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 hiu2/5/201523.4 TCH PHN FOURIER BIN I FOURIERBin i Fourier Cp bin i Fourier dng phc: = 12

= ()

K hiu:

= 1 = () 1()2/5/20153VD: =

= ; <

; Tm bin i Fourier. = = =

2

2

=

/2/2= /2 /2= 2

sin 2VD: = = ; <

; = 2

sin 2

= 2

sin 2= 2

sin 2

2

2= 2

2 sin 2

2= Sinc 2= Sa 2Nhc li, Sinc = sin

S = sin

2/5/20154Tnh Cht Ca Bin i FOURIER Tnh tuyn tnh Tnh cht i xng Thay i t l thi gian Php dch min thi gian Php dch min tn s Vi phn min thi gian Tch phn min thi gian Vi phn min tn s nh l nhn chp tn s nh l moun Parseval nh l nhn chp trongmin thi gian nh l iu ch2/5/20155Tnh Cht Ca Bin i FOURIER

()1) Tuyn tnh

1

1()

2

2()

1

1 +2

2

1

1() +2

2()2) Thay i t l thi gian 1

3) Tnh cht i xng

2()4) Dch min thi gian 0

0

Tnh Cht Ca Bin i FOURIER

()5) Dch min tn s

0 06) Tch phn min thi gian

+ 0 7) Vi phn min thi gian

8) Vi phn min tn s

2/5/20156VD: = =

; < ; Tm bin i Fourier. = = =

= =

+ 2

2

2

sin 2

+ 2

+

2

2

sin 2;

2

2

2

sin 2; = =

2

sin 2

2

2

= 2

2

sin 2

2

2

2= 2

2

sin 2sin 2VD: = =

; < ; 2/5/20157VD: = =

; < ; = 2

2

sin 2sin 2 =

22

sin2 2

=

= 1

22

sin2 2= sin2 2

22= Sa2 2VD: = =

; < ; = 2

2

sin 2sin 2 =

2

cos 1

=

= 1

2

cos 1=

2

2 cos 12/5/20158Tch Chp nh ngha =

1 . 2 = 1 2 = 2 1 nh l Parseval = =

=

2 =12

2 1) Tch chp min thi gian

1 2 12

1 2 2) Tch chp min tn s

1 2

1 2 nh l iu ch

() Tnh cht dch tn s

0

0

0 +0 nh l iu ch cos 0 12 0 + +0 sin 0 12 0 +02/5/201593.5 PHN TCH PH TN HIUWD: Phn Tch Ph Tn Hiu Phng php phn tch ph Tnh X(): = ()

Khi : th biu din bin ca F() theo tn s gi lph bin ca tn hiu f(t). Biu din gc pha ca F() theo cc tn s cachng gi l ph pha ca tn hiu f(t)2/5/201510VD: = . ; > Tm ph tn hiu. = = 0

+ = 1 + + 0=1 + = Ph bin : =1

2 +2 Ph pha: =

Ht chng 3