pass-band data transmission dr. teerasit kasetkasem
TRANSCRIPT
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Pass-band Data Transmission
Dr. Teerasit Kasetkasem
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Block Diagram
Functional model of pass-band data transmission system.
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Signaling
Illustrative waveforms for the three basic forms of signaling binary information. (a) Amplitude-shift keying. (b) Phase-shift keying. (c) Frequency-shift keying with continuous phase.
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What do we want to study?
We are going to study and compare different modulation techniques in terms of Probability of errors Power Spectrum Bandwidth efficiency
B
Rb Bits/s/Hz
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Coherent PSK
Binary Phase Shift Keying (BPSK) Consider the system with 2 basis functions
and
tfT
t cb
2cos2
1
tfT
t cb
2sin2
2
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BPSK
If we want to fix that for both symbols (0 and 1) the
transmitted energies are equal, we have
1
2
s1
We place s0 to minimize probability of error
s0
s0
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BPSK
We found that phase of s1 and s0 are 180 degree difference. We can rotate s1 and s0
1
2
s1
s0
Rotate
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BPSK
We observe that 2 has nothing to do with signals. Hence, only one basis function is sufficient to represent the signals
2
s1s0
1
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BPSK
Finally, we have
tfT
EtEts c
b
bb 2cos
2)(11
tfT
EtEts c
b
bb 2cos
2)(10
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BPSK
Signal-space diagram for coherent binary PSK system. The waveforms depicting the transmitted signals s1(t) and s2(t), displayed in the inserts, assume nc 2.
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BPSK
Probability of error calculation. In the case of equally likely (Pr(m0)=Pr(m1)), we have
0
0
erfc2
1
2erfc
2
1
N
E
N
dP
b
ike
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BPSK
Block diagrams for (a) binary PSK transmitter and (b) coherent binary PSK receiver.
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Quadriphase-Shift Keying (QPSK)
TtitfT
Ets ci
0;
4122cos
2
T is symbol duration E is signal energy per symbol There are 4 symbols for i = 1, 2, 3, and 4
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QPSK
TttiEtiE
tfT
iEtfT
iEts cci
0;4
12sin4
12cos
2sin2
412sin2cos
2
412cos
21
Which we can write in vector format as
412sin
412cos
iE
iEis
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QPSK
i Input Dibit Phase of QPSK
signaling
Coordinate of Message point
si1 si2
1 10
2 00
3 01
4 11
4/
4/3
4/5
4/7
2/E
2/E
2/E
2/E
2/E
2/E2/E
2/E
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QPSK
2
1
s4
s1
s3
s2
(10)(00)
(01) (11)
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QPSK signals
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QPSK
Block diagrams of (a) QPSK transmitter and (b) coherent QPSK receiver.
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QPSK: Error Probability QPSK
Consider signal constellation given in the figure
2
1
s4
s1
s3
s2
(10)(00)
(10) (11)
Z4
Z1
Z3
Z2
2/E
2/E
2/E
2/E
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QPSK
We can treat QPSK as the combination of 2 independent BPSK over the interval T=2Tb
since the first bit is transmitted by 1 and the second bit is transmitted by 2.
Probability of error for each channel is given by
00
12
2erf
2
1
2erfc
2
1
N
Ec
N
dP
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QPSK
If symbol is to be received correctly both bits must be received correctly.
Hence, the average probability of correct decision is given by
Which gives the probability of errors equal to
21 PPc
0
0
2
0
2erfc
2erfc
4
1
2erfc1
N
E
N
E
N
EPP Ce
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QPSK
Since one symbol of QPSK consists of two bits, we have E = 2Eb.
The above probability is the error probability per symbol. The avg. probability of error per bit
Which is exactly the same as BPSK .
0erfcsymbolper
N
EPe b
0erfc
2
1symbolper
2
1bitper
N
EPePe b
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BPSK vs QPSK
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2Power spectrum density of BPSK vs. QPSK
Normalized frequency,fTb
Nor
mal
ized
PS
D,
Sf/2
Eb
BPSKQPSK
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QPSK
ConclusionQPSK is capable of transmitting data twice
as faster as BPSK with the same energy per bit.
We will also learn in the future that QPSK has half of the bandwidth of BPSK.
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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 1 0
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 the message 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 +3/411 -3/410 -/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
2QPSK
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-2
0
2OFFSET QPSK
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-1
0
1D 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 octaphase-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 octaphase-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
babi
lity
of S
ymbo
l 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)
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.
TttfbT
Etfa
T
Ets cicii 0;2sin
22cos
2 00
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M-ary QAM
Again, we have
as the basis functions
TttfT
t c 0;2cos2
1
TttfT
t cb
02sin2
2
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M-ary QAM
QAM square ConstellationHaving even number of bits per symbol,
denoted by 2n. M=L x L possible valuesDenoting 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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16-QAM
Calculation of Probability of errorsSince 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
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
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
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Coherent FSK
FSK = frequency shift keyingCoherent = receiver have information on
where the zero phase of carrier. We can treat it as non-linear modulation
since information is put into the frequency.
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Binary FSK
Transmitted signals are
where
elsewhere ,0
0,2cos2
bib
b
iTttf
T
E
ts
2,1;
iT
inf
b
ci
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Binary FSK
S1(t) represented symbol “1”.
S2(t) represented symbol “0”.
This FSK is also known as Sunde’s FSK.It is continuous phase frequency-shift
keying (CPFSK).
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Binary FSK
There are two basis functions written as
As a result, the signal vectors are
elsewhere ,0
0,2cos2
bibi
TttfTt
b
bE
E 0 and
021 ss
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BFSK
From the figure, we have In case of Pr(0)=Pr(1), the probability of
error is given by
We observe that at a given value of Pe, the BFSK system requires twice as much power as the BPSK system.
bEd 212
022
1
N
EerfcP b
e
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TRANSMITTER
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RECEIVER
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Power Spectral density of BFSK
Consider the Sunde’s FSK where f1 and f2 are different by 1/Tb. We can write
We observe that in-phase component does not depend on mi since
tfT
t
T
Etf
T
t
T
E
T
ttf
T
Ets
cbb
bc
bb
b
bc
b
bi
2sinsin2
2coscos2
2cos2
bb
b
bb
b
T
t
T
E
T
t
T
E cos
2cos
2
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Power Spectral density of BFSK
We have
For the quadrature component
bbb
b
bb
bBI T
fT
fT
E
T
t
T
EFfS
2
1
2
1
2cos
22
Half of the symbol power
bb
b
T
t
T
Etg
sin
2 2222
2
14
cos8
fT
fTTES
b
bbbBQ
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Power Spectral density of BFSK
Finally, we obtain )()()( fSfSfS BQBIB
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Phase Tree of BFSK
FSK signal is given by
At t = 0, we have
The phase of Signal is zero.
bc
b
b
T
ttf
T
Ets
2cos2
0cos20
02cos2
0b
b
bc
b
b
T
E
Tf
T
Es
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Phase Tree of BFSK
At t = Tb, we have
We observe that phase changes by after one symbol (Tb seconds). - for symbol “1” and + for symbol “0”
We can draw the phase trellis as
cos
22cos
2
b
b
b
bbc
b
bb T
E
T
TTf
T
ETs
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Minimum-Shift keying (MSK)
MSK tries to shift the phase after one symbol to just half of Sunde’s FSK system. The transmitted signal is given by
"0"for 02cos2
"1"for 02cos2
2cos2
2
1
tfT
E
tfT
E
ttfT
Ets
b
b
b
b
cb
b
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MSK
Where
Observe that
tT
ht
b
0
bc
bc T
hff
T
hff
2and
2 21
212
1fffc
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MSK
h = Tb(f1-f2) is called “deviation ratio.”
For Sunde’s FSK, h = 1. For MSK, h = 0.5. h cannot be any smaller because the
orthogonality between cos(2f1t) and cos(2f2t) is still held for h < 0.5.
Orthogonality guarantees that both signal will not interfere each other in detection process.
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MSK
Phase trellis diagram for MSK signal 1101000
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MSK
Signal s(t) of MSK can be decomposed into
where
tftstfts
tftT
Etft
T
E
ttfT
Ets
cQcI
cb
bc
b
b
cb
b
2sin2cos
2sinsin2
2coscos2
2cos2
bb
TttT
t 0;2
0
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MSK
Symbol (0) (Tb)
10 /2
-/2
0
0 -/2
/2
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MSK
For the interval –Tb < t 0, we have
Let’s note here that the for the interval -Tb<t 0 and 0< tTb may not be the same.
We know that
0;2
0 tTtT
t bb
bbb T
t
T
t
T
t
2sin0sin
2cos0cos
20cos
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MSK
Since (0) can be either 0 or depending on the past history. We have
“+” for (0) = 0 and “-” for (0) = Hence, we have
bbb T
t
T
t
T
t
2cos
2cos0cos
20cos
bbbb
bI TtT
T
t
T
Ets
;
2cos
2)(
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MSK
Similarly we can write
for 0< tTb and Tb < t2Tb. Note the “+” and “-” may be different between these intervals.
Furthermore, we have that (Tb) can be /2 depending on the past history.
bb
b TtT
Tt 2
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MSK
Hence, we have
we have that (Tb) can be /2 depending on the past history.
22sincos
22cossin
2sincos
2cossin
2sin
bb
bb
b
bb
b
bb
b
bb
T
tT
T
tT
T
TtT
T
TtT
T
TtT
bbb
bb T
t
T
t
T
TtT
2sin
22cos
2sin
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MSK
Hence, we have
“+” for (Tb) = +/2 and “-” for (Tb) = -/2
The basis functions change to
bbb
bQ Tt
T
t
T
Ets 20;
2sin
2)(
bcbb
TttfT
t
Tt
0;2cos
2cos
21
bcbb
TttfT
t
Tt
0;2sin
2sin
22
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MSK
We write MSK signal as
Where and
)()(
)(sinE)(0cosE
2sin2
sinsin2
2cos2
cos0cos2
2sinsin2
2coscos2
2211
2b1b
tsts
tTt
tfT
tT
T
Etf
T
t
T
E
tftT
Etft
T
Ets
b
cb
bb
bc
bb
b
cb
bc
b
b
0cos1 bEs bb TEs sin2
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MSK
Symbol (0) s1 (Tb) s2
10 /2
-/2
0
0 -/2
/2
bE
bE
bE
bE
bE
bE
bE
bE
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02
1
N
EerfcP b
e
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0Phase: /2 /2 0 -/2/2
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MSK
We observe that MSK is in fact the QPSK having the pulse shape
Block diagrams for transmitter and receiver are given in the next two slides.
bT
t
2cos
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b
b
T
T
dtttxx )()( 11
bT
dtttxx2
022 )()(
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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.5
1
1.5
2
2.5
3
3.5
4
Normalized Frequency, fTb
Nor
mal
ized
PS
D,
S(f
)/E
bMSK BPSKQPSK
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MSK
Probability of error of MSK system is equal to BPSK and QPSK
This due to the fact that MSK observes the signal for two symbol intervals whereas FSK only observes for single signal interval.
Bandwidth of MSK system is 50% larger than QPSK.
2
222 116
2cos32)(
fT
fTEfS
b
bbMSK
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Noncoherent Orthogonal Modulation
Noncoherent implies that phase information is not available to the receiver.
As a result, zero phase of the receiver can mean any phase of the transmitter.
Any modulation techniques that transmits information through the phase cannot be used in noncoherent receivers.
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Noncoherent Orthogonal Modulation
cos(2ft)
sin(2ft)
Receiver
cos(2ft)sin(2ft)
Transmitter
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Noncoherent Orthogonal Modulation
It is impossible to draw the signal constellation since we do not know where the axes are.
However, we can still determine the distance of the each signal constellation from the origin.
As a result, the modulation techniques that put information in the amplitude can be detected.
FSK uses the amplitude of signals in two different frequencies. Hence non-coherent receivers can be employed.
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Noncoherent Orthogonal Modulation
Consider the BFSK system where two frequencies f1 and f2 are used to represented two “1” and “0”.
The transmitted signal is given by
Problem is that is unknown to the receiver. For the coherent receiver, is precisely known by receiver.
bi TtitfT
Ets 0,2,1;2cos
2)(
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Noncoherent Orthogonal Modulation
Furthermore, we have
To get rid of the phase information (), we use the amplitude
)()(
2sinsin2
2coscos2
2cos2
)(
2211 tsts
tfT
Etf
T
E
tfT
Ets
ii
ii
i
EEEssts ii 2222
21 sincos
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Noncoherent Orthogonal Modulation
Where
The amplitude of the received signal
TT
i dtttxxdtttss0
110
11 )()()()(
TT
i dtttxxdtttss0
220
22 )()()()(
2/12
0
2
0
2sin)(2cos)(
T
i
T
ii dttftxdttftxl
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Quadrature Receiver using correlators
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Quadrature Receiver using Matched Filter
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Noncoherent Orthogonal Modulation
Decision rule: Let if li > lk for all k. For examples, decide if l1 > l2
This decision rule suggests that if the envelop (amplitude) of the received signal described in term of cos(2f1t) is greater than the envelop of the received signal described in term of cos(2f2t), we say s1(t) was sent.
imm ˆ
1ˆ mm
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Noncoherent Matched Filter
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Noncoherent Orthogonal Modulation
Consider the output of matched filter of cos(2fit).
T
ii dtTfxtTftxty0
2cos)(2cos)(
2sin)()(2in-
2cos)()(2cos)(
0
0
T
ii
T
ii
dfxtTfs
dfxtTfty
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Noncoherent Orthogonal Modulation
Envelope at t=T is
Which is exactly the same as in correlator receiver
2/12
0
2
0
2cos)(2cos)(
T
i
T
ii dfxdfxl
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Generalized binary receiver for noncoherent orthogonal modulation.
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Quadrature receiver equivalent to either one of the two matched filters in part
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Noncoherent Orthogonal Modulation
Probability of Errors
02exp
2
1
N
EPe
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Noncoherent: BFSK
For BFSK, we have
elsewhere ; 0
0;2cos2
bib
b
iTttf
T
E
ts
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Noncoherent: BFSK
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Noncoherent: BFSK
Probability of Errors
02exp
2
1
N
EP b
e
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DPSK
Differential PSK Instead of finding the phase of the signal on
the interval 0<tTb. This receiver determines the phase difference between adjacent time intervals.
If “1” is sent, the phase will remain the same If “0” is sent, the phase will change 180 degree.
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DPSK
Or we have
and
bbcb
b
bcb
b
TtTtfT
E
TttfT
E
ts
2;2cos2
20;2cos2
)(1
bbcb
b
bcb
b
TtTtfT
E
TttfT
E
ts
2;2cos2
20;2cos2
)(2
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DPSK
In this case, we have T=2Tb and E=2Eb
Hence, the probability of error is given by
0exp
2
1
N
EP b
e
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DPSK: Transmitter
11 kkkkk dbdbd
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DPSK
{bk} 1 0 0 1 0 0 0 1 1
{dk-1} 1 1 0 1 1 0 1 0 0
Differential encoded {dk}
1 1 0 1 1 0 1 0 0 0
Transmitted Phase
0 0 0 0 0
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DPSK: Receiver
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DPSK: Receiver
From the block diagram, we have that the decision rule as
If the phase of signal is unchanged (send “1”) the sign (“+” or “-”) of both xi and xQ should not change. Hence, the l(x) should be positive.
If the phase of signal is unchanged (send “0”) the sign (“+” or “-1”) of both xi and xQ should change. Hence, the l(x) should be negative.
0
1say
0say
1010
QQII xxxxl x
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Signal-space diagram of received DPSK signal.