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MON HOC: VI BA SOLp Ky thuat vien
thong
Chng 2:
X LY TN HIEU BANG GOC
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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
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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 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.
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ieu che so
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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.
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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 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
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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:
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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
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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
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
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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?
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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 laAcos(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 Tuynhien cac loai 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 signaltransmitting at 2000 bps. The transmission mode is half-
duplex.
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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.
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Example 4Example 4
Given a bandwidth of 5000 Hz for an ASK signal, whatare the baud rate and bit rate?
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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.
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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.
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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.
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
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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 tanso song mang khac nhau:
( )
=ts( )tfA
12cos
( )tfA2
2cos
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.
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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12/03/0941ThS.Vo Trng Sn
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?
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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?
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.
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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==
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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-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
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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 +=
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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
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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
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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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Example 10Example 10
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
Example 11Example 11
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Example 11Example 11
Compute the bit rate for a 1000-baud 16-QAM signal.
Example 11Example 11
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Example 11Example 11
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
Example 12Example 12
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Example 12Example 12
Compute the baud rate for a 72,000-bps 64-QAM signal.
Example 12Example 12
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Example 12Example 12
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
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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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Ket thuc chng 2