v cc im khng thuc E .

Tp hp cc im bin ca E c gi l bin E , k hiu E . Hnh trn m { }rzzz 0 l cc

tp m c bin ln lt l { }rzzz = 0 v { } { }= rzzz 0 . Hnh trn ng { }rzzz 0 khng phi l tp m v cc im bin rzz = 0

khng phi l im trong.

d. Tp lin thng, min

Tp con D ca mt phng phc hay mt cu phc c gi l tp lin thng nu vi bt k 2 im no ca D cng c th ni chng bng mt ng cong lin tc nm hon ton trong D .

Mt tp m v lin thng c gi l min.

Min D cng bin D ca n c gi l min ng, k hiu DDD = . Min ch c mt bin c gi l min n lin, trng hp ngc li gi l min a lin.

Ta qui c hng dng trn bin ca min l hng m khi ta i trn bin theo hng th min D bn tay tri.

Min D c gi l b chn nu tn ti 0>R sao cho DzRz , . 1.2. HM BIN PHC

1.2.1. nh ngha hm bin phc

nh ngha 1.1: Mt hm bin phc xc nh trn tp con D ca hoc l mt quy lut cho tng ng mi s phc Dz vi mt hoc nhiu s phc w , k hiu ( ) Dzzfw = , .

Nu vi mi z ch cho tng ng duy nht mt gi tr w th ( )zf c gi l hm n tr. Trng hp ngc li f c gi l hm a tr.

Hm s ( ) 32 +== zzfw l mt hm n tr, cn hm s ( ) zzfw == l mt hm a tr.

Tp D trong nh ngha trn c gi l tp xc nh. Ta ch xt tp xc nh D l mt min, v vy D c gi l min xc nh.

Thng thng ngi ta cho hm phc bng cng thc xc nh nh ( )zf , khi min xc nh D l tp cc s phc z m ( )zf c ngha.

Hm s ( )12 +== z

zzfw c min xc nh l { }D z z i= . Ta c th biu din mt hm phc bi hai hm thc ca hai bin ),( yx nh sau:


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iyxz += v ( ) ivuzfw +== th ( )( )

==

yxvvyxuu

,,

(1.24)

Gi ( )yxu , l phn thc, ( )yxv , l phn o ca hm )(zf . Hm s xyiyxiyxzw 2)3(3)(3 2222 ++=++=+= c

=+=

xyvyxu

2322 .

Trng hp min xc nh D th ta c hm phc bin s thc, ta k hiu ( )tfw = c bin s l t thay cho z .

Trng hp min xc nh D l tp s t nhin th ta c dy s phc ( ) = nnfzn , , ta thng k hiu dy s l ( ) nnz hay ( )=1nnz . 1.2.2. Gii hn

nh ngha 1.2: Dy s ( )=1nnz hi t v 000 yxz += , k hiu 0lim zznn

=

, nu

> 0:0,0 zzNnN n (1.25) Dy s ( )=1nnz c gii hn l , k hiu = nn zlim , nu

>>> nzNnN :0,0 (1.26) T (1.17) suy ra rng

==

+==

0

0

000 lim

limlim

yy

xxiyxzz

nn

nn

nn

(1.27)

nh ngha 1.3: Ta ni hm phc ( )zfw = xc nh trong mt ln cn ca 0z c gii hn l L khi z tin n 0z , k hiu ( ) Lzf

zz=

0lim , nu vi mi ln cn ( )LB tn ti ln cn

( )0zB sao cho vi mi ( ) 00 , zzzBz th ( ) ( )LBzf . Trng hp Lz ,0 nh ngha trn c vit di dng c th sau:

( ) ( )


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1.2.3. Lin tc

nh ngha 1.4: Hm phc ( )zfw = xc nh trong min cha im 0z c gi l lin tc ti 0z nu ( ) ( )0

0

lim zfzfzz

=

. Hm phc ( )zfw = lin tc ti mi im ca min D c gi l lin tc trong D .

T (1.29) suy ra rng mt hm phc lin tc khi v ch khi hai hm thc hai bin (phn thc, phn o) xc nh bi (1.24) l lin tc. Do ta c th p dng cc tnh cht lin tc ca hm thc hai bin cho hm phc.

1.2.4. Hm kh vi, iu kin Cauchy-Riemann

nh ngha 1.5: Gi s iyxz += l mt im thuc min xc nh D ca hm phc n tr ( )zfw = . Nu tn ti gii hn

( ) ( )

zzfzzf

z +

0lim (1.33)

th ta ni hm ( )zfw = kh vi (hay c o hm) ti z , cn gii hn c gi l o hm ti z , k hiu ( )zf ' hoc ( )zw' . V d 1.8: Cho 2zw = , tnh ( )zw' .

Gii: ( ) zzzwzzzzzzw +=

+=+= 22 222 ,

Do ( ) ( ) zzzzwzw

zz22limlim'

00=+=

= .

nh l 1.1: Nu hm phc ( ) ( ) ( )yxivyxuzfw ,, +== kh vi ti iyxz += th phn thc ( )yxu , v phn o ( )yxv , c cc o hm ring ti ),( yx v tha mn iu kin Cauchy-

Riemann

( ) ( )( ) ( )

=

=

yxxvyx

yu

yxyvyx

xu

,,

,, (1.34)

Ngc li, nu phn thc ( )yxu , , phn o ( )yxv , kh vi ti ),( yx v tha mn iu kin Cauchy-Riemann th ( )zfw = kh vi ti iyxz += v

( ) ( ) ( ) ( ) ( )yxyuiyx

yvyx

xviyx

xuzf ,,,,'

=

+= . (1.35)

V d 1.8: Hm xyiyxzw 2222 +== V d 1.7 c

==

==

xvy

yu

yvx

xu

2

2 , do hm kh vi

ti mi im v ( ) zyixzw 222' =+= .


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V d 1.9: Hm iyxzw == c 1,1 ==

yv

xu khng tha mn iu kin Cauchy-Riemann,

do hm khng kh vi ti bt k im no.

1.2.5. Hm gii tch

nh ngha 1.6: Hm n tr ( )zfw = kh vi trong mt ln cn ca z c gi l gii tch ti z . Nu ( )zf kh vi ti mi im ca D th ta ni ( )zf gii tch trong D. ( )zf gii tch trong D nu n gii tch trong mt min cha D .

Khi nim kh vi v o hm ca hm phc c nh ngha tng t nh trng hp hm thc. V vy cc tnh cht v quy tc tnh o hm bit i vi hm thc vn cn ng i vi hm phc.

( )( ) ( ) ' '( ) '( )f z g z f z g z = . ( )( ) ( ) ' '( ) ( ) ( ) '( )f z g z f z g z f z g z= + . (1.38)

( )'

2( ) '( ) ( ) ( ) '( ) , ( ) 0( ) ( )

f z f z g z f z g z g zg z g z

= .

( )( ) )(').(')( ' zuufzuf = . 1.2.6. Cc hm phc s cp c bn

1.2.6.1. Hm ly tha nzw = , n nguyn dng 2. Hm s xc nh v gii tch vi mi z , o hm 1= nnzw . Nu ( )+= sincos irz th ( )+= ninrw n sincos . Vy nh ca ng trn Rz = l ng trn nRw = . nh ca tia += 2Arg kz l

tia += 2'Arg knw . nh ca hnh qut nz 2arg0


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1.2.6.2. Hm cn n zw = Hm cn bc n : n zw = l hm ngc ca hm ly tha bc n . Mi s phc khc 0 u c ng n cn bc n, v vy hm cn l mt hm a tr.

1.2.6.3. Hm m zew = M rng cng thc Euler (1.12) ta c nh ngha ca hm m

( )yiyeeew xiyxz sincos +=== + (1.39) +== 2Arg, kywew x .

Hm m gii tch ti mi im v ( )'z ze e= 2121 zzzz eee += , 21

2

1 zzz

ze

ee = , ( )nz nze e= , zikz ee =+ 2 . (1.40)

1,,1 20 ===

ii eiee .

Qua php bin hnh zew = , nh ca ng thng ax = l ng trn aew = , nh ca ng thng by = l tia += 2Arg kbw .

nh ca bng


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iu ny chng t hm lgarit phc l hm a tr. ng vi mi z c v s gi tr ca w , nhng gi tr ny c phn thc bng nhau cn phn o hn km nhau bi s nguyn ca 2 . Vi mi 0kk = c nh ta c mt nhnh n ta tr ca hm zw Ln= .

( )++= 2argln 0kzizw Nhnh n tr ng vi 0=k c gi l nhnh n tr chnh v c k hiu zln . zizz arglnln +=

trong ln v tri l hm bin phc, cn v phi l hm bin thc.

Mt s tnh cht ca hm lgarit.

( ) ( ) ( ) ( ) =+=++= iikki 1ln122)1arg(1ln1Ln ( ) ( ) ( ) ( ) ( ) znzzz

zzzzzz n LnLn,LnLnLn,LnLnLn 212

12121 ==

+= .

Cc nhnh n tr ca hm lgarit gii tch trn na mt phng phc Z b i na trc thc m )0( =>+=

ieenieeni

nnnn.


17

1.2.6.6. Cc hm lng gic hyperbolic phc

zzz

zzzeezeez

zzzz

shchcoth,

chshth,

2sh,

2ch ===+=

(1.43)

Cc hm lng gic hyperbolic phc gii tch trong min xc nh

( ) ( ) ( ) ( )z

zz

zzzzz 2'

2'''

sh1coth,

ch1th,shch,chsh ==== .

zizziizezzezz zz chcos,shsin,shch,shch ====+ . zzzzzzzz 2222 shch2ch,shch22sh,1shch +=== .

1.3. PHP BIN HNH BO GIC

Nhiu vn trong khoa hc v thc tin (v d bi ton n mn, bi ton thit k cnh my bay) a n bi ton: Tm php bin hnh bo gic bin min D thnh min no m ta bit hoc d dng kho st hn. Trong mc ny ta a ra vi nguyn l v phng php tm php bin hnh trong nhng trng hp n gin.

1.3.1. nh ngha php bin hnh bo gic

nh ngha 1.7: Php bin hnh ( )zfw = c gi l bo gic ti z nu tho mn hai iu kin sau:

i. Bo ton gc gia hai ng cong bt k qua im z ( k c ln v hng).

ii. C h s co dn khng i ti z , ngha l mi ng cong i qua im ny u c h s co dn nh nhau qua php bin hnh.

Php bin hnh ( )zfw = c gi l bo gic trong min D nu n bo gic ti mi im ca min ny.

nh l sau y cho iu kin ca php bin hnh bo gic.

nh l 1.2: Nu hm ( )zfw = kh vi ti z v ( ) 0' zf th php bin hnh thc hin bi hm ( )zfw = bo gic ti im z , ng thi ( )zf 'arg l gc quay v ( )zf ' l h s co gin ti im z ca php bin hnh .

T nh l ny ta suy ra rng nu ( )zfw = gii tch trong D v ( ) Dzzf ,0' th n l mt php bin hnh bo gic trong D.

1.3.2. Php bin hnh tuyn tnh 0, += abazw Php bin hnh ny bo gic trong ton min v ( ) zazw = ,0' . Nu = ieaa th bzeaw i += . iu ny chng t php bin hnh tuyn tnh l hp ca

ba php bin hnh sau:

Php v t tm O t s ak = , Php quay tm O, gc quay ,


18

Php tnh tin theo vc t b . Vy php bin hnh tuyn tnh l mt php bin hnh ng dng (hp ca mt php v t,

php quay, php tnh tin). N bin mt hnh bt k thnh mt hnh ng dng vi n. c bit bin mt ng trn thnh mt ng trn, bin mt ng thng thnh mt ng thng, mt a gic thnh mt a gic ng dng.

V d 1.10: Tm php bin hnh bo gic bin tam gic vung cn c cc nh ( )iA 27 + , ( )iB 23+ , ( )iC 45 + thnh tam gic vung cn c cc nh ( )iA 21 , ( )01B , ( )iC +11 .

Gii: Hai tam gic vung cn bt k u ng dng vi nhau nn tn ti mt php ng dng 0, += abazw bin ABC thnh 111 CBA . Php bin hnh ny bin A thnh 1A , bin B thnh 1B , do ba, tha mn h phng trnh

( )( ) iz

iwib

ia

biabiai

231

2231

2230272 =

=

=

++=++=

.

Thay 5 4z i= + ta c 3( 5 4 ) 1 12 2iw i i i= + = + .

1.3.3. Php nghch o z

w 1=

Php bin hnh z

w 1= c th m rng ln mt phng phc m rng bng cch cho nh ca 0=z l v nh ca =z l 0=w .

o hm ( ) = ,0,01' 2 zzzw nn php bin hnh bo gic ti mi im ,0z . Hai im A, B nm trn mt tia xut pht t tm I ca ng trn ( )C bn knh R c gi

l lin hp hay i xng qua ( )C nu 2RIA.IB= .

v

u

1C

1B

1A i2

i

1 x

y

A B

C

7 3

i2

i4

Z W


19

V zzz

ArgArg1Arg == nn z v z

w 1= cng nm trn mt tia xut pht t O.

Ngoi ra 11. =z

z , do z v z

w 1= i xng nhau qua ng trn n v.

Vy php bin hnh nghch o z

w 1= l hp ca php i xng qua ng trn n v v php i xng qua trc thc. Php bin hnh ny bin:

Mt ng trn i qua O thnh mt ng thng. Mt ng trn khng i qua O thnh mt ng trn. Mt ng thng i qua O thnh mt ng thng qua O. Mt ng thng khng i qua O thnh mt ng trn i qua O.

Nu ta xem ng thng l mt ng trn (c bn knh v hn) th php bin hnh z

w 1= bin mt ng trn thnh mt ng trn.

nh ca ng trn R=z l ng trn R1=w , nh ca hnh trn Rw . nh ca M trn tia OB l N trn tia OB', B' l i xng ca B qua trc

thc v 1OM.ON = .

3.4. Php bin hnh phn tuyn tnh 0,0; ++= bcadc

dczbazw

Ta c th m rng hm phn tuyn tnh dczbazw +

+= ln mt phng phc m rng bng

cch cho nh ca cdz = l v nh ca =z l

caw = .

B M

B' x

y

O u

v

O

W Z

N


20

o hm ( ) ( ) += ,,0' 2 c

dzdczbcadzw nn php bin hnh bo gic ti mi im

,cdz .

( )( )

( ) dczcadbc

ca

dczcadbcdcza

dczcbcacz

dczbazw +

+=+++=+

+=++= 1 .

Do php bin hnh phn tuyn tnh l hp ca 3 php bin hnh:

Php bin hnh tuyn tnh: dczz +6 ,

Php nghch o: dcz

dcz ++16 ,

Php bin hnh tuyn tnh: ca

dczcadbc

dcz++

+

11 6 .

V cc php bin hnh tuyn tnh v nghch o bin mt ng trn thnh mt ng trn v bo ton tnh i xng ca 2 im i xng qua ng trn, nn php bin hnh phn tuyn tnh cng c tnh cht .

Php bin hnh 0, ++= c

dczbazw c th vit li

1

11dzbza

cdz

cbz

ca

w ++=

++

= hoc 2

2dzbzkw +

+= (1.44)

v vy ch ph thuc 3 tham s. Do mt hm phn tuyn tnh hon ton c xc nh khi bit nh 321 ,, www ca 3 im khc nhau bt k 321 ,, zzz . xc nh 3 tham s

111 ,, dba ta gii h phng trnh sau y.

13

1313

12

1212

11

1111 ,, dz

bzawdzbzaw

dzbzaw +

+=++=+

+= (1.45)

Hoc hm phi tm c th xc nh bi phng trnh

32

12

3

1

32

12

3

1zzzz

zzzz

wwww

wwww

=

(1.46)

c bit nu ( ) 00 =zw v ( ) =1zw , theo (1.44) ta c

1

0zzzzkw

= (1.47)


21

1.3.5. Cc nguyn l tng qut ca php bin hnh bo gic

a. S tn ti ca php bin hnh

nh l 1.3 (nh l Riemann): Nu D v l hai min n lin (khng phi l mt phng phc m rng hay mt phng phc m rng b i mt im) th tn ti php bin hnh ( )zfw = gii tch, bo gic n tr hai chiu bin D thnh .

Hn na nu cho trc 00 D, wz v 0 th ch c duy nht ( )zfw = tho mn ( )00 zfw = , ( ) 00'Arg =zf .

nh l Riemann ch cho ta bit s tn ti ca php bin hnh ch khng cho ta cch tm c th php bin hnh ny. Trong thc hnh, tm php bin hnh bin min D thnh min ngi ta tm php bin hnh bin D, v hnh trn n v 1


22

Gii: V 0z i xng vi 0z qua Ox , i xng vi 0 qua 1=w , do theo nguyn l tng ng bin ta ch cn tm hm phn tuyn tnh bin trc thc 0Im =z ln 1=w v bo ton chiu.

Hai min cho khng ng dng nn 0c . Mt khc ( ) 00 =zw v tnh cht bo ton tnh i xng nn ( ) =0zw , do theo (1.47) ta c th xt hm phn tuyn tnh dng

0

0zzzzkw

= . Khi = xz th ( ) 1=xw 110

0

0

0 ===

kzxzxk

zxzxk .

= iek . Vy 0

0zzzzew i

= .

V d 1.12: Tm php bin hnh bo gic ( )zfw = bin hnh trn 1


23

Nu ta thm iu kin ( ) iw =0 th ieiiei ii =+

= 00 .

Vy php bin hnh cn tm l iziziw +

= 33

.

V d 1.14: Tm php bin hnh bo gic ( )zfw = bin min

> chn r b ng trn tm a bn knh r : DrC v ( ) ( )


30

Do ( ) 2 2cos cos cos cos

1 1C C C C

z z z zI dz dz dz dzz z z z z

= = + ++ + v v v v . Cc im 0=z v 1=z u nm trong hnh trn gii hn bi C. p dng cng thc

(1.56)' v (1.57)' ta c:

( ) ( )0 0 12 cos 2 cos ' 2 cos 2 1 cos1z z zI i z i z i z i = = == + + = + . 1.4.6. Bt ng thc Cauchy v nh l Louville

T cng thc (1.58) suy ra rng, nu ng trn RazCR =: nm trong D v ( ) Mzf vi mi RCz th

( ) ( )( )( )

1 1! ! 2

2 2R

nn n

C

f zn n M Rf a dzRz a

+ += v

hay

( ) ...,1,0;!)( = nR

Mnaf nn (1.59)

Bt ng thc (1.58) c gi l bt ng thc Cauchy.

nh l 1.12 (nh l Louville): Nu ( )f z gii tch trong ton mt phng v b chn th n l mt hm hng.

Chng minh: Theo gi thit, tn ti 0>M sao cho ( ) Mzf vi mi z . p dng bt ng thc Cauchy (1.58) vi 1=n , ta c ( )' Mf a

R vi mi 0>R suy ra ( )' 0f a =

vi mi a . p dng cng thc Newton - Lepnit, ta c

( ) ( ) ( ) ( ) ( ) === zzfzfdzzfzfzfz

z,0 0

'0

0

.

1.5. L THUYT CHUI PHC

1.5.1. Chui s phc

Cho dy s phc { }= 0nnu , ta nh ngha mt cch hnh thc =0n

nu l mt chui cc s

phc m s hng th n l nu .

Tng nn uuuS +++= "10 c gi l tng ring th n ca chui trn.


31

Nu dy cc tng ring { }=0nnS c gii hn hu hn l S th ta ni chui =0n

nu hi

t v S c gi l tng ca chui, k hiu =

=0n

nuS .

Trong trng hp ngc li, dy { }=0nnS khng c gii hn hoc c gii hn bng th ta ni chui phn k.

Tng t (1.27), mi chui phc =0n

nu hi t khi v ch khi hai chui s thc tng ng

=

= 00,

nn

nn ba hi tu v

=

=

=+=

000 nn

nn

nn biau ; trong n n nu a ib= + .

Vi nhn xt ny, ta c th p dng cc kt qu bit i vi chui s thc cho cc chui s phc. Chng hn:

iu kin cn chui =0n

nu hi t l 0lim = nn u .

Nu chui cc mun 0

nn

u

= hi t th chui

=0nnu cng hi t. Khi ta ni chui

=0n

nu hi t tuyt i. Nu chui =0n

nu hi t nhng chui cc mun 0

nn

u

= khng hi t th

ta ni chui bn hi t.

1.5.2. Chui lu tha

Chui c dng

( )=

0n

nn azc , vi azcn ,, (1.60)

c gi l chui lu tha tm a . Khi cho z mt gi tr c th ta c mt chui s phc, chui s phc ny hi t hoc phn k. Min hi t ca chui (1.60) l tp hp cc gi tr z m chui ny hi t.

R rng rng mi chui lu tha tm a bt k c th a v chui lu tha tm 0 bng cch t az = :

0

nn

nc

= , vi ,nc . (1.61)

V vy n gin, trong cc trng hp sau ta ch xt s hi t ca chui ly tha tm 0.


32

Mt v d c bit ca chui lu tha l chui cp s nhn =0n

nz , c tng ring l tng ca

cc s hng ca cp s nhn z

zzzzSn

nn

=++++=+

111

12 " vi 1z , do

0

1 khi 11

phn k khi 1

n

n

zzz

z

=

.

Chng minh: Chui =0

0n

nn zc hi t suy ra 0lim 0 =

nn

nzc , v vy tn ti 0>M sao cho

,0 Mzcn

n ...,2,1,0= n Do nn

n

nn

nn

nz

zM

zzzczc

000 = .

Chui =

0 0n

n

n

z

zM hi t khi 0zz < . Suy ra chui

=0nn

n zc hi t tuyt i khi

0zz < . Phn 2. ca nh l l h qu ca phn 1. nh ngha 1.8: SR ( R0 ) tha mn mt trong nhng iu kin sau c gi l

bn knh hi t ca chui (1.61):

Nu chui (1.61) hi t ti mi z th ta t =R . Nu chui (1.61) ch hi t ti 0=z th ta t 0=R . Chui (1.61) hi t khi Rz < , phn k khi Rz > .

nh l 1.14: Nu n

nn c

c 1lim += (tiu chun D'Alembert)

hoc n nn

c= lim (tiu chun Cauchy) th

n u

n u 0

n u 0

01R

= < < =

=

(1.62)

l bn knh hi t ca chui (1.61).


33

Nhn xt: nh l trn cho ta cch xc nh bn knh hi t ca chui (1.61). tm min hi t ca chui ny ta ch cn xt thm s hi t ca chui trn ng trn Rz = .

nh l 1.15: a) Nu chui (1.61) c bn knh hi t R th tng ca chui ( ) =

=0n

nn zczf

l mt hm gii tch trong hnh trn hi t Rz < , o hm ( ) =

=1

1'

n

nn znczf .

b) ( ) =

++= 0

11n

nn znc

zF l mt nguyn hm ca )(xf .

c) =

1

1

n

nn znc ,

=+

+01

1nnn z

nc

cng c bn knh hi t l R.

1.5.3. Chui Taylor

nh ngha 1.9: Chui ly tha c dng

( )

=

0

)()(

!nn

naz

naf

(1.63)

c gi l chui Taylor ca hm ( )zf ti a . nh l 1.16: 1) Chui lu tha bt k l chui Taylor ca hm tng ca n trong hnh trn

hi t.

2) Ngc li, mi hm ( )zf gii tch ti a th c th c khai trin thnh chui Taylor trong ln cn Raz


34

=

=+++++=0

2

!!!2!11

n

nnz

nz

nzzze ""

Hm gii tch ti mi im nn bn knh hi t ca chui l =R . b. Hm ( ) zzf sin=

( ) ( )0 0

1sin2 2 ! !

n niz iz

n n

iz ize ezi i n n

= =

= =

( ) ( )

=

+

= +==

0

12

0 )!12()1()1(1

!21

n

nnn

n

n

nz

niz

i.

Hm gii tch ti mi im nn bn knh hi t ca chui l =R . c. Hm ( ) zzf cos= ( )

=

=

+=+

+==0

2

0

12'

)!2()1(

)!12()12()1(sincos

n

nn

n

nn

nz

nznzz .

Hm gii tch ti mi im nn bn knh hi t ca chui l =R . d. Hm ( )

11+= zzf

=

==+ 0)1(

)(11

11

n

nn zzz

.

Bn knh hi t ca chui l 1=R v hm s khng gii tch ti 1 . e. Nhnh chnh ca hm lgarit v hm ly tha

V hm )1ln( z+ l mt nguyn hm ca 1

1+z nn

=

++=+ 0

1

1)1()1ln(

n

nn

nzz .

Bn knh hi t ca chui l 1=R . Hm ly tha m : ( ) "" ++++++=+ nm z

nnmmmzmmmzz

!)1)...(1(

!2)1(11 2

Bn knh hi t ca chui l 1=R .

c bit: ( ) =

=+

+=+=+ 0 222

21

)!(2)!2()1(

!223

21

2111

11

n

nn

nz

nnzzz

z" .

1.5.5. Khng im ca mt hm gii tch, nh l v tnh duy nht

nh ngha 1.10: im a c gi l khng im ca hm gii tch ( )zf nu ( ) 0=af .


35

Khai trin Taylor ca ( )zf ti khng im a c dng ( ) ( ) ( ) ( ) ( ) ( )

=

=++ ==++=

nk

kk

nk

kk

nn

nn azk

afazcazcazczf!

)(1

1 " .

S t nhin n b nht sao cho ( ) 0!

)(=

nafc

n

n th c gi l cp ca khng im a .

Nu n l cp ca khng im a th

( ) ( ) ( )zazzf n= , vi ( ) 0= nca . (1.65) ( )z l tng ca mt chui lu tha c cng bn knh hi t vi chui Taylor ca ( )zf ti a nn

gii tch trong ln cn ca a .

nh l 1.17: Gi s ( )zf gii tch ti a v khng ng nht bng 0 trong bt k ln cn no ca a . Khi , nu a l khng im ca ( )zf th tn ti mt ln cn ca a sao cho trong ln cn ny khng c mt khng im no khc.

Chng minh: V a l khng im ca ( )zf nn c th biu din di dng (1.65) trong hm gii tch ( )z tha mn ( ) 0 a . V vy tn ti mt ln cn ca a trong ln cn ny ( ) 0 z , do ( )zf cng khc 0.

H qu: Nu ( )zf gii tch ti a v tn ti dy khng im { }=0nna c gii hn l a khi n , th ( )zf ng nht bng 0 trong mt ln cn ca a . nh l 1.18 (nh l v tnh duy nht): Nu ( ) ( )zgzf , l hai hm gii tch trong min D

v trng nhau trn mt dy hi t v a trong D th ( ) ( ) Dzzgzf = , . 1.5.6. Chui Laurent v im bt thng

C th xy ra trng hp hm ( )zf khng gii tch ti a nhng gii tch trong mt ln cn ca a b i im a : Raz


36

Tng ( ) ( )=

=0

1n

nn azczf c gi l phn u v ( ) ( )

== 12 n n

n

azczf c gi l

phn chnh ca chui Laurent (1.66).

nh l 1.19 (nh l tn ti v duy nht ca chui Laurent):

1. Mi hm ( )zf gii tch trong hnh vnh khn K: Razr


37

( 1) 1

21

1 1 1 ( 1) ( 1)!0 1( 1)! 2 ( 1)! ( 1)

n n

n nz

nn cn z n

+ ++=

+ = = = + + (theo cng thc

(1.57) nh l 1.11).

Vy ( ) ( ) ( ) =

===

111

n

n

n

nn zzczf .

b. Khai trin Laurent trong min 1 1z > : Chn ng cong kn 2L bao quanh 1 nm trong min ny.

( )( )2 21 1

2 2 1n n

L

c dzi z z += v .

Chn 21, ln lt l 2 ng cong kn nm trong 2L bao quanh 1, 2. p dng cng thc (1.53) h qu 2 ca nh l 1.7 ta c:

( )( )( )( )

( )( )

2 1 2

2

2 2

112 11 1 1 1

2 2 2 22 1 1

n

n n nL

z zc dz dz dz

i i i zz z z +

+ +

= = + v v v

Tng t trn ta c ( )( )1 2

11 2210 12 1 n

nzdz

ni z + = v

nu

nu

( )( ) ( )2

2

2

2

111 1 1

2 2 1

n

n

z

zdz

i z z+

+ =

= = v vi mi n .

Vy

=1120

nn

cn nunu

( ) ( ) ( )

=

= ==

2 111

nn

n

nn

zzczf .

Ta cng c th khai trin Laurent ca hm ( )zf cch phn tch thnh tng ca cc phn thc hu t ti gin

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

f zz z z z

= = .

Trong min 0 1 1z< < th ( ) ( ) ( ) ( )

=

===

= 1011

111

21

n

n

n

n zzfzzz

.

Trong min 1 1z > th ( ) ( ) ( )10 1

1 1 1 112 1 11 1

1

n nn nz z zz

z

+

= == = =


38

( ) ( ) ( )

=

= == 21 1

11

11

1

nn

nn zzz

zf .

1.5.6.2. im bt thng c lp

nh ngha 1.12: Nu hm ( )zf gii tch trong hnh vnh khn Raz


39

1.6.2. Cch tnh thng d

a. T cng thc khai trin Laurent ca hm trong hnh vnh khn RazK


40

a. C l ng trn: 23=z .

b. C l ng trn: 10=z .

Gii: Hm 2)3)(1( + zzez c 1=z l cc im n v 3=z cc im kp.

2 21Res ;1 lim( 1)( 3) ( 3) 16

z z

z

e e ez z z

= = + + ,

3

2 21 1

1 1 1 5Res ; 3 lim lim( 1)( 3) 1! 1 1 ( 1) 16

z zz

z z

e d e eez z dz z z z

= = = +

a. Khi C l ng trn 23=z th trong C hm cho ch c mt cc im 1=z .

Vy 816

2 ieeiI == .

b. Khi C l ng trn. 10=z th trong C hm cho c hai cc im 1=z v 3=z .

Do ( )

3

43

85

165

162

eeieeiI =

=

.

1.6.4. p dng l thuyt thng d tnh cc tch phn thc

1.6.4.1. Tnh tch phn ( )( )

= dx

xQxPI , trong ( ) ( )xQxP , l hai a thc thc.

B : Gi s hm ( )zf gii tch trong na mt phng 0Im z , tr ra ti mt s hu hn cc im bt thng c lp v tho mn:

( ) 0lim;0Im

= zzfzz (1.73)

Khi ( ) 0lim =RC

Rdzzf , trong { }0Im, == zRzzCR .

nh l 1.22: Gi s ( ) ( )zQzP , l hai a thc h s thc bin phc, bc ca ( )zP ln hn bc ca ( )zQ t nht l hai. Nu ( ) xxQ ,0 v naa ,...,1 l cc cc im nm trong na mt phng 0Im >z ca phn thc ( ) ( )( )zQ

zPzR = . Khi

( ) ( )1

2 Res ;n

kk

R x dx i R z a

== (1.74)


41

V d 1.24: Tnh tch phn ( )220 1dxI

x

=

+ .

Gii: Hm ( ) ( ) ( ) ( )2 2 221 1

1R z

z i z iz= = ++

c cc im kp iz = nm trong na mt

phng 0Im >z . Vy

( ) ( )2 2 2 32 21 1 1 1 22 Res ; lim2 2 ( ) (2 ) 41 1 z i

dx dI i i i idz z i ix z

= = = = = + + + .

1.6.4.2. Tch phn dng ( )cosR x xdx , ( )sinR x xdx

Hai tch phn trn l phn thc v phn o ca tch phn ( ) i xR x e dx .

B : Gi s hm ( )zf gii tch trong na mt phng 0Im z , tr ti mt s hu hn cc im bt thng c lp v tho mn:

( ) Rk CzRMzf , ; Mk ,0> l hng s (1.75)

th ( ) 0lim = RC

ziR

dzzfe , vi mi 0> . Trong { }0Im, == zRzzCR . nh l 1.23: Gii s ( )

)()(

zQzPzR = l mt phn thc hu t tho mn cc iu kin sau:

i. )(zR gii tch trong na mt phng 0Im >z ngoi tr ti mt s hu hn cc cc im naa ,...,1 .

ii. )(zR c th c m cc im mbb ,...,1 trn trc thc v ( )i xR x e kh tch ti nhng

im ny.

iii. Bc ca )(zQ ln hn bc ca )(zP t nht l 1.

Khi

( ) ( ) ( )1 1

2 Res ; Res ;n m

i x i z i zk k

k kR x e dx i R z e a i R z e b

= = = + (1.76)

V d 1.25: Tnh tch phn 2 20

cos , ( , 0)xI dx ax a

= >+ . Gii: V hm di du tch phn l hm chn nn


42

2 2 2 2 2 21 cos 1 1Re Re 2 Res ;2 2 2 2

i x i x ax e e eI dx dx i aix a x a x a a

= = = = + + + .

V d 1.26: Tnh tch phn

=0

sin dxx

xI .

Gii: V hm di du tch phn l hm chn nn

==

dx

xedx

xxI

ixIm

21sin

21 .

Hm z

zR 1)( = tho mn cc iu kin ca nh l 1.23, c cc im n duy nht 0=z

trn trc thc. Do ( )1 1Im Res ;0 Im2 2 2

izeI i iz

= = = .

1.6.4.3. Tch phn dng ( )2

0sin,cos dxnxnxR .

t ixez = th izdzdx

izznxzznx

nnnn==+=

,

2sin,

2cos

Khi x bin thin t 20 th ixez = vch ln ng trn n v C theo chiu dng. V vy

( )20

cos ,sin ,2 2

n n n n

C

z z z z dzR nx nx dx Ri iz

+ = v (1.77)

V d 1.27: Tnh tch phn

+=2

0 sin35 xdxI

Gii: V hm s ( )izizziz 3

33

2

13

103

22 +

+

=

+ ch c mt cc im n

3iz = nm

trong ng trn n v C, do

2 2

1 2 22 Res ;3 1 10 10 3 25 3 1 3 12 3 3

C C

dz dz iI ii iizz z z z z

i z

= = = = + + +

v v .

1.7. PHP BIN I Z

Da vo tnh cht xc nh duy nht ca hm s gii tch trong hnh vnh khn Rzr


43

Php bin i Z c rt nhiu ng dng trong l thuyt x l tn hiu v lc s, v ni chung vic kho st cc hm gii tch s thun li v d dng hn so vi kho st cc dy ri rc.

1.7.1. nh ngha php bin i Z

nh ngha 1.13: Bin i Z ca dy tn hiu { } =nnx )( l hm phc ( )

=

= ==

n

n

n

n znxznxzX 1)()()( (1.78)

Min hi t ca chui (1.78) l min xc nh ca bin i Z.

Trng hp dy tn hiu { } =nnx )( ch xc nh vi 0n , ngha l ( ) 0,x n = 0n < , khi bin i Z ca tn hiu ny c gi l bin i mt pha.

V d 1.28: Tm bin i Z ca tn hiu >


44

th )(zX xc nh khi Rzr


45

=

=n

nn zczX )( vi 1

1 ( )2n nC

X zc dzi z += v ,

C l ng cong kn bao quanh gc O v nm trong hnh vnh khn Rzr

nu

nu .


46

c. Min 3 z< : 1

0 0 1 1

1 1 1 1( ) 2 3 2 31 3 22 1 12

n n n n n n n n

n n n nX z z z z z

z zz zz z

= = = =

= + = + = + Vy

1

0 0( )

3 2 1n nn

x nn < =

nu

nu .

TM TT

Dng tng qut ca s phc

z x iy= + , trong ,x y l cc s thc; 12 =i . Dng lng gic, dng m ca s phc

( )cos sinz x iy r i = + = + , iz z e = . Trong 2 2z r OM x y= = = + , Argz 2 ,k k= + . ln cn ca 0z : ( ) { }


47

Nu hm phc ( ) ( ) ( )yxivyxuzfw ,, +== kh vi ti iyxz += th phn thc ( )yxu , v phn o ( )yxv , c cc o hm ring ti ),( yx v tha mn iu kin Cauchy-Riemann

( ) ( )( ) ( )

=

=

yxxvyx

yu

yxyvyx

xu

,,

,,

Ngc li, nu phn thc ( )yxu , , phn o ( )yxv , kh vi ti ),( yx v tha mn iu kin Cauchy-Riemann th ( )zfw = kh vi ti iyxz += v

( ) ( ) ( ) ( ) ( )yxyuiyx

yvyx

xviyx

xuzf ,,,,'

=

+= .

Hm n tr ( )zfw = kh vi trong mt ln cn ca z c gi l gii tch ti z . Nu ( )zf kh vi ti mi im ca D th ta ni ( )zf gii tch trong D. ( )zf gii tch trong D nu n

gii tch trong mt min cha D .

Php bin hnh bo gic

Php bin hnh ( )zfw = c gi l bo gic ti z nu tho mn hai iu kin sau: i. Bo ton gc gia hai ng cong bt k qua im z ( k c ln v hng).

ii. C h s co dn khng i ti z , ngha l mi ng cong i qua im ny u c h s co dn nh nhau qua php bin hnh.

Php bin hnh ( )zfw = c gi l bo gic trong min D nu n bo gic ti mi im ca min ny.

Nu hm ( )zfw = kh vi ti z v ( ) 0' zf th php bin hnh thc hin bi hm ( )zfw = bo gic ti im z , ng thi ( )zf 'arg l gc quay v ( )zf ' l h s co gin ti

im z ca php bin hnh . Nu ( )zfw = gii tch trong D v ( ) Dzzf ,0' th n l mt php bin hnh bo gic trong D.

Tch phn phc

Gi s ( ) ( ) ( )yxivyxuzfw ,, +== xc nh n tr trong min D. L l ng cong (c th ng kn) nm trong D c im mt u l A mt cui l B.

Chia L thnh n on bi cc im BzzzzA n ,...,,, 210 nm trn L theo th t tng dn ca cc ch s. Chn trn mi cung con kk zz ,1 ca ng cong L mt im bt k

kkk i+= . t ,k k kz x iy= + nkzzz kkk ,1;1 == .

( ) ( )=

==

n

kkk

zABzfdzzfI

knk 1

0max1

lim .

( ) ++=ABABAB

udyvdxivdyudxdzzf .


48

Cng thc tch phn Cauchy

Gi s ( )zf gii tch trong min D (c th a lin) c bin l D . Khi , vi mi Da ta c:

( ) ( )12 D

f zf a dz

i z a = v ; ( ) ( )( )( ) 1

!2

nn

C

f znf a dzi z a += v

tch phn c ly theo chiu dng ca D . Chui Taylor

Chui ly tha c dng ( )

=

0

)()(

!nn

naz

naf

c gi l chui Taylor ca hm ( )zf ti a . 1) Chui lu tha bt k l chui Taylor ca hm tng ca n trong hnh trn hi t.

2) Ngc li, mi hm ( )zf gii tch ti a th c th c khai trin thnh chui Taylor trong ln cn Raz


49

Ngc li dy { } =nnx )( xc nh bi cng thc 11( ) ( )2 nCx n z X z dzi= v c gi l

bin i ngc ca bin i Z ca )(zX .

CU HI N TP V BI TP

1.1. Nu hm phc )(zfw = c o hm ti 0z th c o hm mi cp ti 0z . ng Sai .

1.2. Hm phc )(zfw = gii tch ti 0z th c th khai trin thnh tng ca chui ly tha tm 0z .

ng Sai .

1.3. Hm phc )(zfw = c o hm khi v ch khi phn thc v phn o ( )yxu , , ( )yxv , c o hm ring cp 1.

ng Sai .

1.4. Nu 0z l im bt thng c lp ca hm phc )(zfw = th c th khai trin Laurent ca hm s ny ti 0z .

ng Sai .

1.5. Tch phn ca hm phc gii tch )(zfw = trong min n lin D khng ph thuc ng i nm trong D .

ng Sai .

1.6. Tch phn trn mt ng cong kn ca hm phc gii tch )(zfw = trong min n lin D lun lun bng khng.

ng Sai .

1.7. Thng d ca hm phc )(zfw = ti 0z l phn d ca khai trin Taylor ca hm ny ti 0z .

ng Sai .

1.8. Hm phc )(zfw = c nguyn hm khi v ch khi gii tch. ng Sai .

1.9. Tch phn ca mt hm phc )(zfw = ch c mt s hu hn cc im bt thng c lp trn mt ng cong kn C (khng i qua cc im bt thng) bng tng cc thng d ca

)(zfw = nm trong ng C . ng Sai .

1.10. C th tm c mt hm phc b chn v gii tch ti mi im.

ng Sai .

1.11. Rt gn cc biu thc sau


50

a. ( ) ( ) ( )3523352 ++ iii , b. ii 31

131

1+ ,

c. 10

11

+

ii , d.

( )( )( )( ) ( )3

1 2 3 4 21 2 1i i i

i i+ +

+ .

1.12. Gii cc phng trnh sau

a. 012 =++ zz , b. 0423 = zz , 1.13. Tnh: a. 3 1 i+ , b. 3 2424 i+ . 1.14. Tnh qu tch nhng im trong mt phng phc tho mn

a. 243 = iz , b. ( )4

arg = iz ,

c. 622 =++ zz , d. 122 =+ zz . 1.15. Tnh phn thc v phn o ca cc hm s sau

a. 3zw = b. z

w = 11 c. zew 3= .

1.16. Cho z

zw 1+= . Tm o hm )(' zw trc tip t nh ngha. Vi gi tr no ca z th hm s khng gii tch.

1.17. Chng minh hm zzw = khng gii tch ti mi z . 1.18. Chng minh rng hm

a. 4zw = b. izz

w += ,11

2

tho mn iu kin Cauchy-Riemann. Tnh )(' zw trong mi trng hp trn.

1.19. Tm hm phc gii tch ( ) ),(),( yxivyxuzfw +== bit phn thc a. 23 3),( xyxyxu = , b. xyxyxu 2),( 22 += ,

1.20. Tm hm phc gii tch ( ) ),(),( yxivyxuzfw +== bit phn o a. 22)1(

),(yx

yyxv ++= , b. xxyyxv 32),( += ,

1.21. Tm nh ca cc ng cong sau y qua php bin hnh z

w 1= .

a. 422 =+ yx , b. xy = , c. 1,0, , d. 1)1( 22 =+ yx .


51

1.22. Tm nh ca ng thng nm trn tia += kz3

Arg qua php bin hnh zzw

+=11 .

1.23. Cho php bin hnh tuyn tnh 1)1( += ziw a. Tm nh ca on thng ni iz =11 v iz =2 . b. Tm nh ca ng trn 2)1( =+ iz .

1.24. Tm php bin hnh bo gic bin hnh trn 1w sao cho cc im i,1,1 bin ln lt thnh 1,0, . 1.25. Tnh tch phn =

CdzzI trong hai trng hp sau

a. C l on thng ni 2 im 1 v +1. b. C l na cung trn tm 0 nm trong na mt phng trn i t im 1 n im 1.

1.26. Cho C l ng trn 31 =z , tnh cc tch phn sau:

a. cos

C

z dzz v , b. ( 1)

z

C

e dzz z +v .

1.27. Tnh tch phn =C

zdzI trong C l ng gp khc c nh ln lt l ,21,2 i+

2,1 i+ .

1.28. Tnh tch phn 2sin

41C

z

I dzz

= v trong C l ng trn 0222 =+ xyx .

1.29. Tnh tch phn ( ) ( )3 31 1CdzI

z z= + v trong cc trng hp sau:

a. C l ng trn 2,1


52

1.32. Khai trin Laurent ca hm s 2

12 +

+=zz

zw

a. Trong hnh vnh khn 21


53

z .

Chng 2: Cc php bin i tch phn

CHNG II: CC PHP BIN I TCH PHN

GII THIU

54

}Trong chng I chng ta s dng tnh duy nht ca khai trin Laurent ca hm gii tch

trong hnh vnh khn xy dng php bin i Z. Nh php bin i Z ta c th biu din tn hiu s bi hm gii tch . Trong chng ny chng ta s nghin cu hai php bin i tch phn l bin i Laplace v bin i Fourier.

{ )(nx )(zX Nhiu vn trong k thut, trong in t vin thng, trong l thuyt mch, a v gii

cc phng trnh, h phng trnh cha o hm, tch phn ca cc hm no , ngha l phi gii cc phng trnh vi phn, tch phn hay phng trnh o hm ring. Vic gii trc tip cc phng trnh ny ni chung rt kh. K s Heaviside l ngi u tin vn dng php bin i Laplace gii quyt cc bi ton lin quan n mch in.

Php bin i Laplace bin mi hm gc theo bin thnh hm nh theo bin . Vi php bin i ny vic tm hm gc tho mn cc biu thc cha o hm, tch phn (nghim ca phng trnh vi phn, phng trnh tch phn, phng trnh o hm ring) c quy v tnh ton cc biu thc i s trn cc hm nh. Khi bit hm nh, ta s dng php bin i ngc tm hm gc cn tm.

t s

Trong mc ta ny gii quyt hai bi ton c bn ca php bin i Laplace l tm bin i thun, bin i nghch v mt vi ng dng ca n.

Cc hm s trong chng ny c k hiu l thay cho v c k hiu cho cc tn hiu ph thuc vo thi gian

...),(),( tytx ...),(),( xgxf)(),( tytx t .

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 2.57, 2.57'), dng cc (cng thc 2.63) v dng phc (cng thc 2.64, 2.68). Phn 1 ca mc ny s trnh by ba dng ny 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.

Php bin i Fourier ri rc c s dng tnh ton bin i Fourier bng my tnh, khi cc tn hiu c ri rc ho bng cch chn mt s hu hn cc gi tr mu theo thi gian v ph cng nhn c ti mt s hu hn cc tn s. Tuy nhin 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...

NI DUNG

2.1. PHP BIN I LAPLACE

2.1.1. nh ngha bin i Laplace

nh ngha 2.1: Gi s l hm s thc xc nh vi mi . Bin i Laplace ca hm s c nh ngha v k hiu:

)(tx 0>t)(tx

(2.1) { } ==0

)()()( dttxesXtx stL

Php bin i Laplace ca hm s gi l tn ti nu tch phn (2.1) hi t vi gi tr thuc min no . Trng hp ngc li ta ni php bin i Laplace ca hm s khng tn ti. Php bin i Laplace l thc hay phc nu bin s ca hm nh l thc hay phc.

)(tx s)(tx

s )(sX

Theo thi quen ngi ta thng k hiu cc hm gc bng cc ch thng cn cc bin i ca n bng cc ch in hoa . i khi cng c k hiu bi

.

...),(),( tytx...),(),( sYsX

...),(~),(~ sysx

2.1.2. iu kin tn ti

nh ngha 2.2: Hm bin thc c gi l hm gc nu tho mn 3 iu kin sau: )(tx

1) vi mi . 0)( =tx 0M

0,)( 0 > tMetx t . (2.2)

0 c gi l ch s tng ca . )(txR rng 0 l ch s tng th mi s 1 0 > cng l ch s tng. V d 2.1: Hm bc nhy n v (Unit step function)

(2.3)


V d 2.2: Cc hm s cp c bn u lin tc v khng tng nhanh hn hm m. Nhng vn cha phi l hm gc v khng tho mn iu kin 1) ca nh ngha 2.2. Tuy nhin hm s sau:

)(tx

(2.4)

v 0)(lim)Re(

= sXs .

Hn na hm nh gii tch trong min )(sX 0)Re( >s vi o hm

(2.5) =0

)()()(' dttxetsX st

Chng minh: Vi mi s i = + sao cho 0 > , ta c: 0(( ) tstx t e Me ) m

hi t, do tch phn hi t tuyt i. V vy tn ti bin i Laplace

v

0( )

0

te

dt dtetx st 0

)(

)(sX0 0 0

( ) ( ) ( ) ( )st t i t tX s x t e dt x t e e dt x t e dt

= =

( )( )0

0

0 00 0

tt Me MMe dt

= = .

Ngoi ra Re( )0

lim 0 lim ( ) 0s

M X s = = .

Tch phn hi t v tch phn dtetx st 0

)( ( ) dttetxdtetxs

stst =

00

)()()( hi t u

trong min { }1Re( )s s vi mi 1 , 1 0 > (theo nh l Weierstrass), suy ra hm nh c o hm ( dtetx

ssX st

=

0)()(' ) ti mi thuc cc min trn. V vy gii tch trong

min

s )(sX

0Re( )s > .

56


Nhn xt:

1. Theo nh l trn th mi hm gc u c nh qua php bin i Laplace. Tn gi "hm gc" l do vai tr ca n trong php bin i ny.

57

}2. T v d 2.2, cng thc (2.4) suy ra rng mi hm s cp c bn u c bin i

Laplace )(tx

{ )()( ttx L . Tuy nhin, n gin thay v vit ng { })()( ttx L th ta vit tt { )(tx }L . Chng hn ta vit { tsin }L thay cho { }tt sin)(L , { }1L thay cho { })(tL .

3. Ta quy c cc hm gc lin tc phi ti 0. Ngha l )0()(lim0

xtxt

=+ .

V d 2.3: V hm c ch s tng )(t 00 = do bin i

{ }ss

edtest

st 1100===

L vi mi . 0)Re(, >ssV d 2.4: Hm c ch s tng tsin 00 = do bin i

{ } ==0

sin)(sin dttesXt stL tn ti vi mi . 0)Re(, >ss

p dng cng thc tch phn tng phn ta c:

( ) 20 00 0

( ) cos cos 1 sin sinst st st stX s te se t dt se t s e t dt = =

( ) 22 1 1)(1)(1 ssXsXs +==+ . 2.1.3. Cc tnh cht ca php bin i Laplace

2.1.3.1. Tnh tuyn tnh

nh l 2.2: Nu c bin i Laplace th vi mi hng s A, B,)(,)( tytx )()( tBytAx + cng c bin i Laplace v

{ } { } { })()()()( tyBtxAtBytAx LLL +=+ . (2.6) V d 2.5: { } { } { }

145sin415sin45 2 ++=+=+ sstt LLL .

2.1.3.2. Tnh ng dng

nh l 2.3: Nu { )()( txsX }L= th vi mi , 0>a{ }

=

asX

aatx 1)(L . (2.7)

V d 2.6: { } ( ) 222 1/11sin +

=+= sstL .


2.1.3.3. Tnh dch chuyn nh

58

}nh l 2.4: Nu { )()( txsX L= th vi mi a , { } ( )asXtxeat =)(L . (2.8)

V d 2.7: { } { }as

ee atat ==11LL . { } 222ch =

+=

sseet

ttLL ;

{ } 222sh =

=

seet

ttLL . { } 22)(sin + = asteatL .

2.1.3.4. Tnh tr

nh l 2.5: Nu { )()( txsX }L= th vi mi a , { } ( )sXeatxat sa= )()(L . (2.9)

th ca hm c c bng cch tnh tin th ca dc theo trc honh mt on bng . Nu biu din tn hiu theo thi gian th biu din tr n v thi gian ca qu trnh trn.

)()( atxat )()( txta )(tx t )( atx

a

)()( txt )()( atxat

t O

x

t O

x

a

V d 2.8: { }s

eatas

= )(L .

V d 2.9: Hm xung (Impulse) l hm ch khc khng trong mt khong thi gian no .

(2.10)

>


{ } { } { }, ( ) ( ) ( ) as bsa b e et t a t a s = =L L L .

t t

x x

O O a a b b

)(t 1

59

V d 2.10: Tm bin i Laplace ca hm bc thang x


{ } ( )( ) 1 2 ( 1)( ) (0) '(0) (0)n n n n nx t s X s s x s x x = "L . (2.14) V d 2.12: { } 2222

'0sin1sincos +=+

=

=

ss

sstt LL .

H qu: Vi gi thit ca nh l 2.6 th )0()(lim)Re(

xssXs

= .

Chng minh: p dng nh l 2.1 cho o hm ta c . )(' tx 0)0()(lim)Re(

= xssXs2.1.3.6. Bin i Laplace ca tch phn

nh l 2.7: Nu hm gc c )(tx { })()( txsX L= th hm s cng l hm gc v

=t

duuxt0

)()(

( )ssXduux

t=

0)(L . (2.15)

2.1.3.7. o hm nh

nh l 2.8: Gi s l mt hm gc c )(tx { })()( txsX L= th { } ( ) ( )sX

dsdtxt n

nnn 1)( =L . (2.16)

V d 2.13: { } ( ) 1!11 +== nnn

nn

sn

sdsdtL .

t

x

a O

1

V d 2.14: Hm dc

0

0

0

( )

1

t

t a

t a

tx ta

<

=

nu

nu

nu

)()()( 0 attattx a += )()()()()( ata

attatatat

att

at =+= .

t

x

1 O

1

2

{ } 222 11)( ase

ase

astx

asas == L .

V d 2.15: Hm xung tam gic n v

60


>

= tduu utt

c bin i Laplace ss

duu

ut 1arctg1sin

0=

L .

2.1.3.9. Bin i Laplace ca hm tun hon

nh l 2.10: Gi s l mt hm gc tun hon chu k th )(tx 0>T

{ } sT

Tst

e

dttxe

txsX

==

1

)(

)()( 0L . (2.18)

V d 2.17: Tm bin i Laplace ca hm gc tun hon chu k sau: 02 >a

t 1

1

a a2 a3 a4

61


( )222 20 0 0

1( )

a a asa a a st stst st st

a a

ee ee x t dt e dt e dts s s

= = = .

( )( )

22 2

22 2

22

2

sh1 1 1 1 1 1( ) th11 ch

as asas as

as as asas

asas

as

e e e eX ss s s ses e

e e

= = = = = + +.

2.1.3.10. nh ca tch chp

nh ngha 2.3: Tch chp ca hai hm s ( ), ( ); 0x t y t t l hm s c k hiu v xc nh bi cng thc

0

( ) ( ) ( ) ( )t

x t y t x u y t u du = (2.19) Tnh cht:

( ) ( ) ( ) ( )x t y t y t x t = (tch chp c tnh giao hon) Nu l hai hm gc th tch chp ca chng )(),( tytx ( ) ( )x t y t cng l hm gc. nh l 2.11: Nu { })()( txsX L= , { })()( tysY L= th

{ }( ) ( ) ( ) ( )x t y t X s Y s =L (2.20) Ngoi ra nu cng l hm gc th ta c cng thc Duhamel: )('),(' tytx

{ } { }(0) ( ) '( ) ( ) ( ) (0) ( ) '( ) ( ) ( )x y t x t y t x t y x t y t sX s Y s+ = + =L L (2.21) V d 2.17: { } { } { } 2 21 1sin sin 1t t t t s s = = +L L L ( ) { }2 22 2

1 1 1 sin11

t ts ss s

= = = ++ L .

Do tnh duy nht ca bin i ngc (nh l 2.12) ta suy ra: . tttt sinsin* =2.1.4. Php bin i Laplace ngc

T v d 2.17 cho thy cn thit phi gii bi ton ngc: Cho hm nh, tm hm gc. Trong mc ny ta s ch ra nhng iu kin mt hm no l hm nh, ngha l tn ti hm gc ca n, ng thi cng ch ra rng hm gc nu tn ti l duy nht.

nh ngha 2.4: Cho hm , nu tn ti sao cho )(sX )(tx { } )()( sXtx =L th ta ni l bin i ngc ca , k hiu

)(tx

)(sX { })()( 1 sXtx = L . 2.1.4.1. Tnh duy nht ca bin i ngc

nh l 2.12: Nu l mt hm gc vi ch s tng )(tx 0 v { } )()( sXtx =L th ti mi im lin tc t ca hm ta c: )(tx

62


dssXei

txi

i

st+

= )(

21)( (2.22)

trong tch phn v phi c ly trn ng thng =)Re(s theo hng t di ln, vi l s thc bt k ln hn 0 .

Cng thc (2.22) c gi l cng thc tch phn Bromwich.

Cng thc Bromwich cho thy bin i Laplace ngc nu tn ti th duy nht.

2.1.4.2. iu kin mt hm c bin i ngc

nh l 2.1 cho thy khng phi mi hm phc gii tch no cng c bin i ngc. Chng

hn hm khng th l nh ca hm gc no v 2)( ssX = = )(lim)Re( sXs .

nh l sau y cho ta mt iu kin hm gii tch c bin i ngc

nh l 2.13: Gi s hm phc tho mn 3 iu kin sau: )(sX

i. gii tch trong na mt phng )(sX 0)Re( >s , ii. RMsX )( vi mi thuc ng trn s Rs = v 0lim = RR M ,

iii. Tch phn hi t tuyt i. dssXi

i+

)(

Khi c bin i ngc l hm gc cho bi cng thc (2.22). )(sX )(tx

c gi c th tm hiu chng minh nh l 2.12, nh l 2.13 trong Ph lc C ca [2] hoc nh l1 trang 29 ca [5].

2.1.4.3. Mt vi phng php tm hm ngc

a. S dng cc tnh cht ca bin i thun v tnh duy nht ca bin i ngc

T tnh duy nht ca bin i ngc, ta suy ra rng tng ng gia hm gc v hm nh l tng ng 1-1 . V vy ta c th p dng cc tnh cht bit ca php bin i thun tm hm ngc.

V d 2.18: ( ) !51

41 54

614

61 te

se

stt =

=

+ LL

( ) ( )( ) ( ) )3(

!53

44

5345

6

315

6

351 =

+=

+ ttee

see

se tss LL .

b. Khai trin thnh chui ly tha

Nu "+++++= 544332210)( sa

sa

sa

sa

sa

sX th

63


{ } "+++++== !4!3!2

)()(4

43

32

210

1 tatatataasXtx L (2.23)

V d 2.19:

1

2 3 4 2 3 4 51 1 1 1 1 1 1 1 1 1 11

2! 3! 4! 2! 3! 4!se

s s s ss s s s s s s

= + + = + + " "

1 2 3 41

2 2 21( ) 1

(2!) (3!) (4!)s t t tx t e t

s

= = + +

"L

( ) ( ) ( ) ( ) ( )

2 4 6 8

02 2 2 2 2 2 2 2 2 2

2 2 2 21 2

2 2 4 2 4 6 2 4 6 8

t t t tJ t= + + ="

trong l hm Bessel bc 0 (xem chng III). 0J

c. S dng thng d ca tch phn phc

Vi iu kin ca nh l 2.13 th c bin i ngc xc nh bi cng thc Bromwich (2.22).

)(sX)(tx

B

'B

A

RC

1a 2a

na x

y

O

Mt khc gi s hm ch c mt s hu

hn cc im bt thng c lp trong

na mt phng

)(sX

naaa ,...,, 21 .

Chn R ln sao cho cc im bt thng ny u nm trong phn ca mt phng c gii hn bi ng trn tm O bn knh R v ng thng

. Khi RC

=)Re(s

{ } [ ]=

==n

kk

st asXesXtx1

1 ;)(Res)()( L (2.24)

c bit nu )()()(

sQsPsX = , trong bc ca a thc ln hn bc ca a thc .

Gi s ch c cc khng im n l v chng khng phi l khng im ca

th ta c cng thc Heavyside:

)(sQ )(sP

)(sQ naaa ,...,, 21)(sP

11

( )( )( )( ) '( )

kn

a tk

k k

P aP sx tQ s Q a

=

= = L e (2.25)

V d 2.20: Tm hm gc

++++=

)3)(2)(1(53)(

21

ssssstx L .

64


Gii: Hm nh )3)(2)(1(

53)()( 2

++++=

sssss

sQsP c cc cc im n l . 3,2,1

43

)(')(

1=

=ssQsP , 1

)(')(

2=

=ssQsP ,

45

)(')(

3=

=ssQsP ttt eeetx 32

45

43)( += .

V d 2.21: Tm hm gc ( )2

12

3 3 2( )( 2) 4 8

s sx ts s s

+ + = + + L .

Gii: Hm nh ( )2

2

( ) 3 3 2( ) ( 2) 4 8

P s s sQ s s s s

+ += + + c cc cc im n l ii 22,22,2 + .

1)(')(

2=

=ssQsP ,

41

)(')(

22

isQsP

is+=

+=,

41

41

)22(')22(

)(')(

22

iiiQiP

sQsP

is=+=

++=

=.

ittittt eieietx 222224

14

1)( +

+

++=

( ) ( ) +=+++= tteeeeeieeee ttitittitittt 2sin212cos24 222222222 . d. Tm hm gc ca cc phn thc hu t

Mi phn thc hu t c dng )()()(

sQsPsX = , trong bc ca ln hn bc ca

u c th phn tch thnh tng ca cc phn thc ti gin loi I v loi II.

)(sQ )(sP

Cc phn thc hu t loi I: as

1 hay nas )(1 , a c hm gc:

ateas

=

11L ,

)!1()(1 11

=

nte

as

nat

nL . (2.26)

Cc phn thc hu t loi II: ( )2 2( ) nMs N

s a +

+ +, ,,, aNM .

S dng tnh cht dch chuyn nh ta c th a cc phn thc ti gin loi II v mt trong hai dng sau:

( )2 2 ns

s + hoc ( )2 21

ns + (2.27)

Trng hp , t v d 2.6 v v d 2.12 ta c: 1=n

ts

s =

+ cos22

1L , =

+ t

ssin1

221L (2.28)

65


Trng hp : 2=n

( )1

22 2

sin2

s t t

66

s

= +

L ( ), 1

2 32 2

1 sin cos2

t t t

s

= +

L (2.29)

Trng hp : 3=n ( )2

12 32 2

sin cos8

s t t t

s

t

= +

L ,

( )( )2 21

2 32 2

3 sin 3 cos18

t t t

s

t

= +

L (2.30)

V d 2.22: Hm nh v d 2.21. ( )2

2

3 3 2( )( 2) 4 8

s sX ss s s

+ += + + c th phn tch thnh tng

cc phn thc ti gin

4)2(

14)2(

)2(22

184

322

1)( 222 ++++++=++

++= sss

ssss

ssX

( )2

1 2 22

3 3 2 1( ) 2 cos 2 sin 22( 2) 4 8

t t ts s 2x t e es s s

+ + = = + + + L t e t .

V d 2.25: Tm hm gc ca 32

)2)(1(11155)( +

=ss

sssX .

Ta c th phn tch thnh tng cc phn thc ti gin )(sX

323

2

)2(7

)2(4

231

131

)2)(1(11155)(

++++

=+=

sssssssssX

tttt etteeess

sstx 22223

21

274

31

31

)2)(1(11155)( ++=

+= L .

2.1.5. ng dng ca bin i Laplace

2.1.5.1. ng dng ca bin i Laplace gii phng trnh vi phn tuyn tnh

a. Phng trnh vi phn tuyn tnh h s hng

)(0111

1 tyxadtdxa

dtxda

dtxda n

n

nn

n

n =++++

" (2.31)

tha mn iu kin u

1)1(

10 )0(,...,)0(',)0( === nn xxxxxx (2.32)


Ta tm nghim l hm gc bng cch t { })()( txsX L= , { })()( tysY L= . p dng cng thc bin i Laplace ca o hm (2.13), (2.14) vi iu kin u (2.32),

{ } )()( 00 sXatxa =L { } ( )011 )()(' xssXatxa =L .. ... ... ... ... ... ... ... ... ... ... ... ...

{ } ( )( ) 1 0 2( ) ( )n n nn n na x t a s X s s x sx x 1n = "L . (2.33) Thay vo (2.31) ta c

( ) ( )1 11 1 0 0 1( ) ( )n n n nn n n na s a s a s a X s Y s x a s a s a + + + + = + + + +" "2 1 ( )2 31 1 2n nn n n 1 nx a s a s a x a + + + + + +" " . Vy phng trnh nh c dng:

)()()()()()()()(

sAsBsYsXsBsYsXsA +=+= .

nh ngc l nghim cn tm. { )()( 1 sXtx = L }V d 2.27: Tm nghim ca phng trnh: tha mn iu kin u

.

txxx sin"2)4( =++0)0()0(")0(')0( )3( ==== xxxx

Gii: Phng trnh nh: ( ) ( )4 2

2 32

1 12 1 ( ) ( )1 1

s s X s X ss s

+ + = =+ +.

p dng cng thc (2.30) ta c nghim { } ( )21 3 sin 3 cos( ) ( )8

t t tx t X s

= =L t .

V d 2.28: Tm nghim ca phng trnh: tha mn iu kin u texx =+" 1)1( =x , . 0)1(' =x

Gii: Bng cch t ta a iu kin u 1= tu 1=t v iu kin u . 0=ut )()1()( txuxuy =+= . S dng quy tc o hm hm hp ta c:

dtdx

dudt

dtdx

dudx

dudy === , tng t 2

2

2

2

dtxd

duyd = .

Do phng trnh cho c th vit li tng ng: vi iu kin u .

1)()(" +=+ ueuyuy0)0(',1)0( == yy

t . { } { } ssYsuyuysY == )()(")()( 2LLPhng trnh nh: ( ) ( )2 221 ( ) ( )1 1( 1) 1

e es Y s s Y ss ss s

+ = + = + s + + .

67


ueueeeuys

e

sse

s

e

sY u sin2

cos2

12

)(1

212

1)1(

2)( 22 +

+=++

+= .

Vy phng trnh cho c nghim )1sin(2

)1cos(2

121)( +

+= teteetx t .

b. H phng trnh vi phn tuyn tnh h s hng

V d 2.29: Tm nghim ca h phng trnh vi phn:

vi iu kin u .

==

xyyyxx

2'32'

==

3)0(8)0(

yx

Gii: t { })()( txsX L= , { })()( tysY L= { } { } 3)(,8)( == sYtysXtx LL . Thay vo h phng trnh trn ta c h phng trnh nh:

hay

==XYsY

YXsX23

328

=+=+

3)1(283)2(

YsXYXs

Gii h phng trnh nh ta c nghim:

+=+=

++=+=

42

15

)4)(1(223

43

15

)4)(1(178

sssssY

sssssX

=+=

.25)(

35)(4

4

tt

tt

eety

eetx

c. Phng trnh vi phn tuyn tnh h s bin thin

V d 2.31: Gii phng trnh 4 0t x" x' tx+ + =

t { }X( s ) x( t )= L th { }4 4 dXtx( t )ds

= L , { } 0x'( t ) sX x( )= L .

{ } ( )2 20 0 2d dtx"( t ) s X sx( ) x'( ) sX s x( )ds ds= = +L 0X . Phng trnh nh: 22 0 0 4dX dXsX s x( ) sX x( )

ds ds + + = 0 .

Hay 2 24 4dX dX s( s ) sX dsds X s

+ = = + .

Gii phng trnh ny ta c: 2 4CX( s )

s= + .

Nghim ca phng trnh l hm gc 1 02 24Cx( t ) CJ ( t )

s = = + L .

xc nh C ta thay vo 2 v ca ng thc trn: 0t = 00 0x( ) CJ ( ) C= = . Vy nghim ca phng trnh l: 00 2x( t ) x( )J ( t )= . 68


2.1.5.2. ng dng ca bin i Laplace gii phng trnh tch phn

Xt phng trnh tch phn dng tch chp

=+t

tfCduutkuxBtxA0

)()()()( (2.34)

CBA ,, l cc hng s, l cc hm gc. )(),( tktf

Gii phng trnh (2.34) l tm tt c cc hm thc tha mn ng thc vi mi )(tx t thuc mt min no .

Gi s l hm gc. t )(tx { })()( txsX L= , { })()( tfsF L= , { })()( tksK L= . Phng trnh nh

)()(

)()()()()(sKBA

sFCsXsFCsKsXBsXA +==+ .

Nghim

+=

)()(

)( 1sKBA

sFCtx L (2.35)

V d 2.33: Gii phng trnh tch phn Abel:

0

( ) ( ) ; 0 1( )

t x u du f tt u

= <


Gi l hiu in th ca hai u on mch, l cng dng in ca mch ti thi im t . v tha mn cc ng thc sau:

)(tu )(ti)(tu )(ti

)()( 12 tiRuutu == ; dttdiLuu )(23 = ;

+= 0

034 )(

1 qdttiC

uut

. (2.36)

t { })()( tisI L= , { })()( tusU L= th )0()( isIdt

tdi =

L ,

sq

sIqdtti

t0

00

)( +=

+L .

Trong l in lng ban u (0q 0=t ) trn cc thnh t in. Trong cc bi ton ng mch cc iu kin ban u u bng 0: 0)0(,00 == iq . Lc t s gia in th nh v cng nh gi l tr khng nh

IUZ = . Nh vy cc tr khng nh ca in trR , cun dy c h s

t cm L v t in c in dung tng ng l: C

Cs

ZLsZRZ 1;; === (2.37)

Khi tnh ton mt mng gm nhiu mch in kn ta p dng nh lut th nht ca Kirchoff (Kichp) cho tng nt v nh lut th hai cho tng mch kn, sau chuyn cc phng trnh tm c sang phng trnh nh.

p dng hai nh lut Kirchoff ta c th tm tr khng nh tng ng ca mch mc ni tip v mch song song c bn sau:

Tr khng nh tng ng Z ca hai tr khng mc ni tip bng tng hai tr khng ny.

21, ZZ

1Z 2Z

A B C

Gi ln lt l hiu in th gia A, B; B, C v A, C. theo nh lut 1 Kirchoff ta c . Chuyn qua nh

uuu ,, 2121 uuu += IZIZZIUUU 2121 +=+= . Vy

21 ZZZ += (2.38) Nghch o ca tr khng nh tng ng ca hai tr khng mc song song

bng tng nghch o hai tr khng ny. 21, ZZ

1Z

2Z

A B

1I

2I

I

70


Gi ln lt l cng nh trong mch 1, mch 2 v mch chnh. U l in th nh gia A v B.

III ,, 21

p dng nh lut 2 Kirchoff ti nt A ta c 21 III += 21 Z

UZU

ZU += . Vy:

21

111ZZZ

+= (2.39)

V d 2.34: Mt t in c in dung C c np in c in lng . Ti thi im , ta mc n vo 2 mt ca 1 cun dy c in cm

0q0=t L . Tm in lng ca t in v

cng ca dng in trong mch ti thi im . )(tq

)(ti 0t >Gii: p dng nh lut Kirchoff th nht cho mch vng ta c:

C

L

i 01 00

=

++ qdtiCdtdiL

t.

V dtdqti =)( nn phng trnh trn tr thnh

001 22

00

2

2=+=

++ Cqdt

qdLqdtdtdq

CdtqdL

t.

t { )()( tqsQ }L= , v 0)0()0(',)0( 0 === iqqq . Do ta c phng trnh nh: ( )

CLs

sqQCQsqQsL

10

200

2

+==+ .

Vy CLt

CLq

dtdqti

CLtqtq sin)(;cos)( 00 === .

2.2. PHP BIN I FOURIER

2.2.1. Chui Fourier

2.2.1.1. Khai trin Fourier ca hm tun hon chu k 2 nh ngha 2.5: Cho l mt hm tun hon chu k 2)(tx , chui

( )01

cos sin2 n nn

a a nt b n

=+ + t (2.40)

c cc h s xc nh bi

...,2,1;sin)(;cos)(1;)(12

0

2

0

2

00 ====

nntdttxbntdttxadttxa nn (2.41)

71


c gi l chui Fourier ca hm . Cc h s (2.41) gi l h s Fourier. )(tx

C th chng minh c rng nu

=

++=1

0 sincos2

)(n

nn ntbntaatx (2.42)

th cc h s l cc h s Fourier (2.41) ca hm . nn baa ,,0 )(tx

Ngc li mi hm tha mn iu kin Dirichlet th c th khai trin thnh chui Fourier.

nh l 2.14 (nh l Dirichlet): Gi s hm tun hon chu k , n iu tng khc v b chn (gi l iu kin Dirichlet), ti cc im gin on ta k hiu

)(tx 2

2

)0()0()( ++= txtxtx (2.43)

Khi chui Fourier hi t v c ng thc (2.42), trong )0(),0( + txtx ln lt l gii hn phi v gii hn tri ca ti )(tx t .

2.2.1.2. Khai trin Fourier ca hm tun hon chu k lT 20 = Chui Fourier ca hm tun hon chu k c dng: )(tx l2

01

( ) cos sin2 n nn

a n nx t a t bl l

t

== + + (2.44)

Cc h s Fourier c tnh theo cng thc sau:

...,2,1;sin)(1;cos)(1;)(12

0

2

0

2

00 ==== ntdtlntxlbtdtlntxladttxla

l

n

l

n

l (2.45)

Nhn xt:

1. Hm tun hon chu k l mt trng hp c bit ca hm tun hon chu k , v vy cc nhn xt sau y c gi thit l hm tun hon chu k . Ngoi ra do tnh cht tch phn ca hm tun hon nn cc h s Fourier (2.45) cng c th tnh nh sau:

2 l2l2

;cos)(1;)(122

0 ++ ==

cl

cn

cl

ctdt

lntx

ladttx

la

cntdtl

ntxl

bcl

cn ==

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

2 (2.46)

2. Nu l hm l tun hon chu k th )(tx l2 tl

ntx cos)( l hm l v tl

ntx sin)( l hm chn, do cc h s Fourier (2.44) tha mn

...,2,1;sin)(2;00

0 ==== ntdtlntxlbaal

nn (2.47)

72


3. Nu l hm chn tun hon chu k th )(tx l2 tl

ntx cos)( l hm chn v

tl

ntx sin)( l hm l, do cc h s Fourier (2.44) tha mn

...,2,1;cos)(2;)(2;000

0 ==== ntdtlntxladttxlabl

n

l

n (2.48)

4. Gi s 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 ( ba , )abl =2 . Do c th khai trin thnh

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

;2cos)(2;)(20 ==b

an

b

atdt

abntx

abadttx

aba

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

= ntdtab ntxabbb

an (2.49)

5. Gi s 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 (2.48) v nu m rng thnh hm l th cc h s Fourier c tnh theo cng thc (2.47).

)(tx ( l,0 )l2

2.2.1.3. Dng cc ca chui Fourier (Polar Fourier Series)

T cng thc (2.42) nu ta t

2200 ;2 nnnbaAaA +== (2.50)

v gc


=

++

+=1

intint0222 n

nnnn eibaeibaa

Vy ta c th vit chui Fourier di dng phc

(2.53) =

=n

nectxint)(

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

= = = +2/ 2 n

0 02

( )n n n

n n

a ca c cb i c c

= = + = (2.54)

Cc h s Fourier phc (2.54) c th tnh trc tip

cdtetxcc

cn =

+ ,)(21 2 int (2.55)

Hm tun hon chu k c khai trin Fourier dng phc lT 20 =

=

=

n

tl

ninectx )( , cdtetxl

clc

c

tl

nin =

+ ,)(

21 2 (2.56)

Nu k hiu 0

01

Tf = l tn s c bn ca hm tun hon chu k th cng thc (2.68)

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= (2.57) nh l 2.15: i vi mi hm tun hon chu k )(tx lT 20 = tho mn iu kin

Dirichlet th c ng thc Parseval

=

+=

nn

Tc

ccdttx

T22

0

0

)(1 (2.58)

Nhn xt: Cng thc (2.44), (2.52), (2.56) 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 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

nc

74


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

00 T

nnffn == .

2.2.2. Php bin i Fourier hu hn

2.2.2.1. nh ngha php bin i Fourier hu hn

Bin i Fourier hu hn ca tn hiu ri rc { } =nnx )( l (2.59) l { } 2( ) ( ) ( ) i nf

nX f x n x n e

=

= = Fnu chui v phi hi t.

Cng thc bin i ngc

l{ } l110

( ) ( ) ( ) i nf2x n X f X f e = = F df (2.60) V d 2.36: Tm bin i Fourier hu hn ca tn hiu ri rc )(rect)( nnx N= , N l 1 s t

nhin.

Gii: l21

2 22

0

1( ) ( )1

i NfNi nf i nf

i fn n

eX f x n e ee

= =

= = =

ffNe

eeee

ee fNi

fifi

NfiNfi

fi

Nfi

=

=

sinsin)1( .

Nhn xt:

1. Trong cng thc bin i Fourier 2.59, 2.60 i s c k hiu cho tn s. C ti liu khng biu din bin i Fourier qua min tn s m qua min

f nh sau

l { }( ) ( ) ( ) i nn

X x n x n e =

= = F , l{ } l210

1( ) ( ) ( )2

i nx n X X e d

= = F (2.61) 2. Hai cch biu din ny tng ng vi nhau qua php i bin s . f= 23. Mt iu kin tn hiu ri rc { } =nnx )( tn ti bin i Fourier hu hn l


2.2.2.2. 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:

{ } { } { }( ) ( ) ( ) ( )Ax n By n A x n B y n+ = +F F F (2.62) 2. Tr:

l { } { } l020( ) ( ) ( ) ( )i n fX f x n x n n e X= =F F f . (2.63) 3. Dch chuyn nh:

l { } { } l02 0( ) ( ) ( ) ( )i nfX f x n e x n X f f= =F F . (2.64) 4. iu ch:

{ } l l0 02 2 00 ( ) (( ) cos(2 ) ( ) 2 2i nf i nf X f f X f fe ex n nf x n

+ ++= =

F F 0 ) . (2.65)

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

X f x n x n e =

= = F { } l2 2( ) ( ) ( ) ( )i nf i nf

n nx n x n e x n e X

= =

= = = F f (2.66) 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 { } l2 2 ( )( )( ) ( ) ( ) (i nf i n f

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

= =

= = = F f (2.67) 7. Tch chp:

{ } { } { }( ) ( ) ( ) ( )x n y n x n y n = F F F (2.68) 8. Tch chp nh:

{ } { } { }( ) ( ) ( ) ( )x n y n x n y n = F F F (2.69) 9. Bin i ca hm tng quan

= =m

yx nmymxnr )()()(, { } l l, ( ) ( ) ( )x yr n X f Y f=F (2.70) Nu thc th )(),( nynx =

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

10. nh l Weiner-Khinchin:

76


{ } l 2, ( ) ( )x xr n X f=F . (2.71) 11. o hm nh:

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

df= = F F (2.72)

12. 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 . (2.73)

2.2.3. Php bin i Fourier

2.2.3.1. Cng thc tch phn Fourier

nh l 2.16: Nu hm kh tch tuyt i trn ton b trc thc ( ) v

tho mn iu kin Dirichlet th c ng thc

)(tx


4. Cc cng thc tch phn Fourier, nh l 2.16 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

78

25. Nu i bin 2 f d df = = , thay vo cng thc (2.75) 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

= =

(2.78)

2.2.3.2. nh ngha php bin i Fourier

nh ngha 2.6: Gi s hm kh tch tuyt i trn trc thc v tha mn iu kin Dirichlet. Bin i Fourier (vit tt l FT) ca l

)(tx)(tx

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

= = F (2.79) Trong k thut, nu l hm dng sng (waveform) theo thi gian th )(tx t l( )X f c gi

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

T cng thc tch phn Fourier (2.78) ta c cng thc bin i ngc

l{ } l1( ) ( ) ( ) i ft2x t X f X f e

= = F df (2.80) Hm nh qua php bin i Fourier l( )X f c th vit di dng cc

l l ( )( ) ( ) i fX f X f e = (2.81) trong

l l l( ) ( ) ( )X f X f X= f , l( ) ( )f X f = (2.82) c gi dng bin - pha ca php bin i.

Cp l( ), ( )x t X f c gi l cp bin i Fourier. 2.2.3.3. Tnh cht ca php bin i Fourier

a. Tng t cc tnh cht (2.63)-(2.73) ca php bin i Fourier hu hn, php bin i Fourier c cc tnh cht c tng kt trong bng sau:

(2.83)

Tnh cht Hm )(tx Bin i Fourier l( )X f

1. Tuyn tnh )()( 21 tBxtAx + l l1 2( ) ( )AX f BX 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. Dch chuyn nh )(02 txe tfi l 0( )X f f

7. iu ch tftx 02cos)( l l0 01 1( ) (2 2 )X f f X f f + +

8. o hm n

n

dttxd )( ( ) l2 (ni f X f )

9. Tch phn

tduux )( l l1 1( ) (0) ( )

2 2X f X f

i f +

10. o hm nh )(txt n ( ) l( )2 nn nd X fi f df

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

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

12. Tch )()( 21 txtx l l1 2( ) ( )X f X f Hm trong tnh cht 9. l hm Dirac (xem v d 2.40). b. T nh ngha bin i Fourier (2.79) 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

)(tx

)(tx l( )X f cho nhau, ngha l l { } l{ }( ) ( ) ( ) ( )X f x t X t x= =F F f (2.84)

2.2.3.4. 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

= (2.85)

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

79


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

= (2.86)

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

2.2.3.5. Bin i Fourier ca cc hm c bit

V d 2.37: Xung vung n v

(2.87)

>

.

l 20

( ) 2 cos 2t i ft tX f e e dt e ft

= = dt p dng quy tc tch phn tng phn, t

cos 2 sin 2 / 2

t tU e dU e dtdV ftdt V ft f

= = = =

80


l0

0 0

sin 2( ) 2 sin 2 sin 22 2

tt te ftX f e ft dt e

f f f

= + = ft dt

Tip tc t sin 2 cos 2 / 2

t tU e dU e dtdV ftdt V ft f

= = = =

l0

0

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

tte ftX f e ft dt

f f f f f f

= = X f

l2 2

2( )4

X f 2f

= + .

Ngc li 0,4

2222 >=

+ fe

tF .

V d 2.40: Hm Dirac hai pha )(t l hm suy rng, hm chn tha mn tnh cht 0 0

( )0

tt

t = =

vi

vi v (2.90)

= 1)( dtt

1. vi mi hm lin tc ti 0.

= )0()()( fdtttf )(tf

2. . { } { }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

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

)(1)( ta

at = .

5. i bin s ly tch phn ta c

0 0( ) ( ) ( ) ( )0f t f t t t dt f

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

Hm Dirac cn c gi l hm xung kim.

V d 2.41: Hm bc nhy

(2.91)

=

=

td

tt

tu )(0001

)(nu

nu

81


Hm khng kh tch tuyt i trong ton b trc thc nhng t tnh cht A. 9. v )(tu

)()( tdt

tdu = ta c th m rng v xem

{ } )(21

21)()( f

fidtu

t+=

=

FF .

V d 2.42: Hm du

(2.92) )()(0101

)sgn( tututt

t =

=

nu

nu

{ } { } { }fi

ffi

ffi

tutut =

+

+==

1)(21

21)(

21

21)()()sgn( FFF .

2.2.4. Php bin i Fourier ri rc (DFT: Discrete Fourier Tranform)

Vic tnh ton bin i Fourier da vo my tnh phi c ri rc ho bng cch chn mt s hu hn cc gi tr mu theo thi gian v ph c c cng nhn ti mt s hu hn cc tn s. l ni dung ca php bin i Fourier ri rc.

Gi s l mt s t nhin cho trc, cn bc ca 1: 0>N N Nie

=2

E tha mn cc tnh cht sau:

i. . (2.93) nnnN =+ ,EE

ii. nu . nu 01

0=

=

N

k

knE lNn NN

k

kn ==

1

0E lNn = , l nguyn dng (2.94)

iii. Vi mi dy tn hiu { tun hon chu k : })(nx N )()( nxNnx =+ th nk

N

k

N

m

mkmxN

nx EE =

=

=

1

0

1

0)(1)( (2.95)

Da vo (2.95) ta c th nh ngha php bin i Fourier ri rc ca dy tn hiu{ })(nx tun hon chu k . N

nh ngha 2.7: Bin i Fourier ri rc ca dy tn hiu{ })(nx tun hon chu k l N (2.96) l { } 1

0( ) ( ) ( )

Nmk

mX k DFT x n x m

=

= = EBin i Fourier ri rc ngc

l{ } l10

1( ) ( ) ( )N

nk

kx n IDFT X k X k

N

== = E (

toan_vienthong

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