qpsk: change of phase is ±90º or ±180º
TRANSCRIPT
1
PB.51
Offset QPSK
QPSK: Change of phase is ±90º or ±180º
)(1 tφ
)(2 tφ
PB.52
Offset QPSK
Offset QPSK: Change of phase is ±90º only• Envelope variation is reduced.
)(1 tφ
)(2 tφ2/30 2sin2)(
0 2cos2)(
2
1
TttfT
t
TttfT
t
c
c
≤≤=
≤≤=
πφ
πφ
2
PB.53
π/4-shifted QPSK
The carrier phase used for the transmission of successivesymbols is alternately picked from one of the two QPSKconstellation shown below.
)(1 tφ
)(2 tφ
)(1 tφ
)(2 tφ
PB.54
π/4-shifted QPSK
Eight possible phase states for the π/4-shifted QPSK– phase change: ±45º and ±135º– envelope variation is reduced.– Unlike offset QPSK, it can be noncoherently detected.
)(1 tφ
)(2 tφ
3
PB.55
π/4-shifted QPSK
– Can be differentially encoded.– Example:
• π/4-Differential Quaternary Phase Shift Keying(DQPSK) is used in IS-95 (standard for CDMA mobilecommunication)
PB.56
Frequency shifting keying (FSK)
Similar to other passband data transmission system,the function model of FSK is
ModulatorSignal
transmission encoder
im
Carrier signal
)(tsi Channel)(tx
Detectoris x Signal transmission
decoder m̂
4
PB.57
Binary FSK
Symbols 1 and 0 are distinguished from each other bytransmitting one of two sinusoidal wave that differ infrequency.
≤≤=elsewhere0
0)2cos(2)( bi
b
b
iTttf
TE
ts π
where b
i Tinf +
= for some fixed integer and 2,1=i
Note: the two sinusoidal waves must be orthogonal
PB.58
It is a continuous-phase signal as phase continuity isalways maintained, including the inter-bit switchingtimes.
Called continuous-phase frequency-shift keying(CPFSK)
Example:If sTb µ1= , 9=n then MHzf 101 = and MHzf 111 = .
For sending a symbol 1, the sinusoidal wave with 1fcompletes ten cycles.
5
PB.59
The transmitted signal can be written as)()( 11 tEts bφ= and
)()( 22 tEts bφ=
where bib
i TttfT
t <≤= 0 )2cos(2)( πφ
PB.60
Therefore, the constellation of binary FSK is
)(1 tφ
)(2 tφ
bE
bEDecision boundary
bE2
Constellation of binary PSK
)(tφ
Decisionboundary
bEbE−
6
PB.61
GenerationThe binary data sequence is first applied to an on-offlevel encoder, at the output of which symbol 1 isrepresented by a constant amplitude of bE andsymbol 0 is represented by zero volts.
)(1 tφ
X
)(2 tφ
X
On-offlevel
encoder
Inverter
+
PB.62
DetectionThe detector consists of two correlators with acommon input, which are supplied with locallygenerated coherent signals. The correlator outputs arethen subtracted.
∫T
0)(tx
)(1 tφ
X1x
Decisiondevice
0
∫T
0
)(2 tφ
X2x
0 if 00 if 1
<>
yy
+
-
y
7
PB.63
As the decision boundary is )()( 21 tt φφ = ,
choose 1 if 0>ychoose 0 if 0<y
)(1 tφ
)(2 tφ
bE
bEDecision boundary
0=y 0>y
PB.64
Error ProbabilityThe received signal is
∫∫ −=
−=bb TT
dtttxdtttx
xxy
0 20 1
21
)()()()( φφ
The mean of y is
−=
sent was0 Symbolsent was1 Symbol
b
b
EE
y
The variance of y is
o
oo
N
NNxx
=
+=
+=
22
)var()var( 212σ
8
PB.65
Therefore, if symbol 0 was sent,
+−=
−−=
o
b
o NEy
N
yyyf
2)(
exp2
1
2)(exp
21)0|(
2
2
2
π
σσπ
and the error probability is
2)(
exp2
1
)sent was0 symbol0(
0
2
10
dyNEy
N
yPp
o
b
o∫∞
+−=
>=
π
PB.66
Let o
b
NEy
z2+
= ,
[ ]
( )
2/erfc21
exp22/
210
ob
NE
NE
dzzpob
=
−= ∫∞
π
As 1001 pp = ,
( )obe NEP 2/erfc21
=The error probability of binary PSK is
=
o
be N
Ep erfc21 .
i.e., we have to double the bit energy-to-noisedensity ratio, ob NE / , to maintain the samebit error rate as in a binary PSK system.
9
PB.67
Power SpectraThe general form of binary FSK is
≤≤=elsewhere0
0)2cos(2)( bi
b
b
iTttf
TE
tsπ
)/sin()2sin(2)/cos()2cos(2
)/sin()2sin(2)/cos()2cos(2
0 symbolfor 1;- symbolfor )/2cos(2)(
bcb
bbc
b
b
bcb
bbc
b
b
bcb
b
TttfTETttf
TE
TttfTETttf
TE
TttfTEts
ππππ
ππππ
ππ
±=
±−±=
+±=
Consider bTff /121 =− and their arithmetic meanequals to cf
PB.68
• The in-phase component is independent of theinput binary wave. The power spectral densityof this component consists of two deltafunctions. (one delta function if basebandspectrum is considered)
)/sin()2sin(2)/cos()2cos(2)( bcb
bbc
b
b TttfTETttf
TEts ππππ ±=
• The quadrature component is related to theinput binary wave. The power spectral densityis
)14()(cos8
222
2
−fTfTE
b
bb
ππ
10
PB.69
)/sin()2sin(2)/cos()2cos(2)( bcb
bbc
b
b TttfTETttf
TEts ππππ ±=
• The average powers of the delta functionadding up to one-half the total power of thebinary FSK signal.
• The presence of these two discrete frequencycomponents provides a means ofsynchronizing the receiver with the transmitter.
PB.70
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PB.71
M-ary Quadrature Amplitude Modulation (QAM)
Introduction– PSK is usually limited to BPSK, QPSK and 8-PSK.– For further reducing the transmission bandwidth, QAM is
used.
– Example:Consider a voice signal with 3kHz bandwidth,• If analog amplitude modulation (AM) is used, the
transmission bandwidth is 2x3kHz = 6kHz• If analog single sideband modulation (SSB) is used,
the transmission bandwidth is 3kHz.
PB.72
If digital method is used, the minimum sampling rate is2x3kHz = 6kHz. If there are 256 levels for encoding, thedata rate is 48kbps
Referring to the diagram of PB.70, the transmissionbandwidth for BPSK is
Transmission bandwidth for 8-PSK is
Transmission bandwidth for 1024-QAM is
kHz962/2 == bRT
kHz32log/2/2 2 == MRT b
kHz6.9log/2/2 2 == MRT b
12
PB.73
Generation
4-PAM
4-PAMModulator 16-QAM
32-PAMModulator 1024-QAM
32-PAM
I
Q
I
Q
2-PAM
2-PAMModulator
I
Q QPSK
PB.74
M-ary Quadrature Amplitude Modulation (QAM)The M-ary QAM signal is defined by
[ ]
=
±±=≤≤
−=
)()(
,...2.1,00
)2sin(2)2cos(2)(
2
121 t
tss
kTt
tfbTEtfa
TEts ck
bck
bk
φφ
ππ
where
TttfT
t
TttfT
t
c
c
≤≤=
≤≤=
0 )2sin(2)(
0 )2cos(2)(
2
1
πφ
πφ
[ ] [ ]okokkk EbEass =21
13
PB.75
If there are M symbols and ML = , the M-ary squareconstellation can always be viewed as the Cartesian producta one-dimensional L-ary PAM constellation with itself.
Therefore, we have
{ }
+−−+−+−+−+−
−−−+−−+−−−−+−−+−
=
)1,1()1,3()1,1(
)3,1()3,3()3,1()1,1()1,3()1,1(
LLLLLL
LLLLLLLLLLLL
ba ii
L
MLMM
L
L
PB.76
{ }
−−−−−−−−−−−−
−−−−
=
+−−+−+−+−+−
−−−+−−+−−−−+−−+−
=
)3,3()3,1()3,1()3,3()1,3()1,1()1,1()1,3(
)1,3()1,1()1,1()1,3()3,3()3,1()3,1()3,3(
)1,1()1,3()1,1(
)3,1()3,3()3,1()1,1()1,3()1,1(
LLLLLL
LLLLLLLLLLLL
ba ii
L
MLMM
L
L
Example: Consider a 16-QAM, L=4
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PB.77
The signal constellation is
oE oE3 1φ
2φ
PB.78
The encoding of the message is as follows:• Two of the four bits, namely, the left-most two bits,
specify the quadrant in the constellation plane inwhich a message point lies. Thus, starting from thefirst quadrant and proceeding counterclockwise, thefour quadrant are represented by 11, 10, 00, and 01.
• The remaining two bits are used to represent one ofthe four possible lying within each quadrant of theplane.
15
PB.79
oE oE3 1φ
2φ
1111
1101
1110
1100
PB.80
Error Probability
The probability of correct detection for M-ary QAM is2)'1( ec PP −=
where 'eP is the probability of symbol error for thecorresponding L-ary PAM.
The probability of symbol error for M-ary QAM is
−=
≈−−=
−=
o
o
e
e
ce
NE
M
PP
PP
erfc112
'2)'1(1
12
16
PB.81
As the transmitted energy in M-ary QAM depends on theparticular symbol transmitted, the probability of symbolerror is expressed in terms of the average value of thetransmitted energy.
Assuming that the L amplitude level levels of the in-phaseor quadrature component are equally likely, we have
3)1(2 o
avEME −
=
Therefore,
−
−=o
ave NM
EM
P)1(2
3erfc112
PB.82
Noncoherent systems
Introduction– For coherent systems, the receiver is perfectly
synchronized to the transmitter, and the only channelimpairment is noise.
– In certain situations, the receiver cannot follow thechange of the received signal phase.• Noncoherent detection is used.
17
PB.83
Noncoherent binary FSK
The transmitted signal is
bib
bi Tttf
TEts ≤≤= 0)2cos(2)( π
For the nonchoherent detection of this frequency-modulated wave, the receiver consists of a pair of matchedfilters followed by envelope detectors.
PB.84
Filtermatched to
bTttf
≤≤0)2cos( 1π
)(txEnvelopedetector
Comparisondevice
1l
21
21
if 0 if 1
llll
<>
Filtermatched to
bTttf
≤≤0)2cos( 2π
Envelopedetector
2l
Sample atbTt =
Sample at bTt =
18
PB.85
[ ]
[ ] [ ] [ ] [ ] ττπτπττπτπ
ττπτ
dfxtTfdfxtTf
dtTfxty
bb
b
TT
T
∫∫∫
−−−=
+−=
0 110 11
0 1
2sin)()(2sin2cos)()(2cos
)(2cos)()(
For example, if the received signal is)2cos()( 1 θπ += tftx
The upper matched filter output is
convolution
The envelope of the matched filter output is
[ ][ ] [ ][ ] 2/12
0 1
2
0 1 2sin)(2cos)(
+ ∫∫ ττπτττπτ dfxdfx bb TT
PB.86
For example,bT
t
The bit error rate for noncoherent binary FSK is
−=
o
be N
EP2
exp21
19
PB.87
Differential Phase-shift keying
Introduction– the noncoherent version of PSK– differential encoding of the input binary wave
– Provided that the unknown phase contained in thereceived wave varies slowly, the phase different betweenwaveforms received in two successive bit intervals will beindependent of
• To send symbol 0, advance the phase of the currentsignal waveform by 180°.
• To send symbol 1, the phase of the current signalwaveform is unchanged.
θ
θ
PB.88
Suppose the transmitted DPSK signal equals to
bcb
b TttfTE
≤≤0)2cos(2
π ,
the transmitted signal for bTt 20 ≤≤ is
≤≤
≤≤=
bbcb
b
bcb
b
TtTtfTE
TttfTE
ts2)2cos(
2
0)2cos(2)(1
π
π
if symbol 1 was sent
≤≤+
≤≤=
bbcb
b
bcb
b
TtTtfTE
TttfTE
ts2)2cos(
2
0)2cos(2)(2
ππ
π
if symbol 0 was sent
No phase change
Change in phase
20
PB.89
GenerationThe differential encoding process starts with anarbitrary first bit, serving as reference.
Delay
Logicnetwork
{ }kb
bT
DPSKsignal
Productmodulator
Amplitudelevel shifter
{ }kd
{ }1−kd)2cos(2 tf
T cb
π
If the incoming binary symbol kb is 1, leave thesymbol kd unchanged with respect to the previous bit.
PB.90
If the incoming binary symbol kb is 0, change thesymbol kd with respect to the previous bit.
Example,
{ }kb 1 0 0 1 0 0 1 1{ }1−kd 1 1 0 1 1 0 1 1
{ }kd 1 1 0 1 1 0 1 1 1( 1−⊕= kkk dbd )
Output 0 0 π 0 0 π 0 0 0phase
21
PB.91
Detection
)(tx
∫bT
0
)2cos( tfcπ
X
)2sin( tfcπ
∫bT
0X
DelaybT
DelaybT
X
X
+
yDecisiondevice
00 if 00 if 1
<>
yy
Suppose the carrier phase is unknown and the initial phaseis 0, the received signal is
)2cos()( θπ += tfAtx c
PB.92
The output of the In-phase correlator is
θθπθπ cos'cos2
)2(cos)2cos(0
AATdttftfA bc
T
cb
==+∫
and the output of the Quadrature-phase correlator is
θθπθπ sin'sin2
)2(sin)2cos(0
AATdttftfA bc
T
cb
==+∫ .
If the initial phase is 180°, the outputs are θcos'A− andθsin'A− .
22
PB.93
Therefore, the two received signal points are)sin',cos'( θθ AA and )sin',cos'( θθ AA −−
)(1 tφ
)(2 tφ
PB.94
The receiver measures the coordinates ),( 00 QI xx at time
bTt = and ),( 11 QI xx at time bTt 2= .
Therefore,
sent was0 symbol if 0
sent was1 symbol if 0
1011
1011
<+
>+
QQII
QQII
xxxx
xxxx
Error probabilityThe bit error rate for DPSK is
−=
0
exp21
NEP b
e
23
PB.95
Comparison of Digital Modulation Schemes
Probability of Error– Signaling Scheme Bit Error Rate
Coherent BPSKCoherent QPSKCoherent MSK
Coherent FSK
DPSK
Noncoherent binary FSK
( )ob NE /erfc21
( )ob NE 2/erfc21
( )ob NE /exp21
−
( )ob NE 2/exp21
−
PB.96
Comparison of DigitalModulation Schemes
The bit error rate for allthe systems decreasemonotonically withincreasing value of
ob NE /