ch3_01_chuoi_f.pdf

24
2/5/2015 1 Chương 3 CHUỖI FOURIER VÀ BIẾN ĐỔI FOURIER Nội Dung 1. Một số dạng tín hiệu quan trọng 2. Khái niệm hàm tuần hoàn 3. Chuỗi Fourier Khai triển Fourier 4. Tích phân Fourier – Biến đổi Fourier 5. Phân tích phổ tín hiệu

Upload: chi

Post on 17-Aug-2015

219 views

Category:

Documents


0 download

TRANSCRIPT

2/5/20151Chng 3CHUI FOURIER V BIN I FOURIERNi Dung1. Mt s dng tn hiu quan trng2. Khi nim hm tun hon3. Chui Fourier Khai trin Fourier4. Tch phn Fourier Bin i Fourier5. Phn tch ph tn hiu2/5/20152MT S DNG TN HIUQUAN TRNGMt S Dng Tn Hiu Quan Trng1. Tn hiu xung vung gc H()2. Hm dc (Ramp function) ()3. Hm bc nhy n v ()4. Hm xung lc n v o()5. Hm du - Tn hiu ()6. Tn hiu xung tam gic ()7. Hm m suy gim ()8. Hm m tng dn 1 ()9. Xung hm m10. Tn hiu xung cosin11. Cp phn b o() chn l12. Phn b lc13. Dy xung vung lng cc14. Dy xung vung n cc15. Tn hiu Sin suy gim theo hm m16. Tn hiu 17. Tn hiu , 18. Tn hiu 2 , 2 2/5/20153Mt S Dng Tn Hiu Quan Trng1. Xung vung gc H() 2. Hm dc ()212110) ( ) ( t t x H == [ =21t , 121,0) ( ) (tt t xc0) ( . ) (bc ta t xH =ba) ( . ) (bc ta t xH ==0 , 00 ,) (tt tt r= a ta t ta t r, 0,) (0ar(t-a)K0r(t)1Mt S Dng Tn Hiu Quan Trng3. Bc nhy n v ()10 t> 10 0 1 0

0 +1 1 = () 0 = 1 0, 0 10 , 0 > 11-1

0 = () = 1 , > 00 , = 01 , < 02/5/20155Mt S Dng Tn Hiu Quan Trng7. Hm m suy gim 8. Hm m tng dn = . . > 0

0 ()

0 () = 1 . > 0Mt S Dng Tn Hiu Quan Trng9. Xung hm m 10. Xung hm Cos = . .

2

> 0

0 ()

0 ()

20 20 = . cos(0) .

/02/5/20156Mt S Dng Tn Hiu Quan Trng11.Cp phn b o() chn l|()|||() 0

12

12 12 0

12 12 12|| = 12 +12+ 12= 12 +12 12

12Mt S Dng Tn Hiu Quan Trng12.Phn b lc Tnh cht phn b lc Hm chn ||| = ||| Tun hon ||| + = ||| Thay i thang |||

= = Ri rc ||| = =() Lp tun hon ||| = = ||| = = .... ....5 4 3 2 1 0 1 2 3 4 5 ||| 2/5/20157Mt S Dng Tn Hiu Quan Trng13.Dy xung vung lngcc14.Dy xung vung ncc... ... 20

2 ... ...2 0 2

Mt S Dng Tn Hiu Quan Trng15.Tn hiu Sin suy gimtheo hm m16.Tn hiu Gausse = . . sin 0 . > 0 = 22/5/20158Mt S Dng Tn Hiu Quan Trng17.Tn hiu () Tn hiu ()()11 32 43 12 4

= sin

, 01 , = 0 = sin

, 01 , = 0Mt S Dng Tn Hiu Quan Trng18.Tn hiu 2 Tn hiu 2

2()1

2 1 2

2 = sin2

2, 01 , = 0

2 = sin2

2, 01 , = 02/5/20159Tnh Cht Hm Xung Lc1. Quan h vi hm ()

0

= = 2. Thay i thang

0= 0 3. Tnh i xng = 4. Nhn i s = 0 0 = 0 05. Nhn tch phn

= 0

0 = 06. Nhn tch chp = = = 0 = ( 0)3.2 KHI NIM HM TUN HON 2/5/201510Khi Nim Hm Tun Hon = + Mt tn hiu tun hon c thuc tnh l khng i khidch thi gian mt on . Chu k c bn: l gi tr nh dng nh nht mhm cn tun hon. Hm M Phc Tun Hon =

Vi l thc v = 0l s thun o. Bin v pha c v (biu din) tch bit.2/5/2015113.3 CHUI FOURIER KHAI TRIN FOURIERChui Fourier Chui lng gic:

02 + =1

cos +

sin

0

; 1

: h s (l cc hng s); = 1 Chui lng gic m rng:

02 + =1

cos

+

sin

0

; 1

: h s (l cc hng s); = 2

=

2/5/201512nh L Dirichlet Nu l hm tun hon, b chn v c mt sim xc nh khng lin tc trong mt chu k can th khi chui FOURIER ca s hi t n ti tt c nhng im m lin tc. Cn tinhng im m khng lin tc, chuiFOURIER ca n s hi t n gi tr trung bnh cagiihn tri v giihn phica tc nu tiim = 0hm s b gin on th:

= 0 =lim0 + lim0+ 2= 0 + 0+2

02 + =1

cos

+

sin

=

1 = = +2

1 < <

2 =

2 <

3 < < +22/5/201513Khai Trin Fourier Cho hm s f(x) tun hon vi chu k T = 2p tha iukin Dirichlet. Khi hm f(x) c th biu din didng chui Fouier theo cng thc sau: = 02 + =1

cos

+

sin

;

0 = 1

2

= 1

2 cos

= 1

2 sin

, 0,

,

Cng Thc Euler M Rnga. 2cos

= 0; 0b. 2sin

= 0c. 2cos

cos

= 0; d.2cos2

= e. 2cos

sin

= 0f.2sin

sin

= 0; g.2sin2

= ; 02/5/201514VD: = 0, < 0,1, 0 < , Tm khai trin Fourier. Bit () = ( +2) = 2 = 2 = ; 0 = 1

2 = 1

0

=1;

= 1

0

cos = 0

= 1

0

sin

= cos()

0

= 1

1 cos = 2

; 0; = 12+ 2

sin 1+sin 33+sin 55+VD: = , < 0,, 0 < , Tm khai trin Fourier. Bit () = ( +2)Nhn xt: f(x) tha mn nh lut Dirichlet (tun hon, b chn vc mt s im xc nh khng lin tc trong mt chu k ca n), nn c th khai trin Fourier. = 2 = 2 = ;

0 = 1

0 +0

= 1

0+

220

= 322/5/201515VD: = , < 0,, 0 < ,

= 1

0 cos +0

cos = 1

+ = 0 cos =

sin 0= 0 = 0

cos = 0

= 0

=sin

0

0

sin

= cos

20

= 1

2 cos 1 = 1 1

2= 2

2; 0;

= 1 1

2VD: = , < 0,, 0 < ,

= 1

0 sin +0

sin = 1

+ = 0 sin =

cos 0= 1 cos

= 0

sin = 0

= 0

=cos

0

0

cos

= cos

+ sin

20

= cos

= 1

2/5/201516VD: = , < 0,, 0 < ,khai trin Fourier

0 = 32 ;

= 1 1

2;

= 1

= 12

0 + =1

cos

+

sin

= 1232 + =11 1

2cos 1

sin = 34 + 2

112cos 132cos 3 152cos 5 + sin 1+sin 22+sin 33+Chui Fourier Chn Lnh l 1 Nu f(t) l hm tun honchn theo t th

= 0 v

0 = 1

2 =2

= 1

2 cos

= 2

cos

nh l 2 Nu f(t) l hm tun honl theo t th

= 0 v

= 1

2 sin

= 2

sin

2/5/201517VD: = 0, < 0,, 0 < , Tm khai trin Fourier. Bit () = () = Hm chn

= 0; v

0 = 2

0

= 2

220

=

= 2

0

cos = 2

cos

20

= 2

1 1

2= 4

2; 0; VD: = 0, < 0,, 0 < , = ; 0 =

= 4

2; 0; ;

= 0Khai trin Fourier = 12

0 + =1

cos = 2 4

cos 12+cos 332+cos 552+2/5/201518VD: = 0, < 0,, 0 < , Tm khai trin Fourier. Bit = = Hm l

= 0; v

= 2

0

sin = 2

cos

= 2

cos = 2

1 ==1

sin =2 sin 1sin 22+sin 33Cc Dng Chuyn i Ca KhaiTrin Fourier =

02 +=1

cos

+

sin

=

02 +

2 +

2

=1

2 +

2cos

+

2 +

2sin

t

0 = 02

=

2 +

2

= tan1

= 2

2/5/201519Cc Dng Chuyn i Ca KhaiTrin Fourier =

02 +

2 +

2

=1

2 +

2cos

+

2 +

2sin

=0+

=1cos

cos

+cos

sin

(1)=0+

=1sin

cos

+sin

sin

(2)1 = 0 +

=1cos

: 2 = 0 +

=1sin

+

: Chui Fourier Dng M Phc = 02 + =1

cos

+

sin

= 02 + =1

+

2+

2= 02 + =1

2

+

+

2

t C0 = a02 ; Cn =

2 Cn =

+

2 =0+=1

+

=0+=1

+ =1

= =

2/5/201520Chui Fourier Dng M Phc =0+=1

+ =1

= =

= 12

2

= 12

2

0 = 12

2 VD: = , < /,, / , Tm chui Fourier dng m phc. -/2 /2 2 -2/5/201521VD: = , < /,, / , = ;

= 12

2

= 12

2

2

= 12

2

2= 1

2

22= sin

2

=12 sin

2

2=12sinc 2

0 = 12

2 =12VD: = 0, < /2,1, /2 , = =

= 0 + =1

+ =1

= 0 + =1

+= 12+ =1sinc 2 cos 2/5/201522WD: Phn Tch PhTn Hiu Tun HonPhng php phn tch ph Tm Cn Tnh F(e) =2tCn Khi : th biu din bin ca F(e) theo tn sgi l ph bin ca tn hiu f(t). Biu din gc pha ca F(e) theo cc tn s cachng gi l ph pha ca tn hiu f(t)V DCho hm s () tun hon vi chu k T nhhnh bn di. V ph ca ()2/5/201523V D

= 12

2

= 1

201 2

+ 1

0

2

2

= 1

12

2

20+12

2

0

2=12 1 +

+ 1= 122 2

+2=

1 cos V D Tm F(e): = 2

= 2

1 cos = 2

1 cos = 0 = 24

= 2 +12/5/201524V D Ph bin Ph pha- - - --e e 2e 3e 4e 5e -2e -3e -4e -5e44/33/4e.. ..|F(e)|- - - --ee 2e 3e 4e 5e-2e -3e -4e -5et/2e.. ..argF(e)-t/2+ ===1 2k nn42k n0) F(< + => + ===0 1 2k nn40 1 2k n22k n0) ) F( arg(t