Binary Detection

7/27/2019 Binary Detection

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2012/2013 Meixia Tao @ SJ TU 2

Topics to be Covered

Binary digital modulation

M-ary digital modulation

Tradeoff study

data basebandDigital

modulator

Bandpasschannel

Digitaldemodulator

BPFdetector

Noise


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Digital Modulation

The message signal is transmitted by a sinusoidal carrierwave

In digital communications, the modulation processcorresponds to switching or keying the amplitude, frequency,

or phase 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 andperformance analysis


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8.1 Binary Modulation Types

In binary signaling, the modulator produces one oftwodistinct 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 havinga 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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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 ofs2

Message point

Messagepoint

Decision boundary

R1

R2


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Probability 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 betransmitted, the average probability of error forcoherent binary FSKis

3 dB worse than BPSK

i.e. to achieve the same Pe, BFSK needs 3dB moretransmission 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 ifl>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

ofBitError

PSK

ASK/FSK

3dB

Probability of Error Curve for BPSK and FSK/ASK

e.g.


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Example #1

Binary data are transmitted over a microwave linkat the rate of106 bits/sec and the PSD of thenoise at the receiver input is 10-10 watts/Hz.

a) Find the average carrier power required tomaintain an average probability of errorfor coherent binary FSK.

b) Repeat the calculation in a) for noncoherent

binary FSK


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2012/2013 Meixia Tao @ SJ TU 2727

We have discussed

Coherent modulation schemes, .e.g.

BPSK, BFSK, BASK They needs coherent detection,

assuming that the receiver is able todetect and track the carrier wavesphase

Update

We now consider:

Non-coherent detection on binary FSK

Differential phase-shift keying (DPSK)

In many practical situations, strict phasesynchronization is not possible. In thesesituations, 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 withuniform distribution


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Signal Space Representation

No matter what the two phases are, the signalscan be expressed as a linear combination of thefour 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

(selectthelargest)

(For detailed proof, see Section 10.4.2 in the textbook )


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Performance Comparison Between coherent

FSK and Non-Coherent FSK

0 2 4 6 8 10 12 1410

-7

10-6

10

-5

10-4

10-3

10-2

10-1

10

0

Eb/No in [dB]

P

robability

ofBit

Error

BPSK

ASK/FSK

NC FSK

DPSK


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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 signalwaveform 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 twobit intervals), the phase difference betweenwaveforms 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 respect

to 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

Differentiallyencoded

binary data

TransmittedPhase

mi

di___________

1 iii mdd =


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Differential Detection of DPSK Signals

Multiply the received DPSK signal with its delayed version

Output of integrator (assume noise free)

The unknown phase becomes irrelevant

If = 0 (bit 1), the integrator outputy is positive

if = (bit 0), the integrator output y is negative

Decision

device

bT

dt0

Threshold of

zero volts

Choose 1 ifl >0

Otherwise choose 0

Delay

Tb


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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 intervalTb

In M-ary data transmission, send one ofM possible signalsduring each signaling intervalT

In almost all applications, M = 2

n

andT = 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 tocombine different methods of modulation intohybrid forms

For example, we may combine discrete changesin 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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MPSK (contd)



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MPSK Signal Constellations

BPSK QPSK 8PSK 16PSK


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2012/2013 Meixia Tao @ SJ TU

The Euclidean distance between any two signal points in theconstellation is

The minimum Euclidean distance is

dmin plays an important role in determining error performance asdiscussed previously (union bound)

In the case of PSK modulation, the error probability is dominated bythe 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

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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 betweenadjacent signal points

For M=8, determine by how many dB thetransmitted signal energy Es must be increased toachieve the same as M =4.

52

mind

mind


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Error Performance of MPSK

For large M, doubling thenumber of phases requires anadditional 6dB/bit to achievethe same performance

4dB 5dB 6dB

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 theenvelope is constant (circular constellation). If werelax 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:



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MQAM Signal Constellation

Square lattice

Can be related with two L-ary ASK in in-phase andquadrature 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 sameerror performance if the number of bits per symbol is increased

from k to k+1, where k is large.

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0

34

( 1)

b

e

kEP Q

M N

(for )2kM =

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 signalwaveforms, we define the correlation coefficients


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MFSK (contd)

For orthogonality, minimum frequency separationbetween successive frequencies is 1/(2T)

1

0.715/T


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M-ary orthogonal FSK has a geometric presenation as MM-dim orthogonal vectors, given as

The basis functions are

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( )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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Error Performance of MFSK


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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 signalconstellation


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Example #2

The 16-QAM signal constellation shown below is aninternational standard for telephone-line modems (calledV.29).

a) Determine the optimum decision

boundaries for the detectorb) Derive the union bound of the

probability of symbol errorassuming that the SNR issufficiently 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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Bit Error Rate with Gray Coding

Gray coding is a bit-to-symbol mapping When going from one symbol to an adjacent

symbol, only one bit out of the k bits changes

An error between adjacent symbol pairs results inone and only one bit error.


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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 halfthe symbol

error

eb PP2

1

8 4 Comparison of M-ary Modulation


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8.4 Comparison of M ary Modulation

Techniques

Channel bandwidth and transmit power are 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

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 ul tra-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]. J ustify 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.)

Energy Efficiency Comparison


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e gy c e cy Co pa so

MFSK MPSK


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2012/2013 Meixia Tao @ SJ TU71

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

B d idth Effi i C i


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Bandwidth Efficiency Comparison

To compare bandwidth efficiency, we need to know thepower spectral density (power spectra) of a givenmodulation scheme

MPSK/MQAM

If is rectangular, the bandwidth of mainlope is

If it has a raised cosine spectrum, the bandwidth is

Spectrumshaping filter

Inputdata

Signal pointmapper

Spectrumshaping filter

+MPSK/MQAM

signal

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

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= bit rate

(bits/sec/Hz)

B d idth Effi i C i ( td)


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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 separatedby 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, butbandwidth efficiency of MFSK decreases.

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 maximumrate 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

Power-Bandwidth Tradeoff


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Power-Bandwidth TradeoffCapacity boundary

with R = C

UnachievableRegion with R > C

Shannon

limit

N t th F d t l T d ff


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Notes on the Fundamental Tradeoff

In the limits as R/B goes to 0, we get

This value is called the Shannon Limit

Received Eb/N0 must be >-1.6dB for reliable communicationsto be possible

BPSK and QPSK require the same Eb/N0 of 9.6 dB to achievePe=10

-5. However, QPSK has a better bandwidth efficiency,which is why QPSK is so popular

MQAM is superior to MPSK

MPSK/MQAM increases bandwidth efficiency at the cost oflower energy efficiency

MFSK trades energy efficiency at reduced bandwidth efficiency.

S t D i T d ff


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System Design Tradeoff

Power Limited Systems:Powerscarce

but bandwidth available

Bandwidth Limited Systems:

Bandwidth scarce

Power available

Which

Modulation

to Use?

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 ul tra-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]. J ustify 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.)

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 38Mbpsand 54 Mbps)

LTE Cellular Systems

FSK:

Cordless telephone

Paging system

Documents

Binary Detection