no se me ocurre que poner aqui
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONS
Modulaciones de alta velocidad
QAM-OFDM- 1
SIGNAL PROCESSING IN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONS
1. Modulaciones de alta velocidad
Transmiten: voz, imagen y datos
Requieren alta eficiencia espectral
Han de ser robustas a las degradaciones que introduce elcanal
(ej: ecos y desvanecimientos)
Tx. Telefónica por
línea de abonado
Radiodifusión
QAM-OFDM- 2
- Modem V.34
- xDSL
- Modulación OFDM
Todas se apoyan en la modulación
QAM:
Buen compromiso EfBW
en canales AWGN
BER
Bastante robusta a errores de sincron.
SIGNAL PROCESSING IN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONS
Densidad espectralSeñal paso-banda:
{
s (t ) = Re Ac bs (t ) e j ( 2πf ct +θ c )
bs (t ) = is (t ) + jqs (t )
}
información
b (t ) = i (t ) 2 + q (t ) 2
s
s
s
bs (t ) = bs (t ) e jθ b (t )
θ b (t ) = tg −1 qs (t )
is (t )
s (t ) = Ac {is (t ) cos( wc t + θ c ) − qs (t ) sen(wc t + θ c )}
QAM-OFDM- 3
SIGNAL PROCESSING IN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY ANDCOMMUNICATIONS
Rs (τ ) =
Ac2
(R
is
(τ ) + Rqs (τ ) )cos( wcτ )
2
TF
S s (τ ) =
Ac2
4
QAM-OFDM- 4
(S
is
( f − f c ) + Sis ( f + f c ) + S qs ( f − f c ) + S qs ( f + f c ) )
SIGNAL PROCESSING IN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONS
s (t ) =
Modulación M-ASK:
∞
∑a
k = −∞
is (t ) =
∞
∑a
k = −∞
p(t − kT )
k
Sis ( f ) = σ r P( f ) + (ma r )
σ =
2
a
M 2 −1
ma =
12
Si
∞
2
∑
P (nr ) δ ( f − nr )
2
n = −∞
M −1
unipolar
2
t
→ P( f ) = sen πfT = T sinc( fT )
p (t ) = Π
πf
T
Sis ( f ) =
M 2 −1
12r
QAM-OFDM- 5
p (t − kT ) cos( wc t + θ c )
Señal M-PAM
2
2
a
k
sinc ( fT ) +
2
( M − 1) 2
4
δ(f )Ef BW =
rb
BWs
(bits/s/Hz)
SIGNAL PROCESSING IN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONS
Modulación M-APK (QAM):
∞
jθ k
j ( wc t +θ c )
s(t ) = Re Ac ∑ ak e p(t − kT ) e
k = −∞
ASK+PSK
I k + jQk
∞
∞
s(t ) = Ac ∑ I k p(t − kT ) cos( wc t + θ c ) − ∑ Qk p(t − kT ) sen( wc t + θ c )
k = −∞
k = −∞
is (t ) =∞
k = −∞
qs (t ) =
Si
∑I
∞
∑Q
k = −∞
k
M1 = 2
QAM-OFDM- 6
I k = ±1,±3... ± ( M 1 − 1)
n
p (t − kT )
M2 = 2
m
M-APK
Qk = ±1,±3... ± ( M 2 − 1)
p (t − kT )
k
M=M1 M2
M =2
n+m
=2
b
r=
rb
b
baudios
SIGNAL PROCESSING IN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONS
QAM-OFDM- 7
SIGNAL PROCESSINGIN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONS
QAM-OFDM- 8
SIGNAL PROCESSING IN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONS
QAM-OFDM- 9
SIGNAL PROCESSING IN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONS
QAM-OFDM- 10
SIGNAL PROCESSING IN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONSBER
A igual eficiencia espectral que PSK y ASK, QAM permite mejor BER; ya que hay más grados
de libertad en su diseño
M
8
Mejora en SNR de QAM
sobre PSK para igual BER
1.65
16
4.20
32
7.02
64
9.95
Comparación argumentos
BER
QAM-OFDM- 11
SIGNAL PROCESSING IN COMMUNICATIONS GROUP
DEPARTMENT OF SIGNAL THEORY AND COMMUNICATIONS
El límite de Shannon paratransmisión libre de errores en canales limitados en banda y AWGN
C = BWc log 2 [1 + SNR ] (bits / seg )
Ej.: canal telefónico: BWc= 3KHz
SNR=30 dB (1000:1)
C=30 kbps
Para determinar la probabilidad de error de una modulación digital como M-QAM se ha
de recurrir a su constelación. La probabilidad de error pe viene determinada principalmente
por la distancia mínima entre ptos. de...
Regístrate para leer el documento completo.