ly thuyet tin hieu

8/6/2019 Ly Thuyet Tin Hieu

1/46

Chng 1: Mt s khi nim cn bn

1. Tn hiu Tin tc H thng2. Phn lai tn hiu

3. Biu din gii tch tn hiu

1. Tn hiu- Tin tc- H thng

Tn hiu l biu hin vt l ca tin tc m n mang t

ngun tin n ni nhn tin.M hnh l thuyt: hm theo thi gian x(t)

Tin tc l nhng ni dung cn truyn i qua hnh

nh, ting ni, s liu o lng

H thng l nhng thit b hay thut tan, thc

hin nhng tc ng theo mt qui tc no ln tn

hiu to r a mt tn hiu khc

HT

[K]

Tn hiung vo

Tn hiung ra

[K] biu th cho thut tan x l

2. Phn loi

2.1. Tn hiu xc nh v tn hiu ngu nhin

2.2. Tn hiu lin tc v r i rc

2.3. Tn hiu nng lng Tn hiu cng sut

2.4. Cc phn loi khc

2.1.Tn hiu xc nh v tn hiu ngu nhin

Tn hiu xc nh l tn hiu m qu trnh thi gian ca tnhiu c biu din bng mt hm thc hay phc.V d: ( ) 220 2 cos(2 .50 )( )u t t V

......0.01

( )u t 2220

t0.01

x(t)

t

Tn hiu ngu nhin(THNN): l tn hiu m qu trnh thigian ca n khng an trc c. V d: ting ni, hnhnh, m nhc u khng c biu din tan hc. nghincu THNN ta phi tin hnh quan st thng k tm ra quilut phn b ca n.


2/46

2.2. Tn hiu lin tc v r i rc

)(tx

t

Tn hiu tng t (bin ,thi gian lin tc)

)(tx

t

Tn hiu lng t (bin rirc, thi gian lin tc)

)(tx

t

Tn hiu ri rc (bin lintc, thi gian ri rc)

)(tx

t

Tn hiu s (bin , thi gianri rc)

2.3. Tn hiu nng lng TH cng sut

Tn hiu nng lng hu hn gm cc tn hiu c thi hn

hu hn, cc tn hiu qu xc nh v ngu nhin.

Tn hiu cng sut trung bnh hu hn gm cc tn hiu

tun han, tn hiu c thi hn v hn c gi tr tin n

hng s khc khng khi t dn ra v cng

2.4. Cc phn lai khc

Da vo b rng ph ca tn hiu c th phn lai tnhiu nh sau: tn hiu (TH) tn s thp, TH tn s cao,TH di rng, TH di hp.

Da vo bin ca TH c th phn lai thnh TH cbin hu hn, TH c bin v hn.

Da vo bin thi gian ca TH c th phn lai thnh

TH c thi hn hu hn, TH c thi hn v hn.

Tn hiu nhn qu: l tn hiu c gi tr bng khng khit


3/46

3.1. Biu din ri rc3.1.1 Tn hiu trc giao

1 2 1 2. *,x t x t x t x t dt

Tch v hng gia hai tn hiu c nh ngha

Nu tch v hng ny bng khng th ta ni hai tnhiu trc giao

Tn hiu trc chun

)()(2

)(1

txttx

1))(),(( ttx

Nu

v Tn hiu chun ha

21

2121

1

0),(

xx

xxxx

3.1.2 Biu din tn hiu bng chui hm tr c giao

N

n

nn ttx

1

)()(

ni

nini

1

0,

Tp hm c chn, thng l tp hm trc chun, tc l: )(t

n H s khai trin chui c xc nh theo phng trnh

, 1

( ( ), ( )) ( , )N

n i n n i n

x t t

),( ii x Khi

3.1.3 Mt s v d v biu din ri rc

a. Chui Fourier lng gicb. Chui Fourier phc

a. Chui Fourier lng gic

...2,1);2

sin(2

);2

cos(2

;1

)( ntTnTtTnTTtn

Chui Fourier lng gic c to bi tp hm trc chun l tp hm iu ha sau:

T: chu k tn hiu

1

0 )2

sin(2

)2

cos(21

)(

n

nn tT

nT

tT

nTT

tx

Tn hiu x(t) c th biu din bng chui Fourier

nn ,,0

T

dttx

TT

xx

0

00 )(11

,,

dttT

ntxT

tT

nT

x

T

n )2

cos()(2

)2

cos(2

,

0

dttT

ntxT

tT

nT

x

T

n )2

sin()(2

)2

sin(2

,

0

Trong cc h s khai trin c xc nh nh sau:


4/46

a. Chui Fourier lng gic

n

nn

a

barctg

tdtxT

aT

00 )(1

tdtntxT

aT

n 0 0 )(cos)(2

dttntxTb

T

n

00 )(sin)(

2

22nnn bac

T

20

a0, an, bn, cn: h s khai trin chuiFourier.

tn s c bn ca tn hiu

T: chu k ca tn hiu

0 )( ) ( cos sin0 0

1

x t a a n t b n t n n

n

(1)

10

cos0

)(n

ntnncatx (2)

a. Chui Fourier lng gic- V d

2/2/

X

t......

T-T

x(t)

02

Xa

2, 1,5,9...

2sin

2 2

, 3,7,11...

n

Xn

X n na

n X

nn

0nb

1

20

1

2( ) 1 cos

2

n

nn odd

X Xx t n t

n

1

22

1 ,n

n

Xa n odd

n

2T

a. Chui Fourier lng gic- V d

tnn

tt

tttA

0

00

000

cos1

...9cos9

17cos

7

1

5cos5

13cos

3

1cos

4

A

T

t

T

20

Sng vung

n=1n=3

n=1

n=5

n=41

t

b. Chui Fourier phc

...2,1,0;1

)(

2

neTt

t

T

jn

n

Tp hm iu ha phc trc chun c chn:

T: chu k tn hiu

n

tT

jn

n eT

tx

21

)(

2 2

0

1 1, ( )

T jn t jn t

T Tn x e x t e dt

T T

Chui Fourier phc tng ng

n

tjnneXtx

0)(

T

20

Hay:

T

tjnn dtetx

TX

0

0)(1

(3)

nn XC 200 X

2

nnn

jbaX

Chui (1), (2), (3) c quan h vi nhau nh sau:


5/46

a. Chui Fourier phc - V d

2/2/

X

t......

T-T

x(t)

0

2

2

1sin

2

jn t

n

X nX Xe dt

T n

0( ) sin cos

2n

X nx t n t

n

3.2. Biu din lin tc TH3.2.1 Dng tng qut

dtsttxsX ),()()(

dstssXtx ),()()(

)()( sXtx

),( st

),( ts

Bin i thun

Bin i ngc

c gi l nhn bin i

c gi l nhn lin hp

3.2.2 Mt s v d v php bin i lin tc

Bin i Fourier

( ) ( ) ( )j tX F x t x t e dt

( ) ( )x t X

1 1( ) ( )

2

j tx t F X X e d

Bin i Laplace

0( ) ( ) ( )st

X s L x t x t e dt

1

1( ) t 0

2( ) ( )0 t


6/46

Bi tp

1. Tm chui Fourier lng gic v chui Fourier phc cc tn hiu sau

......

2

4sin2( ) tx t

2

2

( )x t

2

4

( )x t

4

Bi tp

2. Tm X() ca cc tn hiu sau:

2. ( ) ta x t e

1 1 0

. ( ) 1 0 1

0 1

t t

b x t t t

t

1 1 0. ( ) 1 0 1

0 1

tc x t t

t

3. Tm x(t) bit cc X() nh sau:

. ( )a X e

2. ( ) 2

0 2

b X


7/46

Chng II: TN HIU XC NH

1. Cc thng s c trng ca tn hiu

2. Tn hiu xc nh thc3. Tn hiu xc nh phc

4. Phn tch tn hiu ra cc thnh phn

5. Phn tch tng quan tn hiu

6. Phn tch ph tn hiu

7. Truyn tn hiu qua mch tuyn tnh

Cc thng s c trng ca tn hiu


1.1 Tch phn tn hiu1.2 Tr trung bnh ca tn hiu

1.3 Nng lng ca tn hiu

1.4 Cng sut trung bnh ca tn hiu

1.1 Tch phn tn hiu

Cho x(t) l tn hiu xc nh, tch phn tn hiu c nhngha nh sau:

2

1

( )

t

t

x x t dt

Vi x(t) tn ti trong khang thi gian hu hn (t1- t2):

( )x x t dt

Vi x(t) tn ti v hn : ,

1.2 Tr trung bnh ca tn hiu2

1

2 1

( )

t

t

x t dt

xt t

Vi tn hiu c thi hn hu hn:

1lim ( )

2

T

T

T

x x t dt T

Vi tn hiu c thi hn v hn:

Vi tn hiu tun han:

0

1( )

T

x x t dt T


8/46

1.3 Nng lng ca tn hiu Ex

2

1

2 2( )

t

x

t

E x x t dt Vi tn hiu c thi hn hu hn:


2( )xE x t dt

tn hiu x l tn hiu nng lng

0 xE Nu

1.4 Cng sut trung bnh ca tn hiu

2

1

2

2 1

( )

t

t

x

x t dt

Pt t

Vi tn hiu c thi hn hu hn:

21lim ( )2

T

xT

T

P x t dt T


Vi tn hiu tun han:2

0

1 ( )T

xP x t dt T

tn hiu x l tn hiu cng sut0 xP Nu








Tn hiu xc nh thc

2. Tn hiu xc nh thc

2.1 Tn hiu nng lng2.2 Tn hiu cng sut

2.3 Tn hiu phn b


9/46

2.1.1 Tn hiu nng lng c thi hn hu hn

2.1.2 Tn hiu nng lng c thi hn v hn

2.1 Tn hiu nng lng

2.1 Tn hiu nng lng c thi hn hu hn

a. Xung vung gc t

21

21

)(tx

b

a

)(tx

0 1/ 21

( ) 1/ 22

1 1/ 2

t

x t t t

t

( )t c

x t ab

1/ 2

1/ 2

1x dt

1/ 2

1/ 2

1xE dt

x ab

2Ex a b

1 1( )0 1

t tx t tt

11

)(tx

0 1

2 2

1 0

(1 ) (1 ) 2 / 3xE t dt t dt

0 1

1 0

(1 ) (1 ) 1x t dt t dt

Tt 0

)(tx

Tt 0 0

t

0( )t t

x t AT

b. Xung tam gic t

2.1 Tn hiu nng lng c thi hn hu hn (tt)

2( ) >0t

Tt

x t Xe T

0 (1 )

T

t TX

x Xe dt e


T0

)(tx

c. Xung hm m

22 2 2

0

(1 )2

T

t T

x

XE X e dt e


10/46

d. Xung cosin

0

0

( ) cost

x t X t

0

0

2

0

0

2

2cos

Xx X tdt

2

02XEx

o

2

)(tx

o

2


2.2 Tn hiu nng lng c thi hn v hn

a. Hm m suy gim

0( ) >0

0 0

tXe tx t

t

0

t Xx Xe dt

T0

)(tx

22 2

02

t

x

XE X e dt

2.2 Tn hiu nng lng c thi hn v hn (tt)

b. Tn hiu sin suy gim theo hm m

0sin 0

( )0 0

tXe t t x t

t

02 20

x X

0

0

)(tx

0

2

0

2 2

2 2 2

04x

XE

0

00

sin 0( )

1 0

t ttx t Sa t

t

0x

c. Tn hiu Sa

0

xE

tx

0

0

2

0

3

0

0

2

0

3



11/46

d. Tn hiu Sa20t

2 0

22

0 0

sin t 0( )

1 t = 0

tx t Sa t t

0

x

0

2

3xE

tx

0

0

2

0

3

0

0

2

0

3


2.2.1 Tn hiu CS khng tun han2.2.2 Tn hiu tun han

2.2 Tn hiu cng sut

2.3 Tn hiu cng sut khng tun han

a. Bc nhy n v 1(t)

0

1 1lim

2 2

T

Tx dt

1 t > 0

( ) 1( ) 1/2 t = 0

0 t < 0

x t t

1

2xP

0

)(tx

0

0( ) .1x t X t t

0t

0

)(tnz

2

1)(1 tZ

)(2 tZ

ntZn

),(

11

2

1 1 1( )

2 2 2

10

2

n

tn

z t nt t n n

tn

b. Hm m tng dn

0

)(tx

( ) 1 1( )tx t X e t

0

1lim (1 ) ;2 2

T

t

T

Xx X e dt T

2

2x

XP

2.3 Tn hiu cng sut khng tun han (tt)

1 t 0( ) > 0

0 t < 0

tX ex t


12/46

1 t > 0

( ) ( ) 0 0

1 t < 0

x t Sgn t t

b. Tn hiu Sgn(t)

0

)(tx

0

2 2

0

1lim ( 1) (1) 12

T

xT

TP dt dt T

0

0

1lim ( 1) (1) 0

2

T

TT

x dt dt T

2.3 Tn hiu cng sut khng tun han (tt)

2.4 Tn hiu tun han

a. Tn hiu iu ha

x(t)

q

X

T

t

tX 0cos

tX 0cos

2

2XPx

0x

2.4 Tn hiu tun han (tt)

x(t)

X

T

t

pha = 0

pha =/4

b. Dy xung vung gc lng cc

0x

2

xP X

2/2/

c. Tn hiu xung vung gc n cc/ 2

/ 2

1;

Xx Xdt

T T

/ 2 2

2

/ 2

1;x

XP X dt

T T

2.3.1 Phn b (t)2.3.2 Phn b lc

2.3 Tn hiu phn b


13/46

2.4 Tn hiu phn b

a. Phn b (t)

)(t

0 0 va t 1

0 -tt dtt

00

va 10 0

-0

t tt t t t dt

t t

Tnh cht

(1) a t dt a t dt a

1( )

(2) ' ' 1( ); ( )d t

t dt t t dt

0 0 0

(3) ( ) (0)

( ) ( ) ( ) ( )

x t t x t

x t t t x t t t

0 0(4) ( ) (0); ( ) ( ) ( )x t t dt x x t t t x t

2.4 Tn hiu phn b

0

0

(5)t

t tt

(6) t t

0 0

(7) ( )

( ) ( ) ( )

x t t x t

x t t t x t t

0 0(4) ( ) (0); ( ) ( ) ( )x t t dt x x t t t x t

2.4 Tn hiu phn b

2.4 Tn hiu phn b

b. Phn b lc |||(t)

Tt

T|||1

|||n

t t n

1 |||n

t t nTT T


14/46

Tnh cht

(1) Tnh cht ri rc

(2) Tnh cht lp tun han

1( ). ||| ( ) ( )

n n

t x t x t t nT x nT t nT

T T

1( ) ||| ( ) ( )n n

t x t x t t nT x t nT T T

2.4 Tn hiu phn b



2. Tn hiu xc nh thc

3. Tn hiu xc nh phc





Tn hiu xc nh phc

3. Tn hiu xc nh phc

Re ( ) Im ( )x t x t j x t

Nng lng ca tn hiu phc:

2

( )x E x t dt

2

1

2

2 1

( )

t

t

x

x t dt

Pt t

21lim ( )

2

T

x TT

P x t dt T

Cng sut trung bnh:

2

0

1( )

T

xP x t dt T



2. Tn hiu xc nh thc

3. Tn hiu xc nh phc





Phn tch tn hiu ra cc thnh phn


15/46


4.1 Thnh phn thc, o

4.2 Thnh phn chn v l

4.3 Thnh phn xoay chiu v mt chiu

4.1 Thnh phn thc, o

Re ( ) Im ( );x t x t j x t

Re ( ) Im ( );x t x t j x t

1

Re [ ( ) ( )]

2

x t x t x t

1

Im [ ( ) ( )]2

x t x t x t j

Re ( ) Im ( ) ;x x t j x t

Re ( ) Im ( ) ;x x t j x t

2

Re Im( ) x x x E x t dt E E

Re Im x x xP P P

4.1 Thnh phn chn, l

( ) ( );ch lx t x t x t

( )ch ch x t x t ( )l l x t x t

1

( ) [ ( )]2

chx t x t x t

1( ) [ ( )]2

lx t x t x t

0

( )chx t

0

( )l

x t

0lx 0lx

x xch xl E E E

x xch xlP P P 1

t0

( )x t

V d: Thnh phn chn v l cax(t) = e-t1(t)

4.1 Thnh phn mt chiu, xoay chiu

( );x t x x t

0x 0x

x x x E E E

x x xP P P

x x

Trong :

:thnh phn mt chiu

:thnh phn xoay chiux

V d: Thnh phn mt chiu v xoay chiu ca TH x(t) :

x ( )x t


16/46








Phn tch tng quan tn hiu


5.1 H s tng quan5.2 Hm tng quan

5.1 H s tng quan

H s tng quan gia hai tn hiu c nh ngha nh sau:

2

( ) ( ) ,

,( )

xy

x t y t dt x y

x x x t dt

2

( ) ( ) ,

,( )

yx

y t x t dt y x

y yy t dt

H s tng quan chun ha

, ,

, , xy yx

x y y x

x x y y

0 1 0

1

khi x v y trc giaokhi x = y

5.2.1 HTQ tn hiu nng lng

5.2.2 HTQ tn hiu cng sut

5.2 Hm tng quan


17/46

( ) ( ) ( ) ( )xy x t y t dt x t y t

( ) ( ) ( ) ( )yx y t x t dt y t x t

Hm tng quan

Hm t tng quan

( ) ( )x x t x t dt

5.2.1 Hm tng quan tn hiu nng lng

5.2.1 Hm tng quan tn hiu nng lng (tt)

Tnh cht:

2

(3) 0 ( )x xx t dt E

(4) 0

(1)

xy xy

xy xy vi tn hiu thc

(2) x x x x vi tn hiu thc

Hm t tng quan ca tn hiu thc l hm chn

Nng lng ca tn hiu = gi tr HTTQ khi = 0

V d 1: Tm hm tng quan ca hai tn hiu sau:

21

21

)(ty

0

)(1)( ttXetx

)(tx

*Xt 1 1

2 2

1/ 2

0

t

xy Xe dt

+1/2-1/2


1/ 21

Xe

*Xt1

2

1/ 2

1/ 2

1/ 2 1/ 2

t

xy Xe dt

Xe e

*Xt1

2

0xy

5.2.1 Hm tng quan tn hiu nng lng

)(tx

+1/2-1/2

)(tx

-1/2 +1/2


18/46

1/21 1/ 2 1/ 2

1/2 1/2 1/ 2

0 1/ 2

Xe

X e exy

1/21 1/ 2 1/ 2

1/2 1/2 1/ 2

0 1/ 2

Xe

X e eyx yx


TC (1)

V d 2: Tm hm t tng quan ca tn hiu xung vung

2T

2T

)(tx

Khi 0 T

2T

2T

)(tx

/ 2

2 2

/ 2

T

xT X dt X T

+T/2-T/2


2T

2T

)(tx

+T/2-T/2

Khi T

0x

V x(t) l tn hiu thc nn HTTQ ca n l hm chn (TC2) nn

T

Khi 0T 2

x X T

0x


Kt qa ta c HTTQ ca xung vung

2 khi 00 khi

x

X T T

T

TT

)(xTX2

Nh vy HTTQ ca xung vung l xung tam gic

2x X TT



19/46

V d : Tm hm t tng quan ca tn hiu sau


X

t

)(tx

T0

2x X TT

5.2.2 Hm tng quan THCS khng tun han

1lim ( ) ( )2

T

xyT

T

x t y t dt T

Hm tng quan

Hm t tng quan

1

lim ( ) ( )2

T

yxT

T

y t x t dt T

1

lim ( ) ( )2

T

xT

T

x t x t dt T

V d 1: Tm hm t tng quan ca x(t) = X1(t)

0

)(tx

0

)(tx

0

2

21lim

2 2

T

xT

XX dt

T

0

22

0

1lim

2 2

T

x T

XX dt

T

2

2x

X

0

)(tx

5.2.2 Hm tng quan THCS khng tun han (tt)

5.2.2 Hm tng quan THCS khng tun han (tt)

V d 2: Tm hm tng quan ca x(t) = X1(t) v y(t) = sgn(t)

)(tx 0

)(ty

0

0

)(tx0

0

1lim

2 2

T

xT

X Xdt Xdt

T

ta cng c kt qa tng t

2

Xx


20/46

5.2.2 Hm tng quan tn hiu tun han

0

1( ) ( )

T

xy x t y t dt T

0

1( ) ( )

T

yx y t x t dt T

0

1( ) ( )

T

x x t x t dt T

Tnh cht

2

(3) 0x xx P

(4) 0

(1) ; xy xy xy xy (i vi TH thc)

(2) ;x x x x (i vi TH thc)

5.2.2 Hm tng quan tn hiu tun han (tt)


21/46



2. Tn hiu xc nh thc

3. Tn hiu xc nh phc





Phn tch ph tn hiu


6.1 Ph ca tn hiu nng lng

6.2 Ph ca tn hiu cng sut

6.3 Mt ph nng lng, mt ph cng sut



6.1.1 nh ngha6.1.2 Cc tnh cht ca ph6.1.3 Ph ca mt s tn hiu thng gp

Ph ca tn hiu nng lng c xc nh bi bin ithun Fourier. Bin i Fourier l mt cng c tan c

nh ngha l mt cp bin i thun ngc nh sau:

6.1.1 nh ngha

( ) ( ) ( ). j t X F x t x t e dt

11

( ) ( ) ( ).2

j t x t F X X e d

x(t) v gi l cp bin i Fourier( )X

( ) ( ) x t X K hiu


22/46

( ) ( ) j X X e P jQ

c im ( )X

trong trng hp tng qut l mt hm phc( )X

( ) , , , X P Q

ph pha, ph thc, ph o.

c tn gi tng ng l ph bin

2 2( ) X P Q

( )Q

arctgP

6.1.2 Cc tnh cht ca ph

. ( ) . ( ) . ( ) . ( )a x t b y t a X bY

1. Nu x(t) l tn hiu thc th P(),|X()| l hm chntheo , Q(),() l hm l theo

3. Tnh cht tuyn tnh

( ) ( )

( ) ( )

( ) ( )

( ) ( )

x t X

x t X

x t X

x t X

2.

( )t

x a X aa

4. Tnh cht i xng

( ) ( ) x t X

5. Tnh cht ng dng

6. Tnh cht dch chuyn trong min thi gian

00( )j t

x t t X e

6.1.2 Cc tnh cht ca ph

00( )j t

x t t X e

( ) 2 X t x

7. Tnh cht dch chuyn trong min tn s (iu ch)

6.1.2 Cc tnh cht ca ph (tt)

0 0( )j t

x t e X

0 0 01

( ) cos2

x t t X X

0 0( )j t

x t e X

0 0 01

( ) sin2

x t t X X j


23/46


9. Vi phn trong min thi gian( )

( ) . ( )n

n

n

d x tj X

dt

( )

( ) 1, 2,3...n

n n

n

d X j t x t n

d

8. Vi phn trong min tn s

( )1: ( )

dXn tx t j

d

22

2

( )2 : ( )

d Xn t x t

d

11. Tch chp trong min thi gian

( ) ( ) ( ) ( )x t y t X Y

12. Tch chp trong min tn s

1( ). ( ) ( ) ( )2

x t y t X Y


10. Tch phn trong min thi gian1

( ) ( )

t

x d X j


13. Ph ca hm tng quan v t tng quan

( ) ( ) ( ) ( ) ( )xy x t y t dt x t y t

Theo nh ngha tac

( ) ( ) ( )xy

F X Y

i vihmt tng quan x(t) = y(t)

2

( ) ( ) ( )x

F X mt ph nng lng

14. nh l Parseval

1( ) ( ) ( ) ( )2

x t y t dt X Y d

Khi x(t) = y(t) 2 21( ) ( )2

x x t dt X d E

l Parseval cho ta mt s lin h gia nng lng cxc nh trong min thi gian v min tn s


2 21( ) ( )

2 x t dt X d


24/46

6.1.3 Ph mt s tn hiu thng gp

( ) 1( ) ( >0)tx t e t

0

( )x t

( )X

( )

1

2

2

11( )te tj

2 21X

1tan

1( )Xj

2 22te

( )x t ( )X 2

( )

tx t e

2 2

2X

6.1.3 Ph mt s tn hiu thng gp (tt)

2T

2T

)(tx

t tx T

( )X

2

T

4

T

2

T

4

T

2

t TTSaT

2T X TSa



0

( ) t x t Sa

0

( )X

0

0

p dng tnh chti xng ta c:

2

TSaTt T

tx

0

0

2

0

3

0

0

2

0

3

0

002

tSa

0 02 Sa t 02

2


25/46


( ) tx t

T1

tTT

)(tx X

2

T

3

T

4

T

2

T

3

T

4

T

( ) 2

x

TSaT

TT

t Tx t

p dng tnh cht ph ca hm ttng quan ta c:

2

2

TF T TSa

T

22Tt TSa

T

20

( ) t x t Sa

tx

0

0

2

0

3

0

0

2

0

3

( )X

02

02

0

2 000

2tSa


2 2/ 2( ) tex t

( )x t1

t

( )X 2

2 2 2 22/ 2 / 22te e





6.3 Mt ph nng lng, mt ph cng sut



26/46


6.2.1 Ph ca tn hiu cng sut khng tun han

6.2.2 Ph ca tn hiu tun han

6.2.1 Ph ca tn hiu cng sut khng tunhan

Cc tn hiu cng sut khng c ph Fourier thng thng. tm ph ca tn hiu cng sut khng tun han, ta c th biudin n bi gii hn ca mt dy tn hiu nng lng.

0

( ) lim ( )x

x t x t

Mi phn t c ph Fourier( )x t

0

( ) lim ( )X X

( ) X F x t

Ph Fourier gii hn

Tn hiu CS x(t) c biu din qua dy tn hiu nng lng sau:

Nu tn ti gii hn ca dy ph th ta s c ph ca

tn hiu x(t):

( )X

a. Tn hiu cng sut khng tun han (tt)

( ) tx t t X

2 2/ 2

20

1lim

2

tt e

2 2 2 2/ 2 / 2

21

2

te e

2 2 / 2

0lim 1X e

Chn dy hm gn ng ca (t) l dy hm Gausse

Cc phn t ca dy c nh Fourier l:

Ph ca (t):

1t


( ) 1x t x t X

2

(tnh ch

t

i x

ng)

( ) gn( )x t S t 0

)(tx

( )X

0

lim sgn( )t x t t e

0

2 20

21 1t j t t j t

j X e e dt e e dt

2 20

2 2lim

jX

j

1 2

2gn( )

jS t


27/46


( ) 1( ) x t t

0

)(tx ( )X

1 1

1 sgn( )2 2

t t

p dng kt qa ca hai v dtrn ta c: 1X

j

1 11cos 1( )0 0( ) ( )0 2 0 0

t tj j

00 00 2

02 2

sin 1( )tj j

t

(p dngnh liu ch cho tn hiu 1(t)

11( )t

j

0 00 2

0

.1( )2 2

cos t t j

6.2.2 Ph ca tn hiu tun han

tm ph ca tn hiu tun han ta biu din chng didng chui Fourier.

0( )jn t

n

n

x t X e

0

0

1( )

T

jn t

n X x t e dt

T

0

2, 0; 1; 2...n

T

0 02 ( )

jn t e nTa c:

Ph Fourier gii hn ca tn hiu tun han

Tn hiu TH x(t) c biu din thnh chui Fourier phc sau:

02 ( )nn

X X n (1)

6.2.2 Ph ca tn hiu tun han (tt)

Cc tn hiu tun han c bit:0( ) cos x t t

0 0( )X

0( ) sin x t t

0 0( ) X j j

0( )j t

x t e

0( ) 2X

(p dng tnhchtiu ch)


V d 1: Ph ca dy xung vung gc n cc

/2/2

5T

Ta c h s khai trin Fourier

0 0

/ 2 / 2

0

/ 2 / 2

1( )

2

T

jn t jn t

n

T

nA A X x t e dt e dt Sa

T T T

02n

A n X Sa n

T T

X

2

T

2

4

2

T

2

4

2T

A

0

2

T


28/46


V d 2: Ph ca phn b lc1|||

t

T T

0/ 2

/ 2

1 1T

jn t

n

T

X t e dt T T

01

2n

X nT

......

0

1

T

0

X

02

02

0


Nhn xt:Gi xT(t) = x(t)(t/T) l phn trung tm ca tn hiu tunhan x(t). THTH x(t) s c biu din bi tch chp caxT(t) v phn b lc.

1( ) ( ) |||T

tx t x t

T T

Vi xT(t) l THNL thi hn hu hn (-T/2,T/2) s c phFourier l XT() = F[xT(t)]v

01 1

||| 2 n

t

nT T T


Theo tnh cht v ph ca tch chp ta c:

01 1||| .2T Tn

t x t X nT T T

Hay

0 02T

n

X nX n

T

(2)

T (1), (2) 0Tn

X nX T


Tnh cht:

( )

( )

( )

n

n

n

x t X

x t X

x t X

2.

( ) n x t X ( ) n y t Y

n n

n n

X X

1.

3. . ( ) . ( ) . .n na x t b y t a X b Y

4. ( ) ; a R(-0)

n

t

x a X a

005. ( )jn t

n x t t X e

06. ( ) jn t n m x t e X

7. ( ) n n x y t X Y


29/46

8. i n ii

x t y t X Y

2x

9. ( )

( )

n n

n

x y t X Y

x x t X

2 2 22

x 0

1

10. ( )

P ( ) 2

n n

n

n n

n n

x t y t X Y

x t X X X


6.3 Mt ph nng ln g Mt phcng sut

6.3.1 Mt ph nng lng

6.3.2 Mt ph cng suta. Tn hiu cng sut khng tun han

b. Tn hiu tun han

6.3.1 Mt ph nng lng

6.3.1 Mt ph nng lng

Mt ph nng lng ca tn hiu nng lng l ilng

2

XTheo tnh cht ca ph(tc 13) ta c:

2

x X

Nh vy v ( l cp bin i Fourier

j

x e d

1

2

jx e d

Vi tn hiu thc, HTTQ chn, do mt ph nnglng cng l hm chn theo .

6.3.1 Mt ph nng lng (tt)

Nh vy nng lng ca TH c th c xc nh theo 3 cchsau:

Khi thay = 0 vo HTTQ ta c:

10

2x xd E

Nng lng ca TH c xc

nh trong min tn s

(1) Tnh trc tip t tch phn bnh phng tn hiu Ex = [x2].

(2) Tnh t hm t tng quan Ex= (0).

0

1 1

2xE d ( khi chn)

(3) Tnh t mt ph nng lng


30/46

Nng lng mt di tn = 2- 1

1 2 2

2 1 1

1 1 1

2 2x xE d d E d

( khi chn)



V d: Tm mt ph nng lng v nng lng ca tnhiu x(t) = e-t1(t) (>0)Ta c:

1Xj

2 21

1 1

2F e

1

2xE

Nng lng tn hiu trong di tn :

3,

3

2 2

3

3

1 1 1 112 6

x xE d E


Mt ph nng lng tng h:

j xy xy xy F e d

1 1

2

j xy xy xy F e d

yx yx F

1

yx yx F

Tng t:

xy yx

Bi v HTQ c tnh cht nn yxxy

6.3 Mt ph nng ln g Mt phcng sut

6.3.1 Mt ph nng lng


b. Tn hiu tun han

6.3.2 Mt ph cng sut

6 3 2 Mt ph cng sut T hi t kh t h


31/46


Ta c HTTQ ca THCS x(t):

/ 2

/ 2

1

lim

T

TT x t x t dt T

/ 2 / 2

/ 2 / 2

1lim

T Tj

TT T

F x t x t dt e d T

/ 2

/ 2

1lim

Tj t

TTT

e dT

1

lim TT T

Ph Fourier gii hn

/ 2 / 2

/ 2 / 2

1lim

T Tj

TT T

x t x t dt e d T

Nh vy HTTQ v mt ph CS l cp bin i Fouriergii hn

trong T() l mt ph nng lng ca tn hiuxT(t) = x(t)(t/T) tc x(t) c xt trong khang thigian T

a. Tn hiu cng sut khng tun han

lim TT T

v


Cng sut ca TH

/ 2

2

/ 2

1lim ( )

T

xT

T

P x t dt T

1

2xP d

Tn hiu xT(t) c nng lng :

/ 2

2

/ 2

1( )

2T

T

x T

T

E x t dt d

Cng sut ca x(t) c xc nh theo biu thc sau:

1 1

lim2

TTd

T

1 1 1

lim

2 2

TTd d

T

Nh vy CS ca tn hiu c t h c xc nh theo cc cchsau:

(1) Tnh trc tip t tr trung bnh bnh phng tn hiu Px =.

(2) Tnh t hm t tng quan Px= (0).(3) Tnh t mt ph cng sut

0

1 1

2xP d d ( khi chn)

1 2 2

2 1 1

1 1 1

2 2xP d d d


b T hi t h


32/46

b. Tn hiu tun han

Theo tnh cht ca ph ta c:

2x nXNh vy, mt ph cng sut ca THTH:

2

0 02 2 x n nn n

X n n

2

n nX l h s khai trin Fourier ca HTTQ

Mt ph cng sut ca THTH l ph ca HTTQ

Cng sut c xc nh t mt ph cng sut :

2 2

0

1

2 x n n

n n

P d X n d X

x nn

P

01

2x n

n

P

Vi tn hiu thc, ph bin l hm chn, do

b. Tn hiu tun han (tt)

Chng II: TN HIU XC NH V d:


33/46



2. Tn hiu xc nh thc

3. Tn hiu xc nh phc




7. Truyn tn hiu qua mch tuyn tnhTruyn tn hiu qua mch tuyn tnh


k(t)K()

x(t)

X()y(t)

Y()

jK F k t K e

*y t k t x t Y K X

Y K X

arg argY X

Quan h gia cc c trng ca tn hiu u vov ra ca h thng tuyn tnh

V d:

Cho tn hiu x(t) = Sa2(2t) qua mch lc nh hnh c png k(t) = Sa2t. Xc nh tn hiu y(t) ng ra.

k(t)x(t) y(t)

Ta c:

Y K X

2 2

2 4 2 4 8 4 8 2

Y

2

8

2

4

Y

2( ) 2 28

y t Sa t Sa t

Hm tng quan v t tng quan ca tn hiunng lng

Mt ph nng lng tng h v mt phnng lng

Quan h gia cc c trng khc


34/46

Mt ph nng lng tng h v mt ph nng lng 7. Truyn tn hiu qua mch tuyn tnh


35/46

Bit rng : xx xx

xy xy

yx yx

yy yy

k K k K

2

yy xx

yy xx

k k

K

xy xx xy xxk K

xxyx yx xx

k K

Nh vy vi tn hiu nng lng ta c mi quan h sau:

2

yy xx

yy xx

k k

K

V c th suy ra cc kt qu tng t i vi tn hiu

cng sut


Vi tn hiu cng sut khng tun han

2

yy xx

yy xx

k k

K

y q y

Vi tn hiu tun han

2

0 0 0

yy xx

yy xx

k k

n K n n

0, 1, 2, .......n

Chng IV: TN HIU IU CH 1.1 S h thng thng tin


36/46

Chng IV: TN HIU IU CH

1. Mt s khi nim c bn

2. Cc h thng iu ch lin tc

3. Ri rc tn hiu

4. iu ch xung

5. Phn knh theo tn s v thi gian

Mt s khi nim c bn

1. 1 S h thng thng tin

1. 2 Mc ch iu ch

1.3 Phn lai iu ch


V d:

- in thai- Truyn hnh- Pht thanh- V tinh

H thng truyn tin tc t ngun n ninhn tin

S h thng thng tin

B bin iMy pht Knh truyn My thu

Ngun tin

ng vo

B bin i

ng ra

Nhn tin

Ngun tin: tng t, sV d: Ting ni, m nhc, hnh nh.

B bin i ng vo: Chuyn tin tc thnh tn hiuphhp cho cc hthng thng tin.

V d: Ting ni Microphone in pMy pht: Khuch i, iu ch

V d: i truyn hnh, i pht thanh, web server

My thu: Gii iu ch, khuch i, lc nhiuV d: TV, radio,

Knh truyn : Mi trng trung gian thc hin vic truyn dn.V d: khng gian, dy dn, cp ng trc, cp quang

1.2 Mc ch iu ch 1.3 Phn loi iu ch


37/46

Chuyn ph ca tn hiu t tn s thplntn scao v bin i thnh dng sng in t lan

truyn trong khng gian

Cho php s dng hu hiu knh truyn

To ra cc tn hiu c k h nng chng nhiu cao

Tn s tn hiu

Cc h thng iu ch

Lin tcXung

Bin Gc Tng t S

AM-SC AM SSB VSBSSB-SC PM FM PCMDeltaPAMPDM PPMAM-SC AM PM FM PAM




3. Ri rc tn hiu

4. iu ch xung


Cc h thng iu ch lin tc

2. Cc h thng iu ch lin tc 2. 2 Tn hiu iu bin


38/46

g

2.1 Sng mang iu ha

2.2 iu ch bin

2.3 iu ch gc

2. 1 Sng mang iu ha

0( ) cos y t Y t trong: Y bin , tn s l hng s

(t) = t + 0gc pha tc thi

Nu tn hiu tin tc x(t) tc ng lm thay i bin casng mang ta c tn hiu iu bin

0( ) ( ) cosy t Y t t Y(t) ng bao bin , l hm ca thi gian bin thin theo quy lut

ca TH x(t).Nu tn hiu tin tc x(t) tc ng lm thay i tn s hocgc pha ca sng mang ta c tn hiu iu ch gc

( ) cos y t Y t

iu bin hai di bn (DSB Double Side band)

iu bin trit sng mang (AM-SC Amplitude

Modulation with Suppressed Carrier)iu bin (AM Amplitude Modulation)

iu bin mt di bn (SSB Single Side band)

iu bin mt di bn trit sng mang (SSB-SC Single Side band with suppressed Carrier)

iu bin mt di bn (SSB Single Side band)iu bin trit mt phn di bn (VSB Vestigal Sideband)

2. 2.1 Tn hiu AM SC

Gi s tn hiu CS x(t) c b rng ph trong khang

(min- max) c c trng bi mt ph CS x()TH x(t) tc ng lm thay i bin ca sng mang tac tn hiu AM-SC nh sau:

( ) ( ) cos AM SC y t x t t

trong: Y(t) = x(t)

0= 0

2. 2.1 Tn hiu AM SC 2. 2.1 Tn hiu AM SC


39/46

tm mt ph CS y() ca tn hiu iu ch AM-SCta xt n trong khang thi gian T hu hn.

( ) ( )cosT T y t x t t Trong : xT(t) = x(t)(t/T) l tn hiu nng lng c phFourier thng thng XT(). Vy yT(t) = xT(t)cost cng ltn hiu nng lng, ph ca n c xc nh theo nhl iu ch

1( ) 2T T TY X X

Mt ph nng lng ca yT(t)

2 2

1( )4

1

4 T T

T T T T

T T

Y X X

X X X X

2. 2.1 Tn hiu AM SC

2 21

4T T TX X

T>> nen X 0m T Do X

Mt ph cng sut ca tn hiu AM-SC theo nh ngha

2 2

1( ) lim lim4

T Ty

T T

X X

T T

1

( )4

y x x

Do 2

( ) limT

xT

XT

Cng sut ca TH AM-SC:

1

2

1 1

2 4

1 1 12 2 2

y y

x x

x x

P d

d

d P

1

2

y xP P

2. 2.1 Tn hiu AM SC

V d Gii iu ch


40/46

t

)(tx

t

)(ty SCAM y

04

1

x0

maxminminmax

1 1

( ) ( ) ( ) cos 22 2

gT T T y t x t x t t

2

( ) cos ( ) cosg AM SC y t y t t x t t

1 1( ) ( )cos 22 2

x t x t t

2 2 21 1

2 24 16

gT T T T X X X

1 1

( ) 2 22 4

gT T T T Y X X X

Gii iu ch

1 12 2

4 16 yg x x x

yg

2

04

1

2

Tn hiu x(t) c th nhn c sau khi lc b cc thnhphn tn hiu c a o tn nh mch lc thng thp

2.2.2 Tn hiu AM

Tn hiu AM c dng :

( ) ( )cos cosAMy t x t t A t

( ) ( ) cosAMy t A x t t trong: Y(t) = A+x(t)0= 0

Lm tng t nh tn hiu AM-SC ta c:

2

( )21

+4

y

x x

A

V d

Gii iu ch tn hiu AM


41/46

t

)(tx

x

0

maxmin

min

max

y

04

1

yc(t)

t

A

A

cy

2

2A

A

A

ty AM

RC tyAM

tx

tA cos

L

( )x t cos t

( )AM

y t

cosA t

S khi to tn hiu AM v mch thc hin

21 2i a u a u

2.2.2 Tn hiu AM

Tn hiu AM c gii iu ch trong mch tch snghnh bao nh sau:

R C tyAM tuc

t

)(tyAM

t

)(tuc

qu iu ch

A

A

ty AMuc(t)

t

Nu ng bao bin c gi tr m:

Nh vy A c chn sao cho ng bao ca TH AM lY(t) = x(t) +A khng m.

iu n y s tha m n nu:

max ( ) : ( ) 0A x t x t

2.2.2 Tn hiu AM

2.2.2 Tn hiu AM 2. Cc h thng iu ch lin tc


42/46

H s hiu sut nng lng : % 100%by

PkP

Pb: Cng sut trung bnh cc di bnPy: Cng sut ca TH AM

AM-SC : % 100%kb yP P

AM : 1

2b xP P

21 12 2

y xP A P

2

% 100%x

x

Pk

A P

B rng ph ca cc TH DSB : max2 AM SC AM B B

2.2.2 Tn hiu AM

V d vi x(t) = acos0t. Tn hiu AM c dng: 0 0( ) cos cos 1 cos cosAMy t A a t t A m t t

0 01

( ) cos cos cos2

AM y t A t mA t t

2

2

222

1

2% 100% 100%1 2

2

mAm

km

A mA

m = a/A: su iu ch ( ) 0 1mVi m = 1 ta c kmax= 33.33% hiu sut nng lng caTH AM khng cao.

2.1 Sng mang iu ha

2.2 iu ch bin

2.3 iu ch gc

2.3 iu ch gc

2.3.1 Tn hiu iu ch gc

2.3.2 Tn hiu iu pha PM

2.3.3 Tn hiu iu tn FM


2. 3.1 Tn hiu iu ch gc 2. 3.1 Tn hiu iu ch gc


43/46

( ) cos y t Y t

Tn hiu iu pha PM (Phase Modulation) txktt

pPM 0

dt

tdxk

dt

tdt pPM

Tn hiu iu tn FM (Frequency Modulation)

tn s sng mang0 gc pha ban ukp hng s t l

dttxktt pFM 0

txktfFM

Tn hiu tin tc c gnvotn s (pha) ca sng mang

max

txkpPM

lch pha v tn s: max

tt

max tPM:

maxdt

tdxkpPM

FM: max

dttxkfFM

maxtxkfFM

nu Tn hiu PM di hp

1max

txkpPM

nu Tn hiu FM di hp 1

max dttxkfFM

2. 3.1 Tn hiu iu ch gc

Quan h gia PM v FM

dttxktt pFM 0

txktt pPM 0

Mchtch

phn

CPM

dttx tx tyFM

Mchvi

phnCFM

dttdx tx tyPM

Sng mangTn hiu

Tn hiu iu ch

2. 3.1 Tn hiu iu ch gc

2.3 iu ch gc Tn hiu PM di hp


44/46






[ ]PM p y t YCos t k x t

Tn hiu PM di hp: 1max txkpPM

( )1p

j k x t j t j t

PM p Z t Ye e Y jk x t e

( )

1p jk x t

pe jk x t 1PMDo nn c th chp nhn

RePM PM p y t Z t YCos t Yk x t Sin t

B rng ph BPM = 2wm

xxp

PMYkY4

2

22

mm

x PM

m m

m

m

m2

RePM PM p y t Z t YCos t Yk x t Sin t

Tn hiu PM di rng (iu ch mc cao):(Rt kh phn tch vi tn hiu x(t) tng qut)

Xt x(t) = Xsinwmt. Ta c: ]sin[ tXktYCosty mpPM )( Xk

p ttj

PMmYetZ

sin)(

tnJYtZtyn

mnPMPM

cos)(Re

c th c khai trin thnh chui Fourier phc nh ngthc Bessel

tj mesin

n

tjn

n

tj mm eJe

sin

Tn hiu PM di rng

Hm Bessell Tn hiu PM di rng


45/46

Jo J1 J2 J3 J4 J5 J6 . . .0 1

0.5 .94 .24 .03

1 .77 .44 .11 .02

2.4 0.0 .52 .43 .20 .06 .02

5.5 0.0 -.34 -.12 .26 .40 .32 .19 . . .

Hm Bessell

0.94 cos 0.24 cos 0.24 cos+ 0.03 cos 2 0.03 cos 2

PM m m

m m

y t Y t Y t Y t Y t Y t

0.5 Vi ta c J0 = 0.94; J1 = 0.24; J2 = 0.03

mm

x

PM

m m

PM

m m

BPM

B rng ph c tnh gn ng theo cng thc Carson

)0.5()1(2 PMmPMPMB

)01(2 PMmPMPM

B

Tn hiu PM di rng

1Vi th b rng ph ca TH PM khng xc nh

2 ( 0.5 )PM m PM

B PM di hp

2.3 iu ch gc

Gii iu ch


46/46



2.3.3 Tn hiu iu tn FM2.3.3 Tn hiu iu tn FM

Vi x(t)=cos(wmt)

][ dttxktYCosty fFM

+m

NBFMB2m

-m

WBFMB2m

.25Y2J2n()

+m-m


BFM

p ng tn s camch cng hng




3. Ri rc tn hiu

4. iu ch xung


Ri rc tn hiu

ly thuyet tin hieu

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