OFFSET QPSK - An-Najah Videos and...OFFSET QPSK To solve the amplitude fluctuation problem, we...
Transcript of OFFSET QPSK - An-Najah Videos and...OFFSET QPSK To solve the amplitude fluctuation problem, we...
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OFFSET QPSK
2
1
s4
s1
s3
s2
(10)(00)
(01) (11)
90 degree shift in phase
180 degree shift in phase
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OFFSET QPSK
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OFFSET QPSK
Whenever both bits are changed simultaneously, 180 degree phase-shift occurs.
At 180 phase-shift, the amplitude of the transmitted signal changes very rapidly costing amplitude fluctuation.
This signal may be distorted when is passed through the filter or nonlinear amplifier.
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OFFSET QPSK
0 1 2 3 4 5 6 7 8-2
-1
0
1
2
0 1 2 3 4 5 6 7 8-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Original Signal
Filtered signal
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OFFSET QPSK
To solve the amplitude fluctuation
problem, we propose the offset QPSK.
Offset QPSK delay the data in quadrature
component by T/2 seconds (half of
symbol).
Now, no way that both bits can change at
the same time.
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OFFSET QPSK
In the offset QPSK, the phase of the signal
can change by 90 or 0 degree only while
in the QPSK the phase of the signal can
change by 180 90 or 0 degree.
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0 1 2 3 4 5 6 7 8-2
-1
0
1
2
0 1 2 3 4 5 6 7 8-1
-0.5
0
0.5
1
0 1 2 3 4 5 6 7 8-1
-0.5
0
0.5
1
OFFSET QPSK
0 1 2 3 4 5 6 7 8-2
-1
0
1
2
0 1 2 3 4 5 6 7 8-1
-0.5
0
0.5
1
0 1 2 3 4 5 6 7 8-1
-0.5
0
0.5
1
0 1 10
1 0 0 0
01 10 10 00
Inphase QPSK
Q phase QPSK
QPSK
0 1 1 0
1 0 0
01 10 1000
Inphase Offset QPSK
Q phase Offset QPSK
Offset QPSK
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Offset QPSK
Possible paths for switching between themessage points in (a) QPSK and (b) offset QPSK.
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OFFSET QPSK
Bandwidths of the offset QPSK and the
regular QPSK is the same.
From signal constellation we have that
Which is exactly the same as the regular
QPSK.
02erfc
N
EPe
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/4-shifted QPSK
Try to reduce amplitude fluctuation by
switching between 2 signal constellation
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/4-shifted QPSK
As the result,
the phase of the
signal can be
changed in
order of /4 or
3/4
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/4-shifted QPSK
Since the phase of the next will be varied in
order of /4 and 3/4, we can designed the
differential /4-shifted QPSK as given below
Gray-Encoded Input Data Phase Change in radians
00 +/4
01 +3/4
11 -3/4
10 -/4
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/4-shifted QPSK:00101001
Step Initial phase
Input Dibit Phase change
Transmitted phase
1 /4 00 /4 /2
2 /2 10 -/4 /4
3 /4 10 -/4 0
4 0 01 3/4 3/4
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/4-shifted QPSK
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-2
0
2
QPSK
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-2
0
2
OFFSET QPSK
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-1
0
1
D OFFSET QPSK
01 10 10 01
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/4-shifted QPSK
Since we only measure the phase different
between adjacent symbols, no phase
information is necessary. Hence, non-
coherent receiver can be used.
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Block diagram of the /4-shifted DQPSK
detector.
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/4-shifted QPSK
Illustrating the
possibility of phase
angles wrapping
around the positive
real axis.
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M-array PSK
At a moment, there are M possible symbol
values being sent for M different phase
values, Mii /12
MiiM
tfT
Ets ci ,,2,1,1
22cos
2
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M-array PSK
Signal-space diagram for
octa phase-shift keying (i.e.,
M 8). The decision
boundaries are shown as
dashed lines.
Signal-space diagram
illustrating the application of
the union bound for octa
phase shift keying.
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M-array PSK
Probability of errors
MEdd /sin21812
4 ;/sinerfc0
MM
N
EPe
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M-ary PSK
0 5 10 15 20 25 3010
-50
10-40
10-30
10-20
10-10
100
Eb/N
0 dB
Pro
babili
ty o
f S
ym
bol err
ors
QPSK
8-ary PSK
16-ary PSK
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M-array PSK
Power Spectra (M-array)
M=2, we have
MfTME
TfEfS
bb
PSK
22
2
2
logsinclog2
sinc2)(
fTEfS bbBPSK2sinc2)(
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M-array PSK
Power spectra of M-ary PSK
signals for M 2, 4, 8.
Tbf
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M-array PSK
Bandwidth efficiency:
We only consider the bandwidth of the main lobe (or
null-to-null bandwidth)
Where B is the bandwidth of the transmitted data
Bandwidth efficiency of M-ary PSK is given by
M
R
MTTB b
b 22 log
2
log
22
MMR
R
B
R
b
bb22 log5.0log
2
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M-ary QAM
QAM = Quadrature Amplitude Modulation
Both Amplitude and phase of carrier change
according to the transmitted symbol, mi
where ai and bi are integers, E0 is the
energy of the signal with the lowest
amplitude
TttfbT
Etfa
T
Ets cicii 0;2sin
22cos
2 00
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M-ary QAM
Again, we have
as the basis functions
There are two QAM constellations, square
constellation and rectangular
constellation
TttfT
t c 0;2cos2
1
TttfT
t c 0;2sin2
2
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M-ary QAM
QAM square Constellation
Having even number of bits per symbol,
denoted by 2n
M=L x L possible values
Denoting ML
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16-QAM
)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(
, ii ba
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16-QAM
L-ary, 4-PAM
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QAM Probability of error
Calculation of Probability of errors
Since both basis functions are orthogonal, we
can treat the 16-QAM as combination of two
4-ary PAM systems.
For each system, the probability of error is
given by
0
0
0
11
2
11
N
Eerfc
MN
derfc
LPe
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16-QAM Probability of error
A symbol will be received correctly if data
transmitted on both 4-ary PAM systems are
received correctly. Hence, we have
Probability of symbol error is given by
21 ec PsymbolP
eee
ece
PPP
PsymbolPsymbolP
2211
111
2
2
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16-QAM Probability of error
Hence, we have
But because average energy is given by
We have
0
0112
N
Eerfc
MsymbolPe
3
1212
22 0
2/
1
20 EMi
L
EE
L
iav
012
3112
NM
Eerfc
MsymbolP av
e
32