ch3_01_chuoi_f.pdf
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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