comm ch08 digmod en
TRANSCRIPT
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Principles of ommunicationsMeixia Tao
Shanghai Jiao Tong University
Chapter 8: Digital Modulation TechniquesTextbook: Ch 8.4 8.5, Ch 10.1-10.5
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Topics to be Covered
Binary digital modulation
M-ary digital modulation
Tradeoff study
data basebandDigital
modulator
Bandpass
channel
Digital
demodulatorBPFdetector
Noise
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Digital Modulation
The message signal is transmitted by a sinusoidal carrierwave
In digital communications, the modulation process
corresponds to switching orkeying the amplitude, frequency,
orphase of the carrier in accordance with the incomingdigital data
Three basic digital modulation techniques Amplitude-shift keying (ASK) - special case of AM
Frequency-shift keying (FSK) - special case of FM Phase-shift keying (PSK) - special case of PM
Will use signal space approach in receiver design and
performance analysis
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8.1 Binary Modulation Types
In binary signaling, the modulator produces one of twodistinct signals in response to 1 bit of source data at atime.
Binary modulation types
Binary PSK (BPSK)
Binary FSK
Binary ASK
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Binary Phase-Shift Keying (BPSK)
Modulation
, , bit duration
: carrier frequency, chosen to be for some fixedinteger or
: transmitted signal energy per bit, i.e.
The pair of signals differ only in a relative phase shift of 180degrees
0 1 1 0 1 0 0 1
1
0
1 /c bf T>>
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Clearly, there is one basis function of unit energy
Then
A binary PSK system is therefore characterized by having
a signal space that is one-dimensional (i.e. N=1), and with
two message points (i.e. M = 2)
Signal Space Representation for
BPSK
s1s20
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Decision Rule of BPSK
Assume that the two signals are equally likely, i.e. P(s1) =P(s2) = 0.5. Then the optimum decision boundary is themidpoint of the line joining these two message points
Decision rule:
Guess signal s1(t) (or binary 1) was transmitted if thereceived signal point r falls in region R1
Guess signal s2(t) (or binary 0) was transmitted otherwise
s1s2
Region R1Region R2
0
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Proof of the Decision Rule
Observation scalar (output of the demodulator) r is
where n represents the AWGN component, which hasmean zero and variance
Thus, the likelihood function of r is
If s1 is transmitted
If s2 is transmitted
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Recall ML decision criterion:
Thus
And
Finally
Choose s1> r
2
When r1< r
2, r falls inside region R2 and the receiver decides in
favor of s2
Message point
Message
point
Decision boundary
R1
R2
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Probabil ity of Error for Binary FSK
Given that s1 is transmitted,
Since the condition r1< r
2corresponds to the receiver
making a decision in favor of symbol s2, the conditional
probability of error given s1 is transmitted is given by
Define a new random variable
Since n1 and n2 are i.i.d with
Thus, n is also Gaussian with
and
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By symmetry
Since the two signals are equally likely to be
transmitted, the average probability of errorfor
coherent binary FSK is
3 dB worse than BPSK
i.e. to achieve the same Pe, BFSK needs 3dB more
transmission power than BPSK
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Binary FSK Transmitter
On-off signalling form
0
1
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Coherent Binary FSK Receiver
bT
dt0
bT
dt0
Decision
Device+
Choose 1 if l>0
Choose 0 otherwise
+
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Binary ASK
Modulation
Average energy per bit
0 1 1 0 1 0 0 1
(On-off signalling)
1
0
s1s2
Region R1Region R2
0
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Probability of Error for Binary ASK
Average probability of error is
Exercise: Prove Pe
Identical to that of coherent binary FSK
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Probability of Error and the Distance
Between Signals
These expressions illustrate the dependence of the error
probability on the distance between two signal points. Ingeneral,
BPSK BFSK BASK
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0 2 4 6 8 10 12 1410
-7
10-6
10-5
10-4
10-3
10-2
10-1
100
Eb/No in [dB]
Probability
ofB
itError
PSK
ASK/FSK
3dB
Probability of Error Curve for BPSK and FSK/ASK
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Example #1
Binary data are transmitted over a microwave linkat the rate of 106 bits/sec and the PSD of the
noise at the receiver input is 10-10 watts/Hz.
a) Find the average carrier powerrequired tomaintain an average probability of error
for coherent binary FSK.
b) Repeat the calculation in a) for noncoherent
binary FSK
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We have discussed
Coherent modulation schemes, .e.g.
BPSK, BFSK, BASK They needs coherent detection,
assuming that the receiver is able to
detect and track the carrier waves
phase
Update
We now consider:
Non-coherent detection on binary FSK
Differential phase-shift keying (DPSK)
In many practical situations, strict phase
synchronization is not possible. In these
situations, non-coherent reception is required.
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8.2: Non-coherent scheme: BFSK
Consider a binary FSK system, the two signals are
Where and are unknown random phases with
uniform distribution
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Signal Space Representation
No matter what the two phases are, the signalscan be expressed as a linear combination of the
four basis functions
Signal space representation
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Correlating the received signal r(t) with the four basis
functions produces the vector representation of thereceived signal
Detector
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Decision Rule for Non-coherent FSK
ML criterion, assume P(s1) = P(s2):
Conditional pdf
Similarly,
Choose s1> envelop detector
Carrier phase is irrelevant in decision making
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Structure of Non-Coherent Receiver for
Binary FSK
It can be shown that
Comparator
(selectthe
largest)
(For detailed proof, see Section 10.4.2 in the textbook )
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Differential PSK (DPSK)
DPSK can be viewed as the non-coherent versionof PSK.
Phase synchronization is eliminated using
differential encoding Encoding the information in phase difference
between successive signal transmission
In effect:
to send 0, we phase advance the current signal
waveform by 1800 ;
to send 1, we leave the phase unchanged
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DPSK (contd)
Provided that the unknown phase contained inthe received wave varies slowly (constant over two
bit intervals), the phase difference between
waveforms received in two successive bit interval
will be independent of
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Generation of DPSK signal
We can generate DPSK signals by combining two basicoperations
Differential encoding of the information binary bits
Phase shift keying
The differential encoding process starts with an arbitraryfirst bit, serving as reference
Let {mi} be input information binary bit sequence, {di} bethe differentially encoded bit sequence
If the incoming bit miis 1, leave the symbol d
iunchanged
with respect to the previous bit di-1
If the incoming bit mi is 0, change the symbol di with respectto the previous bit di-1
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Illustration
The reference bit is chosen arbitrary, here taken as 1
DPSK transmitter diagram
1 0 0 1 0 0 1 1Binary data
1 1 0 1 1 0 1 1 1
0 0 0 0 0 0 0
Initial bit
Differentially
encoded
binary data
TransmittedPhase
mi
di___________
1 iii mdd =
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Error Probability of DPSK
The differential detector is suboptimal in the senseof error performance
It can be shown that
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Summary of Pe for Different Binary
Modulations
Coherent PSK
Coherent ASK
Coherent FSK
Non-Coherent FSK
DPSK
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0 2 4 6 8 10 12 1410
-7
10-6
10-5
10-4
10
-3
10-2
10-1
100
Eb/No in [dB]
Probability
of
BitError
BPSK(QPSK)
ASK/FSK
NC FSK
DPSK
Pe Plots for Different Binary Modulations
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We have discussed binary case
Coherent modulation techniques:
BPSK, BFSK, BASK
Noncoherent modulation techniques:
Non-coherent FSK, DPSK
Update
We now consider:
M-ary modulation techniques
MPSK
MQAM
MFSK
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8.3 M-ary Modulation Techniques
In binary data transmission, send only one of two possiblesignals during each bit interval Tb
In M-ary data transmission, send one of M possible signalsduring each signaling interval T
In almost all applications, M = 2
n
and T = nTb, where n isan integer
Each of the M signals is called a symbol
These signals are generated by changing the amplitude,
phase or frequency of a carrier in M discrete steps. Thus, we have M-ary ASK, M-ary PSK, and M-ary FSK
digital modulation schemes
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Binary is a special case of M-ary
Another way of generating M-ary signals is to
combine different methods of modulation into
hybrid forms
For example, we may combine discrete changes
in both the amplitude and phase of a carrier to
produce M-ary amplitude phase keying.A specialform of this hybrid modulation is M-ary QAM
(MQAM)
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M-ary Phase-Shift Keying (MPSK)
The phase of the carrier takes on M possible values:
Signal set:
= Energy per symbol
Basis functions
1c
fT
>>
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MPSK (contd)
Signal space representation
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MPSK Signal Constellations
BPSK QPSK 8PSK 16PSK
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The Euclidean distance between any two signal points in the
constellation is
The minimum Euclidean distance is
plays an important role in determining error performance as
discussed previously (union bound)
In the case of PSK modulation, the error probability is dominated by
the erroneous selection of either one of the two signal points adjacent
to the transmitted signal point. Consequently, an approximation to the symbol error probability is
2 ( )
2 1 cosmn m n sm n
d E M
= =
s s
min
22 1 cos 2 sin
s sd E EM M
= =
min
0
/ 22 2 2 sin
/ 2MPSK s
dP Q Q E
MN
=
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Exercise
Consider the M=2, 4, 8 PSK signal constellations.All have the same transmitted signal energy Es.
Determine the minimum distance between
adjacent signal points For M=8, determine by how many dB the
transmitted signal energy Es must be increased to
achieve the same as M =4.
mind
mind
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Error Performance of MPSK
For large M, doubling the
number of phases requires an
additional 6dB/bit to achieve
the same performance
4dB 5dB 6dB
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M ary Quadrature Amplitude Modulation
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M-ary Quadrature Amplitude Modulation
(MQAM)
In an M-ary PSK system, in-phase and quadraturecomponents are interrelated in such a way that the
envelope is constant (circular constellation). If we
relax this constraint, we get M-ary QAM.
Signal set:
E0 is the energy of the signal with the lowest amplitude
ai, bi are a pair of independent integers
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MQAM (contd)
Basis functions:
Signal space representation
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MQAM Signal Constellation
Square lattice
Can be related with two L-ary ASK in in-phase and
quadrature components, respectively, where M = L2
1 3 5 7
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Error Performance of MQAM
It can be shown that the symbol error probability ofMQAM is tightly upper bounded as
Exercise: From the above expression, determine the increasein the average energy per bit Eb required to maintain the same
error performance if the number of bits per symbol is increased
from k to k+1, where k is large.
0
34
( 1)
b
e
kEP Q
M N
(for )2kM=
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M ary Frequency Shift Keying (MFSK) or
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M-ary Frequency-Shift Keying (MFSK) or
Multitone Signaling
Signal set:
where
As a measure of similarity between a pair of signal
waveforms, we define the correlation coefficients
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MFSK (contd)
For orthogonality, minimum frequency separationbetween successive frequencies is 1/(2T)
The minimum correlation is at
-0.217
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M-ary orthogonal FSK has a geometric presenation as MM-dim orthogonal vectors, given as
The basis functions are
( )0 , 0, 0, , 0sE=s
( )1 0, , 0, , 0sE=s
( )1 0, 0, , 0,M sE =s
( )2
cos2m c
f m f tT
= +
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Notes on Error Probabil ity Calculations
Pe is found by integrating conditional probability oferror over the decision region
Difficult for multi-dimensions
Can be simplified using union bound (see ch07)
Pe depends only on the distance profile of signal
constellation
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Example #2
The 16-QAM signal constellation shown below is aninternational standard for telephone-line modems (called
V.29).
a) Determine the optimum decision
boundaries for the detectorb) Derive the union bound of the
probability of symbol error
assuming that the SNR is
sufficiently high so that errors
only occur between adjacentpoints
c) Specify a Gray code for this 16-
QAM V.29 signal constellation
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Symbol Error versus Bit Error
Symbol errors are different from bit errors When a symbol error occurs, all bits could
be in error
In general, we can find BER using
is the number bits which differ between and
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Example: Gray Code for QPSK
0001
11 10
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Bit Error Rate for MPSK and MFSK
For MPSK with gray coding An error between adjacent symbol will most likely occur
Thus, bit error probability can be approximated by
For MFSK When an error occurs anyone of the other symbols may result
equally likely.
On average, therefore, half of the bits will be incorrect. That is k/2bits every k bits will on average be in error when there is a symbol
error Thus, the probability of bit error is approximately half the symbol
error
eb PP2
1
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8 4 Comparison of M-ary Modulation
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8.4 Comparison of M ary Modulation
Techniques
Channel bandwidth and transmit powerare twoprimary communication resources and have to beused as efficient as possible
Power utilization efficiency (energy efficiency):
measured by the required Eb/No to achieve acertain bit error probability
Spectrum utilization efficiency (bandwidthefficiency): measured by the achievable data rate
per unit bandwidth Rb/B It is always desired to maximize bandwidth
efficiency at a minimal required Eb/No
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Example # 3
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Example # 3
Suppose you are a system engineerin Huawei, designing a part of thecommunication systems. You are required to design three systems as follow:
I. An ultra-wideband system. This system can use a large of amount ofbandwidth to communicate. But the band it uses is overlaying with the othercommunication system. The main purpose of deploying this system is toprovide high data rates.
II. A wireless remote control system designated for controlling devicesremotely under unlicensed band.
III. A fixed wireless system. The transmitters and receivers are mounted in afixed position with power supply. This system is to support voice and dataconnections in the rural areas. This system works under licensed band.
You are only required to design a modulation scheme for each of the abovesystems. You are allowed to use MFSK, MPSK and MQAM only. You alsoneed to state the modulation level. For simplicity, the modulation level shouldbe chosen from M=[Low, Medium, High]. Justify your answers.
(Hints: Federal Communications Commission (FCC) has a power spectraldensity limit in unlicensed band. It is meant that if your system works underunlicensed band, the power cannot be larger than a limit.)
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Energy Efficiency Comparison
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e gy c e cy Co pa so
MFSK MPSK
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Energy Efficiency Comparison (contd)
MFSK: At fixed Eb/No, increase M can provide an improvement
on Pb
At fixed Pb increase M can provide a reduction in theE
b/N
orequirement
MPSK
BPSK and QPSK have the same energy efficiency
At fixed Eb/No, increase M degrades Pb
At fixed Pb, increase M increases the Eb/Norequirement
MFSK is more energy efficient than MPSK
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B d idth Effi i C i ( td)
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Bandwidth Efficiency Comparison (contd)
In general, bandwidth required to pass MPSK/MQAM signalis approximately given by
But
Then bandwidth efficiency may be expressed as
= bit rate
(bits/sec/Hz)
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B d idth Effi i C i ( t d)
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MFSK:
Bandwidth required to transmit MFSK signal is
Bandwidth efficiency of MFSK signal
Bandwidth Efficiency Comparison (contd)
(Adjacent frequencies need to be separated
by 1/2T to maintain orthogonality)
(bits/s/Hz)
M 2 4 8 16 32 64
1 1 0.75 0.5 0.3125 0.1875(bits/s/Hz)
As M increases, bandwidth efficiency of MPSK/MQAM increases, but
bandwidth efficiency of MFSK decreases.
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Fundamental Tradeoff :
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Fundamental Tradeoff :Bandwidth Efficiency and Energy Efficiency
To see the ultimate power-bandwidth tradeoff, we need touse Shannons channel capacity theorem:
Channel Capacity is the theoretical upper bound for the maximum
rate at which information could be transmitted without error
(Shannon 1948)
For a bandlimited channel corrupted by AWGN, the maximum rateachievable is given by
Note that
Thus
)1(log)1(log0
22BN
PBSNRBCR s+=+=
R
BSNR
BRN
BP
RN
P
N
TP
N
Esssb
====0000
)12( /
0
= BRbR
B
N
E
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S t D i T d ff
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System Design Tradeoff
Power Limited Systems:Power scarce
but bandwidth available
Bandwidth Limited Systems:
Bandwidth scarce
Power available
Which
Modulation
to Use?
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Example # 3
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p
Suppose you are a system engineerin Huawei, designing a part of thecommunication systems. You are required to design three systems as follow:
I. An ultra-wideband system. This system can use a large of amount ofbandwidth to communicate. But the band it uses is overlaying with the othercommunication system. The main purpose of deploying this system is toprovide high data rates.
II. A wireless remote control system designated for controlling devicesremotely under unlicensed band.
III. A fixed wireless system. The transmitters and receivers are mounted in a
fixed position with power supply. This system is to support voice and dataconnections in the rural areas. This system works under licensed band.
You are only required to design a modulation scheme for each of the abovesystems. You are allowed to use MFSK, MPSK and MQAM only. You alsoneed to state the modulation level. For simplicity, the modulation level shouldbe chosen from M=[Low, Medium, High]. Justify your answers.
(Hints: Federal Communications Commission (FCC) has a power spectraldensity limit in unlicensed band. It is meant that if your system works underunlicensed band, the power cannot be larger than a limit.)
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Practical Applications
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Practical Applications
BPSK:
WLAN IEEE802.11b (1 Mbps)
QPSK:
WLAN IEEE802.11b (2 Mbps, 5.5 Mbps, 11 Mbps)
3G WDMA
DVB-T (with OFDM)
QAM
Telephone modem (16QAM)
Downstream of Cable modem (64QAM, 256QAM)
WLAN IEEE802.11a/g (16QAM for 24Mbps, 36Mbps; 64QAM for 38Mbps
and 54 Mbps)
LTE Cellular Systems
FSK:
Cordless telephone
Paging system