MOÂN HOÏC: VI BA SOÁ Lôùp Kyõ Thuaät Vieãn Thoâng

8/14/2019 MON HOC: VI BA SO Lp Ky Thuat Vien Thong

1/81

12/03/091DH GTVT BM KTVT

MON HOC: VI BA SOLp Ky thuat vien

thong

Chng 2:

X LY TN HIEU BANG GOC


2/81

12/03/092ThS.Vo Trng Sn

NOI DUNG CHNG 2

ieu che so la g? Cac loai ieu che so:

ASK

FSK PSK QAM

Chuyen oi ma

Ngau nhien hoa Ma hoa vi sai Tao khung vo tuyen


3/81


ieu che so

e co the truyen dan cac thong tin so bang song ien t,can phai tien hanh ieu che so.

ieu che so la ky thuat gan thong tin so vao dao ong hnhsine (song mang), lam cho song mang co the mang thong tin cantruyen i.

Ta cung co the hieu: ieu che so la s dung thong tin sotac ong le cac thong so cua song mang, lam cho cac thongso cua song mang bien thien theo quy luat cua thong tin.


4/81


ieu che so


5/81


ieu che so

Song mang hnh sine co dang:

x(t) =Acos(2fct + )

Co ba thong so cua song mang co the mang

tin:la bien o (A), tan so (fc) va goc pha (). Do o, ta co the tac ong len mot trong 3

thong so cua song mang e co cac phngphap ieu che tng ng.

Ngoai ra, ta cung co the tac ong len motluc 2 thong so cua song mang e co phngphap ieu che ket hp.


6/81


Co cac phng phap ieu che sau: Amplitude-shift keying (ASK): ieu che khoa dch

bien o. Frequency-shift keying (FSK) : ieu che khoa dch

tan so. Phase-shift keying (PSK) : ieu che khoa dch pha. Quadrature Amplitude Modulation (QAM): ieu che

bien o cau phng. ay la phng phap kethp gia ASK va PSK.

Cac phng phap ieuche so


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Cac phng phap ieuche so


8/8112/03/098ThS.Vo Trng Sn

Bit rate is the number of bits per

second.Baud rate is the number ofsignal units per second.Baud rate is

less than or equal to the bit rate.

Note:Note:


9/8112/03/099ThS.Vo Trng Sn

Example 1Example 1

An analog signal carries 4 bits in each signal unit. If 1000

signal units are sent per second, find the baud rate and the

bit rate


10/8112/03/0910ThS.Vo Trng Sn

Example 1Example 1

An analog signal carries 4 bits in each signal unit. If 1000

signal units are sent per second, find the baud rate and the

bit rate

SolutionSolution

Baud rate = 1000 bauds per second (baud/s)Baud rate = 1000 bauds per second (baud/s)

Bit rate = 1000 x 4 = 4000 bpsBit rate = 1000 x 4 = 4000 bps


11/8112/03/0911ThS.Vo Trng Sn

Example 2Example 2

The bit rate of a signal is 3000. If each signal unit carries

6 bits, what is the baud rate?


12/8112/03/0912ThS.Vo Trng Sn

Example 2Example 2

The bit rate of a signal is 3000. If each signal unit carries

6 bits, what is the baud rate?

SolutionSolution

Baud rate = 3000 / 6 = 500 baud/sBaud rate = 3000 / 6 = 500 baud/s


13/8112/03/0913ThS.Vo Trng Sn

ieu che ASK (2 ASK)

Mc thap nhat la ASK hai mc (2 ASK) Bit 1 nh phan c bieu dien bang mot

song mang co bien o la hang so. Bit 0 nh phan: khong xuat hien song mang.

Vi tn hieu song mang laAcos(2fct)

( )

=ts( )tfA c2cos

0

1bit

0bit


14/8112/03/0914ThS.Vo Trng Sn

Dang tn hieu 2-ASK

Tn hieu ASK hai mc

Ta co the tao ra c 4ASK, 16 ASK Tuynhien cac loai ieu che nay co kha nang

chong nhieu kem.


15/8112/03/0915ThS.Vo Trng Sn

Figure 5.3 ASK


16/8112/03/0916ThS.Vo Trng Sn

Figure 5.4 Relationship between baud rate and bandwidth in ASK


17/8112/03/0917ThS.Vo Trng Sn

Example 3Example 3

Find the minimum bandwidth for an ASK signaltransmitting at 2000 bps. The transmission mode is half-

duplex.


18/8112/03/0918ThS.Vo Trng Sn

Example 3Example 3

Find the minimum bandwidth for an ASK signaltransmitting at 2000 bps. The transmission mode is half-

duplex.

SolutionSolution

In ASK the baud rate and bit rate are the same. The baud

rate is therefore 2000. An ASK signal requires aminimum bandwidth equal to its baud rate. Therefore,

the minimum bandwidth is 2000 Hz.


19/8112/03/0919ThS.Vo Trng Sn

Example 4Example 4

Given a bandwidth of 5000 Hz for an ASK signal, whatare the baud rate and bit rate?


20/8112/03/0920ThS.Vo Trng Sn

Example 4Example 4

Given a bandwidth of 5000 Hz for an ASK signal, whatare the baud rate and bit rate?

SolutionSolutionIn ASK the baud rate is the same as the bandwidth,

which means the baud rate is 5000. But because the baud

rate and the bit rate are also the same for ASK, the bit

rate is 5000 bps.


21/8112/03/0921ThS.Vo Trng Sn

Example 5Example 5

Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz),draw the full-duplex ASK diagram of the system. Find

the carriers and the bandwidths in each direction. Assume

there is no gap between the bands in the two directions.


22/8112/03/0922ThS.Vo Trng Sn

Example 5Example 5

Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz),draw the full-duplex ASK diagram of the system. Find

the carriers and the bandwidths in each direction. Assume

there is no gap between the bands in the two directions.

SolutionSolution

For full-duplex ASK, the bandwidth for each direction is

BW = 10000 / 2 = 5000 Hz

The carrier frequencies can be chosen at the middle of

each band (see Fig. 5.5).

fc (forward) = 1000 + 5000/2 = 3500 Hz

fc (backward) = 11000 5000/2 = 8500 Hz


23/8112/03/0923ThS.Vo Trng Sn

Figure 5.5 Solution to Example 5


24/8112/03/0924ThS.Vo Trng Sn

ieu che FSK (FSK hai mc)

Mc thap nhat la FSK hai mc (2 FSK, BFSK)

Ca hai bit nh phan 0 va 1 c bieu dien hai tanso song mang khac nhau:

( )

=ts( )tfA

12cos

( )tfA2

2cos

1bit

0bit


25/8112/03/0925ThS.Vo Trng Sn

Dang tn hieu 2-FSK

Tn hieu FSK hai mc


26/8112/03/0926ThS.Vo Trng Sn

Figure 5.6 FSK


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Figure 5.7 Relationship between baud rate and bandwidth in FSK


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Example 6Example 6

Find the minimum bandwidth for an FSK signal

transmitting at 2000 bps. Transmission is in half-duplex

mode, and the carriers are separated by 3000 Hz.

SolutionSolutionFor FSK

BW = baud rate + fBW = baud rate + fc1c1 ffc0c0

BW = bit rate + fc1BW = bit rate + fc1 fc0 = 2000 + 3000 = 5000 Hzfc0 = 2000 + 3000 = 5000 Hz


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Example 7Example 7

Find the maximum bit rates for an FSK signal if the

bandwidth of the medium is 12,000 Hz and the difference

between the two carriers is 2000 Hz. Transmission is in

full-duplex mode.

SolutionSolution

Because the transmission is full duplex, only 6000 Hz is

allocated for each direction.

BW = baud rate + fc1BW = baud rate + fc1 fc0fc0Baud rate = BWBaud rate = BW (fc1(fc1 fc0 ) = 6000fc0 ) = 6000 2000 = 40002000 = 4000

But because the baud rate is the same as the bit rate, the

bit rate is 4000 bps.


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MFSK FSK M mc

S dung M tan so song mang Hieu qua hn ve s dung bang thong

nhng loi nhieu hn.

fi= f

c+ (2i 1 M)f

d

fc= Tan so song mang

fd= o di tan so

M = So trang thai ieu che = 2 L

L = So bt trong moi trang thai ieu che

( ) tfAtsii

2cos=Mi

1


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Multiple Frequency-Shift Keying(MFSK)

e tng thch vi toc o d lieu cualuong bit vao, moi mot phan t tn hieuau ra co khoang thi gian la:

Ts=LTseconds

Trong o T chu ky bit (toc o d lieu = 1/T)

V vay, moi phan t tn hieu ieu chemang lng tin la L bits


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Multiple Frequency-Shift Keying(MFSK)


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PSK

PSK hai mc (BPSK) S dung hai goc pha bieu dien cho 2 bit nh phan

( )=ts ( )tfA c2cos( ) +tfA c2cos

1bit0bit

=

( )tfA c2cos

( )tfA c2cos1bit

0bit


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Tn hieu PSK 2 mc

Tn hieu PSK hai mc


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2PSK

Cac v trpha

0 1

2PSK2PSK ra

y(t)

NRZ vao

b(t)

Songmang

c(t)

S o nguyenly

Figure 5 8 PSK


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Figure 5.8 PSK

Figure 5 9 PSK constellation


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Figure 5.9 PSK constellation


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Phase-Shift Keying (PSK)

PSK 4 mc (QPSK) Moi trang thai song mang mang thong tin 2 bit

( )

=ts

+ 42cos

tfA c 11

+

4

32cos

tfA c

432cos tfA c

42cos

tfA c

01

00

10

l


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4PSK S o nguyen lyieu che

4PSK ra

y(t)

NRZ vao

b(t)

c(t)

2PSK

2PSK

S

P 90o

I

Q

b1(t)

b2(t)


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4PSK - Cac trang thai phase

1

0

I

Q

10

11

10

00

01


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4PSK S o nguyen ly giaiieu che

LPF

So pha

LPF

9004 PSKvao

Bo chiacongsuat

Luong bit ra

b(t)

c(t)

Q

I

Figure 5.10 The 4-PSK method


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Figure 5.10 The 4 PSK method

Figure 5.11 The 4-PSK characteristics


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Figure 5.11 The 4 PSK characteristics

Figure 5.12 The 8-PSK characteristics


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Figure 5.12 The 8 PSK characteristics

Figure 5.13 Relationship between baud rate and bandwidth in PSK


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g p


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Example 8Example 8

Find the bandwidth for a 2-PSK signal transmitting at2000 bps. Transmission is in half-duplex mode.


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Example 8Example 8

Find the bandwidth for a 2-PSK signal transmitting at2000 bps. Transmission is in half-duplex mode.

SolutionSolution

For 2-PSK the baud rate is the same as the bandwidth,

which means the baud rate is 2000.


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Example 9Example 9

Given a bandwidth of 5000 Hz for an 8-PSK signal, whatare the baud rate and bit rate?


50/81


Example 9Example 9

Given a bandwidth of 5000 Hz for an 8-PSK signal, whatare the baud rate and bit rate?

SolutionSolutionFor 2PSK the baud rate is the same as the bandwidth,

which means the baud rate is 5000. But in 8-PSK the bit

rate is 3 times the baud rate, so the bit rate is 15,000 bps.


51/81


Phase-Shift Keying (PSK)

Multilevel PSK Using multiple phase angles with each angle having more

than one amplitude, multiple signals elements can beachieved

D = modulation rate, baud R = data rate, bps M = number of different signal elements = 2L

L= number of bits per signal element

M

R

L

R

D2log==


52/81


Performance

Bandwidth of modulated signal (BT)

ASK, PSK BT=(1+r)R

FSK BT=2DF+(1+r)R

R = bit rate 0 < r < 1; related to how signal is filtered

DF = f2-f

c=f

c-f

1


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Performance

Bandwidth of modulated signal (BT)

MPSK

MFSK

L = number of bits encoded per signal element

M = number of different signal elements

RM

rR

L

rBT

+=

+=

2log

11

( ) RM

MrBT

+=

2log

1

Bang thong cua tn hieu


54/81


Bang thong cua tn hieuieu che

V du: Tm bang thong cua tn hieu ieuche PSK 2 mc, 4 mc, 8 mc, 16 mc cho1 luong E1, E3, E4.


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So sanh ba loai ieu che


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QAM

QAM la s ket hp cua ASK va PSK Ve ban chat ay la 2 tn hieu c phat i

tren cung mot tan so song mang.

( ) ( ) ( ) tftdtftdts cc 2sin2cos 21 +=


57/81


Quadrature amplitude modulation is a

combination ofASKandPSKso that amaximum contrast between each

signal unit (bit, dibit, tribit, and so on)

is achieved.

Note:Note:

Cac trang thai cua 4 QAM & 8


58/81


Cac trang thai cua 4-QAM & 8QAM


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Dang tn hieu 8-QAM


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S DUNGPHO BIENNHAT

16-QAM

16QAM S o nguyen ly


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16QAM S o nguyen lyieu che

16QAM ra

y(t)

NRZ vao

b(t)

c(t)

4PAM

4PAM

S

P 90o

I

Q

b1(t)

b4(t)

b2(t)

b3(t)

16QAM - Cac trang thai


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16QAM - Cac trang thaiphase

10

00

I

Q

1000

1010

01 11

01

11 1011

100

1

1000

Figure 5.17 Bit and baud


63/81


Table 5 1 Bit and baud rate comparison


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Table 5.1 Bit and baud rate comparison

ModulationModulation UnitsUnits Bits/BaudBits/Baud Baud rateBaud rate Bit Rate

ASK, FSK, 2-PSKASK, FSK, 2-PSK Bit 1 N N

4-PSK, 4-QAM4-PSK, 4-QAM Dibit 2 N 2N

8-PSK, 8-QAM8-PSK, 8-QAM Tribit 3 N 3N

16-QAM16-QAM Quadbit 4 N 4N

32-QAM32-QAM Pentabit 5 N 5N

64-QAM64-QAM Hexabit 6 N 6N

128-QAM128-QAM Septabit 7 N 7N

256-QAM256-QAM Octabit 8 N 8N


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Example 10Example 10


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A constellation diagram consists of eight equally spacedpoints on a circle. If the bit rate is 4800 bps, what is the

baud rate?

SolutionSolutionThe constellation indicates 8-PSK with the points 45

degrees apart. Since 23 = 8, 3 bits are transmitted with

each signal unit. Therefore, the baud rate is4800 / 3 = 1600 baud



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Compute the bit rate for a 1000-baud 16-QAM signal.



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Compute the bit rate for a 1000-baud 16-QAM signal.

SolutionSolution

A 16-QAM signal has 4 bits per signal unit sincelog216 = 4.

Thus,

(1000)(4) = 4000 bps



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Compute the baud rate for a 72,000-bps 64-QAM signal.



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Compute the baud rate for a 72,000-bps 64-QAM signal.

SolutionSolution

A 64-QAM signal has 6 bits per signal unit sincelog2 64 = 6.

Thus,

72000 / 6 = 12,000 baud

Ch i


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Chuyen oi ma

Tx

Rx

Tx

Rx

M

OD

M

UX

D

E

MO

D

D

E

MU

X

D

E

M

O

D

D

E

M

U

X

M

O

D

M

U

X

HDB3/CMI

NRZ

HDB3 t NRZ


72/81


HDB3 to NRZ

Tre T

NRZ

X1

X2

f

Y1

+L1

-L2

Y2

+V

OP-Amp 1

OP-Amp 2

a

b

c

R1

R2

OR1 OR2

d

eHDB-3

RZ

CMI t NRZ


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CMI to NRZ

Tre bit

D Q

C Q

RNZ

CLK

CMIa

b

D

EX-OR

N hi h (S bl )


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Ngau nhien hoa (Scrambler)

Trong truyen dan vi ba so, khoi giai ieuche co c s ong bo ong ho vikhoi ieu che nh vao viec khoi phuong ho t tn hieu thu c.

Neu tn hieu truyen dan co tnh ngaunhien th pho nang lng cua tn hieu taptrung tai tan so bang toc o bit (Rb) do

o ong ho khoi phuc c la Rb.

N hi h (S bl )


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Ngau nhien hoa (Scrambler)

Neu tn hieu truyen dan khong co tnh ngaunhien, v du tn hieu: 1010101010101010101010101010

110110110110110110110110110110

11001100110011001100110011001100

th pho nang lng cua tn hieu tap trung tai tanso lap lai cua tng nhom bit (Rb/2, Rb/3, Rb/4) doo ong ho khoi phuc c la Rb/2, Rb/3, Rb/4.

V vay au thu khong co c ong bo vi au

phat. Ngau nhien hoa nham muc ch bien oi luong

tn hieu trc khi phat i mang tnh ngau nhien.

N l hi h


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aD lieu vao

D lieu nh phan2

EX-ORb Moi

trngtruyen

dan

EX-ORb

c cy

D lieura

Mach ma hoa Mach giai mahoa

Nguyen ly ngau nhien hoa

a c b c y

0 0 0 0 0

0 1 1 1 0

1 0 1 0 1

1 1 0 1 1

T t hi hi


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Tao tn hieu ngau nhien

Cac bo tao tn hieu ngau nhien (xem tailieu, trang 228,229, 230)

M h i i


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Ma hoa vi sai

c s dung e phuc vu cho ieu che visai.

Xem tai lieu, trang 231, 232, 233, 234, 235.

Tao kh n o t en


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Tao khung vo tuyen

He thong truyen dan vo tuyen can cothem cac loai thong tin khac nhau ephuc vu cho giam sat, quan ly va baodng he thong (goi la cac bit nghiep

vu). Cac thong tin nay c ghep vao luong

so trc ieu che va lay ra sau khi giaiieu che.

e co the ghep va tach cac bt nghiepvu chnh xac can phai tao mot cau truckhung mi, c goi la khung vo tuyen.

Cau truc khung vo tuyen


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Cau truc khung vo tuyen

Mot khung vo tuyen gom co cac thanhphan sau: Cac bit d lieu cua luong so can truyen Cac bit nghiep vu

Cac bit kiem tra loi Cac bit ong bo khung.

Moi he thong truyen dan khac nhau nh

ngha cau truc khung vo tuyen khac nhau.


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MOÂN HOÏC: VI BA SOÁ Lôùp Kyõ Thuaät Vieãn Thoâng

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