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ET 2060H thng thng tin
TS. ng Quang Hiuhttp://ss.edabk.org
Trng i hc Bch Khoa H NiVin in t - Vin thng
2011-2012
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Outline
H thng thng tin v iu ch bin
Khng gian tn hiu v h thng thng tin s
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Khi nim h thng thng tin
x(t) y(t) yr (t) x(t)
iu ch gii iu ch
knh h(t)
My pht - my thu (im - im). Knh h(t) (fading, Doppler, v.v.) v
nhiu Gauss n(t). Signal-to-Noise Ratio (SNR). Ghp tin x(t) vo sng
mang ti pha pht sao cho ph hp
vi mi trng truyn dn (iu ch - modulation). Tch tin x(t) ra khi
sng mang ti pha thu (gii iu ch -
demodulation). tin cy: x(t) x(t).
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iu ch / gii iu ch
iu ch l qu trnh thay i cc thuc tnh ca sng mang c(t)theo tn hiu
thng tin x(t).
c(t) = Ac cos(ct + c)
iu bin (AM) iu tn (FM) iu pha (PM)
Mt s u im khi thc hin iu ch: Dch di tn hot ng ca tn hiu v trung
tm bng tn
c cp php. Cho php truyn tin khong cch xa hn, kh nng chng
nhiu, chng giao thoa tt hn, v.v. Ph hp hn vi tng ng dng, tng hon
cnh c th.
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Khi nim iu bin (AM) DSB-SC
x(t)
cos(c t)
y(t)
t
y(t)
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Ph ca tn hiu iu bin
y(t) = x(t) cos(ct)
=12x(t)[ejc t + ejc t ]
= X (j) =12[X (j( c)) + X (j( + c))]
X (j)
1
Y (j)
12
cc
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Gii iu bin ng b pha (coherent detection)
y(t)
cos(c t)
12x(t)LPF
w(t)
w(t) = y(t) cos(ct) = x(t) cos2(c t)
=12x(t) +
12x(t) cos(2ct)
W (j)
12
2c2c
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Trng hp khng ng b pha sng mang
w(t) = y(t) cos(ct + 2) = x(t) cos(ct + 1) cos(ct + 2)
=12x(t) cos(2 1) +
12x(t) cos(2c t + 2 + 1)
Tn hiu thu c sau khi lc thng thp:
x(t) = x(t) cos(2 1)
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Trng hp khng ng b pha sng mang
w(t) = y(t) cos(ct + 2) = x(t) cos(ct + 1) cos(ct + 2)
=12x(t) cos(2 1) +
12x(t) cos(2c t + 2 + 1)
Tn hiu thu c sau khi lc thng thp:
x(t) = x(t) cos(2 1)
Nu (2 1) thay i theo thi gian? Vng kha pha (PLL)
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Cc phng php iu bin khc
y(t) = [B + x(t)] cos(ct)
su iu ch (modulation depth): h = max{x(t)}B
t
y(t)h = 0.25
t
y(t)h = 0.75
Gii iu ch dng mch tch ng bao (envelop detector), kocn ng b pha
nhng lng ph cng sut pht vo sng mang.
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QAM (Quadrature Amplitude Modulation)
xI (t)
xQ(t)
cos(c t)
pi2
b
y(t)
12xI (t)
12xQ(t)
cos(c t)
pi2
b
by(t)
LPF
LPF
Chng minh? V ph tn hiu? Tng gp i hiu qu s dng di tn!
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iu ch bin xung (PAM)
y(t) =
n=
x(nTs)h(t nTs)
trong ,
h(t) =
{1, 0 < t < T00, t cn li
v Ts < 12B .
Ghp knh phn chia theo tn s (FDM) - dng AM Ghp knh phn chia theo
thi gian (TDM) - dng PAM
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Bi tp
Vit chng trnh Matlab minh ha iu ch AM trng hpDSB-SC.
(a) V trn min thi gian cc tn hiu x(t),y(t),w(t) v x(t)trong
khong thi gian [0, 1] giy, khi x(t) = cos(2 10t),c(t) = cos(2
100t)
(b) V ph cc tn hiu trn
(c) V dng tn hiu ti my thu x(t) khi SNR = 10 dB.
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Outline
H thng thng tin v iu ch bin
Khng gian tn hiu v h thng thng tin s
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S h thng thng tin s
u vo
u ra
m ha ngun m ha knh iu ch
knh
gii m ngun gii m knh gii iu ch
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Cc khi nim trong thng tin s
rng bng thng B [hertz] Dung lng knh C = B log2(1+ SNR) Tc truyn
d liu
(i) Tc k hiu (symbol / baud rate) Rs(ii) Tc bit (bit rate) R =
Rs log2 M
T s nng lng bit trn nhiu Eb/N0. T l li bit BER
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Nguyn l thng tin s
m my pht my thu ms(t) r(t)
n(t)
{mi}, {P[mi ]} {si (t)} {mi}
Pht i dng sng s(t) = si (t) khi u vo l m = mi . Di tc ng ca nhiu
l: r(t) = s(t) + n(t). Nu bit trc {P [mi ]} (xc sut pht i mi trong
tp hu
hn cc gi tr {m0,m1, . . . ,mM1}) v cho trc cc dngsng {s0(t),
s1(t), . . . , sM1(t)}; my thu c nhim v x ltn hiu thu c r(t) m sao
xc sut li Pe = P [m 6= m]l nh nht.
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V d v dng sng (1) BPSK: m {0, 1}, hoc {1, 1}.
s(t) =
s0(t) =
EbT
cos(2fc t), m = 0
s1(t) =
EbT
cos(2fct), m = 1
vi fc = nT .
0 1 2 3 4 5 6 7 8 9 10
1
0
1
0 1 2 3 4 5 6 7 8 9 10
1
0
1
4-ASK: m {0, 1, 2, 3}, u[n] {3d/2,d/2, d/2, 3d/2}
s(t) =
n
u[n]g(t nT )
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V d v dng sng (2)QPSK: m {0, 1, 2, 3} hoc {00, 01, 11, 10},
s(t) =
s0(t) =
EsTcos(2fc t + /4), m = 0
s1(t) =
EsTcos(2fc t + 3/4), m = 1
s2(t) =
EsTcos(2fc t + 5/4), m = 2
s3(t) =
EsTcos(2fc t + 7/4), m = 3
0 1 2 3 4 5 6 7 8 9 10
1
0
1
0 1 2 3 4 5 6 7 8 9 100
1
2
3
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Khng gian tn hiu
Tp hp cc dng sng s(t) (hm thc / phc) c nnglng hu hn v php nhn,
php cng thng thng khng gian vector N-chiu
+ Tch trong (inner product) v ton t 2-norm khnggian Hilbert
H c s trc chun {k(t)}
k(t)(t)dt =
{1, k = 0, k 6=
vi mi 0 k, (N 1).
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V d v h trc chun Tp cc xung dch theo thi gian
k(t) = g(t k), k = 0, 1, . . . , (N 1)
vi g(t) l xung c nng lng n v
g(t) =
{1, 0 t
0, t cn li
Tp cc xung dch trn min tn s, vi k = 0, 1, . . . , (N 1).
k(t) =
{ 2Tcos(2
Tkt), 0 t T
0, t cn li
Hai hm hnh sin lch pha 90 .
0(t) =
{ 2Tcos(2f0t), 0 t T
0, t cn li
1(t) =
{ 2Tsin(2f0t), 0 t T
0, t cn li
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Chm sao tn hiu
Biu din si (t) theo c s
si(t) =N1j=0
sijj (t), i = 0, 1, . . . , (M 1)
Mi dng sng si(t) c xc nh bi vector:
si = [si0, si1, . . . , si(N1)]
Tp hp M im si = [si0, si1, . . . , si(N1)] trong khng gianN-chiu
gi l chm sao tn hiu (signal constellation).
Mi im c gi l mt k hiu (symbol) si . Truyn tn hiu M-mc (M-ary
signaling)
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V d v chm sao tn hiu 64-QAM (N=2,M=64)
b
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I
Q
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My thu khi khng c nhiu
r(t) b
s0
s1
sN1
0(t)
1(t)
N1(t)
b
b
b
b
b
b
b
b
b
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My thu khi c nhiu
Tm im si trn chm sao tn hiu sao cho gn vi[s0, s1, . . . , sN1]
nht. iu kin: D liu u vo {mi} phn phi u Nhiu trng Gauss n(t) vi gi
tr trung bnh bng khng
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S b thu pht s
b
cos(2fc t)
cos(2fc t)
b
m ha p(t)
gii m matched filter LPF
s baseband x(t)
x(t)s
Ts
knh
Trn thc t hay dng s QAM!!!
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M Gray
b m ha s
M ha lung bit u vo b thnh cc k hiu s sao cho hai khiu cnh nhau
(trn chm sao) ch khc nhau duy nht 1 bit.
b b b b b b b b
000 001 011 010 1001011111108-ASK
b
b
b
b
I
Q
00 01
1110QPSK
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To dng xung
Xung vung
p(t) =
{ 1T, 0 t T
0, t cn li
gy ra ISI. Xung hm sinc, cos nng (raised cosine), Gauss. T
c!!!
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Matched filter (MF)
s
n(t)
p(t) h(t) s
r(t)
T
Tm h(t) sao cho u ra c SNR ln nht? Chng minh c khi h(t) = p(T
t).
r(t)
p(t)
T s
T
Hnh: Cch tip cn khc i vi MF
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Bi tp
1. Vit chng trnh Matlab thc hin m Gray
2. Vit chng trnh minh ha iu ch BPSK, QPSK, 16-QAM(a) V dng tn
hiu baseband ti my pht v my thu khi c
nhiu / khng c nhiu, vi cc dng xung khc nhau(b) V dng tn hiu ti u
ra b matched filter.(c) Khi phc li tn hiu, so snh vi u vo.
H thng thng tin v iu ch bin Khng gian tn hiu v h thng thng tin
s
