Thong Tin So

7/27/2019 Thong Tin So

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B CNG THNGTRNG I HC CNG NGHIP H NI

B mn in t Vin thng-Khoa in t

------------------------

BI GiNG MN:

1GV: B Quc Bo BM TVT - H CNHN

TH NG TIN S

Ging vin: Th.s. B Quc Bo

H ni 2007


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Ch-ng 7:Ch-ng 7: iu ch tn hiu siu ch tn hiu s

2


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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


,

thong tin can truyen i.

Ta cung co the hieu: ieu che so la s dung thong tin so

tac ong len cac thong so cua song mang, lam cho cac

thong so cua song mang bien thien theo quy luat cua

thong tin.


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ieu che so



5/60

ieu che so

Song mang hnh sine co dang:

x(t) = A cos(2fct + )

Co ba thong so cua song mang co the mang tin:la


, c .

Do o, ta co the tac ong len mot trong 3 thong socua song mang e co cac phng phap ieu chetng ng.

Ngoai ra, ta cung co the tac ong len mot luc 2 thongso cua song mang e co phng phap ieu che kethp.


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Co cac phng phap ieu che sau:

Amplitude-shift keying (ASK): ieu che khoa dch bien o.

Frequency-shift keying (FSK) : ieu che khoa

Cac phng phap ieu che so


dch tan so.

Phase-shift keying (PSK) : ieu che khoa dchpha.

Quadrature Amplitude Modulation (QAM): ieuche bien o cau phng. ay la phng phap kethp gia ASK va PSK.


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Cac phng phap ieu che so



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Bit rate is the number of bits per

second.Baud rate is the number of

Note:Note:


signal units per second.Baud rate is

less than or equal to the bit rate.


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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


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


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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


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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 la Acos(2fct)

ts tfA c2cos

0

1bit

0bit


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Dang tn hieu 2-ASK


Tn hieu ASK hai mc

Ta co the tao ra c 4ASK, 16 ASK Tuy nhien cacloai ieu che nay co kha nang chong nhieu kem.


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Figure 5.3 ASK



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Figure 5.4 Relationship between baud rate and bandwidth in ASK



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Example 3Example 3

Find the minimum bandwidth for an ASK signal

transmitting at 2000 bps. The transmission mode is half-

duplex.


o u ono u on

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.


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Example 4Example 4

Given a bandwidth of 5000 Hz for an ASK signal, what

are the baud rate and bit rate?

SolutionSolution


In 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.


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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.


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

BW = 10000 / 2 = 5000 Hz

The carrier frequencies can be chosen at the middle ofeach band (see Fig. 5.5).

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

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


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Figure 5.5 Solution to Example 5



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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 tan sosong mang khac nhau:


ts

tfA 12cos

tfA 22cos 1bit

0bit


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Dang tn hieu 2-FSK


Tn hieu FSK hai mc


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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.


SolutionSolution

For FSK

BW = baud rate + fBW = baud rate + fc1c1

ffc0c0BW = 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.


o u ono u on

Because the transmission is full duplex, only 6000 Hz is

allocated for each direction.

BW = baud rate + fc1BW = baud rate + fc1 fc0fc0

Baud 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 nhieuhn.

tfAts ii 2c o s Mi 1


fi= fc+ (2i 1 M )fd 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


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Multiple Frequency-Shift Keying (MFSK)

e tng thch vi toc o d lieu cua luong bit vao,moi mot phan t tn hieu au ra co khoang thi gianla:

Ts=LTseconds


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

V vay, moi phan t tn hieu ieu che mang lngtin 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

tfA c2cos 1bit


tfA c2cos 0bit

tfA c2cos

tfA c2cos1bit

0bit


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Tn hieu PSK 2 mc


Tn hieu PSK hai mc


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2PSK

0 1

2PSK2PSK ra

y(t)

NRZ vao

b(t)


Cac v tr pha

Song mangc(t)

S o nguyen ly


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Figure 5.8 PSK



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Figure 5.9 PSK constellation



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Phase-Shift Keying (PSK)

PSK vi sai (DPSK)

S dch phase cua trang thai hien tai da tren stham khao pha cua trang thai trc o.

Bit 0 cum tn hieu hien tai ong phase vi


cum n eu rc o.

Bit 1 cum tn hieu hien tai ngc pha vi cumtn hieu trc o.


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

2cos

tfA c 11


ts

4

32cos

tfA c

432cos tfA c

42cos

tfA c

01

00

10


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4PSK S o nguyen ly ieu che

2PSKIb1(t)


4PSK ra

y(t)

NRZ vao

b(t)

c t

2PSK

P 90o

Qb2(t)


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4PSK - Cac trang thai phase

1

I

Q

10

1101


0 1000


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4PSK S o nguyen ly giai ieu che

LPF

4 PSK Luong bit rac(t)

I


LPF

900

vao

Bo chiacong suat

b(t)

Q


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Figure 5.10 The 4-PSK method



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Figure 5.11 The 4-PSK characteristics



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Figure 5.12 The 8-PSK characteristics



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Figure 5.13 Relationship between baud rate and bandwidth in PSK



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Example 8Example 8

Find the bandwidth for a 2-PSK signal transmitting at

2000 bps. Transmission is in half-duplex mode.

SolutionSolution


- ,

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, what

are the baud rate and bit rate?

SolutionSolution


For 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.


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Multilevel PSK Using multiple phase angles with each angle having more than

one amplitude, multiple signals elements can be achieved

RRD


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

L = number of bits per signal element

2og


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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-fc=fc-f1


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Performance

Bandwidth of modulated signal (BT)

MPSK

RM

rR

L

rB

T

2log

11


MFSK

L = number of bits encoded per signal element M= number of different signal elements

RM

BT

2log


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Bang thong cua tn hieu ieu che

V du: Tm bang thong cua tn hieu ieu che PSK 2mc, 4 mc, 8 mc, 16 mc cho 1 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.


s cc s ncos 21


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Quadrature amplitude modulation is a

combination ofASKandPSKso that a

Note:Note:


maximum contrast between each

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

is achieved.


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Cac trang thai cua 4-QAM & 8 QAM



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Dang tn hieu 8-QAM



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S DUNG PHOBIEN NHAT

16-QAM



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16QAM S o nguyen ly ieu che

4PAMIb1(t)

b2(t)


16QAM ra

y(t)

NRZ vao

b(t)

c t

4PAM

P 90o

Qb4(t)

b3(t)


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16QAM - Cac trang thai phase

10

Q

1010

11 1011


00

I1000 01 11

01 1001

1000


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Figure 5.17 Bit and baud



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Table 5.1 Bit and baud rate comparison

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

ASK, FSK, 2ASK, FSK, 2--PSKPSK Bit 1 N N

44--PSK, 4PSK, 4--QAMQAM Dibit 2 N 2N

88--PSK, 8PSK, 8--QAMQAM Tribit 3 N 3N


1616--QAMQAM Quadbit 4 N 4N

3232--QAMQAM Pentabit 5 N 5N

6464--QAMQAM Hexabit 6 N 6N

128128--QAMQAM Septabit 7 N 7N

256256--QAMQAM Octabit 8 N 8N


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Example 10Example 10

A constellation diagram consists of eight equally spaced

points on a circle. If the bit rate is 4800 bps, what is the

baud rate?

SolutionSolution


The 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.

SolutionSolution

A 16-QAM signal has 4 bits per signal unit since


log216 = 4.

Thus,

(1000)(4) = 4000 bps


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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 since

60

log2 64 = 6.

Thus,

72000 / 6 = 12,000 baud

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